kinematic position analysis

8/13/2019 Kinematic position Analysis

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Chapter 4Position Analysis

Theory is the distilled essence of practice


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4.0 Introduction

Kinematic AnalysisDetermine the accelaration of all the moving parts in the

assembly.

Calculates Stresses of componentsStatic or Dynamic Forces on the parts

AccelerationsPosition / Velocity / Acceleration


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Graphical approach None of the information obtained graphically for the

first position will be applicable to the second position. It is useful fro checking the analytical results.

Derive the general equations of motion Solve analytical expressions Once the analytical solution is derived for a particular

mechanism, it6 can be quickly solved (with acomputer) for all positons.


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Graphical Position Analysis Is more simple then the algebraic approach

Graphical Velocity and Acceleration analysis Becomes quite complex and difficult then the

algebraic approachGraphical analysis is a tedious exercise and was the only

practical method available in the day B.C.(Before

Computer) , not so long ago.


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4.1 Coordinate Systems Global (Absolute) Coordinate System

Local Coordinate System


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Inertial reference frame : A system which itself has no acceleration

Local Coordinate system: Typically attached to a link at some point of interest.

This might be a pin joint, a center of gravity, or a lineof centers of a link. These local coordinate systemmay be either rotating or non-rotating as we desire.


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4.2 Position and Displacement Position Vector : The position of a point in the plane

can be defined by the use of a position vector as abovefigure.

Polar coordinated / Cartesian coordinate

A position vector Polar form : a magnitude and angle of vector Cartesian form : X and Y components of the vector


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=

+=

X

Y

Y X A

R

R R R R

arctan

22

R AR X

R Y


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cossin

sincos

y xY

y x X

R R R

R R R

+=

=

In matrix form

=

y

x

Y R R

R

R

cossin

sincosX


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Displacement

Displacement of a point

Is the change in its position and can be defined as thestraight line between the initial and final position of apoint which has moved in the reference frame.


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The position vector R BA defines the displacement ofthe point B with respect to point A . This can beexpressed as the position difference equations .

R BA= R B - R A

The position of B with respect to A is equal to the(absolute) position of B minis the (absolute) position ofA, where absolute means with respect to the origin of

the global reference frame.

R BA= R B - R A

R BA= R BO - R AO


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Case 1 One body in two successive position

position difference

Case 2 Two bodies simultaneous in separate position

relative position


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4.3 TRANSLATION, ROTATION, ANDCOMPLEX MOTION

Translation All points on the body have the same

displacement .

B B A A R R =


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Rotation Difference points in the body undergo

difference displacements and thus there is adisplacement difference between any twopoints chosen.

BA A B B B R R R =


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Complex motion The sum of the translation and rotation

components .total displacement = translation component + rotation component


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Referred to the point B

B B B B B B R R R +=

Referred to the originat A

A B A A A B R R R +=


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Theorems

Eulers theorem

The general displacement of a rigid body withone point fixed is a rotation about some axis.

Chasles theorem Any displacement of a rigid body is equivalent

to the sum of a translation of any one point onthat body and a rotation of the body about anaxis through that point.


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4.4 GRAPHICAL POSITION ANALYSISOF LINKAGES


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4.5 ALGEBRAIC POSITION ANALYSISOF LINKAGES

2

2sin

cos

a A

a A

y

x=

=

( ) ( )( ) 222

222

y x

y y x x

Bd Bc

A B A Bb

+=

+=

The coordinates of point A

The coordinates of point B

Above equation provides a pair of simultaneousequations in B

x and B

y .


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=

=

d B

B

A B

A B

x

y

x x

y y

14

13

tan

tan

The link angles for this position can then befound and a two argument arctangent functionmust be used to solve following equation


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Vector Loop Representation of Linkage

An alternate approach to linkage position

analysis creates a vector loop (or loops)around the linkage. The links arerepresented as position vectors .


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Multi-Loops Mechanism

Cam Follower Mechanism


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Complex Numbers as Vectors Vectors may be defined in polar coordinates

by their magnitude and angle , or in cartesian coordinates as x and y components .

Polar form

R

re j

Cartesian form

sincos

sincos

jr r

jr ir +

+)


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Complex number notationX direction component real protion

Y direction component imaginary portion

This imaginary number is used in a complexnumber as operator , not as a value.

This complex number notation to representplanar vectors comes from the Euler identity :

sincos je j =


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The Vector Loop Equation for a FourbarLinkage

01432 =+ RRRR

01432 =+ j j j j decebeae


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Above equation can be solved for two unknowns .There are four variables in above equation andonly one independent variable ( 2). We needto find the algebraic expressions which define 3and 4 as functions of the constant link lengthsand the one input angle 2.

{ }{ }24

23

,,,,

,,,,

d cbag

d cba f =

=


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Substituting the Euler identity into the complexform equation:

( ) ( ) ( )

( ) 0sincos

sincossincossincos

11

443322

=+

++++

jd

jc jb ja


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+

=

+

=

2tan1

2tan1

cos,

2tan1

2tan2

sin42

42

442

4

4

Half angle identities


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Vector loop equation:

01432 = RRRR

01432 = j j j j decebeae

Complex form equation

Real part

0coscoscos

;00coscoscoscos

432

1

1432

=

=

=

d cba

sod cba

0sinsinsin 432 = cbaImaginary part


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015432 =+ RRRRR

015432 =+ j j j j j fedecebeae

+= 25

( ) 012432 =+ + j j j j j fedecebeae

Vector Loop Equation

Complex polar notation (Three unkonwns 3

4

5)

Two factors determine the relationship between the

two geared links. The gear ratio and the phaseangle

New equation


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Step 4 Square both equations and add them to eliminateone unknown


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Step 5 Substitute the tangent half angle identities forthe sin and cosine terms


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Step 6 Repeat steps 3 to 5 for the other unknown


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Any point on link 1

Any point on link 4

Any point on link 3


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Extreme Values of the Transmission Angle

For a Grashof crank rocker four-bar linkage theminimum value of the transmission angle will occurwhen the crank is collinear with the ground link asfollowing figure

+==

bc

ad cb

2

)(cos

2221

11

++==

bcad cb

2

)(cos

2221

22

For the overlapping

case

For the extendedcase


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To find the maximum and minimum values ofinput angle 2 we differentiate previousequation, from the derivate of 2 with respectto , and set it equal to zero.

0sin

sin

2

2 ==

ad bc

d d

The link length a, b, c, d are never zero, so thisexpression can only be zero when sin is zero.

This is consistent with the definition of toggleposition. If is zero or 180 then cos will be 1.


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4.12 CIRCUITS AND BRANCHES IN

LINKAGESCircuit: all possible orientations of the links that

can be realized without disconnecting any of the joints.Branch: a continuous series of positions of the

mechanism on a circuit between two stationaryconfigurations.The stationary configurations divide a circuit into a

series of branches.

kinematic position analysis

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