ME 3230 Chapter 6 Plannar Linkage Design -1

7/29/2019 ME 3230 Chapter 6 Plannar Linkage Design -1

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Lecture Slides

ME 3230 Kinematics and Mechatronics

Chapter 6 Planar Linkage Design

By Dr. Debao Zhou


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Scope: Design/SynthesisWe are about to begin the study of the means and

method to design ( create ) linkages 4-bar linkage Slide-crank and its inversion Other mechanisms

Find joint locations and link length Based on the following constraints

Motion generation Function generation

Methods Graphical approach Analytical approach


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Design Goals: Find joint locations and link length basedon design requirements

Motion Generation coupler as a wholefollows a desiredtrajectory

Function Generation Input / output

relationship


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Review: 4-bar linkage - Grashof Law The fundamental 4-bar linkage design law:

s + l < p + q s is the shortest link, l is the longest link, p and q are the other two

links

This inequality must be true, if continuous relative motion exists between any 2 links.

Grashof linkages Type 1: the inequality holds

We state that these linkages (type 1) have two joints that performcomplete (360 ) rotation and they are located at either end of the

shortest link Nomenclature: Fixed link, Turning Links and Coupler

Type 2: the inequality is not held They have no fully rotating joints, all 4 joints oscillate between limits May not be able to assemble


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Review: Grashof Type 1 Mechanisms:

s is connected to thebase (a, b) this is acrank rocker

s is the base link (c) this is double crankor drag-linemechanism

s is the coupler link this is a double

rocker where thecoupler can performa complete rotationrelative to the base


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Some concepts The coupler - moving plane

Moving Pivots : End points (or any point) on the chosen line-segment selected as joints

Circle point = moving pivot

Fixed Pivot : is located at thecenter of the circle on whichall three equivalent points onthe moving coupler line lie

A1, A2, A3 which are the samemoving center at the 3locations

Center point = fixed pivot

Pole : P ij 6

Ai

Bi

B j

A j

P ij

Bisector line


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4-bar Linkage Design: 2 crank positions Two fixed pivots, one

moving pivot (one length) Or start with a known base

length (O 2 to O 4) and thelength of the Output Link(O 4B)

We desire that themechanisms 2 rockers bein precisely two locations(at some time) they areseparated by the angle with respect to link 2 and

wrt Link 4

B1

B2

O 4O 2

A1 A2


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Graphical Method To solve, lay out 2 points

(O2

O4)

From O 2 strike two lineseparated by

From O4 strike Output Link(O4B) at two positionsseparated by

Then strike a line from O 2 to B 2 about O 2 by - (theinversion!)

Connect B 1 and B 2 Form a perpendicular

bisector to this line Where it strikes O 2 A1 line

locates A 1 Sets Link 2 Length and a

line from A 1 to B 1 establishes Coupler length

B1

B2

O 4O 2

A1 A2

B*


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Analytical Method

1 1 1 1 2 2 2 2

1 1 1 1 2 2 2 2

2 2 2 2

3

2 2 2 2

2 2

2 1 1 4 1 2 1 1 4 1

2

2 2 1 4 2 2 2 1

The distance between A & B is:

r

substituting from earlier:

r r

r r

A B A B A B A B

A B A B A B A B

x x y y x x y y

x x y y x x y y

Cos r r Cos Sin r r Sin

Cos r r Cos Sin r

2

4 2r Sin

22113

12

12

32

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:Equations

, :Unkown, :Given

,,, :Given

B A B Ar

r r

r r

r 1

r 4

B2

B1

A1 A2

O 4

O 2

1

1


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2-Position Design Analytical Methods

1 4 2 12

4 1 1 1 2 1 4 2 2

2 2

3 2 1 1 4 1 2 1 1 4 1

2 2

2 2 1 4 2 2 2 1 4 2

:

os

os

r r Cos Cosr

r Cos r Cos Cos r Cos

now

r r C r r Cos r Sin r r Sin

r C r r Cos r Sin r r Sin


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Motion Generation: 2-Position Coupler

Moving pivots as A/B or A/C:

Construct perpendicular bisectors for each (or any 2)pairs of points on the coupler

Crank Pivots can be placed

anywhere along them onealong As perpendicular Bisector and one along Bs (Cs)perpendicular Bisector : isocelestriangle

Notice, If the Perpendicular Bisectors are extended far enough they will intersect at thesingle pivot position theDisplacement pole P 12 (wecould build a 3-bar solution!)

