東京工業大学工学院機械系 教授

イワツキ ノブユキ

岩附 信行

大学院講義

「先端機械要素」

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

Lecture for Graduate Students

Advanced Mechanical Elements

Lecture syllabus

Course title Advanced Mechanical ElementsAcademic major Mechanical EngineeringOffered quarter 2QDayperiod Thursday 3-4Credits 1-0-0Lecture room ZOOMRegistration number MECHH431Lecturer Prof Nobuyuki Iwatsuki

Office I1-305E-mail iwatsukinaamtitechacjp

Course description and aimsThis course offers the knowledge on kinematic and dynamic

analysis of planar and spatial link mechanisms and methods to control redundant and overunderactuated mechanisms

The characteristic of motion mechanisms affects the performance of advanced mechanical systems such as robots It is thus required to quantitatively evaluate the characteristic and to design and control mechanisms based on the evaluation criteria

For this purpose students will understand how to analyze the kinematics and dynamics of planar and spatial closed-loop link mechanisms by utilizing the systematic kinematic analysis method and will experience to apply them to kinematic and dynamic analyses and motion control of redundant mechanisms overactuated mechanisms of which actuator inputs are more than mobility of mechanism and underactuated mechanisms constrained by elastic elements or gravitational force

Student learning outcomesBy the end of this course students will be able to

(1)Explain mobility of mechanism and relation between inputoutput motion of mechanism

(2)Analyze displacement velocity and acceleration of planarspatialclosed-loop link mechanisms with the systematic kinematic analysis method

(3)Analyze the dynamics of planarspatial closed-loop link mechanisms utilizing the systematic kinematic analysis method

(4)Explain the optimum motion control of redundant link mechanisms

(5)Explain motion control of overactuated or underactuatedmechanisms with elastic elements

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

Lecture for Graduate Students

Advanced Mechanical Elements

Lecture syllabus

Course title Advanced Mechanical ElementsAcademic major Mechanical EngineeringOffered quarter 2QDayperiod Thursday 3-4Credits 1-0-0Lecture room ZOOMRegistration number MECHH431Lecturer Prof Nobuyuki Iwatsuki

Office I1-305E-mail iwatsukinaamtitechacjp

Course description and aimsThis course offers the knowledge on kinematic and dynamic

analysis of planar and spatial link mechanisms and methods to control redundant and overunderactuated mechanisms

The characteristic of motion mechanisms affects the performance of advanced mechanical systems such as robots It is thus required to quantitatively evaluate the characteristic and to design and control mechanisms based on the evaluation criteria

For this purpose students will understand how to analyze the kinematics and dynamics of planar and spatial closed-loop link mechanisms by utilizing the systematic kinematic analysis method and will experience to apply them to kinematic and dynamic analyses and motion control of redundant mechanisms overactuated mechanisms of which actuator inputs are more than mobility of mechanism and underactuated mechanisms constrained by elastic elements or gravitational force

Student learning outcomesBy the end of this course students will be able to

(1)Explain mobility of mechanism and relation between inputoutput motion of mechanism

(2)Analyze displacement velocity and acceleration of planarspatialclosed-loop link mechanisms with the systematic kinematic analysis method

(3)Analyze the dynamics of planarspatial closed-loop link mechanisms utilizing the systematic kinematic analysis method

(4)Explain the optimum motion control of redundant link mechanisms

(5)Explain motion control of overactuated or underactuatedmechanisms with elastic elements

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Lecture syllabus

Course title Advanced Mechanical ElementsAcademic major Mechanical EngineeringOffered quarter 2QDayperiod Thursday 3-4Credits 1-0-0Lecture room ZOOMRegistration number MECHH431Lecturer Prof Nobuyuki Iwatsuki

Office I1-305E-mail iwatsukinaamtitechacjp

Course description and aimsThis course offers the knowledge on kinematic and dynamic

analysis of planar and spatial link mechanisms and methods to control redundant and overunderactuated mechanisms

The characteristic of motion mechanisms affects the performance of advanced mechanical systems such as robots It is thus required to quantitatively evaluate the characteristic and to design and control mechanisms based on the evaluation criteria

For this purpose students will understand how to analyze the kinematics and dynamics of planar and spatial closed-loop link mechanisms by utilizing the systematic kinematic analysis method and will experience to apply them to kinematic and dynamic analyses and motion control of redundant mechanisms overactuated mechanisms of which actuator inputs are more than mobility of mechanism and underactuated mechanisms constrained by elastic elements or gravitational force

Student learning outcomesBy the end of this course students will be able to

(1)Explain mobility of mechanism and relation between inputoutput motion of mechanism

(2)Analyze displacement velocity and acceleration of planarspatialclosed-loop link mechanisms with the systematic kinematic analysis method

(3)Analyze the dynamics of planarspatial closed-loop link mechanisms utilizing the systematic kinematic analysis method

(4)Explain the optimum motion control of redundant link mechanisms

(5)Explain motion control of overactuated or underactuatedmechanisms with elastic elements

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Course description and aimsThis course offers the knowledge on kinematic and dynamic

analysis of planar and spatial link mechanisms and methods to control redundant and overunderactuated mechanisms

The characteristic of motion mechanisms affects the performance of advanced mechanical systems such as robots It is thus required to quantitatively evaluate the characteristic and to design and control mechanisms based on the evaluation criteria

For this purpose students will understand how to analyze the kinematics and dynamics of planar and spatial closed-loop link mechanisms by utilizing the systematic kinematic analysis method and will experience to apply them to kinematic and dynamic analyses and motion control of redundant mechanisms overactuated mechanisms of which actuator inputs are more than mobility of mechanism and underactuated mechanisms constrained by elastic elements or gravitational force

Student learning outcomesBy the end of this course students will be able to

(1)Explain mobility of mechanism and relation between inputoutput motion of mechanism

(2)Analyze displacement velocity and acceleration of planarspatialclosed-loop link mechanisms with the systematic kinematic analysis method

(3)Analyze the dynamics of planarspatial closed-loop link mechanisms utilizing the systematic kinematic analysis method

(4)Explain the optimum motion control of redundant link mechanisms

(5)Explain motion control of overactuated or underactuatedmechanisms with elastic elements

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Student learning outcomesBy the end of this course students will be able to

(1)Explain mobility of mechanism and relation between inputoutput motion of mechanism

(2)Analyze displacement velocity and acceleration of planarspatialclosed-loop link mechanisms with the systematic kinematic analysis method

(3)Analyze the dynamics of planarspatial closed-loop link mechanisms utilizing the systematic kinematic analysis method

(4)Explain the optimum motion control of redundant link mechanisms

(5)Explain motion control of overactuated or underactuatedmechanisms with elastic elements

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

KeywordsClose-loop link mechanism kinematics dynamics systematic kinematic analysis motion control optimum control redundancy overactuated mechanism underactuated mechanism wire-driven mechanism

Class flowImportant issues are summarized at the end of lecture every week Because sample software to analyze mechanisms which are explained in this lecture will be offered to students students are expected to try to analyze various mechanisms with the software by themselves so as to understand what they learn

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

ScheduleClass 1 Kinematic analysis of planar link mechanism with the (June 25) systematic kinematic analysis method

- Displacement velocity and acceleration analyses of planar closed-loop link mechanism with the systematic kinematic analysis method -

Class 2 Kinematic analysis of spatial link mechanism with the(July 2) systematic kinematic analysis method

