MENG 372
Chapter 5
Analytical Position Synthesis
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
1
5.1 Types of Kinematic Synthesis
• 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
2
5.2 Precision Points
• The points, positions prescribed for successive
locations of the output (coupler or rocker) link in the
plane.
• In graphical synthesis:
P1
move from C1D1 to C2D2
a2
• In analytical synthesis:
P2
move from P1 to P2 while
rotating coupler a2
(note: angles are measured
anticlockwise)
3
Precision Points
• Can define vectors
Z and S from the
attachment points
E and F to P
• Note: the coupler
is not triangular,
but 3 points are
defined on the
coupler
• Points E and F are
called A and B
P1
a2
Z1
S1
A
P2
B
4
5.3 Two Position Synthesis
• Want to move from P1 to
P2 while coupler rotates a2
• Given P21, d2 and a2
• Design each half
separately
• Write vector loop
equation(s) to include
given values, find free
choices to make problem
easy to solve.
5
Problem Statement
Y
Design a 4-bar linkage which will move P1 to P2
while coupler rotates thru a2. P lies on coupler.
Find the lengths and angles of all links.
R2 R
1
d2
1. Choose any coordinate system X-Y
P2 P P1
2. Draw vector P21 inclined at d2
21
3. Define position vectors R1 and R2
Z1
a
Z2 2
4. Draw an arbitrary vector Z1. Then
form vector Z2 with same
magnitude but angle a2 with Z1.
5. Draw vectors W1 and W2 to meet at
O2.
W1
6. Write vector loop equation.
W2
O2
X
6
Two Position Synthesis
• Vector loop equation
W2 + Z2 - P21 - Z1 - W1 = 0
• Write complex vectors
we
i    2 
 ze
i   a 2 
 p 2 1e
id 2
 ze
i
 we
i
0
• Expand exponents
i
we e
i 2
i
 ze e
ia 2
 p 21 e
id 2
 ze
i
 we
i
0
• Combine terms
we
i

e
i 2

 1  ze
i

e
ia 2

 1  p 21 e
id 2
7
we
i
e
i 2
Two Position Synthesis
 1   ze  e  1   p e
i
ia 2
id 2
21
• Variables
w, , 2, z, , a2, P21, d2 = 8
• Given
P21, d2, a2
=-3
• Complex equations: 1
can solve for 2 unknowns
=-2
• Free Choices
=3
8
we
i
e
Two Position Synthesis
 1   ze  e  1   p e
i 2
i
ia 2
id 2
21
• Choose (, 2, )
S  e  e  1
T  e  e  1
i
i 2
i
ia 2
U  p 21 e
id 2
Gives 2 simultaneous eqns.
wS  zT  U
w S  zT  U
9
we
i
e
i 2
Two Position Synthesis
 1   ze  e  1   p e
ia 2
i
id 2
21
• Choose (2, z, )
we
i

p 21 e
id 2
 ze
e
i 2
i

e
1
ia 2
1

Q
from which the magnitude and
angle can be calculated
w=abs(Q), =angle(Q)
• The other side can be calculated
similarly
10
Two Position Synthesis
• Once both sides have been
solved, the coupler and
ground can be calculated
using
V1  Z 1  S 1
G1  W1  V1  U 1
v=abs(V1)
g=abs(G1)
11
Two Position Synthesis Comparison
• For graphical, position of attachment points A and B
relative to P in x and y directions (4) and points of O2 and
O4 along the perpendicular bisectors (2) gives 6 total
• For analytical, 3 free choices each side * 2 sides=6 total
12
5.6 Three Position Synthesis
• Want to move from
P1 to P2 while
coupler rotates a2
and from P1 to P3
while coupler
rotates a3
• Given P21, d2, P31,
d3, a2 and a3.
13
Three Position Synthesis
• Vector loop equations
W2 + Z2 - P21 - Z1 - W1 = 0
W3 + Z3 - P31 - Z1 - W1 = 0
• Write complex vectors
w. e
we
i    2 
i    3 
 ze
 ze
i   a 2 
i   a 3 
 p 21e
id 2
 ze  w e
 p 31e
id 3
 ze  w e
i
i
0
i
i
0
• Combine terms
we
we
i
i

e
e
i 2
i3


 1   ze  e
 1  ze
i
i
e
ia 2
ia 3

 1 
 1  p 21 e
p 31 e
id 2
id 3
14
we
we
i
i

e
e
i 2
i3
Three Position Synthesis


 1   ze  e
 1  ze
i
i
e
ia 2
ia 3

 1 
 1  p 21 e
p 31 e
id 2
id 3
• Variables
w,,2,3,z,,a2,a3,P21, P31,
d2 ,d3
= 12
• Given
P21,P31,d2,d3,a2,a3
=-6
• Complex equations *2
2*2
=-4
• Free Choices
=2
15
Three Position Synthesis




w e  e  1   ze  e  1  
• Choose (2, 3 )
we
i
e
 1  ze
i3
i
W  we
Z  ze
i 2
i
i
i
i
e
ia 2
 1  p 21 e
ia 3
p 31 e
id 2
id 3
S2  e
i 2
1
T2  e
ia 2
 1 U 2  p 21e
id 2
S3  e
i 3
1
T3  e
ia 3
1
id 3
U 3  p 31 e
• Two linear equations
.
WS 2  ZT 2  U 2
WS 3  ZT 3  U 3
• Gives solution
w=abs(W), =angle(W)
z=abs(Z), =angle(Z)
16
WS 2  ZT 2  U 2
WS 3  ZT 3  U 3
Solution
Eliminate W to get:
Z 
U 2S3  U 3S 2
T 2 S 3  T3 S 2
Then solve for W:
W 
U 3  Z T3
S3
(USE MATLAB)
17
ue
j
e
j 2
 1   se
j
e



