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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.
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