- Expansion of the systematic kinematic analysis method to spatial mechanisms -

Class 3 Dynamic analysis of planarspatial link mechanisms (July 9) - Driving force and joint force analysis using the

systematic kinematic analysis -Class 4 Motion control of redundant link mechanisms(July 16) - Optimum motion control to utilize redundancy -

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Class 5 Kinematic analysis and motion control of overactuated(July 23) mechanisms

- Motion control of overactuated mechanisms using relaxation with elastic elements -

Class 6 Kinematic analysis and motion control of underactuated(July 30) mechanisms with elastic elements

- Motion control of underactuated mechanisms constrained by elastic elements -

Class 7 Kinematic analysis and motion control of underactuated(August 6) wire-driven mechanisms

- Motion control of wire-driven underactuatedmechanisms under gravitational force-

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Textbook reference book course materialNo textbook is required Several handouts will be distributedvia OCW-i Students should download and print the handouts before class The following book is recommended as a reference book

JSME Textbook series Kinematics of Machineryedited by the Japan Society of Mechanical Engineers (2007)

Assessment criteria and methodsStudents knowledge of kinematic and dynamic analyses of planarspatial closed-loop link mechanisms with the systematic kinematic analysis method and their application to analyses and motion controls of redundant and underoveractuated mechanisms is assessed via submitted reports on several issues

Related coursesSynthesis of robotic mechanisms Design of robot controller

PrerequisitesStudents must have successfully completed Mechanical elements and draftings or have equivalent knowledge

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

June 25 2020

Advanced Mechanical Elements(Lecture 1)

Kinematic analysis of planar link mechanism with the systematic kinematic analysis method

Tokyo Institute of TechnologyDept of Mechanical EngineeringSchool of Engineering

Prof Nobuyuki Iwatsuki

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

1 Introduction of lecturer

Please visit my WEB-sitehttpwwwdynamicsmeptitechacjpindexphp

orSearch by Google with a keyword lsquoIwatsuki labrsquo

Nobuyuki Iwatsuki Professor Dr EngAffiliation

Dynamics Engineering LaboratoryDepartment of Mechanical EngineeringSchool of EngineeringTokyo Institute of Technology

OfficeRoom 305 Ishikawadai-1 Building Ohokayama Campus

Contact Tel +81-3-5734-2538Fax +81-3-5734-3917 (Department office)E-mail nobmeptitechacjp

Several artistic works

Research Themes(A)Synthesis and Control of Robotic Mechanisms

Especially focused on lsquoMechanism Design and Motion Control of Hyper RedundantUnderactuatedRobotsrsquo

(B)Silent EngineeringEspecially focused on lsquoEstimation of Sound Power Radiating from Vibrating Structure and Structural Optimization to Reduce the Sound Powerrsquo

(C)Functional Material ActuatorsEspecially focused on lsquoDevelopment of Micro CiliaryActuators in Grouprsquo

Several Examples of My Research Works(1)Biped Walking Machine

ndash While undergraduatePhD student (-1987)

Straight Walk of Biped Walking Machine(1982)

Knee joint

Ankle joint

Toe

Heel

Hip joint

Leg Mechanism with 1 DOF

Ankle path generator(Planar 6-bar link mech)

Foot driving mechanism(Planar 8-bar link mech)

Driving links

Planar 8-bar link mechanism

Counter weightDrawing 8 figure

Plane-cylinder joint

Slanting crank shaft

Attitude Stabilizing Mechanism

950m

m

A Biped Waking Machine with only 1 DOF (Mechanical synthesis for simple control)

Walking speed 120 stepsmin (World record）

2nd prototype which can startand stop walking

Improvement for non steady walkingChange strideIndependent drive of foot driving mechanismArbitrary control of attitude stabilizing mechanismA new prototype with 5 DOF

Items Prototype-1 Prototype-2Degree-of-freedom 1 5Weight 206kgf 488kgfStraight walk Max 120stepmin Max15 stepsminStartstop to walk - Sucessful

Dynamics is important not only to stabilize biped machine but also to drive actuators

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Items		Prototype-1		Prototype-2
Degree-of-freedom		1		5
Weight		206kgf		488kgf
Straight walk		Max 120stepmin		Max15 stepsmin
Startstop to walk		-		Sucessful

Several artistic works

Research Themes(A)Synthesis and Control of Robotic Mechanisms

Especially focused on lsquoMechanism Design and Motion Control of Hyper RedundantUnderactuatedRobotsrsquo

(B)Silent EngineeringEspecially focused on lsquoEstimation of Sound Power Radiating from Vibrating Structure and Structural Optimization to Reduce the Sound Powerrsquo

(C)Functional Material ActuatorsEspecially focused on lsquoDevelopment of Micro CiliaryActuators in Grouprsquo

Several Examples of My Research Works(1)Biped Walking Machine

ndash While undergraduatePhD student (-1987)

Straight Walk of Biped Walking Machine(1982)

Knee joint

Ankle joint

Toe

Heel

Hip joint

Leg Mechanism with 1 DOF

Ankle path generator(Planar 6-bar link mech)

Foot driving mechanism(Planar 8-bar link mech)

Driving links

Planar 8-bar link mechanism

Counter weightDrawing 8 figure

Plane-cylinder joint

Slanting crank shaft

Attitude Stabilizing Mechanism

950m

m

A Biped Waking Machine with only 1 DOF (Mechanical synthesis for simple control)

Walking speed 120 stepsmin (World record）

2nd prototype which can startand stop walking

Improvement for non steady walkingChange strideIndependent drive of foot driving mechanismArbitrary control of attitude stabilizing mechanismA new prototype with 5 DOF

Items Prototype-1 Prototype-2Degree-of-freedom 1 5Weight 206kgf 488kgfStraight walk Max 120stepmin Max15 stepsminStartstop to walk - Sucessful

Dynamics is important not only to stabilize biped machine but also to drive actuators

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Items		Prototype-1		Prototype-2
Degree-of-freedom		1		5
Weight		206kgf		488kgf
Straight walk		Max 120stepmin		Max15 stepsmin
Startstop to walk		-		Sucessful

Research Themes(A)Synthesis and Control of Robotic Mechanisms

Especially focused on lsquoMechanism Design and Motion Control of Hyper RedundantUnderactuatedRobotsrsquo

(B)Silent EngineeringEspecially focused on lsquoEstimation of Sound Power Radiating from Vibrating Structure and Structural Optimization to Reduce the Sound Powerrsquo

(C)Functional Material ActuatorsEspecially focused on lsquoDevelopment of Micro CiliaryActuators in Grouprsquo

Several Examples of My Research Works(1)Biped Walking Machine

ndash While undergraduatePhD student (-1987)

Straight Walk of Biped Walking Machine(1982)

Knee joint

Ankle joint

Toe

Heel

Hip joint

Leg Mechanism with 1 DOF

Ankle path generator(Planar 6-bar link mech)

Foot driving mechanism(Planar 8-bar link mech)

Driving links

Planar 8-bar link mechanism

Counter weightDrawing 8 figure

Plane-cylinder joint

Slanting crank shaft

Attitude Stabilizing Mechanism

950m

m

A Biped Waking Machine with only 1 DOF (Mechanical synthesis for simple control)

Walking speed 120 stepsmin (World record）

2nd prototype which can startand stop walking

Improvement for non steady walkingChange strideIndependent drive of foot driving mechanismArbitrary control of attitude stabilizing mechanismA new prototype with 5 DOF

