Supplement 3011

7/29/2019 Supplement 3011

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Mechanism Kinematics & Dynamics and

Vibrational Modeling

Dr. Robert L. Williams IIMechanical Engineering, Ohio University

NotesBook Supplement forME 3011 Kinematics & Dynamics of Machines

2013 Dr. Bob Productions

[email protected]/williar4

These notes supplement the ME 3011 NotesBook by Dr. Bob

This document presents supplemental notes to accompany the ME 3011 NotesBook. The outlinegiven in the Table of Contents on the next page dovetails with and augments the ME 3011 NotesBook

outline and hence is incomplete here.

mk

x(t)

k

x (t)x (t)


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ME 3011 Supplement Table of Contents

1. INTRODUCTION ........................................................................................................................................................................................................................ 3

1.3 VECTORS. CARTESIAN RE-IM REPRESENTATION (PHASORS) ............................................................................................................................................... 31.6 MATRICES .............................................................................................................................................................................................................................. 5

2. KINEMATICS ANALYSIS ...................................................................................................................................................................................................... 14

2.1 POSITION KINEMATICS ANALYSIS....................................................................................................................................................................................... 142.1.1 Four-Bar Mechanism Position Analysis ..................................................................................................................................................................... 14

2.1.1.1 Tangent Half-Angle Substitution Derivation and Alternate Solution Method .................................................................................................. 142.1.1.3 Four-Bar Mechanism Solution Irregularities ...................................................................................................................................................... 192.1.1.4 Grashofs Law and Four-Bar Mechanism Joint Limits ...................................................................................................................................... 20

2.1.2 Slider-Crank Mechanism Position Analysis ............................................................................................................................................................... 282.1.3 Inverted Slider-Crank Mechanism Position Analysis ................................................................................................................................................ 312.1.4 Multi-Loop Mechanism Position Analysis ................................................................................................................................................................. 37

2.2 VELOCITY KINEMATICS ANALYSIS ..................................................................................................................................................................................... 412.2.2 Three-Part Velocity Formula Moving Example ......................................................................................................................................................... 412.2.5 Inverted Slider-Crank Mechanism Velocity Analysis ................................................................................................................................................ 432.2.6 Multi-Loop Mechanism Velocity Analysis ................................................................................................................................................................ 47

2.3 ACCELERATION KINEMATICS ANALYSIS............................................................................................................................................................................. 512.3.2 Five-Part Acceleration Formula Moving Example .................................................................................................................................................... 51

2.3.4 Slider-Crank Mechanism Acceleration Analysis........................................................................................................................................................ 532.3.5 Inverted Slider-Crank Mechanism Acceleration Analysis ......................................................................................................................................... 542.3.6 Multi-Loop Mechanism Acceleration Analysis.......................................................................................................................................................... 60

2.4 OTHERKINEMATICS TOPICS................................................................................................................................................................................................ 642.4.1 Link Extensions Graphics ........................................................................................................................................................................................... 64

2.5 JERKKINEMATICS ANALYSIS .............................................................................................................................................................................................. 662.5.1 Jerk Analysis Introduction .......................................................................................................................................................................................... 662.5.2 Mechanism Jerk Analysis ........................................................................................................................................................................................... 69

2.6 BRANCH SYMMETRY IN KINEMATICS ANALYSIS................................................................................................................................................................ 702.6.1 Four-Bar Mechanism .................................................................................................................................................................................................. 702.6.2 Slider-Crank Mechanism ............................................................................................................................................................................................ 72

3. DYNAMICS ANALYSIS .......................................................................................................................................................................................................... 73

3.1 DYNAMICS INTRODUCTION.................................................................................................................................................................................................. 733.2 MASS,CENTER OF GRAVITY, AND MASS MOMENT OF INERTIA......................................................................................................................................... 743.4 FOUR-BARMECHANISM INVERSE DYNAMICS ANALYSIS................................................................................................................................................... 853.5 SLIDER-CRANKMECHANISM INVERSE DYNAMICS ANALYSIS ........................................................................................................................................... 873.6 INVERTED SLIDER-CRANKMECHANISM INVERSE DYNAMICS ANALYSIS.......................................................................................................................... 893.7 MULTI-LOOP MECHANISM INVERSE DYNAMICS ANALYSIS................................................................................................................................................ 97

3.8 BALANCING OF ROTATING SHAFTS ................................................................................................................................................................................... 102

4. GEARS AND CAMS ............................................................................................................................................................................................................... 106

4.1 GEARS ................................................................................................................................................................................................................................ 1064.1.1 Gear Introduction ...................................................................................................................................................................................................... 1064.1.2 Gear Ratio ................................................................................................................................................................................................................. 1124.1.3 Gear Trains ................................................................................................................................................................................................................ 1154.1.4 Involute Spur Gear Standardization .......................................................................................................................................................................... 1174.1.5 Planetary Gear Trains ................................................................................................................................................................................................ 126

4.2 CAMS.................................................................................................................................................................................................................................. 1344.2.1 Cam Introduction ...................................................................................................................................................................................................... 134

4.2.2 Cam Motion Profiles ................................................................................................................................................................................................. 1374.2.3 Analytical Cam Synthesis ......................................................................................................................................................................................... 142

5. MECHANICAL VIBRATIONS INTRODUCTION .............................................................................................................................................................. 149

5.2 MECHANICAL VIBRATIONS DEFINITIONS.......................................................................................................................................................................... 149

6. VIBRATIONAL SYSTEMS MODELING ............................................................................................................................................................................. 152

6.1 ZEROTH-ORDERSYSTEMS ................................................................................................................................................................................................. 1526.2 SECOND-ORDERSYSTEMS ................................................................................................................................................................................................. 162

6.2.1 Translational m-c-k System Dynamics Model ......................................................................................................................................................... 1626.2.3 Pendulum System Dynamics Model ......................................................................................................................................................................... 1636.2.4 Uniform Circular Motion .......................................................................................................................................................................................... 165

6.4 ADDITIONAL 1-DOF VIBRATIONAL SYSTEMS MODELS..................................................................................................................................................... 1666.5 ELECTRICAL CIRCUITS MODELING.................................................................................................................................................................................... 1976.6 MULTI-DOF VIBRATIONAL SYSTEMS MODELS.................................................................................................................................................................. 205


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1. Introduction

1.3 Vectors . Cartesian Re-Im Representation (Phasors)

Here is an alternate vector representation.

i

P Pe

The phasoriPe is a polar representation for vectors, where P is the length of vector P , e is the

natural logarithm base, 1i is the imaginary operator, and is the angle of vector P . ie gives thedirection of the length P, according to Eulers identity.

cos sinie i

ie is a unit vector in the direction of vector P .

PhasorRe-Im representation of a vector is equivalent to CartesianXYrepresentation, where the real (Re)

axis is alongX(or i ) and the imaginary (Im) axis is along Y(or j ).

cos(cos sin )

sin

cos (cos sin )

sin

Re i

Im

X

Y

P PP P i Pe

P P

P PP P i j

P P

A strength of CartesianRe-Im representation using phasors is in taking time derivatives of vectors the

derivative of the exponential is easy ( ( )s sd ds e e ).2 2

2 2

2

2

22 2

2

22

2

22

2

( )

2

cos 2 sin sin cos

sin 2 cos cos

i

i i

i i i i i

i i i i

d P d Pe

dt dt

d P dPe iP e

dt dt

d P

Pe iP e iP e iP e i P edt

d PPe iP e iP e P e

dt

P P P Pd P

dt P P P

2sinP


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Where we had to use extensions of Eulers identity

2

2 2

cos sin

cos sin sin cos

sin cos cos sin

i

i

i

e i

ie i i i

i e i i i

Compare this double-time-derivative with theXYapproach.

2 2

2 2

2

2

22

2 2

2

2

cos

sin

cos sin

sin cos

cos sin sin sin cos

sin cos cos cos sin

cos 2 sin

Pd P d

Pdt dt

P Pd P d

dt dt P P

P P P P Pd P

dt P P P P P

P P Pd P

dt

2

2

sin cos

sin 2 cos cos sin

P

P P P P

We obtain the same result, but theRe-Im phasor time differentiation is made in compact vector notationalong the way.

Above we used the product andchain rules of time differentiation.

product rule( ) ( )

( ) ( ) ( )( )( ( ) ) ( ) ( ) ( )i t i t

i t i t i t d dP t de deP t e e P t P t e P t dt dt dt dt

chain rule( ) ( )

( )( ) ( )( )

i t i t i tde de d t

ie tdt d t dt

The result for this example is

( ) ( ) ( )( ( ) ) ( ) ( ) ( )i t i t i t d

P t e P t e P t ie t dt


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Matrix Addition add like terms and keep the results in place

a b e f a e b f

c d g h c g d h

Matrix Multiplication with Scalar multiply each term and keep the results in place

a b ka kbk

c d kc kd

Matrix Multiplication

C A B

In general, A B B A

The row and column indices must line up as follows.

