Analytical Linkage Synthesis

Analytical Linkage

SynthesisChapter 5

Types of Kinematic Synthesis Types

Function Generation

Correlation of an input function with an output function in a mechanism

Path Generation

Control of a point in the plane such that it follows some prescribed path

Motion Generation

Control of a line in the plane such that it assumes some sequential set of prescribed

positions

Chapter 3 - Example 1 (revisited) Rocker Output- Two Position with

Angular Displacement (Function)

Design a four bar Grashof crank-rocker speed motor input to give 45 of rocker

motion with equal time forward and back,

from a constant speed motor input.

Two-Position Synthesis for Rocker Output

Generic annotation

Link 4 is the output link to be driven by a dyad consisting

of link 2 and 3.

To be determine: links 1, 2, 3, and O2.

Defined: link 4, O4, 4 and

Procedure Choose a location on link 4 to

attached link 3, B1 and B2 in its

extreme locations. This defined R4.


4441 cos ROB xx 4441 sin ROB yy

4442 cosROB xx 4442 sinROB yy

12 BBRRM

uMuRuL B 1

Procedure Place the crank pivot O2 suitable far

from B1 along line L

The length of the crank must be half the length of M


3 Class ,

2 Class ,

1 Class ,

21

21

21

MOB

MOB

MOB

32 12

KKMRR BO

2/sin5.0 42 RMR

Procedure Link 3 can be found by

subtracting R2 from the

magnitude of RB1-RO2

Link 1 is found by subtracting RO2 from RO4

Grashof crank-rocker /no quick return


23 21RRRR OB

241 OORRR

Two-Position Motion Generation AL

Two-Position Motion Generation AL Problem Statement

Design a fourbar linkage which will move a line on its coupler link such that

a point P on that line will be first at P1and later at P2 and will also rotate the

line through an angle 2 between those two positions. Find the lengths and

angles of the four links and the coupler

link dimension A1P1 and B1P1.

Two-Position Motion Generation AL Procedure

Define the two desired precision positions

The dyad W1Z1 (red)defines the left half of the linkages.

U1S1 (black)the right.

The pin-to-pin length and angle of link 3 is define in

terms of vector Z1and S1.

1221 RRP

111 SZV

Two-Position Motion Generation AL Procedure

The ground links is define in terms of the two dyads

First solve for the left side of the linkage

To solve for W1 and Z1write the vector loop

equation

1111 UVWG

0112122 WZPZW

Two-Position Motion Generation AL0112122 WZPZW

0

0

222

222

21

21

jjjjjjj

jjjjj

wezeepezeewe

wezeepzewe

222 2111 jjjjj epezeewe

22122

22

cossinsin1coscos

sinsin1coscos

pzz

ww

Real part:

Imaginary part:

22122

22

sinsincos1cossin

sincos1cossin

pzz

ww


22122 and,,,,,,,

:ableseight vari are There

pzw

2212 and,,:statement

problem in the defined are Three

p

,, :assumed are Three 2

equations two with thesolve are Two

g,simplifyin

221

22

22

cos

sinsin1coscos

sinsin1coscos

pC

B

A

Strategy 1


g,simplifyin

221

22

22

sin

sincos1cossin

sincos1cossin

pF

E

D

then,

FEzDw

CBzAw

usly,simultaneo solving

BDAE

BFCEw

BDAE

CDAFz


Strategy Second



p

2

1

2link

,, Zvector :assumed are Three

z

1W vector for the solveThen

22122 and,,,,,,,


pzw


11 Z and W vectorsof componentsy and x The

2212121

2121

cossin1cos

sin1cos

pZZ

WW

yx

yx

Real part:

Imaginary part:

sin sin

cos cos

11

11

zZwW

zZwW

yy

xx

2212121

2121

sinsin1cos

sin1cos

pZZ

WW

xy

xy

Two-Position Motion Generation ALg,simplifyin

2212212

222

sin cos sin

1cos sin 1cos

pFpED

CBA

substituting:

FDZCZBWAW

EDZCZBWAW

xyxy

yxyx

1111

1111

the solution is;

A

FDZCZBEDZCZAW

xyyx

x 2

1111

1

A

EDZCZBFDZCZAW

yxxy

y 2

1111

1


0112122 USPSU

222 2111 jjjjj epeseeue

22122

22

cossinsin1coscos

sinsin1coscos

pss

uu

For the right hand dyad, U1S1

Imaginary part:

22122

22

sinsincos1cossin

sincos1cossin

pss

uu

Real part:


22122 and,,,,,,,


psu



p

,, :assumed are Three 2

equations two with thesolve are Two

g,simplifyin

221

22

22

cos

sinsin1coscos

sinsin1coscos

pC

B

A


g,simplifyin

221

22

22

sin

sincos1cossin

sincos1cossin

pF

E

D

then,

FEsDu

CBsAu

usly,simultaneo solving

BDAE

BFCEu

BDAE

CDAFs


Strategy Second



p

2

1

3link

,, Svector :assumed are Three

s

1U vector for the solveThen

22122 and,,,,,,,


psu


11 Z and W vectorsof componentsy and x The

2212121

2121

cossin1cos

sin1cos

pSS

UU

yx

yx

Real part:

Imaginary part:

sin sin

cos cos

11

11

sSuU

sSuU

yy

xx

2212121

2121

sinsin1cos

sin1cos

pSS

UU

xy

xy

Two-Position Motion Generation ALg,simplifyin

2212212

222

sin cos sin

1cos sin 1cos

pFpED

CBA

substituting:

FDSCSBUAU

EDSCSBUAU

xyxy

yxyx

1111

1111

the solution is;

A

FDSCSBEDSCSAU

xyyx

x 2

1111

1

A

EDSCSBFDSCSAU

yxxy

y 2

1111

1

Comparison Graphical/Analytical

Example 5-1 Design a fourbar linkage to

move the link APB shown

from position A1P1B1 to

A2P2B2.

Solution

1. Draw the link APB in its two desired positions to

scale

2. Measure or calculate the values of the magnitude and

angle of vector P21

2.165

416.2

2

21

p

Example 5-1 3. Measure or calculate the

values of the change in

angle, 2, of vector Z from position 1 to 2,

4. Assume three additional free choice. Using strategy 2,

3.432

4.38

5.26

298.1

z

Example 5-1 5. Substitute these values in

the corresponding equation

A-F, W1x .. and obtain,

6. Compare to graphical,

This vector W1 is link 2

6.71

467.2

w

71

48.2

w

Example 5-1 7. Repeat the procedure for

the link-4 side, the free

choices,

8. Substitute these values along with the original

values,

9. Compare the graphical solution

6.85

1.104

035.1

2

s

3.432.165416.2 2221 p

4.15

486.1

u

1453.1 u Vector U1 is link 4

Example 5-1 10. Line A1B1 is link 3,

Line O2O4 is link 1,

11. Check the grashofcondition

12. Construct a model in CAD. Check for limiting

conditions

13. Check transmission angles

111 SZV

1111 UVWG

Analytical Linkage Synthesis

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