Items Prototype-1 Prototype-2Degree-of-freedom 1 5Weight 206kgf 488kgfStraight walk Max 120stepmin Max15 stepsminStartstop to walk - Sucessful

Dynamics is important not only to stabilize biped machine but also to drive actuators

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Items		Prototype-1		Prototype-2
Degree-of-freedom		1		5
Weight		206kgf		488kgf
Straight walk		Max 120stepmin		Max15 stepsmin
Startstop to walk		-		Sucessful

Several Examples of My Research Works(1)Biped Walking Machine

ndash While undergraduatePhD student (-1987)

Straight Walk of Biped Walking Machine(1982)

Knee joint

Ankle joint

Toe

Heel

Hip joint

Leg Mechanism with 1 DOF

Ankle path generator(Planar 6-bar link mech)

Foot driving mechanism(Planar 8-bar link mech)

Driving links

Planar 8-bar link mechanism

Counter weightDrawing 8 figure

Plane-cylinder joint

Slanting crank shaft

Attitude Stabilizing Mechanism

950m

m

A Biped Waking Machine with only 1 DOF (Mechanical synthesis for simple control)

Walking speed 120 stepsmin (World record）

2nd prototype which can startand stop walking

Improvement for non steady walkingChange strideIndependent drive of foot driving mechanismArbitrary control of attitude stabilizing mechanismA new prototype with 5 DOF

Items Prototype-1 Prototype-2Degree-of-freedom 1 5Weight 206kgf 488kgfStraight walk Max 120stepmin Max15 stepsminStartstop to walk - Sucessful

Dynamics is important not only to stabilize biped machine but also to drive actuators

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Items		Prototype-1		Prototype-2
Degree-of-freedom		1		5
Weight		206kgf		488kgf
Straight walk		Max 120stepmin		Max15 stepsmin
Startstop to walk		-		Sucessful

Walking speed 120 stepsmin (World record）

2nd prototype which can startand stop walking

Improvement for non steady walkingChange strideIndependent drive of foot driving mechanismArbitrary control of attitude stabilizing mechanismA new prototype with 5 DOF

Items Prototype-1 Prototype-2Degree-of-freedom 1 5Weight 206kgf 488kgfStraight walk Max 120stepmin Max15 stepsminStartstop to walk - Sucessful

Dynamics is important not only to stabilize biped machine but also to drive actuators

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Items		Prototype-1		Prototype-2
Degree-of-freedom		1		5
Weight		206kgf		488kgf
Straight walk		Max 120stepmin		Max15 stepsmin
Startstop to walk		-		Sucessful

(2)Dexterous Motion of Hyper Redundant Robot- Optimum learning control of snake-like robot

with 10DOF to avoid obstackles

Dexterity is set as an objective function togenerate optimum trajectory

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

(3)Control of Structural Flexibility and Its application

Output point

Motion control of a flexible link (coil spring)taking account of its reaction force

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

x z

y

Four legged walking machine with spring leg

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

(4)Estimation of Sound Radiation from Thin Platesubjected to Acoustic Excitation

Thin rectangular plate

Acoustic excitation

Transmitted sound power

Transmitted sound power should be estimated to designsilent structure

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Pseudo random noise Impulsive signal

Transmitted sound power can be estimatedwith an adequate accuracy

Due to mode of vibration

Input sound pressure Input sound pressure

Transmitted sound powerTransmitted sound power

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

(5)Estimation of Sound Radiation from Frog-type Guiro

The calculated acceleration and sound pressureExpanded

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

(6)Development of Micro Ciliary Actuators in GroupBase plate

Micro machine driven by many artificial micro cilium actuators

Artificial cilium motion

Piezoelectric pipemorph actuator

Yy

2θ

L do

φdi

Elastic bar(Internal electrode)

Piezoelectric ceramics

External electrode

~

Piezoelectric Pipemorph actuator

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Partial Ni plating with Watt solution bath

125μm

100μm

Fabrication of PZT film

~

Ni electrodes ~20kV

Internal electrodes

Vacuum desiccator Heater for soldering

Sn-Pb SolderNi plate

Several fabrication methods

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

Convey of a small sheet with micro cilium actuators in group

Microscope

Micro cilium actuators in group

A paper sheet 6mmtimes6mm

8μm

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

I hope you will visit my WEB-site

httpwwwdynamicsmeptitechacjpindexphp

if you are interested in other research works

Thank you

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Itrsquos easy to calculate output motion of themanipulator

2 Kinematics of planar serial link mechanisms

Forward kinematics

+++++minusminus

=

)cos()(sin)sin()(sin

10101000

10101000

θθθθθθθθθθθθ

LLLL

YX

P

P

By differentiating the displacement the output velocity can be calculated as

Linear relationship between output velocity and joint velocity

It can be deformed as

++++minus+minusminus

=

1

0

10110100

10110100

)cos()cos(cos)sin()sin(sin

θθ

θθθθθθθθθθ

LLLLLL

YX

P

P

Jacobian matrix J(θ) Joint velocityvector θ

Output velocityvector r

θθJr )(=

θθJθθJ

)()(

)]sin()()cos()([

)sincos()]cos()()sin()([

)cossin(

102

1010101

020000

102

1010101

020000

+=

++minus+++

minus+++++minus

+minus

=

　　　

　

　　　　　　　　　　

　

θθθθθθθθ

θθθθθθθθθθθθ

θθθθ

L

LL

L

YX

P

P

Output acceleration can also be calculated as

I t is easy to analyze output motion with knowledge at high school

For general planar serial nR manipulatorhellip

+

+=

sum sum

sum summinus

= =

minus

= =

0

1

0 0

0

1

0 0

sin

cos

YL

XL

YX

n

i

i

kki

n

i

i

kki

P

P

θ

θ

minus

=

sum sumsum

sum sumsumminus

= ==

minus

= ==1

0 00

1

0 00

cos

sinn

i

i

kk

i

kki

n

i

i

kk

i

kki

P

P

L

L

YX

θθ

θθ

minus

+minus=

sum sum sum sumsum

sum sum sum sumsumminus

= = = ==

minus

= = = ==1

0 0 0 0

2

0

1

0 0 0 0

2

0

]sin)(cos[

]cos)(sin[n

i

i

k

i

k

i

kkkk

i

kki

n

i

i

k

i

k

i

kkkk

i

kki

P

P

L

L

YX

θθθθ

θθθθ

Itrsquos also easy to analyze

Inverse kinematics

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

To calculate joint input motion forthe specified output motion

Can θ0θ1 be calculated as the function of XPYP

It will be a little bit complicated

By using the prementioned equation for forward kinematics

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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+++++minusminus

=

)cos()(sin)sin()(sin

10101000

10101000

θθθθθθθθθθθθ

LLLL

YX

P

P

By differentiating the displacement the output velocity can be calculated as

Linear relationship between output velocity and joint velocity

It can be deformed as

++++minus+minusminus

=

1

0

10110100

10110100

)cos()cos(cos)sin()sin(sin

θθ

θθθθθθθθθθ

LLLLLL

YX

P

P

Jacobian matrix J(θ) Joint velocityvector θ

Output velocityvector r

θθJr )(=

θθJθθJ

)()(

)]sin()()cos()([

)sincos()]cos()()sin()([

)cossin(

102

1010101

020000

102

1010101

020000

+=

++minus+++

minus+++++minus

+minus

=

　　　

　

　　　　　　　　　　

　