( x ) ( x )( x )

C A B

m n m p p n

That is, in a matrix multiplication product, the number of columns p in the left-hand matrix must equalthe number of rowsp in the right-hand matrix. If this condition is not met, the matrix multiplication isundefined and cannot be done.

The size of the resulting matrix [C] is from the number of rows m of the left-hand matrix and thenumber of columns n of the right-hand matrix, m x n.

Multiplication proceeds by multiplying like terms and adding them, along the rows of the left-hand matrix and down the columns of the right-hand matrix (use your index fingers from the left andright hands).

Example

(2x1) (2x3)(3x1)

ga b c ag bh ci

C hd e f dg eh fi

i

Note the inner indices (p = 3) must match, as stated above, and the dimension of the result is dictated bythe outer indices, i.e. m x n = 2x1.


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Matrix Multiplication Examples

1 2 3

4 5 6A

7 8

9 8

7 6

B

7 8

1 2 39 8

4 5 67 6

7 18 21 8 16 18 46 42

28 45 42 32 40 36 115 108

C A B

(2x2) (2x3)(3x2)

7 81 2 3

9 84 5 6

7 6

7 32 14 40 21 48 39 54 69

9 32 18 40 27 48 41 58 75

7 24 14 30 21 36 31 44 57

D B A

(3x3) (3x2)(2x3)


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Matrix Inversion

Since we cannot divide by a matrix, we multiply by the matrix inverse instead. Given

C A B , solve for [B].

C A B

1 1A C A A B

I B

B

1

B A C

Matrix [A] must be square (m = n) to invert.

1 1

A A A A I

where [I] is the identity matrix, the matrix 1 (ones on the diagonal and zeros everywhere else). Tocalculate the matrix inverse use the following expression.

1 adjoint( )A

AA

where A is the determinant of [A].

adjoint( ) cofactor( )T

A A

cofactor(A) ( 1)i jij ij

a M

minor minorMij is the determinant of the submatrix with row i andcolumnj removed.

For another example, given C A B , solve for [A]

C A B

1 1C B A B B

A I

A

1

A C B

In general the order of matrix multiplication and inversion is crucial and cannot be changed.


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Matrix DeterminantThe determinant of a square n x n matrix is a scalar. The matrix determinant is undefined for a

non-square matrix. The determinant of a square matrix A is denoted det(A) or A . The determinant

notation should not be confused with the absolute-value symbol. The MATLAB function for matrixdeterminant is det(A).

If a nonhomogeneous system ofn linear equations in n unknowns is dependent, the coefficientmatrix A is singular, and the determinant of matrix A is zero. In this case no unique solution exists tothese equations. On the other hand, if the matrix determinant is non-zero, then the matrix is non-singular, the system of equations is independent, and a unique solution exists.

The formula to calculate a 2 x 2 matrix determinant is straight-forward.

a b

Ac d

A ad bc

To calculate the determinant of 3 x 3 and larger square matrices, we can expand about any one

row or column, utilizing sub-matrix determinants. Each sub-determinant is formed by crossing out thecurrent row and its column and retaining the remaining terms as an n1 x n1 square matrix, each ofwhose determinant must also be evaluated in the process. The pivot term (the entry in the cross-out rowand column) multiplies the sub-matrix determinants, and there is an alternating + / / + / etc. signpattern. Here is an explicit example for a 3 x 3 matrix, expanding about the first row (all other optionswill yield identical results).

a b c

A d e f

g h k

( ) ( ) ( )

e f d f d eA a b c

h k g k g h

a ek hf b dk gf c dh ge

For a 3 x 3 matrix only, the determinant can alternatively be calculated as shown, by copying columns 1and 2 outside the matrix, multiplying the downward diagonals with + signs and multiplying the upwarddiagonals with signs (clearly the result is the same as in the above formula).

( ) ( ) ( )

a b c a b

A d e f d e

g h k g h

aek bfg cdh gec hfa kdb a ek hf b kd fg c dh ge

A common usage of the 3 x 3 matrix determinant is to calculate the cross product 1 2P P .

1 2 1 2

1 1 1 11 1

1 2 1 1 1 1 2 1 2

2 2 2 22 2

2 2 2 1 2 1 2

y z z y

y z x yx z

x y z x z z x

y z x yx z

x y z x y y x

i j k p p p pp p p pp p

P P p p p i j k p p p pp p p pp p

p p p p p p p


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System of Linear Equations

We can solve n linear equations in n unknowns with the help of a matrix. Below is an exampleforn = 3.

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

a x a x a x b

a x a x a x ba x a x a x b

Where aij are the nine known numerical equation coefficients, xi are the three unknowns, andbi are thethree known right-hand-side terms. Using matrix multiplication backwards, this is written as

A x b .

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x b

a a a x b

a a a x b

where

11 12 13

21 22 23

31 32 33

a a a

A a a a

a a a

is the matrix of known numerical coefficients

1

2

3

xx x

x

is the vector of unknowns to be solved and

1

2

3

b

b b

b

is the vector of known numerical right-hand-side terms.

There is a unique solution 1

x A b

only if [A] has full rank. If not, 0A (the determinant ofcoefficient matrix [A] is zero) and the inverse of matrix [A] is undefined (since it would require dividingby zero; in this case the rank is not full, it is less than 3, which means not all rows/columns of [A] arelinearly independent). Gaussian Elimination is more robust and more computationally efficient thanmatrix inversion to solve the problem A x b for {x}.


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Matrix Example solve linear equations

Solution of 2x2 coupled linear equations.

1 2

1 2

2 5

6 4 14

x x

x x

1

2

1 2 5

6 4 14

x

x

1 2

6 4A

1

2

xx

x

5

14b

1

x A b

1 4 2 6 8A

The determinant of [A] is non-zero so there is a unique solution.

1 4 2 1/ 2 1/ 41

6 1 3/ 4 1/ 8A

A

check 1 1

2

1 0

0 1A A A A I

1

2

1/ 2 1/ 4 5 1

3/ 4 1/ 8 14 2

x

x

answer.

Check this solution by substituting the answer {x} into the original equations A x b and ensuring

the required original {b} results.

1 2 1 1(1) 2(2) 5

6 4 2 6(1) 4(2) 14


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Same Matrix Examples in MATLAB

%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -% Mat r i ces. m - mat r i x exampl es% Dr . Bob, ME 3011%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

cl ear ; cl c;

A1 = di ag( [ 1 2 3] ) % 3x3 di agonal mat r i xA2 = eye( 3) % 3x3 i dent i t y mat r i x

A3 = [ 1 2; 3 4] ; % mat r i x addi t i onA4 = [ 5 6; 7 8] ;Add = A3 + A4

k = 10; % mat ri x- scal ar mul t i pl i cat i onMul t Sca = k*A3

Tr ans = A4' % mat r i x t r anspose ( swap rows and col umns)

A5 = [ 1 2 3; 4 5 6] ; % def i ne t wo mat r i cesA6 = [ 7 8; 9 8; 7 6] ;A7 = A5*A6 % mat r i x- mat r i x mul t i pl i cat i onA8 = A6*A5

A9 = [ 1 2; 6 4] ; % mat r i x f or l i near equat i ons sol ut i onb = [ 5; 14] ; % def i ne RHS vector dA9 = det ( A9) % cal cul at e det er mi nant of Ai nvA9 = i nv( A9) % cal cul at e t he i nver se of Ax = i nvA9*b % sol ve l i near equat i onsx1 = x( 1) ; % ext r act answer sx2 = x( 2) ;Check = A9*x % check answer shoul d be bxG = A9\ b % Gaussi an el i mi nat i on i s mor e ef f i ci ent

who % di spl ay t he user - creat ed var i abl eswhos % user - cr eat ed var i abl es wi t h di mensi ons

The first solution of the linear equations above uses the matrix inverse. To solve linearequations, Gaussian Elimination is more efficient (more on this in the dynamics notes later) and morerobust numerically; Gaussian elimination implementation is given in the third to the last line of theabove m-file (with the back-slash).

Since the equations are linear, there is a unique solution (assuming the equations are linearlyindependent, i.e. the matrix is not near a singularity) and so both solution methods will yield the sameanswer.