Choose (2, 3 )
ue
j
e
j 3
 1  se
j
e
ja 2
ja 3
 1   p 21e

 1  p 31e
jd 2
jd 3
REPEAT FOR RIGHT-HAND SIDE
OF LINKAGE
U  ue
j
S2  e
j 2
1
T2  e
ja 2
1
U 2  p 21e
jd 2
S  se
j
S3  e
j 3
1
T3  e
ja 3
1
U 3  p 31e
jd 3
Two linear equations
U S 2  ST 2  U 2
U S 3  ST3  U 3
(USE MATLAB)
18
Three Position Synthesis Comparison
• For graphical, position of attachment points A and B
relative to P in x and y directions (4)
• For analytical, 2 free choices each side * 2 sides=4 total
19
Example
Design a 4-bar
linkage to move A1P1
to A2P2 to A3P3
20
21
3 Position Synthesis with Specified
Fixed Pivots
.
• Want to move from P1 to
P2 while coupler rotates a2
and from P1 to P3 while
coupler rotates a3 and
attach to ground at O2 and
O4
• Given R1,R2,R3,z1,z2, z3,
a2 and a3
• Note: if R1 and R2 are
satisfied, P21 is satisfied,
and R1 and R3 give P31

22
3 Position Synthesis with Specified
Fixed Pivots
.
• Vector loop equations
W1+Z1=R1
W2+Z2=R2
W3+Z3=R3
• Use relationships
W 2  W1 e
i 2
, W 3  W1 e
Z 2  Z 1e
ia 2
, Z 3  Z 1e
to get

i 3
ia 3
W1  Z 1  R 1
e
e
i 2
W1  e
ia 2
Z1  R 2
i 3
W1  e
ia 3
Z1  R 3
23
3 Position Synthesis with Specified Fixed Pivots
.
W1  Z 1  R 1
e
e
we
i 2
W1  e
ia 2
Z1  R 2
e
i 3
W1  e
ia 3
Z1  R 3
e
i 2
i 3
we
we
i
 ze
i
 Re
iz 1
i
 Re
iz 2
i
 Re
iz 2
i
e
ia 2
i
e
ia 3
ze
ze
• Variables
w,,2,3,z,,a2,a3 ,R,z1,z2,z3
= 12
• Given
R,z1,z2,z3,a2,a3
=-6
• Complex equations *2
3eqn*2
=-6
• Free Choices (Sub)
=0

This makes the problem hard
24
3 Position Synthesis with Specified Fixed Pivots
.
W1  Z 1  R 1
e
e
we
i 2
W1  e
ia 2
Z1  R 2
e
i 3
W1  e
ia 3
Z1  R 3
e
i 2
we
i 3
we
i
 ze
i
 Re
iz 1
i
 Re
iz 2
i
 Re
iz 2
i
e
ia 2
i
e
ia 3
ze
ze
From 1st equation: Z 1  R 1  W1
Use this to eliminate Z1
e  e  W  R  e R
i 2
e
i3
ia 2
e
ia 3
ia 2
W
1
2
1
1
 R3  e
ia 3
R1
Divide 2 eq’ns to eliminate W1
e
e
i 2
e
ia 2
i 3
e
ia 3



R2  e
ia 2
R3  e
ia 3
R1
R1
Cross Multiply
e
i 2
e
ia 2
R
3
e
ia 3
 
R1  e
i3
e
ia 3
R
2
e
ia 2
R1

25
3 Position Synthesis with Specified Fixed Pivots
.

e
i 2
e
ia 2

R3  e
ia 3
 
R1  e
i3
e
ia 3

R2  e
ia 2
R1

Arrange into form
A  Be
i 2
 Ce i  0
3
where
B e
ia 3
ia 2
R3  e
ia 3
R1  R 3
A e
C  R2  e
ia 2
R2
R1
using s and t:
te
gives
i 2
and s  e
i3
A  Bt  Cs  0
26
3 Position Synthesis with Specified Fixed Pivots
.
A  Bt  Cs  0
(a)
Taking conjugate
A  Bt  Cs  0
Since s and t represent angles
A
B
t

C
0
s
Multiplying by st
A st  B s  C t  0 (b)
A  Bt
From (a) s  
C
• Substituting into (b) gives a
quadratic function of only t
27
3 Position Synthesis with Specified Fixed Pivots
.
at
where
2
 bt  c  0
a  A B , b  A A  B B  C C , and c  A B
Solving gives
t 1, 2 
b
b  4 ac
2
 e
j 2
2a
Only one of the t will be valid.
s can be solved using
s
A  B t1, 2
e
i 3
C
Any 2 of the first eqns can be
used to solve for W1 and Z1
 1
 i 2
e
1  W 1   R 1 


ia 2  
e   Z1   R 2 
28
3 Position Synthesis with Specified Fixed Pivots
.
Summary of calculations (for MATLAB implementation)
B e
ia 3
ia 2
R3  e
ia 3
R1  R 3
A e
C  R2  e
ia 2
R2
R1
a  A B , b  A A  B B  C C , and c  A B
t 1, 2 
b
s  
b  4 ac
2
 e
j 2
2a
A  B t 1, 2
e
j 3
C
 1
 j 2
e
1  W 1   R 1 
 

ja 2  
e
  Z 1  R 2 
w=abs(W1), =angle(W1), z=abs(Z1), =angle(Z1)
29
Example Problem
• Move from C1D1 to C2D2 to C3D3 using attachment
points O2 and O3
• Call point C, P
3
2
1
30