θθθθθθθθ

θθθθθθθθθθθθ

θθθθ

L

LL

L

YX

P

P

Output acceleration can also be calculated as

I t is easy to analyze output motion with knowledge at high school

For general planar serial nR manipulatorhellip

+

+=

sum sum

sum summinus

= =

minus

= =

0

1

0 0

0

1

0 0

sin

cos

YL

XL

YX

n

i

i

kki

n

i

i

kki

P

P

θ

θ

minus

=

sum sumsum

sum sumsumminus

= ==

minus

= ==1

0 00

1

0 00

cos

sinn

i

i

kk

i

kki

n

i

i

kk

i

kki

P

P

L

L

YX

θθ

θθ

minus

+minus=

sum sum sum sumsum

sum sum sum sumsumminus

= = = ==

minus

= = = ==1

0 0 0 0

2

0

1

0 0 0 0

2

0

]sin)(cos[

]cos)(sin[n

i

i

k

i

k

i

kkkk

i

kki

n

i

i

k

i

k

i

kkkk

i

kki

P

P

L

L

YX

θθθθ

θθθθ

Itrsquos also easy to analyze

Inverse kinematics

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

To calculate joint input motion forthe specified output motion

Can θ0θ1 be calculated as the function of XPYP

It will be a little bit complicated

By using the prementioned equation for forward kinematics

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

θθJθθJ

)()(

)]sin()()cos()([

)sincos()]cos()()sin()([

)cossin(

102

1010101

020000

102

1010101

020000

+=

++minus+++

minus+++++minus

+minus

=

　　　

　

　　　　　　　　　　

　

θθθθθθθθ

θθθθθθθθθθθθ

θθθθ

L

LL

L

YX

P

P

Output acceleration can also be calculated as

I t is easy to analyze output motion with knowledge at high school

For general planar serial nR manipulatorhellip

+

+=

sum sum

sum summinus

= =

minus

= =

0

1

0 0

0

1

0 0

sin

cos

YL

XL

YX

n

i

i

kki

n

i

i

kki

P

P

θ

θ

minus

=

sum sumsum

sum sumsumminus

= ==

minus

= ==1

0 00

1

0 00

cos

sinn

i

i

kk

i

kki

n

i

i

kk

i

kki

P

P

L

L

YX

θθ

θθ

minus

+minus=

sum sum sum sumsum

sum sum sum sumsumminus

= = = ==

minus

= = = ==1

0 0 0 0

2

0

1

0 0 0 0

2

0

]sin)(cos[

]cos)(sin[n

i

i

k

i

k

i

kkkk

i

kki

n

i

i

k

i

k

i

kkkk

i

kki

P

P

L

L

YX

θθθθ

θθθθ

Itrsquos also easy to analyze

Inverse kinematics

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

To calculate joint input motion forthe specified output motion

Can θ0θ1 be calculated as the function of XPYP

It will be a little bit complicated

By using the prementioned equation for forward kinematics

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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For general planar serial nR manipulatorhellip

+

+=

sum sum

sum summinus

= =

minus

= =

0

1

0 0

0

1

0 0

sin

cos

YL

XL

YX

n

i

i

kki

n

i

i

kki

P

P

θ

θ

minus

=

sum sumsum

sum sumsumminus

= ==

minus

= ==1

0 00

1

0 00

cos

sinn

i

i

kk

i

kki

n

i

i

kk

i

kki

P

P

L

L

YX

θθ

θθ

minus

+minus=

sum sum sum sumsum

sum sum sum sumsumminus

= = = ==

minus

= = = ==1

0 0 0 0

2

0

1

0 0 0 0

2

0

]sin)(cos[

]cos)(sin[n

i

i

k

i

k

i

kkkk

i

kki

n

i

i

k

i

k

i

kkkk

i

kki

P

P

L

L

YX

θθθθ

θθθθ

Itrsquos also easy to analyze

Inverse kinematics

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

To calculate joint input motion forthe specified output motion

Can θ0θ1 be calculated as the function of XPYP

It will be a little bit complicated

By using the prementioned equation for forward kinematics

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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Inverse kinematics

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

To calculate joint input motion forthe specified output motion

Can θ0θ1 be calculated as the function of XPYP

It will be a little bit complicated

By using the prementioned equation for forward kinematics

++++++

=

010100

010100

)sin(sin)cos(cos

YLLXLL

YX

P

P

θθθθθθ

Planar serial 2R manipulator

Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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Let draw an auxiliary line

By defining angles αβ

βαθ minus=0

where

0

01tanXXYY

P

P

minusminus

= minusα

How to derive β X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

α

β

Planar serial 2R manipulator

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1

β

Let consider PJ0J1

L

20

20 )()( YYXXL PP minus+minus=

By using side lengths L0L1 anddistance

0

21

20

2

2cos

LLLLL minus+

=β

We can apply cosine theorem

0

21

20

21

2cos

LLLLL minus+

= minusβTherefore

Planar serial 2R manipulator

0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
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0

21

20

21

0

01

0

2costan

LLLLL

XXYY

P

P minus+minus

minusminus

=

minus=

minusminus　　

βαθ

θ1 can then be calculated

X

J0 ( X0 Y0 )

θ 0

X

Y

O

J1 ( X1 Y1)

P ( XP YP )

θ 1

L0

L1 L

Let consider PJ0J1 and an angle γ

γ

θ1=πminusγ

10

221

201

2cos

LLLLL minus+

= minusγ

From cosine theorem

10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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10

221

201

1

2cos

LLLLL minus+

minus=

minus=

minusπ

γπθ

　　

Joint input velocity and acceleration can be derived as

γθ

γθ

βαθ

βαθ

minus=

minus=

minus=

minus=

1

1

0

0

γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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γγ

sin10LLLL

=

γγγγγ 2

10

2

sincossin)(

LLLLLLL

minus+

=

LYYXXL PPPP

+

=

2

22 )()(L

LYYXXLYYYXXXL PPPPPPPPPP

+minus+++=

They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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They can also be derived as

+minusminus+minusminus

++=

++++minus+minusminus

=

=minus

minus

P

P

P

P

YX

LLLLLL

LL

YX

LLLLLL

)sin(sin)cos(cos)sin()cos(

sin1

)cos()cos(cos)sin()sin(sin

)(

1010010100

101101

110

1

10110100

10110100

1

θθθθθθθθθθ

θ

θθθθθθθθθθ

rθJθ

++minus++minusminus++minus++minusminus

minus

times

+minusminus+minusminus

++=

minus= minus

1

0

1010110101000

1010110101000

1010010100

101101

110

1

)sin()()sin()(sin)cos()()cos()(cos

)sin(sin)cos(cos)sin()cos(

sin1

])([)(

θθ

θθθθθθθθθθθθθθθθθθθθ

θθθθθθθθθθ

θ

LLLLLL

YX

LLLLLL

LL

P

P　　

　

θθJrθJθ

Anyway it is easy to analyze kinematics of serial mechanisms

3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
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3 Kinematics of planar closed-loop link mechanisms

The simplestclosed-loop mechanism

Planar 4-bar link mechanism

θ1L3L 2L

)( YX出力

Input link

X

Y φ

0L

Follower link

Coupler link

ξ

η

ξ

η

J0

J1

J2

J3

P

O

Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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Crank-rocker motion