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Output of Matrices.m

A1 =1 0 00 2 00 0 3

A2 =

1 0 00 1 00 0 1

Add =6 8

10 12

Mul t Sca =10 2030 40

Tr ans =

5 76 8

A7 =46 42

115 108

A8 =39 54 6941 58 7531 44 57

dA9 = - 8

i nvA9 =- 0. 5000 0. 25000. 7500 - 0. 1250

x = 12

Check = 514

xG = 12


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2. Kinematics Analysis

2.1 Position Kinematics Analysis

2.1.1 Four-Bar Mechanism Posi tion Analys is

2.1.1.1 Tangent Half-Angle Substitution Derivation and Alternate Solution Method

Tangent half-angle substitution derivation

In this subsection we first derive the tangent half-angle substitution using ananalytical/trigonometric method. Defining parametertto be

tan2

t

i.e. the tangent of half of the unknown angle , we need to derive cos and sin as functions ofparametert. This derivation requires the trigonometric sum of angles formulae.

cos( ) cos cos sin sin

sin( ) sin cos cos sin

a b a b a b

a b a b a b

To derive the costerm as a function oft, we start with

cos cos2 2

The cosine sum of angles formula yields

2 2cos cos sin2 2

Multiplying by a 1, i.e. 2cos2

over itself yields

2 2

2 2 2

2

cos sin

2 2cos cos 1 tan cos2 2 2

cos2

The cosine squared term can be divided by another 1, i.e. 2 2cos sin 12 2

.


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2

2

2 2

cos2

cos 1 tan2

cos sin2 2

Dividing top and bottom by2

cos 2

yields

2

2

1cos 1 tan

21 tan

2

Remembering the earlier definition fort, this result is the first derivation we need, i.e.

2

21cos1

tt

To derive the sinterm as a function oft, we start with

sin sin2 2

The sine sum of angles formula yields

sin sin cos cos sin 2sin cos2 2 2 2 2 2

Multiplying top and bottom by cosine yields

2 2

sin2

sin 2 cos 2 tan cos2 2 2

cos

2

From the first derivation we learned

2

2

1cos

21 tan

2


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Substituting this term yields

2

1sin 2 tan

21 tan

2

Remembering the earlier definition fort, this result is the second derivation we need, i.e.

2

2sin

1

t

t

The tangent half-angle substitution can also be derived using a graphical method as in the figure

below.


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Alternate solution method

The equation form

cos sin 0E F G

arises often in the position solutions for mechanisms and robots. It appeared in the 4 solution for thefour-bar mechanism in the ME 3011 NotesBook and was solved using the tangent half-anglesubstitution.

Next we present an alternative and simpler solution to this equation. We make two simpletrigonometric substitutions based on the figure below.

Clearly from this figure we have

2 2cos E

E F

2 2sin FE F

In the original equation we divide by 2 2E F and rearrange.

2 2 2 2 2 2cos sin

E F G

E F E F E F

The two simple trigonometric substitutions yield

2 2cos cos sin sin

G

E F

Applying the sum-of-angles formula cos( ) cos cos sin sina b a b a b yields

2 2cos( )

G

E F


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And so the solution foris

1

1,22 2

cosG

E F

where

1tanF

E

and the quadrant-specific inverse tangent function atan2 must be used in the above expression for.

There are two solutions for, indicated by the subscripts 1,2, since the inverse cosine function isdouble-valued. Both solutions are correct. We expected these two solutions from the tangent-half-anglesubstitution approach. They correspond to the open- and crossed-branch solutions (the engineer must

determine which is which) to the four-bar mechanism position analysis problem.

For real solutions forto exist, we must have

2 21 1

G

E F

or

2 21 1

G

E F

If this condition is violated for the four-bar mechanism, this means that the given input angle 2 isbeyond its reachable limits (see Grashofs Law).


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2.1.1.3 Four-Bar Mechanism Solution Irregularities

Four-bar mechanism position singularity 0G E 4 1 1 2 2

2 2 2 2

1 2 3 4 1 2 1 2

2 ( )

2 cos( )

E r r c r c

G r r r r r r

For simplicity, let 1 = 0 (just rotate the entire four-bar mechanism model for zero ground link angle).

2 2 2 2

1 2 3 4 1 4 2 4 1 22 2 ( ) 0G E r r r r r r r r r c

I have encountered two example four-bar mechanisms with this 0G E singularity.

Case 1

When 1 4r r and 2 3r r ,2 2 2 2 2

1 2 2 1 1 2 1 1 22 2 ( ) 0G E r r r r r r r r c ALWAYS, regardless

of2.

ExampleGiven 1 2 3 410, 6, 6, 10r r r r ; this mechanism is ALWAYS singular. To fix this let

1 2 3 410, 5.9999, 6.0001, 10r r r r and MATLAB will be able to calculate the positionanalysis reliably at every input angle.

Case 2

When 1 32r r and 4 22r r , and furthermore 3 23 5r r ,

2 2 2 2

3 2 3 2 2 3 2 2 3 2

2 2 2 2 2 2

2 2 2 2 2 2 2

2

2 2

4 4 8 4 ( )c

100 25 40 84 c

9 9 3 3

8c

3

G E r r r r r r r r r

r r r r r r

r

This 0G E occurs only when 2 90 . Case 2 is much less general than case 1.

Example

Given 1 2 3 410, 3, 5, 6r r r r ; this mechanism is singular when 2 90 . To fix this ignore

2

90 or set your2 array to avoid these values.


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2.1.1.4 Grashofs Law and Four-Bar Mechanism Joint Limits

Grashofs Law

Grashofs Law was presented in the ME 3011 NotesBook to determine the input and output linkrotatability in a four-bar mechanism. Applying Grashofs Law we determine if the input and output

links are a crank (C) or a rocker (R). A crank enjoys full 360 degree rotation while a rocker has arotation that is a subset of this full rotation. This section presents more information on Grashofs Lawand then the next subsection presents four-bar mechanism joint limits.

Grashof's condition states "For a four-bar mechanism, the sum of the shortest and longest linklengths should not be greater than the sum of two remaining link lengths". With a given four-bar

mechanism, the Grashof Condition is satisfied if L S P Q where SandL are the lengths of theshortest and longest links, andP andQ are the lengths of the other two intermediate-sized links. If theGrashof condition is satisfied, at least one link will be fully rotatable, i.e. can rotate 360 degrees.

For a four-bar mechanism, the following inequalities must be satisfied to avoid locking of the

mechanism for all motion.

2 1 3 4

4 1 2 3

r r r r

r r r r

With reference to the figure below, these inequalities are derived from the fact that the sum of

two sides of a triangle must be greater than the third side, for triangles 4 1 1O A B and 2 2 2O A B , respectively.

Note from our standard notation, 1 2 4r O O , 2 2r O A , 3r AB , and 4 4r O B .

A

B

A

O

2

2 O4

A1

B1

B2


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Four-Bar Mechanism Joint Limits

IfGrashof's Law predicts that the input link is a rocker, there will be rotation limits on the inputlink. These joint limits occur when links 3 and 4 are aligned. As shown in the figure below, there willbe two joint limits, symmetric about the ground link.

To calculate the joint limits, we use the law of cosines.

2 2 2

3 4 1 2 1 2 2

2 2 21 1 2 3 4

2

1 2

( ) 2 cos

( )cos

2

L

L

r r r r r r

r r r r

r r

with symmetry about r1.


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Joint Limit Example 1 Given 1 2 3 410, 6, 8, 7r r r r

L S P Q (10 6 8 7 )

so we predict only double rockers from this Non-Grashof Mechanism.

2 2 2

1 1

2

10 6 (8 7)cos cos 0.742 137.92(10)(6)

L

This method can also be used to find angular limits on link 4 when it is a rocker. In this case links 2 and3 align.

2 2 2

1 1

4

10 7 (6 8)cos cos 0.336 109.6

2(10)(7)

180 70.4L

In this example, the allowable input and output angle ranges are:

2137.9 137.9 470.4 289.6

This example is shown graphically in the ME 3011 NotesBook, in the Grashofs Law section (2. Non-Grashof double rocker, first inversion).

CautionThe figure on the previous page does not apply in all joint limit cases. For Grashof

Mechanisms with a rocker input link, one link 2 limit occurs when links 3 and 4 fold upon each otherand the other link 2 limit occurs when links 3 and 4 stretch out in a straight line. See Example 4 (andExample 3 for a similar situation with the output link 4 limits).


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L S P Q (10 4 8 7 )

Since the S link is adjacent to the fixed link, we predict this Grashof Mechanism is a crank-rocker.

Therefore, there are no 2 joint limits.

2 2 2

1 1

2

10 4 (8 7)cos cos 1.3625

2(10)(4)L

which is undefined, thus confirming there are no 2 joint limits.

There are limits on link 4 since it is a rocker. For 4min, links 2 and 3 are stretched in a straight line(their absolute angles are identical).

