162a 2 - position and displacement

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162A. Position andDisplacement in

MechanismsChapter 2: Position &

Displacement



Chapter 2

Position and Displacement



Notation Consider a fixed reference

coordinate system Oxyz or simplyOxy if we have a 2-d referencesystem. O is the origin fromwhich positions are measured

The position of a point P in thecoordinate s stem is re resented

as RPO which is a vectorverbalized as “R P to O”.Remember: tip-to-tail. This RPOvector is sometimes simply writtenas RP, when it is w.r.t. the origin.

Vector RP is written with R in bold(or with a line above or below theletter) to connote it is a vector.Without bold (or a line), it

connotes just the magnitude



Terms Absolute Position of a point

is its position with respect to

the origin of the coordinatesystem Relative Position between

two points P and Q can beexpressed as position

two pointsRPQ = RPO - RQO

Apparent Position of apoint P is the position of thepoint P relative to theobserver

Absolute and RelativePositions are related as:

RPO = RQO + RPQ



We’ll employ two methods forposition analysis

Graphical Method

Intuitive

Inexact Complex Algebra Method

Analytic

Exact



Before that….The Loop Closure Equation

Loop Closure Equation: A vector equationrepresenting all or part of mechanismformulated accordin to the fact: The

vector sum of a closed polygon equalszero.

That is, the sum of vectors to close theloop is 0, or,



Loop-Closure Equation

Applying the loop closure equation to thehand-operated clamp

R + R + R + R = 0



The above loop closure equationcan be derived…

y

X

RB = RA + RBA

RC = RB + RCB = RA + RBA + RCB

RD = RC + RDC = RA + RBA + RCB + RDC

RA = RD + RAD = RA + RBA + RCB + RDC + RAD

RA

O

RBA + RCB + RDC + RAD = 0



Graphical Method for PositionAnalysis

Suited for solving planar motion problems Points on the mechanism move in a single or parallel

planes

Loo Closure vector e uation can be broken down into 2

scalar equations:C = A + B => Cx = Ax + Bx and Cy = Ay + By

Can solve for 2 unknowns

Note: Vectors in bold



4 Cases for 2 unknowns in solvingplanar position equations

Case 1: Magnitude and direction of the

Case 2: Magnitude of one and direction ofthe other vector are unknown

Case 3: Magnitudes of 2 vectors unknown

Case 4: Directions of 2 vectors unknown



Graphical Method Steps

Identify the loop closure equation Choose a coordinate system and scale factor Draw the known vectors

not the magnitude

Draw a circular arc representing the magnitude ifmagnitude is known but not the direction

There could be more than one solutions…a longline can intersect an arc in 2 places or 2 arcscan intersects in 2 places

Measure and report the unknowns



Example: The Slider-CrankMechanism

Given: Lengths oflinks 2 and 3 andposition of slider B

determine thepositions of alllinks

Unknowns: θ2 andθ3

(Slider crank in IC engines versus air compressors)



Example: The Slider-CrankMechanism (contd.)

Identify Loop ClosureEquation:

RB = RA + RBA

o a ona pos onConvention: Draw a line at the tail of the

vector going towards right

The rotational position willbe the angle starting fromthe line to the link in theCCW direction



Example: The Slider-CrankMechanism (contd.)

Draw vector RB

With O as center, draw arcequal in length to link 2

t as center, raw arcequal in length to link 3

Two points where the 2

arcs intersect Measure and report 2 sets

of solutions for θ2 and θ3



Another Example: 4-BarMechanism Given: the lengths of

bars, crank angle, θ2

and a coupler point

link 3

Goal: To performpositional analysisand determine θ3 andθ4



Example: 4-Bar Mechanism(contd.) 1: Identify Loop Closure

Equation: RA + RBA = RC +RBC

2: Draw vectors RC and RA

3: With A as center, draw

distance between points Aand B. With C as center,draw an arc equal in lengthto link 4

4: Measure off the 2

unknowns for the (two)points of intersection ofarcs

Draw point P and measureoff RP and θθθθP



In-class exercise

Given currentposition with link2 vertical (θ2 =

Reposition link2 to 30 degreesrotated CCW

Find movementof link 6



Approach:

Don’t forget to check the mobility first



Complex Algebra Method Suitable for Planar problems

Position of a point A can be represented as below:



Position vector of a point A can berepresented as a complex number



Properties of vector “e jθ”

e jθis a unit vector

jθ

counterclockwise by 90 degrees

Proofs:



Complex Algebra Method

Identify a triangular loop for loop closure If one is not readily available, improvise a



Sliding Block Mechanism

Given O4O2 = 9”,O2A=4.5”, andθ =135o

Find θ4 and O4A



Sliding Block Mechanism (contd.) Identify the loop closure equation:

RA = RO2 + RAO2

RA = RA θ4 = RA cos θ4 + j RA sin θ4

O2 = 90o

= coso

+ s no

= + RAO2 = RAO2 135o = RAO2 cos 135o + jRAO2

sin135o = -3.182 + j3.182”

RA = (0 + j9) + (-3.182 + j3.182)

RA = [(-3.182)2 + 12.1822] = 12.59”

θ4 = tan-1(12.182/-3.182) = 104.64o



Four Bar Linkage

Note: Remember to use a triangular loop closure

How do you get a triangular loop in a 4-bar?

Draw a diagonal “S”



Four Bar Linkage (contd.)



Four Bar Linkage (contd.)

Using Cosine and/or Sine Laws, you can set 2 equations and solvefor the 2 unknowns:



Position analysis of any arbitrary pointusing Complex Algebra method

Once the position analysis of the linkage is,

any arbitrary point on any linkage



Absolute Positions of A, B and Cwill be:



Displacement

Displacement of a Point P when ittraverses a path:

∆∆∆∆RP = RP’ – RP

Displacement =new absolute position minus old

absolute position



Euler Theorem for Rigid BodyMotion

Any displacement of a rigid body isequivalent to the sum of a net translationof one oint and a net rotation of the bod

about that point



In-class Exercise

162a 2 - position and displacement

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