3 degree of freedom manipulator

7/28/2019 3 Degree of Freedom Manipulator

1/5

3 Degree of Freedom Manipulator

S G Krishnan and R Harilal

May 2, 2013

Position Analysis

Using (x,y ,) as inputwe first calculate the positions of A,B,C.

xCi = x lcicos(i)yCi = y lcisin(i)

1 = + (/6)2 = + (/2)3 = + (5/6)

D = (xC xA)2 + (yC yA)2

= cos1((lA)

2+D2(lB)2)

2lAD)

= tan1 yCyAxCxA

=

In the same way we can calculate the orientation of all the three legs.

1


2/5

Velocity Analysis

For velocity analysis we have inputs as (x,y,, x, y, ). We obtain 1, 2, 3 by position analysis.

J X =

J is the jacobian. J is obtained by differentiating the contraints of the legs.

(xBi xCi)2

+ (yBi yCi)2

= (lbi)2

xC = x lccos1xB = xA lacos1yC = y lcsin1yB = yA lasin1

Differentiating with respect to time

2(xB xC)( xB xC) + 2(yB yC)( yB yC) = 0

On simplifying

1 la[(y yA)cos1 (x xA)sin1 + lccos(1 1)] =(x xA lccos1 lacos1) x + (y yA lcsin1 lasin1) y lc[(y yA)cos1 (x xA)sin1 + lasin(1 1)] 1

J can be represented as

p1/s1 q1/s1 r1/s1p2/s2 q2/s2 r2/s2p3/s3 q3/s3 r3/s3

pi = (x xAi lcicosi laicosi)qi = (y yAi lcisini laisini)ri = lci[(y yAi)cosi (x xAi)sini + laisin(i i)]si = lai[(y yAi)cosi (x xAi)sini + lcicos(i i)]

Thus given X, we can find

Acceleration Analysis

Accelerations are obtained by differentiating the velocity equation

J

X = on differentiation

JX+ JX = X = J1( JX)J can be represented as

P1 Q1 R1P2 Q2 R3P3 Q3 R3

2


3/5

Pi =si pipisi

s2iQi =

siqiqisis2i

Ri =siririsi

s2i

pi = x + laisini + lbisiniqi = y laicosi lbicosiri = lci[(x xa)cosi + (y ya)sini] + lc[xsini ycosi] + lalc(i i)cos(i i)si = lai[(x xa)cosi + (y ya)sini] la[xsini ycosi] + lalc(i i)cos(i i)

Hence acceleration inversion is completed

Force and Torque Analysis

Force and torque analysis is done to determine the amount of torque needed by the motor tobring about the required motion.

Free Body Diagrams

Platform

Link

3


4/5

Crank

Force equation for platform

Fx1 + Fx2 + Fx3 = MPlatformx

Fy1 + Fy2 + Fy3 = MPlatformy

(Fx1sin1 Fy1cos1)lc1+ (Fx2sin2 Fy2cos2)lc2+ (Fx3sin3 Fy3cos3)lc3 = Iplatform

Iplatform =Mplatforml

2

ci

4

xci = x lcicosixci = x + lcisinixci = x + lcisini + lcicosi

2

yci = y lcisiniyci = y lcicosi

yci = y lcicosi + lcisini2

xBi = laicosixBi = laisiniixBi = laisinii laicosii2

yBi = laisiniyBi = laicosii

yBi = laicosi

i laisini

i

2

= tan1[ yciybixcixbi

]

xcom =xBi+ xCi

2

ycom =yBi+ yCi

2

com =2( xCi xBi)sin+2( yci yBi)cos

lbi

4


5/5

Force Equations of a link

Fx1 Fx4 = Mlinkxcom

Fy1 Fy4 = Mlinkycom

(Fy1cos Fx1sin+ Fy4cos Fx4sin) lbi2 = Ilinkcom

Ilink =Mlinkl

2

bi

12

Solving for Fx1 to Fx6 , Fy1 to Fy6

Torque of the motor is obtained by

1 = (Fx4sin1 Fy4cos1)la1

Similarly we can calculate torque for the other two motors.

5

3 degree of freedom manipulator

Documents