2 2 21 1

4min

10 7 (4 8)cos cos 0.036 88.02(10)(7)

180 92.0

For4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

2 2 2

1 1

4min

10 7 ( 4 8)cos cos 0.95 18.2

2(10)(7)

180 161.8

In this example, the output angle range is

492.0 161.8

and2 is not limited. This example is shown graphically in the ME 3011 NotesBook, in the Grashofs

Law section (1a. Grashof crank-rocker).


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Joint Limit Example 3 Given 1 2 3 411.18, 3, 8, 7r r r r (in) and 1 10.3

L S P Q (11.18 3 8 7 )

This is the four-bar mechanism from Term Example 1 and it is a four-bar crank-rocker Grashof

Mechanism. There are no limits on 2 since link 2 is a crank.

The 4 limits are

4 120.1L (links 2 and 3 stretched in a line)

4 172.5L (links 2 and 3 folded upon each other in a line)

The output angle range is

4120.1 172.5

and2 is not limited. This example is NOT shown graphically in the ME 3011 NotesBook Grashofs

Law section. However, these 4 limits are clearly seen in the F.R.O.M. plot for angle 4 in TermExample 1 in the ME 3011 NotesBook.


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L S P Q (10 4 8 7 )

so we predict this Grashof Mechanism is a double-rocker (Sopposite fixed link). The 2 joint limitsare no longer symmetric about the ground link, as was the case in the Non-Grashof Mechanism double

rocker (Example 1). For 2min, links 3 and 4 are folded upon each other (their absolute angles areidentical).

2 2 2

1 1

2min

10 8 (7 4)cos cos 0.969 14.4

2(10)(8)

For2max, links 3 and 4 are instead stretched in a straight line (their absolute angles are different by asin Example 1).

2 2 21 1

2max

10 8 (7 4)

cos cos 0.269 74.42(10)(8)

2min Diagram 2max Diagram0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5

6

7

8

X (m)

Y(m)

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y(m)


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26

This behavior reverses for the 4 joint limits. For4min, links 2 and 3 are stretched in a straight line(their absolute angles are identical).

2 2 2

1 1

4min

10 7 (8 4)cos cos 0.036 88.0

2(10)(7)

180 92.0

For4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

2 2 2

1 1

4min

10 7 (8 4)cos cos 0.95 18.2

2(10)(7)

180 161.8

4min Diagram 4max Diagram0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5

6

7

8

X (m)

Y(m)

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y(m)


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In this plot we can see the minimum and maximum values we just calculated for links 2 and 4.

Note at 2min 14.4

, 3 138.6

and 3 221.4

are the same angle.

Again, this example is NOT shown graphically in the ME 3011 NotesBook Grashofs Law section.However, a similar case with the same dimensions, in different order, is shown in the ME 3011

NotesBook ( 1 2 3 47, 10, 4, 8r r r r , 1d. Grashof double rocker).

Grashofs Law only predicts the rotatability of the input and output links; it says nothing aboutthe rotatability of the coupler link 3 in this case, what is the rotatability of the coupler link? (In thiscase the coupler linkS rotates fully, proving that the relative motion is the same amongst all four-barmechanism inversions, though the absolute motion with respect to the possible 4 ground links is very

different.)

For more information, see:

R.L. Williams II and C.F. Reinholtz, 1987, Mechanism Link Rotatability and Limit PositionAnalysis Using Polynomial Discriminants, Journal of Mechanisms, Transmissions, and Automationin Design, Transactions of the ASME, 109(2): 178-182.

10 20 30 40 50 60 70 80-250

-200

-150

-100

-50

0

50

100

150

200

2

(deg)

(deg)

3

1st

3

2nd

4

1st

4

2nd


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2.1.2 Slider-Crank Mechanism Posi tion Analys is

Step 6.Solve for the unknowns alternate solution

Here are the same slider-crank mechanism position analysisXYcomponent equations, rearranged

to isolate the 3 terms.

3 3 2 2

3 3 2 2

r c x r c

r s h r s

We can square and add to eliminate 3, similar to the four-bar mechanism solution approach.

2 2 2 2 2

3 3 2 2 2 2

2 2 2 2 2

3 3 2 2 2 2

2

2

r c x xr c r c

r s h hr s r s

2 2 2 2

3 2 2 2 2 22 2r x h r xr c hr s

This quadratic equation inx has the following form:

2 0ax bx c 2 22 2 2

2 3 2 2

1

2

2

a

b r c

c r r h hr s

There are two solutions forx, corresponding to the right and left branches.

2 2 2 2

1,2 2 2 3 2 2 2 22x r c r h r s hrs

Then 3 is found from a ratio of the YtoXequations.

1,23 2 2 1,2 2 2atan2( , )h r s x r c

This alternate solution yields identical results as the earlier solution approach in the ME 3011NotesBook for the right (

13 1,x ) and left (

23 2,x ) branches.


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Slider-Crank Mechanism Snapshot and F.R.O.M. MATLAB m-files

No sample m-files are given in the ME 3011 NotesBook for the slider-crank mechanism sinceyou can readily adapt the snapshot and F.R.O.M. m-files given for the four-bar mechanism previously.

However, below we include a partial m-file to show how to draw the slider and fixed pistonwalls for the slider-crank mechanism graphics, since this was not required for the four-bar mechanism.

Outside the loop:

Lp =put a number here; % length of piston (slider link)Hp =put a number here; % height of pistonXp = [-1 -1 1 1]*Lp/2;Yp = [-1 1 1 -1]*Hp/2;

This establishes the rectangular corner coordinates for the slider link, centered at the origin of yourcoordinate frame. It can be done once, outside the loop. Instead of typing numbers forLp andHp, I

scale them to a fraction of r2, for generality in different-sized slider-crank mechanisms. Note I onlyincluded the four corner points MATLABpatch (below) closes the rectangular figure, i.e. back to

the starting point.

Inside the loop (right after theplot command where links 2 and 3 are drawn to the screen)

patch(Xp+x(i),Yp+h,'g'); % draw piston to screen

where x(i) is the variable horizontal slider displacement andh is the constant vertical offset. These

position parameters shift the piston coordinates from the origin to the correct location in each loop. Youcan use any piston color you like (I show green here, 'g').

Further, to draw the horizontal lines representing the piston walls:

Outside the loop

Xpt = [-1000 1000]; % fixed piston wallsYpt = [h+wall/2 h+wall/2];Xpb = [-1000 1000];Ypb = [h-wall/2 h-wall/2];

Inside the loop (right after theplot command where links 2 and 3 are drawn to the screen)

line(Xpt,Ypt,'LineWidth',2); line(Xpb,Ypb,'LineWidth',2);

Set the piston wall width wall to allow a small clearance between the piston and the walls. Again, it

can be scaled to a small fraction ofr2 for generality. The -1000 and1000 coordinates used above are

to extend the piston wall lines off the screen to the left and to the right.


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MATLAB subplot feature

In a slider-crank mechanism full-range-of-motion (F.R.O.M.) simulation you will need to plot

both 3 andx vs. the independent variable 2. Since the units of3 (deg) andx (m) are dissimilar, theymay not fit clearly on the same plot. In this situation you should use a sub-plot arrangement.

Outside the F.R.O.M. loop you can do the subplot in this way:

subplot(211); % 2x1 arrangement of plots, first plotplot(th2/DR,th3/DR);subplot(212); % 2x1 arrangement of plots, second plotplot(th2/DR,x);

Now, you can use the standard axis labels, linetypes, titles, axis limits, grid, etc., for each plot within asubplot (repeat these formatting commands after eachplot statement above to use similar formatting

for each). These options are not shown, for clarity.

The generalized usage ofsubplot is shown below.

subplot(mni); % m x n arrangement of plots, ith plotplot( . . . );

As seen in the example syntax above, the integers need not be separated by spaces or commas.However, I believe they may be so separated if you desire.


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2.1.3 Inverted Slider-Crank Mechanism Posit ion Analysis

This slider-crank mechanism inversion 2 is an inversion of the standard zero-offset slider-crankmechanism where the sliding direction is no longer the ground link, but along the rotating link 4.Ground link length r1 and input link length r2 are fixed; r4 is a variable. The slider link 3 is attached tothe end of link 2 via an R joint and slides relative to link 4 via a P joint. This mechanism converts rotaryinput to linear motion and rotary motion output. Practical applications include certain doors/windowsopening/damping mechanisms. The inverted slider-crank is also part of quick-return mechanisms.

