four point synthesis crank constraint equationsmoreno/kinematics/me 5150 kinematics spring... · 1...

1

Design Synthesis

Four Point Synthesis – Crank Constraint Equations

4 unknowns a0x, a0y, a1x, a1y

3 Constraint equations will be non-linear, require iterative method for solution

)()(

0)()()()(

11

010100

aDa

aaaaaaaa

jj

Tj

Tj

http://www.uconn.edu/index.php

2

Design Synthesis Newton-Raphson Method, finding root of non linear equation (1 variable)

dxxdf

xfxxxx

dxxdf

xfxxxx

dxxdf

xfx

rootxxf

xdx

xdfxfxxf

)()(

)()(

)()(

0)(

)()()(

1

1112

0

0001

Taylor Series

correction

Sensitive to initial guess

A is stable

B is not


3

Design Synthesis In kinematic analysis often have n non linear equations with n unknowns

solutionaatxf

ninx

fix

fix

fixfxf

xxxxf

xxxxf

xxxxf

i

nii

nn

n

n

0)(

2,1)()(

0),,,(

0),,,(

0),,,(

22

11

321

3212

3211

Expand Taylor series


4

Design Synthesis

nnn

n

i

f

f

f

xfn

xfn

xf

xf

xf

equationslinearofset

fx

fxf

2

1

2

11

2

1

1

1

1

0)(

Solve for δs and update, repeat

0

1

k

kkk xx

Correction vector


5

Design Synthesis Constant length for guiding cranks

Length constraint equation

0)()()()(

0)()()()(

0)()()()(

)()(

4,3,20)()()()(

01010404

01010303

01010202

1

010100

aaaaaaaa

aaaaaaaa

aaaaaaaa

aDa

jaaaaaaaa

TT

TT

TT

ijj

Tj

Tj 1)


6

Design Synthesis Constant length for guiding cranks

Length constraint equation

0)()()()(

0)()()()(

0)()()()(

010101140114

010101130113

010101120112

aaaaaaDaaD

aaaaaaDaaD

aaaaaaDaaD

TT

TT

TT

3 non linear equations with 4 unknowns a0x,a0y,a1x,a1y

Assume 1 value (say a0x) and solve for other 3

Based on initial assumption, get

A series of fixed and moving crank points (initial position)

Any pair should work but need to check mobility

All possible fixed points = center point curve

All possible moving points = circle point curve

1)

0a

x

1a1a

0a

y


7

Design Synthesis Example - 4 bar linkage with 4 precision points

90)0.2,0.2(

45)5.1,0.3(

0)5.0,0.2(

)0.1,0.1(

144

133

122

1

P

P

P

P

Added 4 th point

P4

3P

2P1P

y

x12

13

14


8

Design Synthesis

Solving 1)

Center point (fixed)

solutions

Circle point (fixed)

solutions


9

Design Synthesis

100

101

310

100

086.707.707.

3707.707.

100

5.010

101

100

)(

)(

14

13

12

1111211

1111211

12

D

D

D

cpsppcs

spcppsc

D yxy

yxx

Displacement Matrices

for 4 precision points


10

Design Synthesis

0.2,5.0

1.2,2.1

25

4.0,5.1

25.1,5.2

30

:intint

11

00

11

00

yx

yx

yx

yx

ccmoving

ccfixed

solution

aamoving

aafixed

solution

curvespocirclepocenterfrompick

30

25


11

Design Synthesis

1

0.2

5.0

100

101

310

1

5.1

0.1

1

4.0

5.1

100

101

310

1

5.2

6.2

1

0.2

5.0

100

086.0707.0707.0

3707.0707.0

1

85.1

94.1

1

4.0

5.1

100

086.0707.0707.0

3707.0707.0

1

43.1

78.3

1

0.2

5.0

100

5.010

101

1

5.1

5.1

1

4.0

5.1

100

5.010

101

1

1.0

5.2

11441144

11331133

11221122

cDcaDa

cDcaDa

cDcaDa

Other positions


12

Design Synthesis


• Start with constraint equations

– Crank length or slider slope

• Calculate D matrix for defined point/angles of

rigid body (1 to 2, 1 to 3, 1 to j)

• Substitute displacement equation constraint

eqns

• Get linear equations for initial crank/slider

position (e.g.a1x,a1y)

• Determine initial position

• Determine additional positions using D

equations 13

Design Synthesis Process Summary


14

Design Synthesis –Function Generation

Function Generation:

Previously: Considered position synthesis (points/angles)

Now: Consider mechanisms where output motion is specified function

of input motion. (Crank and sliders)

Approach: Function generator synthesis problem converted to equivalent

rigid body guidance problem (last section) using

principle of inversion

Motion of guided rigid body described by the relative motion

of the input member with respect to the output member


15


Function generation requires consideration of error (error curve)

Δ between mechanical output φ and theoretical function is called

structural error

Actual error curve is dependent on precision point spacing.