Bisectors is constructed usingonly the moving pivots

B1

B2

O 4O 2

A1

A2

Bisector line

O 2 A1B1O 4O 2 A2B2O 4

B1

A1

C 1

A2

B2

C 2

O 2

O 4


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Motion Generation: 3-Position Coupler Task: Finding fixed pivot and length of

each link

1. Moving Pivots: Points A, B We find these points at

the intersection of thePerpendicular Bisectors of each moving pivot takenin pairs

The Crank(s) are drawnfor the Fixed Pivot to the

1st

pose of the movingPivot

2. Moving Pivot: Pointsother than A, B, such aspoints C, D.

O 2

O 4

A1

B1 A2 B2

A3

B3


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Motion Generation: 3-Position Coupler: After Design

Check Grashof Criteria for type

s =1.131 (Base) (Checkactual values)

l = 3.708 (Crank 1) p = 1.75 (coupler), q = 2.785

(Crank 2)

This is a Type II device notcapable of full rotation(p+q


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3: nominated fixed pivots One of points A, B is not moving pivot Using Nominated Fixed Pivots to

define moving pivot It relies on Apparent rather than

true positions that lie at vertices of congruent triangles with the desiredlamina positions for the fixed pivot of design interest.

Once the series of apparent fixedpivot positions are found as above,the position of the moving pivot on the1 st coupler position is set at the center of the circle that include all of theapparent fixed pivot locations

Congruent triangles

For the second crank we need to performa similar relationship with a differentnominated fixed Pivot

A2

B1

A1

A3

B2

B3

C *

C *2

C *3

B*

C 1

D

E

C 2

C 3


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A2

B1

A1

A3

B2

B3

C *

B*

C 1

C 2

C 3

B1

A1

A3

B2

B3

C *

C *2

C *3

A2


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3: nominated fixed pivots One of points A, B is not moving pivot Using Nominated Fixed Pivots to

define moving pivot It relies on Apparent rather than

true positions that lie at vertices of congruent triangles with the desiredlamina positions for the fixed pivot of design interest.

Once the series of apparent fixedpivot positions are found as above,the position of the moving pivot on the1 st coupler position is set at the center of the circle that include all of theapparent fixed pivot locations

Congruent triangles

For the second crank we need to performa similar relationship with a differentnominated fixed Pivot

A2

B1

A1

A3

B2

B3

C *

B*

C 1

C 2

C 3


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Starts Here:

We will prepare Apparent PositionsC*2 and C* 3 bymaking CongruentTriangles upon

A1-B1 workingwith triangles

A 2B2C* and A 3B3C*

The true position of the crank starting out is C* to C1where C1 is the center of the circle containing C* -- C 2*

-- C3* developed in the earlier steps


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Getting C2* -- used intersecting Arcs equal in length to A2-C* and B2-C*


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Now After the Position of C1 determined: The Crank is C*-C1

Found by making the 2congruent trianglesand then drawing a 3-point circle thru C*,C2* & C3* -- true

position C1 is thecenter of this circle


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If a 2nd

Fixed Pivot is desired! Make Congruent triangles to find apparentpositions of the Fixed Pivot

True Position of first instance of the moving

pivot is the center of circle containing D* - D2*- D3*

Then we would check for Grashof type andultimately ability to solve our problem.


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Change of Branch: Will it Work? We must determine if the linkage

can be assembled and travelwithout being disassembled andreassembled in the other SolutionBranch

Method 1: animation check Method 2:

For 4-bar linkages, we can define aangle -- an angle between thecoupler and the Driven (longer) crank in

all positions If the angle changes sign or direction,

we are looking at an impracticalsolution one that cant be operated


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3 Coupler Positions: Slider Crank Design Note the Coupler

Positions The Circle of Sliders is

formed by finding poles of positional pairs 1-2, 1-3 &2-3 and Image Pole P 23

note this image pole is areflection of the Truepole P 23 about a lineconnecting poles 1-2 and1-3