Step 1. Draw the Kinematic Diagram

r1 constant ground link length 2 variable input angle

r2 constant input link length 4 variable output angler4 variable output link length L4 constant total output link length

Link 1 is the fixed ground link. Without loss of generality we may force the ground link to behorizontal. If it is not so in the real world, merely rotate the entire inverted slider-crank mechanism so it

is horizontal. Both angles 2 and4 are measured in a right-hand sense from the horizontal to the link.

Step 2. State the Problem

Given r1, 1 = 0, r2; plus 1-dof position input 2

Find r4 and4

2

1

4

3


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Step 3. Draw the Vector Diagram. Define all angles in a positive sense, measured with the right handfrom the right horizontal to the link vector (tail-to-head; your right-hand thumb is located at the vectortail).

Step 4. Derive the Vector-Loop-Closure Equation. Starting at one point, add vectors tail-to-head untilyou reach a second point. Write the VLCE by starting and ending at the same points, but choosing adifferent path.

2 1 4r r r

Step 5. Write the XY Components for the Vector-Loop-Closure Equation. Separate the one vectorequation into its twoXandYscalar components.

2 2 1 4 4

2 2 4 4

r c r r c

r s r s

Step 6.Solve for the Unknowns from theXYequations. There are two coupled nonlinear equations in

the two unknowns r4, 4. Unlike the standard slider-crank mechanism, there is no decoupling ofXandY. However, unlike the four-bar mechanism, there is only one unknown angle so the solution is easierthan the four-bar mechanism. First rewrite the aboveXYequations to isolate the unknowns on one side.

4 4 2 2 1

4 4 2 2

r c r c r

r s r s

A ratio of the YtoXequations will cancel r4 and solve for4.

4 4 2 2

4 4 2 2 1

r s r s

r c r c r

4 2 2 2 2 1atan2( , )r s r c r

Then square and add theXYequations to eliminate 4 and solve forr4.

2 2

4 1 2 1 2 22r r r rr c

2

1

4


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Note the same r4 formula results from the cosine law. Alternatively, the same r4 can be solved from

either theXorYequations after is 4 known.

X) 2 2 144

r c rr

c

Y) 2 24

4

r sr

s

Both of these r4 alternatives are valid; however, each is subject to a different artificial mathematicalsingularity ( 4 90

and 4 0,180 , respectively), so only the former square-root formula should be

used forr4, which has no artificial singularity. The Xalgorithmic singularity 4 90 never occurs

unless 2 1r r , which is to be avoided (see below), but the Yalgorithmic singularity occurs twice per fullrange of motion.

Technically there are two solution sets the one above and2 2

4 1 2 1 2 22r r r rr c , 4 .However, the negative r4 is not practical and so only the one solution set (branch) exists, unlike mostplanar mechanisms with two or more branches.

Full-rotation condition

For the inverted slider-crank mechanism to rotate fully, the fixed length of link 4, L4, must begreater than the maximum value of the variable r4.

Slider Limits

The slider reaches its minimum and maximum displacements when 2 = 0 and, respectively.

Therefore, the slider limits are 1 2 4 1 2r r r r r . Thus, the fixed lengthL4 must be greater than 1 2r r .In addition we require 1 2r r for full rotation.


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Graphical Solution

The Inverted Slider-Crank mechanism position analysis may be solvedgraphically, by drawingthe mechanism, determining the mechanism closure, and measuring the unknowns. This is an excellentmethod to validate your computer results at a given snapshot.

Draw the known ground link (points O2 andO4 separated by r1 at the fixed angle 1 = 0). Draw the given input link length r2 at the given angle 2 (this defines pointA). Draw a line from O4 to pointA. Measure the unknown values ofr4 and4.


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Inverted Slider-Crank Mechanism Position Analysis: Term Example 3

Given:

1

2

4

1

0.20

0.10

0.32

0

r

r

L

m

Snapshot Analysis (one input angle)

Given this mechanism and 2 70 , calculate 4 andr4.

4 (deg) r4(m)

150.5 0.191

This Term Example 3 position solution is demonstrated in the figure below.

Term Example 3 Position Snapshot

-0.1 -0.05 0 0.05 0.1 0.15 0.2-0.1

-0.05

0

0.05

0.1

0.15

0.2

X (m)

Y(m)


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Full-Range-Of-Motion (F.R.O.M.) Analysis: Term Example 3

A more meaningful result from position analysis is to report the position analysis unknowns for

the entire range of mechanism motion. The subplots below gives r4 (m) and 4 (deg), for all

20 360 , for Term Example 3.

Term Example 34and r44 varies symmetrically about 180

, being 180 at 2 0,180,360

. r4 varies like a negative cosine

function with minimum displacement 1 2 0.1r r at 2 0,360 and maximum displacement 1 2 0.3r r

. Since r1 is twice r2 in this example, whenever 2 60,300 , a perfect 30 60 90 triangle is

formed; the relative angle between links 2 and 4 is 90 which corresponds to the max and min of

4 150,210 , respectively. At these special points, 4 ( 3 2)0.2 0.173r m.

There is another right triangle that shows up for 2 90 ; in these cases

2 2

4 0.2 0.1 0.224r m and 4 153.4,206.6 , respectively. Check all of these special values in the

F.R.O.M. plot results.

0 50 100 150 200 250 300 350150

160

170

180

190

200

210

4

(deg)

0 50 100 150 200 250 300 3500.1

0.15

0.2

0.25

0.3

2

(deg)

r4

m


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2.1.4 Multi-Loop Mechanism Position Analysis

Thus far we have presented position analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The position analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms. A good rule ofthumb is to look for four-bar (or slider-crank) parts of the multi-loop mechanism as we already knowhow to solve the complete position analyses for these.

This section presents position analysis for the two-loop Stephenson I six-bar mechanism shownbelow as an example multi-loop mechanism. This is one of the five possible six-bar mechanisms shownin the on-line Mechanisms Atlas.

Stephenson I 6-Bar Mechanism

We immediately see that the bottom loop of the Stephenson I six-bar mechanism is identical toour standard four-bar mechanism model. Since we number the links the same as in the four-bar, and ifwe define the angles identically, the position analysis solution is identical to the four-bar presentedearlier. With the complete position analysis of the bottom loop thus solved, we see that points CandDcan be easily calculated. Then the solution for the top loop is essentially another four-bar solution:graphically, the circle of radius r5 about point Cmust intersect the circle of radius r6 about pointD toform pointE(yielding two possible intersections in general). The analytical solution is very similar tothe standard four-bar position solution, as we will show.

For multi-loop mechanisms, the number of solution branches for position analysis increasescompared to the single-loop mechanisms. Most single-loop mechanisms mathematically have twosolution branches. For multi-loop mechanisms composed of multiple single-loop mechanisms, thenumber of solution branches is 2n, where n is the number of mechanism loops. For the two-loopStephenson I six-bar mechanism, the number of solution branches for the position analysis problem is 4,two from the standard four-bar part, and two for each of these branches from the upper loop.


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Now let us solve the position analysis problem for the two-loop Stephenson I six-bar mechanismusing the formal position analysis steps presented earlier. Assume link 2 is the input link.

Step 1. Draw the Kinematic Diagram this is done in the figure above.

r1 constant ground link length 1 constant ground link angle

r2

constant input link length to pointA 2

variable input link angle

r2a constant input link length to point C 2 constant angle on link 2

r3 constant coupler link length, loop I 3 variable coupler link angle, loop I

r4 constant output link length, loop I 4 variable output link angle, loop I

r4a constant input link length to pointD 4 constant angle on link 4

r5 constant coupler link length, loop II 5 variable coupler link angle, loop II

r6 constant output link length, loop II 6 variable output link angle, loop II

As usual, all angles are measured in a right-hand sense from the absolute horizontal to the link, as shownin the kinematic diagram.


Given r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 2, 4; plus 1-dof position input 2

Find 3, 4, 5, 6

Step 3. Draw the Vector Diagram. Define all angles in a positive sense, measured with the right handfrom the right horizontal to the link vector (tail-to-head with the right-hand thumb at the vector tail andright-hand fingers towards the arrow in the vector diagram below).


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Step 4. Derive the Vector-Loop-Closure Equations. One VLCE is required for each mechanism loop.Start at one point, add vectors tail-to-head until you reach a second point. Write each vector equation bystarting and ending at the same points, but choosing a different path.

2 3 1 4

2 5 1 4 6a a

r r r r

r r r r r

Note an alternative to the second vector loop equation is 2 5 3 4 6b br r r r r . See if you can identify

2br and 4br , plus their angles 2b and4b.

Step 5. Write the XYComponents for each Vector-Loop-Closure Equation. Separate the two vectorequations into fourXYscalar component equations.