Planar 4-bar link mechanismX

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

Planar 4-bar link mechanismX

Y

X

Y

)( YXOutput coupler point

X

Y

O

Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
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• スライド番号 27
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• スライド番号 33
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• スライド番号 35
• スライド番号 36
• スライド番号 37
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Planar 4-bar link mechanism

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O

Let analyze output angular motion

)( 00 YX)( 03 YX

By eliminating θ1θ2 φ＝f(θ) may be derived

Complicated

πφθθθθ

θθθ

+=+++=

+++=

2

2312100

2312103

sinsinsincoscoscos

LLLYYLLLXX

(1)From closed-loop equation

θ1

θ2

Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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Planar 4-bar link mechanism

Let analyze output angular motion

θ1L3L 2L

)( YX

X

Yφ

0Lξ

η

ξ

η

J0

J1

J2

J3

P

O)( 00 YX)( 03 YX

By deforming equation with t=tan φ 2 and solving the equation to obtain t then φ＝f(θ) may be derived

Complicated

（2）From length of coupler link

　

　　

　　

+=+=

++

=

=minus+minus

332

332

01

01

1

1

22

221

221

sincos

sincos

)()(

YLYXLX

YLXL

YX

LYYXX

φφ

θθ

θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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θ1L

3L 2L

X

Y

φ

0L J0

J1

J2

J3

)( 11 YX

)( 22 YX

α

β

βαφ +=Let derive α and β respectively

(3)Let draw an auxiliary line

)( 00 YX)( 03 YXPlanar 4-bar link mechanism

Input link

Follower link Coupler link

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YX

θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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θ1L

J1

J1

J3

)( 11 YX

α

θθ 221

2301 sin)cos( LXXLL +minus+=

3L2L

J1

J2

J3

)( 11 YX

)( 22 YX

β

301

11

cossintan

XXLL

minus+= minus

θθα

Easy

=βLet consider J3J1J2

Cosine theorem

3

22

23

21

2cos

LLLLL minus+

= minusβ

)( 00 YX)( 03 YX

)( 03 YXθ

θ

θθφ

cos)(2)(2cos)(2)(cos

cossintan

3012

30212

30123

22

230

211

301

11

XXLXXLLXXLLLXXL

XXLL

minus+minus+

minus++minusminus++

minus+=

minus

minus

　　

Resultantly

φ can be represented as function of θ

Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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• スライド番号 33
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• スライド番号 35
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Displacement of coupler point

3L

)( YXCoupler point

X

Y φ

Follwerlink

Coupler linkξ

η

ξ

η

J1

J2

J3

ψ

++

=

01

01

1

1

sincos

YLXL

YX

θθ

++

=

03

33

2

2

)(sin)(cos

YLXL

YX

θφθφAngle ψ

3031

311

21

211

)(coscos)(sinsintan

tan

XXLLLL

XXYY

minus+minusminus

=

minusminus

=

minus

minus

θφθθφθ

ψ

　　

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

P

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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• スライド番号 39
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• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

+++minus

=

+

minus=

2

2

2

2

cossinsincos

cossinsincos

)()(

YX

YX

YX

ψηψξψηψξ

ηξ

ψψψψ

θθ

　　　　

)( YXCoupler point

X

YCoupler link

ξ

η

ξ

η

J1

J2ψ

Coordinate transformation

PResultantlyDisplacement of coupler point can be derived

HoweverhellipPeculiar to a mechanismIt will be complicated to analyze velocity and

acceleration

ldquoGeneral and systematic method is strongly requiredrdquo

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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４ Systematic kinematics analysis method

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

sCrank

Two adjacent linkswith revolute pairs

Two adjacent links withprismatic and revolute pairs

Coupler point

Crank input-Slider output planar 6-bar link mechanism

A mechanism is divided into several units and each of them will be analyzed with systematic calculation

Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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Feature of the systematic kinematics analysis(1)It is easy to understand because motion of kinematic

pair can be calculated in order(2)Applicable to various planar mechanisms with

revolute and prismatic pairs(3)Velocity and acceleration can be calculated(4)Fundamentals to analyze dynamics of mechanism

ldquoGeneral systematic and available methodrdquo

rarrCalculation program can be offered

41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
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41 Derivation of systematic kinematic analysis (1)Motion of crank linkcrank_input

++++

=

0

0

1

1

)sin()cos(

YLXL

YX

φθφθ

To consider the postureof front link

++++++minus

=

0

0

1

1

)cos()()sin()(

YLXL

YX

φθφθφθφθ

+++minus++++++++minus

=

02

02

1

1

)]sin()()cos()[()]cos()()sin()[(

YLXL

YX

φθφθφθφθφθφθφθφθ

Differentiating

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φ

θL

Crank link

Input parameters

Output parameters

Mechanism constants

Position and posture of pair J0 rarrCrank angle rarrPosition of pair J1

（X0Y0）φ θ （X1Y1）

From a pair to a pairVelocity and acceleration

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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• スライド番号 27
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• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

(2)Motion of input sliderslider_input

++

=

L

L

S

S

YsXs

YX

ψψ

sincos

+++minus

=

L

L

S

S

YssXss

YX

ψψψψψψ

cossinsincos

+minus+++minusminusminus

=

L

L

S

S

YssssXssss

YX

ψψψψψψψψψψψψψψ

sincoscos2sincossinsin2cos

2

2

+minus+minusminus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψψψψψ

sincoscossin

++minusminus+minusminus+minus

=

SCC

SCC

C

C

YyxXyx

YX

)cossin()sincos()sincos()cossin(

22

22

ψψψψψψψψψψψψψψψψ

Position and posture of slider axisrarrTranslational input rarrPosition of pair JC

（XLYL）ψ s （XCYC）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

Input slider link

Posture of slider axis

+++minus

=

SCC

SCC

C

C

YyxXyx

YX

ψψψψ

cossinsincos Position of pair JC

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

φL

(3)Angular motion of linklink_angle

XY

∆∆1tanminus=φ

2LXYXY

∆∆∆∆ minus=φ

2LXYXY

∆∆∆∆ minus=φ

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Posture angleIs represented with positionof pairs

Position of pairs at both ends rarr Posture of link（X0 Y0） （X1 Y1） φ

Angular motion of link

O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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O X

Y

X

J1( X1 Y1 )

J0 ( X0 Y0 )

αL

(4)Angular motion of segmentsegment_angle

Length varies

Segment connecting two points

XY

∆∆1tanminus=α Let consider

change of distance

αα 22 cos

XXYXY

∆∆∆∆∆

minus

=

3cossin)(2

cos)](2)([

XXXYXY

XYXYXXYXYX

∆sdot

∆∆∆minus∆∆minus

∆∆minus∆∆∆minus

∆∆minus∆∆∆=

α

αα

α

α

　　　　

　　　

　　　

010101

010101

YYYYYYYYYXXXXXXXXX

minus=∆minus=∆minus=∆

minus=∆minus=∆minus=∆where

Position of two points rarr Posture of segment （X0 Y0） （X1 Y1） α

O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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O X

Y

X

J1 ( X1 Y1 )

J0 ( X0 Y0 )

JC ( XC YC ) xy

yCxCφ

(5)Motion of coupler pointcoupler_point

Point on a link

+++minus

=

0

0

cossinsincos

YyxXyx

YX

CC

CC

C

C

φφφφ

+minus++minus

=

0

0

)sincos()cossin(

YyxXyx

YX

CC

CC

C

C

φφφφφφ

++minusminus+minusminus+minus

=

02

02

)cossin()sincos()sincos()cossin(

YyxyxXyxyx

YX

CCCC

CCCC

C

C

φφφφφφφφφφφφ

φ can be calculated with link_angle

Position of pairs J0J1 rarr Position of pair JC（X0Y0） （X1Y1 ） （XCYC）

Coordinate transformation due to rotation

(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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(6)Motion of two adjacent links with only revolute pairsRRR_links