We need to define a fixedpivot (A* or B*) andconnect the crank from the* fixed pivot andappropriate end of thecoupler (the one used to

find fixed pivot)


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A1 B1 A2

B2 A3

B3

B*


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Generating P 12 :

Similar work forother two poles


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After Generating all three poles:


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With Image Pole P 23


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After Adding the Slider Circle & Slider line

The Triangle of Coupler position

to points on theslider line are allcongruent herethe lowest vertexis C markingthe Slider Line the Circle is 3-ptincluding I 12 , I 13 & I23


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Change of Branch?It looks like there is

no change of Branch ! A fix A i areall CCW;


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Will the Slider Crank Work?

We need to check order of points The individual C is should be traversed in the

positional order else the design fails assketched (like the one seen earlier as it turns

out!)


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Analytical Method : Techniques #1 and #2

#1: Finding poles ( P ij ) where apole is a point from which twolines can be repositioned by asimple rotation see EQN 6.16 itcontains the solution point locations assumed known in

Frame space We must be aware of matrix

singularity issues

#2: Find Circle Centers on which 3

points lie Treat Bi in Eqn 6.16 as A j (themiddle point of the 3)


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Analytical Method: Technique #3

Technique #3 : Coordinate Transformations: oxy, AXY Rotational transformations (Rs) as matrices Translational vectors relating origins of coor. systems

Relating Coupler (link AP) to Frame ( oxy ) is seen byvectors:

Typically we would know positions of the coupler inFrame Space ( a x , a y )

We can compute the Angle:


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Technique #3: Coordinate Transformations

Coordinate Transformation from Coupler toFrame is (given X, Y):

X, Y are Coupler coor. frame

x, y are base coor. frame

where is the angle X to x

x

y

a x X R

a y Y

Cos Sin RSin Cos

-


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Transforming Frame (oxy) to Coupler (AXY)

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/ /

/ /

1

and notice:

x P A P A

y P A P A

T

a X x A

aY y

Cos Sin A

Sin Cos

A R R

Technique #3: Inverse Transformation

-O

O


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Technique #4: Imaging Pole

223'23223'232121321213

23'23121323

'231213

23'2323

'2312131213

'23

'23

'23231312

'23231312

0

0

0

thenunlnownes,areandframe,fixedIn

y y x x y y x x

r r

y y y y x x x x

j y yi x x j y yi x x

r r

y x

P P P P

P P P P


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x

y

o

Motion Generation Design: Analytical Models

Crank design using circlepoints moving points oncoupler Requires circle point (joint on

coupler) coordinates betransformed to base frame or from ( X,Y ) to ( x, y ) if necessary

Find poles of the 3positions to define fixedpivot ( x*, y* )

Crank is ( x*, y* ) to ( x 1, y 1)from transformed circle point1

Locate 2 nd crank similarlyand build the linkage


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Given fixed pivotcoordinates in baseframe Transform coordinates of

( x*, y* ) to coupler space(for each coupler position)using A matrices

Find coupler pole of the 3*ed transformed position

it becomes ( X,Y ), nowtransform this pole to

frame coordinate for pos i (chosen one and use its Rmatrix)

Crank in pos i is located by( x*, y* ) to ( x i , y i )

x

y

o

X 1

Y 1 X 2

X 3


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Design of Slider: Follow the steps on page282 283

We use all the tools:

Transforming coor. betweenframe and coupler Finding poles Finding image poles Finding circles (center and

radius) Defining slider line points and

ultimately the slider line parametrically

Determining if slider order iscorrect

x

y

o


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Design Topics

Function generation 4-bar linkages

Input-output angle function Two precise positions

Three precise position

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Function Generation Using 4-Bar Linkage

The goal is to design a linkage whoscoupler or cranks follow the motion definedas a function between an input variable anda desired output result

Finding Precision Points Constraint equation: f ( , , a 1, a 2, a 3, a 4) = 0

, : Input and output a

1, a

2, a

3, a

4: design variables defining the linkage,

such as r 1, r 2, r 3, r 4, unknowns . Input/output function g ( , ) = 0: define , to

give precision points. If we have 4 a i s then itis 4 ( , ) precision points!