2 2 3 3 1 1 4 4

2 2 3 3 1 1 4 4

r c r c rc r cr s r s r s r s

2 2 5 5 1 1 4 4 6 6

2 2 5 5 1 1 4 4 6 6

a a a a

a a a a

r c r c rc r c r c

r s r s r s r s r s

where2 2 2

4 4 4

a

a

Step 6.Solve for the Unknowns from the fourXYEquations. The four coupled nonlinear equations in

the four unknowns 3, 4, 5, 6 can be solved in two stages, one for each mechanism loop.

Loop I. This solution is identical to the standard four-bar mechanism solution for 3, 4,

summarized here from earlier. From the first twoXYscalar equations above, isolate 3 terms, square

and add both equations to obtain one equation in one unknown 4. This equation has the form

4 4cos sin 0E F G , where termsE, F, andG are known functions of constants and the input angle

2. Solve this equation for two possible values of4 using the tangent half-angle substitution. The two

4 values correspond to the open and crossed branches. Then return to the originalXYscalar equations

with 3 terms isolated, divide the Y by the X equations, and solve for 3 using the atan2 function,substituting the solved values for4. One unique 3 will result for each of the two possible 4 values.

Loop II. The method is analogous to the Loop I solution above. Since 4 is now known, we also

know 4 4 4a . From the second two XYscalar equations above, isolate 5 terms, square and add

both equations to obtain one equation in one unknown 6. This equation is of the form

2 6 2 6 2cos sin 0E F G , where termsE2, F2, andG2 are known functions of constants and the known

angles 2 2 2a and 4 4 4a . Solve this equation for two possible values of6 using the tangent


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half-angle substitution. The two 6 values correspond to open and crossed branches, for each of the

Loop I open and crossed branches. Then return to the original XY scalar equations with 5 terms

isolated, divide the Yby the Xequations, and solve for 5 using the atan2 function, substituting the

solved values for6. One unique 5 will result for each of the two possible 6 values from each 4.

Branches. There are two 5, 6 branches for each of the two 3, 4 branches, so there are four

overall mechanism branches for the two-loop Stephenson I six-bar mechanism.

Full-rotation condition

The range of motion of a multi-loop mechanism may be more limited than that of single-loopmechanisms. One can perform a compound Grashof analysis when four-bars are the componentmechanisms. For the two-loop Stephenson I six-bar mechanism, the second loop may constrain the firstloop (e.g. it may change an expected crank motion of the input link to a rocker). This is an importantissue in design of multi-loop mechanisms if the input link must still rotate fully.

Graphical Solution

The two-loop Stephenson I six-bar mechanism position analysis may readily be solvedgraphically, by drawing the mechanism, determining the mechanism closure, and measuring theunknowns. This is an excellent method to validate your computer results at a given snapshot.

Loop I (this part is identical to the standard four-bar graphical solution)

Draw the known ground link (points O2 andO4 separated by r1 at the fixed angle 1). Draw the given input link length r2 at the given angle 2 to yield pointA. Draw a circle of radius r3, centered at pointA. Draw a circle of radius r4, centered at point O4. These circles intersect in general in two places to yield two possible pointsB. Connect the two branches and measure the unknown angles 3 and4.

Loop II (this part is a modification of the standard four-bar graphical solution). Start with the end of theprocedure above, on the same drawing. For each solution branch above, perform the following steps.

Draw the linkr2a at angle 2 2 2a from point O2. Draw the linkr4a at angle 4 4 4a from point O4. Draw a circle of radius r5, centered at point C. Draw a circle of radius r6, centered at pointD. These circles intersect in general in two places to yield two possible pointsE. Connect the two branches and measure the unknown values 5 and6.

In general, there are four overall position solution branches.


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2.2 Veloci ty Kinematics Analysis

2.2.2 Three-Part Velocity Formula Moving Example

Given initial positions 0 0 0 0 0 0x yP P L (m, rad) and constant velocities

0 0 0.2 0.1 0.3 0.4x yV V V (m/s, rad/s), simulate this motion and determine PV at each

instant. tf = 5 and 0.1t sec was used. The initial and final snapshots, with their three-part velocitydiagrams, are shown below.

Initial Kinematic Diagram Initial Three-Part Velocity Diagram

Final Kinematic Diagram Final Three-Part Velocity Diagram

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y(m)

-0.5 0 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Vx

(m/s)

Vy

(m/

s)

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y(m)

-0.5 0 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Vx

(m/s)

Vy

(m/s)

V0

V x L

VP


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Three-Part Velocity Moving Example Plots

Position Terms Velocity Terms

What is the relationship between these plots?

Point P Translational Velocity Sliding and Tangential Veloci ty Components

Constant velocity terms 0 0 0.2 0.1 0.3 0.4x yV V V lead to non-constant point P velocity(due to the nonlinear position kinematics).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

time (sec)

PositionTerms

(ma

ndrad)

P0x

P0y

L

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

time (sec)

VelocityTerms(m/sandrad/s)

V0x

V0y

V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time (sec)

PointPVelocity(m/s)

VPx

VPy

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time (sec)

SlidingandTangentialVelocities(

m/s)

Vsx

Vsy

Vtx

Vty


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2.2.5 Inverted Slider-Crank Mechanism Velocity Analysis

Again, link 2 (the crank) is the input and link 4 is the output. Remember r4 is a variable so

4 0r in this problem.

Step 1. The inverted slider-crank mechanism Position Analysis must first be complete.

Given r1, 1 = 0, r2, and2, we solved forr4 and4

Step 2. Draw the inverted slider-crank mechanism Velocity Diagram.

where i (i = 2,4) is the absolute angular velocity of link i. 4r is the slider velocity along link 4.

3 4 since the slider cannot rotate relative to link 4.


Given the mechanism r1, 1 = 0, r2

the position analysis 2, r4, 4

1-dof of velocity input 2

Find the velocity unknowns 4r and4

2

1

4

3


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Step 4. Derive the velocity equations. Take the first time derivative of the vector loop closureequations from position analysis, inXYcomponent form.

Here are the inverted slider-crank mechanism position equations from earlier.

2 1 4

2 2 1 4 4

2 2 4 4

r r r

r c r r c

r s r s

The first time derivative of the position equations is given below.

2 2 2 4 4 4 4 4

2 2 2 4 4 4 4 4

r s r c r s

r c r s r c

These two linear equations in two unknowns can be written in matrix form.

4 4 4 4 2 2 2

4 4 4 4 2 2 2

c r s r r s

s r c r c

Step 5. Solve the velocity equations for the unknowns 4 4,r .

2 2 2 4 2 4

4 4 4 4 4 2 2 2

2 2 2 4 2 4

4 4 4 2 2 24

4

2 2 4 24

2 2 4 2

4

4

( )1

( )

sin( )cos( )

r s c c sr r c r s r s

r s s c cs c r cr

r

rrr

r

where we have used the trigonometric identities:

cos cos cos sin sin

sin sin cos cos sin

a b a b a b

a b a b a b

sin( ) sin( )

cos( ) cos( )

a a

a a

The units are all correct in the solution above, m/s andrad/s, respectively.


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Inverted slider-crank mechanism singularity conditionWhen does the solution fail? This is an inverted slider-crank mechanism singularity, when the

determinant of the coefficient matrix goes to zero. The result is dividing by zero, resulting in infinite

4 4,r .

4 4 4

4 4 4

c r sA

s r c

2 2 2 2

4 4 4 4 4 4 4 4( ) ( ) 0A r c r s r c s r

Physically, assuming1 2r r as in the full rotation condition from the inverted slider-crank mechanism

position analysis, this is impossible, i.e. r4 never goes to zero.

This matrix determinant 4A r was used in the solution of the previous page.

Inverted slider-crank mechanism velocity example Term Example 3 continued

Given

1

2

4

1

0.20

0.10

0.32

0

r

r

L

(m)

2

4

4

70

150.5

0.191r

(deg andm)

Snapshot AnalysisGiven this mechanism position analysis plus 2 25 rad/s, calculate 4 4,r for this snapshot.

4

4

0.870 0.094 2.349

0.493 0.166 0.855

r

4

4

2.47

2.18

r

(m/s andrad/s)

Both are positive, so the slider link 3 is currently traveling up link 4 and the link 4 is currently rotating in

the ccw direction, which makes sense from the physical problem.


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Full-Range-Of-Motion (F.R.O.M.) Analysis Term Example 3 continued

A more meaningful result from velocity analysis is to report the velocity analysis results for the

entire range of mechanism motion. The subplot below gives 4 (top, rad/s) and 4r (bottom, m/s), for all

20 360 , for Term Example 3. Since 2 is constant, we can plot the velocity results vs. 2 (since it

is related to time tvia 2 2t ).