++

=

000

000

sincos

YLXL

YX

B

B

φφ

Two adjacent links with only revolute pair

By giving posture angle φ0 position of pair JB can be calculated

J1( X1 Y1 )

O X

Y

XJ0 ( X0 Y0 )

L

αβ

β

L0

L1

L0

L1

JB ( XB YB )

JBrsquo

Φ0

φ0

+minus++minus

=

002

0000

002

0000

)sincos()cossin(

YLXL

YX

B

B

φφφφφφφφ

βαφβαφβαφ plusmn=plusmn=plusmn= 000 　　where

++minus

=

0000

0000

cossin

YLXL

YX

B

B

φφφφ

Calculated withsegment angle

22 YXL ∆∆ +=

From cosine theorem

LLLLL

0

21

2201

2cos minus+

= minusββ

βsin2

)(2

0

21

220

LLLLLL

minusminus=

)sin2(cos)(

sin]2)2)([(23

12

122

0

22221

220

βββ

ββ

LLLLLLL

LLLLLLLL

minusminusminus

minusminusminusminus=

　　

By differentiating

Sign corresponds to mechanical inversion

Position of pairs J0J1rarrPosition of pair JB（X0Y0） （X1Y1 ） （XBYB）

O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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O X

Y

XPL ( XL YL )

ψ

s

JC ( XC YC )

x

y

yC

xC

JR ( XR YR )L

Two adjacent links with prismatic and revolute pairs

P Displacement of sliderDxBAs C plusmnminusminusminus= ψψ sincos

where

RLC YYyB minus+= ψcosRLC XXyA minus+minus= ψsin

22 )cossin( ψψ BALD minusminus=where

DDBBAAs

2cossinsincos

plusmnminusminus+minus= ψψψψψψ

DDDDDBBBB

AAAAs

42sincoscos2sin

cossinsin2cos2

2

2

minusplusmn+minusminusminus

+++minus=

ψψψψψψψ

ψψψψψψψ

　　

)cossin)(sincoscossin(2 ψψψψψψψψ BABBAAD minus+minus+minus=

RLC XXyA minus+minus= ψψ cos RLC YYyB minus+minus= ψψ sin hellipVelocity and acceleration can be calculated with slider_input

Position and posture ofslider axis position of pair JR（XLYL）ψ（XRYR）

Position of pair JCDisplacement of slider

（XCYC）s

++++minus+

=

LCC

LCC

C

C

YyxsXyxs

YX

ψψψψ

cossin)(sincos)(

222 )()( LYYXX RCRC =minus+minus

Pair JC is located at the intersectionbetween straight line P and a circle with radius L

(7)Motion of two adjacent links with prismatic and revolute pairsPRR_links

C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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C language program of systematic kinematics analysis

------------------------------------------------------------ Kinematics Calculation lt mechcpp gt 2002 814 Created by D Matsuuura 20031013 Improved by D Matusuura 2008 112 Improved by D Matsuuura 2008 114 Improved by N Iwatsuki 2010 821 Improved by N Iwatsuki 20111010 Improved by N Iwatsuki 2018 415 Improved by N Iwatsuki------------------------------------------------------------

include ltstdafxhgtinclude ltstdiohgtinclude ltmathhgtinclude mechhinclude matrixh

----------------------------------------------------------- Input slider motion pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide s Displacement velocity and acceleration of an input slider gsi eta Moving coordinate of a pair on a slider jp Displacement velocity and acceleration of a slider jc Displacement velocity and acceleration of a pair on a slider-----------------------------------------------------------

void crank_input( POINT j1 double l ANGLE theta ANGLE phi POINT j2 )

double thephi dthephi ddthephidouble cs

thephi = thetathe + phithedthephi = thetadthe + phidtheddthephi = thetaddthe + phiddthe

c = cos( thephi )s = sin( thephi )

j2-gtPx = j1Px + lcj2-gtDPx = -lsdthephij2-gtDDPx = -l( sddthephi + cdthephidthephi )

j2-gtPy = j1Py + lsj2-gtDPy = lcdthephij2-gtDDPy = l( cddthephi - sdthephidthephi )

----------------------------------------------------------- Input crank motion j1 Displacement velocity and acceleration of root pair l Link length theta Angular displacement velocity and acceleration of crank phi Angular displacement velocity and acceleration of root link j2 Displacement velocity and acceleration of tip pair-----------------------------------------------------------

void slider_input( POINT pg double l ANGLE psig VARIABLE sdouble gsi double eta POINT jp POINT jc )

double cp sp

cp = cos( psigthe )sp = sin( psigthe )

jp-gtPx = svcp + pgPxjp-gtDPx = sdvcp - svsppsigdthe + pgDPxjp-gtDDPx = sddvcp - 20sdvsppsigdthe - svsppsigddthe

- svcppsigdthepsigdthe + pgDDPx

jp-gtPy = svsp + pgPyjp-gtDPy = sdvsp + svcppsigdthe + pgDPyjp-gtDDPy = sddvsp + 20sdvcppsigdthe + svcppsigddthe

- svsppsigdthepsigdthe + pgDDPy

jc-gtPx = gsicp - etasp + jp-gtPxjc-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjc-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )

- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jc-gtPy = gsisp + etacp + jp-gtPyjc-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPyjc-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe )

- eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

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----------------------------------------------------------- Angular motion of a segment connecting two points j1 j2 Displacement velocity and acceleration of each point angle Angular displacement velocity and acceleration of a segment connecting two points Function value Completeincomplete(Two points are quite same）-----------------------------------------------------------

int segment_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble c sdouble absx absydouble q1 q2double r dr

get_diff( j1 j2 ampdiff )absx = fabs( diffPx )absy = fabs( diffPy )if( absx lt 10e-20 ampamp absy lt 10e-20 )

angle-gtthe = 00angle-gtdthe = 00angle-gtddthe = 00printf(ldquo++ Error in segment_angle yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )

c = cos( angle-gtthe )s = sin( angle-gtthe )

if( absy gt 10e-20 ampamp absxabsy lt 10e-10 )r = sqrt( diffPxdiffPx + diffPydiffPy )dr = ( diffDPxdiffPx + diffDPydiffPy )rif( s gt= 00 )

angle-gtdthe = -diffDPxrangle-gtddthe = ( -diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

angle-gtdthe = diffDPxrangle-gtddthe = ( diffDDPx - 20drangle-gtdthe )rreturn SUCCESS

else

q1 = diffDPydiffPx - diffPydiffDPxq2 = diffDDPydiffPx - diffPydiffDDPxangle-gtdthe = q1cc(diffPxdiffPx)angle-gtddthe = ( (diffPxq2 - 20diffDPxq1)c - 20diffPxangle-gtdtheq1s)c( diffPxdiffPxdiffPx )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links Connecting Revolute Pairs j1 j2 Displacement velocity and acceleration of pairs at both ends l1 l2 Link lengths minv Indicator for mechanical inversion jb Displacement velocity and acceleration of a pair which connects two links Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int RRR_links( POINT j1 POINT j2 double l1 double l2 int minv POINT jb )