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Function Generation Design Procedure for 4-bar linkage

Target: a 4-bar linkage

Given: Function g : input/output relationship Range of input variable and/or output variable

1. Build constrain equation f by scaling one link length(such as r 1) as 1

2. Select number of (maybe 3) input positions: Chebyshev3. Using g to get three output positions4. Put steps 2, 3 into step 1 to get number of ( such as

three) equations5. Solve the equations to get design variables, such asr 2, r 3, r 4

6. Scale back to get the final design7. Check your design using Grashof Inequality

40


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The Actual Goal Is Linkage Design

We start with 3 values for and related to thedesired function Then we will develop x -, y - Component Models

for the links (with a unit base)

d l l k


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Models: linkage constraints

2 3 4

2 3 4

2 2 23 4 2 4 2 4 2

we will plug in our 3 pairs of values ,

to build 3 equations. & solve for t

We will eliminate leading to:

1

1 2 2 2

1, 2 3i i

r Cos r Cos r Cosr Sin r Sin r Sin

r r r r Cos r Cos r r Cos

i or

he 'si

r


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Our Text Suggests a Change of Variable as

This last equation canbe easily solved forthe Z is and then r isusing Coeff INV and

Mathematica or Matlab

2 2 22 4 3

1

2 4

22

34

1 2 1 3 1 1 1

1 2 2 3 2 2 2

1 2 3 3 3 3 3

1 1

2 2

3 3

Leads to modified Equations:z

z

z

into Matrix Form:

1

21

1

111

r r r z

r r z r

z r

z Cos z Cos Cos

z Cos z Cos Cos

z Cos z Cos Cos

Cos CosCos CosCos Cos

1 1 1

2 2 2

3 3 3

z Cos z Cos z Cos

S l b f i i i


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Select number of input positions:

The number is based on how many links areunknown1.Give three input values (three input posture)2.If input range is give, then we use Chebyshev formula to get output3.If no direct relationship between input and output( and ) is given, however, we are given both input

an output range and a second function ( x and y ),then we use linear approximation

44

Ch b h P l i l


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Chebyshev Polynomial

A tool to reduce approximation error: N = # of precision Points, i = 1 to N x 0 is the starting point and x f is the final point

x i are the roots of the polynomial used to solvethe model for precision points at minimumerror.

45

0 0 *2 2 2

f f i

x x x x i x Cos

N N

+

Ch b h P l i l ( )


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Chebyshev Polynomial, (cont.)

Here N = 2 thus we will find x 1 [ 1] & x 2 [ 2]

0 0

0 0

are the roots of the polynomial used to solve the

generator model for precision points at min. error

*2 2 2

* 2 12 2 2

i

f f i

f f i

x

x x x x i x Cos N N

x x x x x Cos i N

+

+

A li i li k i P bl 6 24


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Application - a linkage in Problem 6.24

Get 3 Precision Points using Ch ebyshevsPolynomial: (generating function: = 2 between .5and 1 radian)

NOTE: we neveruse end pointsas precisionpoints but themiddle is used if an odd numberof precisionpoints is needed

0

1

2

3

21 1

22 2

23 3

.5 ; 1.0

1.5 .5 2 1 1 .53352 2 2 3

1.5 .5 2 2 1 0.752 2 2 3

1.5 .5 2 3 1 .96652 2 2 3.2846

.5625

.9341

f r r

r r Cos r

r r Cos r

r r Cos r

r

r

r

R i S f M l b


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Resorting to Software: Matlabtheta1=0.5335;

theta2=0.75;theta3=0.9665;

fei1=0.2846;fei2=0.5625;fei3=0.9341;

A=[1 cos(fei1) -cos(theta1);1 cos(fei2) -cos(theta2);1 cos(fei3) -cos(theta3)];b=[cos(theta1-fei1); cos(theta2-fei2);cos(theta3-fei3)];

z= inv(A)*b

z =

1.0580-0.00190.1011

D i P


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Design Parameters:

r 2 = 1/z 2 = 1/-.00188 = -531.91 r 4 = 1/z 3 = 1/.1011 = 9.89 r 3 = (1+r 2^2 + r 4^2 - 2r 2r 4z1) = 521.44 Scaling back to Base size of 2 (not 1!)