Optimum spacing results in equal error between successive pairs of precision

Points (over range x0 to xf)

f

f

f

f

rangewith

xfyxf

xxxfor

yxftoalproportionisoutput

xtoalproportionisinputwhere

xfythatsuchlinkageasynthesize

statementoblem

0

0

0

0

:

)()(

:

)()(

)(:

)(:

Pr


16


11

11

00

00

0

0

0

0

)]()([

)(

:)(int

))()((

)(

)(

)(

:

,

jj

jj

jj

jj

f

f

f

f

positionfirsttoreferencedgenerally

xfxfk

xxk

jspo

xfxfyk

xxxk

define

yxthatinsureTo

0xfx

0 f

jx

j

0 j f

)( fxf

0 j f


17


Chebyshev Spacing – good first approximation

n

spoprecisionn

xxx

njjxxx

f

j

180

int#

,2,1)2

cos(12

0

0

xxx

xxx

xxx

n

933.0

500.0

067.0

3

03

02

01

xxx

xxx

xxx

xxx

n

962.0

692.0

309.0

038.0

4

04

03

02

01


18


0a0b

1a

ja

1b

jb

j1

j1

Four Bar Linkage Function Generator

jj and 11

Specified crank rotations

input output

Convert specified rotations into equivalent rigid body motion

of input crank a0a with respect to output crank b0b – kinematic inversion


19


0a0b

1a

ja

1b

jb

j1

j1

First consider input crank - Function Generator

z

y

x

jj

jj

jz

jy

jx

a

a

a

cs

sc

a

a

a

1

1

1

11

11

100

0

0

Displacement Matrix for rotation of a1 about a0 (0,0)

100

0

0

100

11

11

1111211

1111211

1 jj

jj

jyjxyjj

jyjxxjj

cs

sc

cpsppcs

spcppsc

Dj


20


0a 0b

1a

ja

1b

jb

j1

j1

Four Bar Linkage Function Generator

0b

0a

1a

ja

1b

jb

j1

0'a

ja'

j1

Rigid body rotation about bo

-angles are rigid

- bring b0bj Coincident with b0b1

a0 a’0, aj a’j

inversion

Next


21


100

)()()(

)(1)()(

100

)()()(

)(1)()(

100

)cossin(cossin

)sincos(sincos

111

111

111

111

112

112

1

1

1

jjj

jjj

jjj

jjj

yxy

yxx

scs

csc

D

scs

csc

D

ppp

ppp

D

j

j

j

Displacement matrix form

about b0=(1.0,0.0)

Rigid body rotation about b0

0b


22

100

)()(

1)()(

)()(

100

0

0

100

)1(

11111

11111

11

11

111

111

11

jjjjj

jjjjj

R

jj

j

jj

jjj

R

jjR

scs

csc

D

sAsBcAcBBAccAsBsAcBBAs

cs

jsc

jscs

csc

D

DDD


Total relative Displacement matrix

Rigid body rotation Input crank rotation

using


23


0b

In the inversion

Coupler link (ab) acts as a guiding crank

for the input crank –rigid body

Motion of input crank defined by [Dr]

a0b0 can also be considered

a guiding crank

inversion

0a

1a

ja

1b

jb

j1

0'a

ja'

j1


24

yx

jjyjxjj

xjyjjjjjj

yjxjjjjjj

jyjxj

aaforsolve

definedspoprecisionwith

ddadadC

adadddddB

adadddddA

jCaBaA

11

232

132

023013

01202223221312

02101123211311

11

,

int2

)(2

1

)1(

)1(

3,2


)()()()( 01010 aaaaaaaa Toj

Tj

Solution Approach – consider previous crank synthesis


25

)(2

1

)1(

)1(

3,2

232

132

123113

11212223221312

12111123211311

11

jjyjxjj

xjyjjjjjj

yjxjjjjjj

jyjxj

drdrbdrbdrC

bdrbdrdrdrdrdrB

bdrbdrdrdrdrdrA

jCaBaA

Use Dr elements (drikj) for dikj

Specified b1x and b1y replaces a0x and a0y

Specified co-ordinates of first position of output crank

Center of relative motion of a relative to bob1


)()()()( 01010 aaaaaaaa Toj

Tj

0b

Use same solution approach as position synthesis


26

Example: Synthesis 4 bar linkage with

)1875.0,375.0(),0,1(),0,0(:

4560

15:30:

10

1313

1212

bbawith

outputinput

o


0a

1b

0b

?1a

x

y

Defined Initial position


27


100

707.259.966.

293.966.259.

100

)45()4560()4560(

))45(1()4560()4560(

45,60

100

259.707.707.

034.707.707.

100

)15()1530)1530(

))15(1()1530()1530(

15,30

3

2

scs

csc

D

scs

csc

D

jR

jR

100

)()(

1)()(

11111

11111

jjjjj

jjjjj

R scs

csc

D


28


)(2

1

)1(

)1(

3,2

232

132

123113

11212223221312

12111123211311

11

jjyjxjj

xjyjjjjjj

yjxjjjjjj

jyjxj

drdrbdrbdrC

bdrbdrdrdrdrdrB

bdrbdrdrdrdrdrA

jCaBaA

523.0,344.0

050.026.106.