POINT diffdouble q2double q dq ddq

double cb sb ssANGLE angledouble phi dphi ddphidouble bet dbet ddbetdouble cp spint ier

get_diff( j1 j2 ampdiff )q2 = diffPxdiffPx + diffPydiffPy

if( q2 lt= 10e-30 )printf(++ Error in RRR_links (２対偶が一致してます) yenn)return ERROR

q = sqrt( q2 )dq = ( diffDPxdiffPx + diffDPydiffPy )qddq = ( ( diffDDPxdiffPx + diffDPxdiffDPx

+ diffDDPydiffPy + diffDPydiffDPy )q- ( diffDPxdiffPx + diffDPydiffPy )dq )q2

cb = ( l1l1 + q2 - l2l2 )( 20l1q )if( fabs( cb ) gt 10 )

printf(ldquo++ Error in RRR_links (連鎖が構成できません)yennrdquo)return ERROR

sb = sqrt( 10 - cbcb )ss = l1l1 - l2l2 - q2

ier = segment_angle( j1 j2 ampangle )if( ier = SUCCESS )

printf(ldquo++ Error in RRR_links (２対偶が一致してます) yennrdquo)return ERROR

bet = acos( cb )

dbet = ssdq( 20l1q2sb )

C language program of systematic kinematics analysis

ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
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ddbet = ( ( ss( ddqq - 20dqdq ) - 20( dqq )( dqq) )sb - ssdqqdbetcb )( 20l1q2qsbsb )

if( minv gt 0 ) phi = anglethe + betdphi = angledthe + dbetddphi = angleddthe + ddbet

else

phi = anglethe - betdphi = angledthe - dbetddphi = angleddthe - ddbet

cp = cos( phi )sp = sin( phi )

jb-gtPx = j1Px + l1cpjb-gtPy = j1Py + l1spjb-gtDPx = j1DPx - l1dphispjb-gtDPy = j1DPy + l1dphicpjb-gtDDPx = j1DDPx - l1( ddphisp + dphidphicp )jb-gtDDPy = j1DDPy + l1( ddphicp - dphidphisp )

return SUCCESS

----------------------------------------------------------- Motion of Adjacent Two links with a Prismatic Pair and Two Revolute Pairs jr Displacement velocity and acceleration of revolute pair at end l Link legth between two revolute pairs pg Displacement velocity and acceleration of a point on a linear guide to describe position of the linear guide psig Angular displacement velocity and acceleration of linear guide gsi eta Moving coordinate of a pair on a slider minv Indicator for mechanical inversion s Displacement velocity and acceleration of a slider on moving coordinate system jp Displacement velocity and acceleration of a slider on fixed coordinate system jb Displacement velocity and acceleration of a revolute pair connecting two linksr Function value Completeincomplete(Link chain cannot be formed)-----------------------------------------------------------

int PRR_links( POINT jr double l POINT pg ANGLE psig double gsi double etaint minv VARIABLE s POINT jp POINT jb )

double cp spdouble a da dda b db ddb d dd ddd q dq ddq s0 ds0 dds0

cp = cos( psigthe )sp = sin( psigthe )

a = -etasp + pgPx - jrPxb = etacp + pgPy - jrPy

d = ll - ( asp - bcp )( asp - bcp )if( dlt00 ) return ERROR

da = -etacppsigdthe + pgDPx - jrDPxdda = -eta( cppsigddthe - sppsigdthepsigdthe ) + pgDDPx - jrDDPx

db = -etasppsigdthe + pgDPy - jrDPyddb = -eta( sppsigddthe + cppsigdthepsigdthe ) + pgDDPy - jrDDPy

dd = -20( dasp + acppsigdthe - dbcp + bsppsigdthe )( asp - bcp )ddd = -20( ( ddasp + 20dacppsigdthe + acppsigddthe

- asppsigdthepsigdthe- ddbcp + 20dbsppsigdthe + bsppsigddthe+ bcppsigdthepsigdthe )( asp - bcp ) + ( dasp + acppsigdthe - dbcp + bsppsigdthe )

( dasp + acppsigdthe - dbcp + bsppsigdthe ) )

q = sqrt( d )dq = dd( 20q )ddq = ( 20dddd - dddd )( 40dq )

s0 = -acp - bsp - gsids0 = -dacp + asppsigdthe - dbsp - bcppsigdtheC language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
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• スライド番号 19
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• スライド番号 21
• スライド番号 22
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• スライド番号 30
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• スライド番号 33
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• スライド番号 35
• スライド番号 36
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• スライド番号 46
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• スライド番号 49
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• スライド番号 51
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• スライド番号 53
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• スライド番号 56
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• スライド番号 60
• スライド番号 61
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dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- 節の角変位（節上の２対偶間を結ぶ線分の成す角） j1 j2 位置ベクトルとその微分 angle 線分の成す角 関数値 0 正常終了 1 異常終了（２点が一致）-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
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• スライド番号 49
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• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
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dds0 = -ddacp + 20dasppsigdthe + asppsigddthe+ acppsigdthepsigdthe - ddbsp - 20dbcppsigdthe - bcppsigddthe + bsppsigdthepsigdthe

if( minvgt0 ) s-gtv = s0 + qs-gtdv = ds0 + dqs-gtddv = dds0 + ddq

else

s-gtv = s0 - qs-gtdv = ds0 - dqs-gtddv = dds0 - ddq

jp-gtPx = s-gtvcp + pgPxjp-gtPy = s-gtvsp + pgPy

jp-gtDPx = s-gtdvcp - s-gtvsppsigdthe + pgDPxjp-gtDPy = s-gtdvsp + s-gtvcppsigdthe + pgDPy

jp-gtDDPx = s-gtddvcp - 20s-gtdvsppsigdthe - s-gtvsppsigddthe - s-gtvcppsigdthepsigdthe + pgDDPx

ｊp-gtDDPy = s-gtddvsp + 20s-gtdvcppsigdthe + s-gtvcppsigddthe - s-gtvsppsigdthepsigdthe + pgDDPy

jb-gtPx = gsicp - etasp + jp-gtPxjb-gtPy = gsisp + etacp + jp-gtPy

jb-gtDPx = -gsisppsigdthe - etacppsigdthe + jp-gtDPxjb-gtDPy = gsicppsigdthe - etasppsigdthe + jp-gtDPy

jb-gtDDPx = -gsi( sppsigddthe + cppsigdthepsigdthe )- eta( cppsigddthe - sppsigdthepsigdthe ) + jp-gtDDPx

jb-gtDDPy = gsi( cppsigddthe - sppsigdthepsigdthe ) - eta( sppsigddthe + cppsigdthepsigdthe ) + jp-gtDDPy

return SUCCESS

----------------------------------------------------------- Angular Motion of Link j1 j2 Displacement velocity and acceleration of two points on a link angle Angular displacement velocity and acceleration of link Function value Completeincomplete(Two points are quite same)-----------------------------------------------------------

int link_angle( POINT j1 POINT j2 ANGLE angle )

POINT diffdouble q

get_diff( j1 j2 ampdiff )

q = diffPxdiffPx + diffPydiffPyif( q lt= 10e-30 )

printf(ldquo++ Error in link_angle （２対偶が一致してます）yennrdquo)return ERROR

angle-gtthe = atan2( diffPy diffPx )angle-gtdthe = ( diffDPydiffPx - diffPydiffDPx )qangle-gtddthe = ( diffDDPydiffPx - diffPydiffDDPx )q

return SUCCESS

C language program of systematic kinematics analysis

You can download these programs fromWEB-sitehttpwwwdynamicsmeptitechacjpjapanesedownloadhtml