Term Example 3 F.R.O.M., 4and 4r As expected, 4 is zero at the max and min for 4 (at 2 60

); also, 4 has a large range of nearly

constant positive velocity near the middle of motion this can be seen in a MATLAB animation.4

r is

zero at the beginning, middle, and end of motion and is max at 2 60 .

0 50 100 150 200 250 300 350-25

-20

-15

-10

-5

0

5

10

4

(rad/s)

0 50 100 150 200 250 300 350

-2

-1

0

1

2

2

(deg)

r4d

(m/s)


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47

2.2.6 Multi-Loop Mechanism Velocity Analysis

Thus far we have presented velocity analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The velocity analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms.

This section presents velocity analysis for the two-loop Stephenson I six-bar mechanism shownbelow as an example multi-loop mechanism. It follows the position analysis for the same mechanismpresented earlier.


The bottom loop of the Stephenson I six-bar mechanism is identical to the standard four-bar

mechanism model and so the velocity analysis solution is identical to the four-bar presented earlier.With the complete velocity analysis of the bottom loop thus solved, the solution for the top loop isessentially another four-bar velocity solution.

As in all velocity analysis, the velocity solution for multi-loop mechanisms is a linear analysisyielding a unique solution (assuming the given mechanism position is not singular) for each positionsolution branch considered. The position analysis must be complete prior to the velocity solution.

Now let us solve the velocity analysis problem for the two-loop Stephenson I six-bar mechanismusing the formal velocity analysis steps presented earlier. Again, assume link 2 is the input link.


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Step 1. The Stephenson I six-bar mechanism Position Analysis must first be complete.

Given r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 2, 4, and2 we solved for3, 4, 5, 6.

Step 2. Draw the Stephenson I six-bar mechanism Velocity Diagram.This should include all the information from the position diagram, plus the new velocity

information. For clarity, we show only the new velocity information here. Refer to the previousStephenson I position kinematics diagram for complete information.

where i (i = 2,3,4,5,6) is the absolute angular velocity of linki. Triangular links 2 and 4 each have asingle angular velocity for the whole link. 0ir for all links since all links are of fixed length (nosliders).


Given the mechanism r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 2, 4,

the position analysis 2, 3, 4, 5, 6,

1-dof velocity input 2

Find the velocity unknowns 3, 4, 5, 6


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Step 4. Derive the velocity equations. Take the first time derivative of each of the two vector loopclosure equations from position analysis, inXYcomponent form.

Here are the Stephenson I six-bar mechanism position equations.

Vector equations

2 3 1 4

2 5 1 4 6a a

r r r r

r r r r r

XYscalar equations

2 2 3 3 1 1 4 4

2 2 3 3 1 1 4 4

r c r c rc r c

r s r s r s r s

2 2 5 5 1 1 4 4 6 6

2 2 5 5 1 1 4 4 6 6

a a a a

a a a a

r c r c rc r c r c

r s r s r s r s r s

where2 2 2

4 4 4

a

a

The first time derivatives of the Loop I position equations are identical to those for the standardfour-bar mechanism.

2 2 2 3 3 3 4 4 4

2 2 2 3 3 3 4 4 4

r s r s r s

r c r c r c

These equations can be written in matrix form.

3 3 4 4 3 2 2 2

3 3 4 4 4 2 2 2

r s r s r s

r c r c r c

The first time derivative of the Loop II position equations is

2 2 2 5 5 5 4 4 4 6 6 6

2 2 2 5 5 5 4 4 4 6 6 6

a a a a a a

a a a a a a

r s r s r s r s

r c r c r c r c


5 5 6 6 5 2 2 2 4 4 4

5 5 6 6 6 2 2 2 4 4 4

a a a a

a a a a

r s r s r s r s

r c r c r c r c

where we have used2 2

4 4

a

a

since 2 and4 are constant angles.


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50

Step 5. Solve the velocity equations for the unknowns 3, 4, 5, 6.

The two mechanism loops decouple so we find 3 and4 from Loop I first and then use 4 to

find5 and6 from Loop II. The solutions are given below.

Loop I (identical to the standard four-bar mechanism)

1

3 3 4 43 2 2 2

3 3 4 44 2 2 2

r s r s r s

r c r c r c

Loop II (similar to the standard four-bar mechanism)

1

5 5 5 6 6 2 2 2 4 4 4

6 5 5 6 6 2 2 2 4 4 4

a a a a

a a a a

r s r s r s r s

r c r c r c r c

Remember, Gaussian elimination is more efficient and robust than the matrix inverse. Also, theseequations may be solved algebraically instead of using matrix methods to yield the same answers.

Stephenson I six-bar mechanism singularity condition

The velocity solution fails when the determinant of either coefficient matrix above goes to zero.The result is dividing by zero, resulting in infinite angular velocities for the associated loop.

For the first loop, the singularity condition is identical to the singularity condition of the standardfour-bar mechanism, i.e. when links 3 and 4 either line up or fold upon each other, causing a link 2 jointlimit. For the second loop, the singularity condition is similar, occurring when links 5 and 6 either lineup or fold upon each other. These conditions also cause angle limit problems for the position analysis,so the velocity singularities are known problems.


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51

2.3 Acceleration Kinematics Analysis2.3.2 Five-Part Acceleration Formula Moving Example

Given initial positions 0 0 0 0 0 0x yP P L (m, rad), initial velocities

0 0 0 0 0 0x yV V V (m/s, rad/s) and constant accelerations 0 0x yA A A

0.2 0.05 0.15 0.1 (m/s2, rad/s2), simulate this motion and determine PA at each instant. tf = 5

and 0.1t sec was used; the initial and final snapshots, with their five-part acceleration diagrams, areshown below.

Initial Kinematic Diagram Initial Five-Part Acceleration Diagram

Final Kinematic Diagram Final Five-Part Acceleration Diagram

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y(m)

-0.8 -0.2 0.40

0.2

0.4

0.6

0.8

1

1.2

A (m/s2)

Ay

(m/s2)

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y(m)

-0.8 -0.2 0.40

0.2

0.4

0.6

0.8

1

1.2

Ax

(m/s2)

Ay

(m/s2)

A0

A

2 x V

x L

x ( x L)

AP


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Five-Part Acceleration Moving Example Plots

Position Terms Velocity Terms

Acceleration Terms Point P Translational Acceleration

What is the relationship amongst the first three plots?

Constant acceleration terms 0 0 0.2 0.05 0.15 0.1x yA A A lead to non-constant point Pacceleration (due to nonlinear position kinematics and centripetal acceleration).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

time (sec)

PositionTerms(man

drad)

P0x

P0y

L

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time (sec)

VelocityTerms(m/sandrad/s)

V0x

V0y

V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

time (sec)

AccelerationTerms(m/s2

andrad/s2)

A0x

A0y

A

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

time (sec)

PointPAcceleration(m/s2)

APx

APy


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53

2.3.4 Slider-Crank Mechanism Acceleration AnalysisDerivative/Integral Relationships

When one variable is the derivative of another, recall the relationships from calculus. For example:

( )( )

dx tx t

dt

0( ) ( )x t x x t dt ( )

( )dx t

x tdt

0( ) ( )x t x x t dt

Term Example 2 F.R.O.M. Slider Results,x, x ,x

The value ofx at any point is the slope of thex curve at that point. The value ofx at any point is

the integral of the x curve up to that point (the value ofx at any point is the area under the x curve up

to that point plus the initial valuex0). A similar relationship exists forx andx .

These graphs are plotted vs. 2, but the same type of relationships hold when plotted vs. time t

since 2 is constant. This is the Term Example 2 F.R.O.M. result.

Note these curves should be plotted vs. time tinstead of2 in order to see the true slope and areavalues accurately.

0 50 100 150 200 250 300 350

0.1

0.2

x(m)

0 50 100 150 200 250 300 350-2

-1

0

1

2

3

xd(m/s)

0 50 100 150 200 250 300 350-50

0

50

2

(deg)

xdd(m/s2)


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54

2.3.5 Inverted Slider-Crank Mechanism Acceleration Analysis

Again, link 2 (the crank) is the input and link 4 is the output.

Step 1. The inverted slider-crank mechanism Position and Velocity Analyses must first be complete.

Given r1, 1 = 0, r2, and2 we solved forr4 and4; then given 2 we solved for4

r and4.

Step 2. Draw the inverted slider-crank mechanism Acceleration Diagram.

where i (i = 2,4) is the absolute angular acceleration of linki. 4r is the slider acceleration along link 4.

3 4 since the slider cannot rotate relative to link 4.