027.003.083.

050.3

026.3

106.3

027.)259.034(.5.0)1875)(.259(.)375((.034(.

003.)375)(.707(.)1875)(.707.1()259)(.707(.)034)(.707(.

083.)1875)(.707.()375)(.707.1()259)(.707.()034)(.707(.

11

11

11

22

2

2

2

yx

yx

yx

aa

aa

aa

C

B

A

C

B

A

Initial position for a1


29


For other points, use displacement matrix for crank rotation

Rotation about a0 (0,0)

1

036.

625.

1

523.

344.

100

05.866.

0866.5.

60

1

281.

559.

1

523.

344.

100

0866.5.

05.866.

30

100

0

0

13

13

12

12

11

11

13

12

1

aDa

input

aDa

input

cs

sc

D jj

jj

j


30


100

)()()(

)(1)()(

100

)cossin(cossin

)sincos(sincos

111

111

1111211

1111211

1

1

jjj

jjj

jyjxyjj

jyjxxjj

scs

csc

D

ppp

ppp

D

j

j

Displacement matrix form Rigid body rotation about b0=(1.0,0.0)

with


31


For b2 and b3

1

309.

425.

1

1875.

375.

100

707.707.707.

293.707.707.

45

1

019.

347.

1

1875.

375.

100

259.966.259.

034.259.966.

15

100

1

100

010

011

13

13

12

12

111

111

111

111

13

12

1

bDb

output

bDb

output

scs

csc

scs

csc

D jjj

jjj

jjj

jjj

j


32


4560

15:30:

1313

1212

outputinput


33


Slider Function Synthesis

Input crank θ1j ~ x

Output slide displacement d1j ~ y

Assume slide slope is α relative to x axis

Assume a0=(0,0) and b1 =(1,0) 1st position of output slider

y=f(x)

(0,0)

(1,0) a0

a1 aj

b1

θ1j

α

d1j

Inversion of input crank

about b1 b1j


34


Slider Function Synthesis

Input crank θ1j ~ x

Output slide displacement d1j ~ y

Assume slide slope is α relative to x axis

Assume a0=(0,0) and b1 =(1,0) 1st position of output slider

y=f(x)

(0,0)

(1,0) a0

a1 aj

b1

θ1j

α

d1j


about b1 b1j


35


100

0

0

:

11

11

1 jj

jj

cs

sc

D

crankinputofrotation

j

For a0=(0,0)


36



about b1

(1,0) a0

a1 aj

b1

θ1j

d1j

b1j

100

1111211

1111211

1

1 jyjxyjj

jyjxxjj

d cpsppcs

spcppsc

D

baboutlinkagerigidofinversion

j


37


bj

b1

Θ =0 for constant slope of slider

Inversion about b1, going from b2 to b1

(1,0)

100

100

0

0

100

10

01

100

10

01

100

)()(

)()(

0

1

111

111

11

11

1

1

1

1

111

111

1112

1112

11

1

sdcs

cdsc

D

cs

sc

sd

cd

DDD

sd

cd

csdbscdbbcs

ssdbccdbbsc

D

sdsdbb

cdcdbb

jjj

jjj

R

jj

jj

j

j

dR

j

j

ijyijxy

ijyijxx

d

jjyy

jjxx

jj

j

α

-d1j

x

y

For a0=(0,0), b1=(1,0)


38

100

758.707.707.

4375.707.707.

100

433.866.5.

25.5.866.

60.875.045

)0,0.1(),0,0(.5.030

.

1312

1313

101212

RR DD

ind

baindoutputinput

withCrankSliderPositionThreeaSynthesizeEx


100

111

111

1

sdcs

cdsc

Dr jjj

jjj

j


39


71.6,348.4

820.0480.0552.0

375.0250.0299.0

820.0480.0,552.0

375.0,250.0,299.0

11

11

11

333

222

yx

yx

yx

aa

aa

aa

CBA

CBA

Initial position of crank pivot

)(2

1

)1(

)1(

3,2

232

132

023013

01202223221312

02101123211311

11

jjyjxjj

xjyjjjjjj

yjxjjjjjj

jyjxj

ddadadC

adadddddB

adadddddA

jCaBaA


40


For positions a2 and a3

100

0707.707.

0707.707.

100

0866.5.

05.866.

100

0

0

13

12

1 11

11

D

D

cs

sc

D jj

jj

j

1

81.7

67.1

1

894.7

410.0

13

12

13

12

aDa

aDa


41


a1

(-4.348,-6.71)

a2

(-0.410,-7.984)

a3

(+1.67,-7.81)

-5.0


four point synthesis crank constraint equationsmoreno/kinematics/me 5150 kinematics spring... · 1...

Documents