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
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• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
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• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

(1)Give position of fixed pairsGive position of fixed revolute pairs and position and posture of fixed prismatic pairs

(3)Calculate motion of moving pair oninput links

By using coupler_point motion of other moving pairs on input links can be calculated

(2)Calculate motion of input linksBy using crank_input or slider_input motion of input link can be calculated

(4)Calculate motion of two adjacent linksBy using RRR_links or PRR_links motion of two adjacent links with revoluteprismatic pair can becalculated

(5)Calculate motion of pairs on coupler linksBy using coupler_point motion of moving pairs on coupler link can be calculated(6)Repeat (4) and (5) and then calculate all moving pairs

END

Ex Crank input-Slider outputplanar 6-bar link mechanism

ldquoIt is important to find two adjacentlinks to be analyzedrdquo

42 Procedure of the systematic kinematics analysis

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Examples of analysis(1)Crank input-Slider output planar 6-bar link mechanism

(0000)

(0000)

(0800)

03

0805

(0302)(0002) 08

135deg

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

X

Y

O

J1J6

J2

J0 ( X0 0 )J3( X3 0 )

J4 ( X4 Y4 )

J5 ( 0 yC5 )

y2

x2

xC4

yC4

θψ

L0

L1L2

L3

s

Coupler point coupler_point

Two adjacent linkswith only revolute pairsRRR_links

Crank input linkcrank_input

Two adjacent links with prismaticrevolute pairsPRR_links

1)Crank input-Slider output planar 6-bar link mechanismProcedure hellip

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

Locus of J4 of the 6-bar mechanism

Y 4m

X4 m

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

Displacement velocity and acceleration of output slider of the 6-bar mechanism

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

(2)A planar closed-loop manipulator with 2 DOF

Forward kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

Output motion

J1

J0J4

J3

J2 Crank input linkcrank_input

Two adjacent links with only revolute pairsRRR_links

Moving point on coupler linkcoupler_point

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

Inverse kinematics

θ1

r

X

Y

O

θ2 L0L3

L2

L1

J1

J0J4

J3

J2 Two adjacent links with only revolute pairsRRR_links

Moving point on coupler link

coupler_pointTwo adjacent links with only revolute pairsRRR_links

All motion can easilybe calculated

Input angle link_angle

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

X

X

X

X

X

Y

O

xy

φ

θ0

θ1

θ2

L0L1

L0

L1

L0

L1

J0

J1

J2

J3

J4

J5J6

J7

J8

P

By specifying the position and motion of output platform

0606 crank_input

coupｌer_pointRRR_links

link_angle

Angular displacement of input link can be calculated

(3)Inverse kinematics of a planar closed-loop manipulator with 3 DOF

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

O X

Y

X

( X Y )

θ1

L1

θ2L2

M

M

(X0Y0)

(4)Planar serial 2R manipulator

J0

J1

J2

Serially connected twocrank input linkcrank_input

Only crank_input

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

XJ0 (X0Y0=(18 03)

Y

O

J3

J4

J1(X1Y1)=(10 02 )

J5 ( 02 04)

J6

η0

θ

φ

04

07

J2(X2Y2)=(05 03)

ξ0

0710

J7 ( 04 05)

ξ0

η1

J4

10

L0

L1

L2L3

L4

ξ1

08

L5

(4)Planar 6-bar link mechanism type-E

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

XJ0 (X0Y0=(12 00)

Y

O

J3

J4

J1(X1Y1)=(03 02 )

J5 ( 06 06)J6

η0

θ

φ

05

07

J2(X2Y2)=(00 18)

ξ0

04

10

J7 ( 04 -07)

ξ1

η1

J4

10

L0

L1L2

L3

L4

(5)Planar 6-bar link mechanism type-B

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

XJ0 (X0Y0)=（05 00)

Y

O

J2

J4

J5 ( 00 10)

J6 η0

θ

φ

03

04

05

ξ0

J7 ( 10 10)

ξ1

η1

J3

05

η2

ξ2

J4 (00 05)

14

L0

L1L2

J1 (X1Y1)=（00 00)

L4

L3

(6)Planar 6-bar link mechanism type-F

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

(7)Planar closed-loop leg-mechanism with 1 DOF

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

2 legs with only 1 DOF

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
• スライド番号 47
• スライド番号 48
• スライド番号 49
• スライド番号 50
• スライド番号 51
• スライド番号 52
• スライド番号 53
• スライド番号 54
• スライド番号 55
• スライド番号 56
• スライド番号 57
• スライド番号 58
• スライド番号 59
• スライド番号 60
• スライド番号 61
• スライド番号 62
• スライド番号 63
• スライド番号 64
• スライド番号 65
• スライド番号 66
• スライド番号 67
• スライド番号 68
• スライド番号 69
• スライド番号 70
• スライド番号 71
• スライド番号 72
• スライド番号 73
• スライド番号 74
• スライド番号 75
• スライド番号 76
• スライド番号 77
• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

(8)Planar closed-loop leg-mechanism with 1 DOFldquoTheo Jansenrsquos mechanismrdquo

Planar 8-bar mechanism with only 1 DOF

2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
• スライド番号 7
• スライド番号 8
• スライド番号 9
• スライド番号 10
• スライド番号 11
• スライド番号 12
• スライド番号 13
• スライド番号 14
• スライド番号 15
• スライド番号 16
• スライド番号 17
• スライド番号 18
• スライド番号 19
• スライド番号 20
• スライド番号 21
• スライド番号 22
• スライド番号 23
• スライド番号 24
• スライド番号 25
• スライド番号 26
• スライド番号 27
• スライド番号 28
• スライド番号 29
• スライド番号 30
• スライド番号 31
• スライド番号 32
• スライド番号 33
• スライド番号 34
• スライド番号 35
• スライド番号 36
• スライド番号 37
• スライド番号 38
• スライド番号 39
• スライド番号 40
• スライド番号 41
• スライド番号 42
• スライド番号 43
• スライド番号 44
• スライド番号 45
• スライド番号 46
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2 legs with only 1 DOF

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
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• スライド番号 80
• スライド番号 81
• スライド番号 82

A quadruped walking machineswith only 1 DOF

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
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• スライド番号 80
• スライド番号 81
• スライド番号 82

Slippage between 2 legsdarr

It can walk on sand beach

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
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• スライド番号 78
• スライド番号 79
• スライド番号 80
• スライド番号 81
• スライド番号 82

5 Concluding remarksAs an available tool to analyze or design link mechanisms is proposed formulated and coded(1)Various planar mechanisms can be systematically

analyzed(2)Not only displacement but also velocity and

acceleration can be calculated and will be applied to dynamics calculation

(3)Students are expected to experience to analyze some mechanisms with the offered programs

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
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• スライド番号 80
• スライド番号 81
• スライド番号 82

Homework 1Calculate time history of output displacementvelocity and acceleration of some planar linkmechanism by using the offered program orself-coded program It is OK to use the offered program as is

The result will be summarized in A4 size PDFwith less than 5 pages and sent to Prof Iwatsukivia OCW-i by July 6 2020

Of course students should describe comments for the calculated results

• スライド番号 1
• スライド番号 2
• スライド番号 3
• スライド番号 4
• スライド番号 5
• スライド番号 6
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