Given the mechanism r1, 1 = 0, r2

the position analysis 2, r4, 4

the velocity analysis 2, 4r , 4

1-dof acceleration input 2

Find the acceleration unknowns 4r and4

2

1

4

3


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Step 4. Derive the acceleration equations. Take the first time derivative of the inverted slider-crankmechanism velocity equations from velocity analysis, inXYcomponent form.

Here are the inverted slider-crank mechanism velocity equations.

4 4 4 4 4 2 2 2

4 4 4 4 4 2 2 2

r c r s r s

r s r c r c

The first time derivative of the velocity equations is given below.2 2

4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2

2 2

4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2

2

2

r c r s r s r c r s r c

r s r c r c r s r c r s

These equations can be written in matrix form.2 2

4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2 24 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2

2

c r s r r s r c r s r c

s r c r c r s r c r s

Step 5. Solve the acceleration equations for the unknowns 4 4,r .

2 24 4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2 24 4 44 2 2 2 2 2 2 4 4 4 4 4 4

2 2

4 2 2 4 2 2 2 4 2 4 4

24 2 2 4 2 2 2 4 2 4 4

4

21

2

sin( ) cos( )

cos( ) sin( ) 2

r r c r s r s r c r s r c

s cr r c r s r c r s

r r r r

r r r

r

where we have again used the trigonometric identities:

cos cos cos sin sin

sin sin cos cos sin

a b a b a b

a b a b a b

sin( ) sin( )

cos( ) cos( )

a a

a a

A major amount of algebra and trigonometry is required to get the final analytical solution for

4 4,r above. Interestingly, the link 4 Coriolis term 4 42r cancelled in 4r while the link 4 centripetal

term 24 4r cancelled in 4. The units are all correct, m/s2 andrad/s2, respectively.


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56

Inverted slider-crank mechanism singularity condition

The acceleration problem has the same coefficient matrix [A] as the velocity problem, so thesingularity condition is identical (see the singularity discussion in the inverted slider-crank mechanism

velocity section the only singularity is when r4 goes to zero; this will never to occur if 1 2r r ).

Inverted slider-crank mechanism acceleration example Term Example 3 continued

Given

1

2

4

1

0.20

0.10

0.32

0

r

r

L

(m)

2

4

4

70

150.5

0.191r

(deg andm)

2

4

4

25

2.47

2.18

r

(rad/s andm/s)

Snapshot Analysis

Given this mechanism position and velocity analyses, plus 2 0 rad/s2, calculate 4 4,r for this

snapshot.

4

4

0.870 0.094 16.873

0.493 0.166 48.957

r

4

4

9.46

267.14

r

(m/s2

andrad/s2)

4r is negative, so the slider link 3 is currently slowing down its positive velocity up link 4 and 4 is

positive, so the link 4 angular velocity is currently increasing in the ccw direction.


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57

Full-Range-Of-Motion (F.R.O.M.) Analysis Term Example 3 continued

A more meaningful result from acceleration analysis is to report the acceleration analysis results

for the entire range of mechanism motion. The subplot below gives 4 (top, rad/s2) and 4r (bottom,

m/s2), for all20 360

, for Term Example 3. Again, since 2 is constant, we can plot the

acceleration results vs. 2 (since it is related to time tvia 2 2t ).

Term Example 3 F.R.O.M., 4and 4r As expected, 4 is zero at the beginning, middle, and end since the 4 curve flattens out at those

points. The maximum (and minimum) 4 values correspond to the greatest slopes for 4. 4r is

maximum (and minimum) at the beginning, middle, and end since the 4r curve is steepest at thosepoints; 4r is zero when the 4r curve is flat, i.e. 2 60

.

0 50 100 150 200 250 300 350-1000

-500

0

500

1000

4

(rad/s2)

0 50 100 150 200 250 300 350-50

0

50

100

2

(deg)

r4dd

(m/s2)


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58

Derivative/Integral Relationships

When one variable is the derivative of another, recall the relationships from calculus (thederivative is the slope of the above curve at each point; the integral is the area under the curve up to thatpoint, taking into account the initial value). For example:

44

( )( )

d tt dt

4 40 4( ) ( )t t dt

44

( )( )

d tt

dt

4 40 4( ) ( )t t dt

0 50 100 150 200 250 300 350

160

180

200

4

(deg)

0 50 100 150 200 250 300 350

-20

-10

0

10

4

(rad/s)

0 50 100 150 200 250 300 350-1000

-500

0

500

1000

2

(deg)

4

(rad/s2)


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59

44

( )( )

dr tr t

dt 4 40 4( ) ( )r t r r t dt

44

( )( )

dr tr t

dt

4 40 4( ) ( )r t r r t dt

These plots are all from Term Example 3.

0 50 100 150 200 250 300 3500.1

0.15

0.2

0.25

r4

(m

)

0 50 100 150 200 250 300 350

-2

-1

0

1

2

r4d

(m

/s)

0 50 100 150 200 250 300 350-50

0

50

100

2

(deg)

r4dd

(m

/s2)


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60

2.3.6 Multi-Loop Mechanism Acceleration Analysis

Thus far we have presented acceleration analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The acceleration analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms.

This section presents acceleration analysis for the two-loop Stephenson I six-bar mechanismshown below as an example multi-loop mechanism. It follows the position and velocity analyses for thesame mechanism presented earlier.


The bottom loop of the Stephenson I six-bar mechanism is identical to the standard four-bar

mechanism model and so the acceleration analysis solution is identical to the four-bar presented earlier.With the complete acceleration analysis of the bottom loop thus solved, the solution for the top loop isessentially another four-bar acceleration solution.

As in all acceleration analysis, the acceleration solution for multi-loop mechanisms is a linearanalysis yielding a unique solution (assuming the given mechanism position is not singular) for eachsolution branch considered. The position and velocity analyses must be complete prior to theacceleration solution.


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61

Now let us solve the acceleration analysis problem for the two-loop Stephenson I six-barmechanism using the formal acceleration analysis steps presented earlier. Again, assume link 2 is theinput link.

Step 1. The Stephenson I six-bar mechanism Position and Velocity Analyses must first be complete.

Given r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 2, 4, 2, and2, we solved for3, 4, 5, 6, 3, 4, 5, 6.

Step 2. Draw the Stephenson I six-bar mechanism Acceleration Diagram.This should include all the information from the position and velocity diagrams, plus the new

acceleration information. For clarity, we show only the new acceleration information here. Refer to theprevious Stephenson I position and velocity kinematics diagrams for complete information.

where i (i = 2,3,4,5,6) is the absolute angular acceleration of linki. Triangular links 2 and 4 each have

a single angular acceleration for the whole link. 0ir for all links since all links are of fixed length (nosliders).


Given the mechanism r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 2, 4,the position analysis 2, 3, 4, 5, 6,

the velocity analysis 2, 3, 4, 5, 6,

1-dof acceleration input 2

Find the acceleration unknowns 3, 4, 5, 6


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62

Step 4. Derive the acceleration equations. Take the first time derivative of both sides of each of thefour scalarXYvelocity equations.

The Stephenson I six-bar mechanism velocity equations are given below.

XYscalar velocity equations

2 2 2 3 3 3 4 4 4

2 2 2 3 3 3 4 4 4

r s r s r s

r c r c r c

2 2 2 5 5 5 4 4 4 6 6 6

2 2 2 5 5 5 4 4 4 6 6 6

a a a a a a

a a a a a a

r s r s r s r s

r c r c r c r c

where2 2 2

4 4 4

a

a

The first time derivative of the Loop I velocity equations is identical to that for the standard four-bar mechanism.

2 2 2

2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

2 2 2

2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

r s r c r s r c r s r c

r c r s r c r s r c r s


2 2 23 3 4 4 3 2 2 2 2 2 2 3 3 3 4 4 4

2 2 2

3 3 4 4 4 2 2 2 2 2 2 3 3 3 4 4 4

r s r s r s r c r c r c

r c r cr c r s r s r s

The first time derivatives of the Loop II velocity equations are:

2 2 2 2

2 2 2 2 2 2 5 5 5 5 5 5 4 4 4 4 4 4 6 6 6 6 6 6

2 2 2 2

2 2 2 2 2 2 5 5 5 5 5 5 4 4 4 4 4 4 6 6 6 6 6 6

a a a a a a a a a a a a

a a a a a a a a a a a a

r s r c r s r c r s r c r s r c

r c r s r c r s r c r s r c r s


2 2 2 25 5 6 6 5 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

2 2 2 25 5 6 6 6 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

a a a a a a a a

a a a a a a a a

r s r s r s r c r c r s r c r c

r c r c r c r s r s r c r s r s

where we have used2 2

4 4

a

a

and

2 2

4 4

a

a

since 2 and4 are constant angles.

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