r1 = 2 (given in problem) r2 = 1063.8 r3 = 1042.9 r4 = 19.8

Looking at these values: since they are sodifferent (link ratios are large) this is likely not avery good design (structurally unsound)

Additionally, it is a type II Grashof design

L t T P 6 23


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Lets Try Pr. 6.23: Problem:

Input-output constraints : = 2 a 1 + a 2Sin with a 1 anda 2 unknowns Function we wish to generate: = 2 2 (angles in

radians) To get a 1 and a 2 over the range: 0< < /2

Analysis : We need to select , to decide u nknowns a 1 &a 2 Need 2 precision points ( 1, 1) and ( 2, 2) Using the Chebyshev Polynomial

Fi di [ ] d [ ]


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1

1

2

2

0 02 2 * 2*1 12 2 2*2

0.7854 0.7854* ( / 4) 0.2300 13.180

0 02 2 * 2*2 12 2 2*2

0.7854 0.7854* (3 / 4) 1.3408 76.820

Cos

Cos r

Cos

Cos r

2 21 1

2 22 2

2 2*.2300 0.1058 6.06392 2*1.3408 3.5953 205.993

r r

Finding xs [ s] and ys [ s]:

or or

or

or

S l i f &


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Solving for a 1 & a 2

)sin(67956.4)48052.0(2

areThere

,for equationsthesolevcanWe

)sin(2

)sin(2

areThere,getcanwe,,Given

21

2212

1211

2121

aa

aa

aa

D t i i g d l t 45


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Determining model error at = 45

21 2

2

2 2

2 .96104 4.679564 4

1.2337 2.34791 1.11421

act est

g g

e

e a a Sin

e Sin

e

N di t l ti b t d d


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No direct relation between and x, and y:

In the previous slides we have explored thetechniques using precision points

Usually, however, we are not given output andinput relationship directly In this case, we normally use a linear relationship

between and x and another linear relationshipbetween and y as follows:

0

0

0

0

0

0

0

0

f f

f f

y y y y

x x

x x

00

0

0

000

0

f f

i

f f

i

i

y y y y

x x x x

i

Gi en F nction m + b


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Given a Function: y = mx + b

Example Y from 0 to 1.0 while X is .5 to 2.3 Therefore: b = -0.179 and m = 0.357

Y = 0.357 x -0.179

We want to change by -45 , (110 to 65 ) Rad: -0.78539, (1.9200, 1.1345)

while changes by -25 , (135 to 110 ) Rad: -0.4361, (2.3562, 1.9200)

Lets use: linear model relationships between X/ and Y/ 3 precision points due to r 1=1, r 2, r 3, r 4

Modeling: Chebyshev Polynomial


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Modeling: Chebyshev Polynomial

0 01

02

0 03

1

2

3

30 0.620582 2

1.42

150 2.179422 2

0.357*.62058 .179 0.042550.357*1.4 .179 0.32080

0.357* 2.17942 .179 0.59905

f f

f

f f

x x x x x Cos

x x x

x x x x x Cos

y y

y

Angles for the linkage:


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Angles for the linkage:With theseangular inputs,we enter the

Matrix Design Models of earlier.

Solve for Zisconvert to r2, r3,r4 (with unitbase length)

00

0

00

0

1

2 3

1

2 3

:

.62058 .5 .7854 1.9200 1.86742.3 .5

:

1.5273 ; 1.1872

0.04255 0 * .4363 2.3562 2.33761 0

:

2.2162 ; 2.0948

ii

f

ii

f

x x

x x y y y y

here

r

similarly

r r

r

similarily

r r

r means the unitis radian

D

D

If r = 1 then:


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If r 1 = 1 then:

r 2 = 1/ z 2 = 1/3.77042 = 0.2652 r 4 = 1/ z 3 = 1/.68066 = 1.4692 r 3 = (1+r 22 + r 42 2r 2r 4z 1) =1.9143

This is a Type I Grashof linkage ( s+l < p+q ) and r

2is the crank in a crank rocker

type of linkage

Graphical Approach for Function Generation 4 bar Linkage


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Graphical Approach for Function Generation 4 bar Linkage

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ME 3230 Chapter 6 Plannar Linkage Design -1

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