MENG 372
Chapter 4
Position Analysis
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
1
Coordinate Systems
• Cartesian (Rx, Ry)
• Polar (RA, q)
• Converting between the two
R A  Rx 2  R y 2
q  arctan R y Rx 
Rx  RA cosq
Ry  RA sin q
• Position Difference, Relative position
– Difference (one point, two times)
– relative (two points, same time)
RBA=RB-RA
Y
A
RBA
B
RA
RB
2
X
4.3 Translation, Rotation, and
Complex motion
• Translation: keeps the same angle
• Rotation: one point does not move
• Complex motion: a combination of rotation
and translation
3
Graphical Position Analysis of
Linkages
Given the length of the links (a,b,c,d), the ground
position, and q2. Find q3 and q4
b
B
A
c
b
q3
c
a
q2
O2
d
q4
O4
4
Graphical Linkage Analysis
• Draw an arc of radius b,
centered at A
• Draw an arc of radius c,
centered at O4
• The intersections are the
two possible positions for
the linkage, open and
crossed
b
c
B1
b
q3
A
c
a
q2
d
O2
q4
O4
B2
5
Algebraic Position Analysis
Obtain coordinates of point A:
Ax  a cos q 2
Ay  a sin q 2
Obtain coordinates of point B:
2
2
b 2  Bx  Ax   B y  Ay 
c 2  B x  d   B y
2
2
These are 2 equations in 2 unknowns: Bx and By
See solution in textbook pages 171, 172.
6
Complex Numbers as Vectors
• We can plot complex numbers on the realimaginary plane
• Euler identity e±iq=cos q ± i sin q
• Cartesian form: RAcos q + i RAsin q
Imaginary
• Polar form: RAeiq
• Multiplying by eiq
corresponds to rotating
by q
Real
7
Analytical Position Analysis
• Given: link lengths a,b,c and d, q1, q2 (the motor
position)
• Find: the unknown angles q3 and q4
8
Analytical Position Analysis
Write
 the
 vector
 loop
 equation:
R2  R3  R4  R1  0
(Positive from tail to tip)
Substitute with complex vectors
aeiq 2  beiq3  ceiq 4  deiq1  0
Take knowns on one side, unknowns on the other.
Call the knowns Z
be
iq 3
 ce
iq 4
Unknowns
 ae
iq 2
 de
iq1
Z
Knowns
9
Fourbar Linkage Analysis
beiq3  ceiq 4  aeiq 2  deiq1  Z
Define: s  eiq3 , t  eiq4
bs  ct  Z
Take conjugate to get a second equation:
bs  ct  Z
For the conjugate of s we have (only true for
eiq)
s e
 iq3
1
1
 iq3 
e
s
So our second equation is
b c
 Z
s t
Note:
1
Z 
Z
10
Fourbar Linkage Analysis
bs  ct  Z
b c
 Z
s t
Use algebra to eliminate one of the unknowns
b
c
Z 
s
t
bs  Z  ct
Multiplying the two gives:
c
b  ZZ  Z  Z ct  c 2
t
2
Multiplying by t and collecting terms gives:
2
2
2
Quadratic equation in t
0  Z ct  ZZ  c  b t  Zc
From the quadratic formula t 

 ZZ  c
 ZZ  c  b 
2
2
2
b

2 2
2Z c
s
 4c 2 Z Z
Z  ct
b
11
Fourbar Linkage Analysis
t

 ZZ  c
 ZZ  c 2  b 2 
2
b2

2
Z  ct
s
b
 4c 2 Z Z
2Z c
• In MATLAB,
0  Z ct  ZZ  c  b t  Zc
Zc=conj(Z)
B1
b
t=roots([Zc*c,Z*Zc+c^2-b^2,Z*c])
A
q3
• q4=angle(t), q3=angle(s)
c
• Two solutions relate to the a
q4
d
open and crossed
q2
positions
O2
O
2
2
2
4
B2
12
MATLAB
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13
>> a=2; b=3; c=4; d=5;
>> th1=0; th2=60*pi/180;
>> z=-a*exp(i*th2)+d*exp(i*th1)
>> th4=angle(t)*180/pi
th4 =
z=
4.0000 - 1.7321i
>> zc=conj(z)
zc =
4.0000 + 1.7321i
B
114.7975
-161.6240
A
>> th3=angle(s)*180/pi
th3 =
c
a
>> t=roots([zc*c,z*zc+c^2-b^2,z*c])
t=
-0.4194 + 0.9078i
-0.9490 - 0.3153i
b
q3
q2
39.2750
-86.1015
O2
q4
d
O4
>> s=(z+c*t)/b
s=
0.7741 + 0.6330i
0.0680 - 0.9977i
14
Inverted Crank Slider linkage
• Given: link lengths a, c and d, q1, q2 (the motor
position), and g the angle between the slider and rod
• Find: the unknown angles q3 and q4 and length b
15
Inverted Crank Slider linkage
• Write the vector loop equation




R2  R3  R4  R1  0
• Substitute with complex vectors
aeiq 2  beiq3  ceiq 4  deiq1  0
• Geometry keeps
q3 q4 g
• so
aeiq 2  bei q 4 g   ceiq 4  deiq1  0
16
Inverted Crank Slider
iq 2
i q 4 g 
iq 4
iq1
ae  be
 ce  de  0
• Grouping knowns and unknowns
bei q 4 g   ceiq 4  aeiq 2  deiq1  Z
iq 4
ig
• Calling
s  e and t  e
• Gives bst  cs  Z  s (bt  c)
• Taking the conjugate to get the second equation
1 1 
s (bt  c)  Z   b  c 
s t

• Multiplying the two gives
 1 2
b  bc t    c  ZZ
 t
2
17
Inverted Crank Slider
 1 2
b  bc t    c  ZZ
 t
2
• The solution is a quadratic
equation in b
 1
2
2
0  b  c  t   b  c  ZZ
t

• Which has a solution of
b

c t  1
t
 

t 1
t
2

2

c 2  4 c 2  ZZ

• b=roots([1 c*(t+1/t),c^2-Z*Zc])
• Once b is known, s can be found using s 
Z
bt  c
18
Crank Slider Mechanism
• Given: link lengths a, b and c, q1, q2 (the motor
position)
• Find: the unknown angle q3 and length d
19
4.8 Linkages of More than Four Bars
• Geared fivebar linkage
• vector loop equation





R2  R3  R4  R5  R1  0
• Complex vectors
ae
iq 2
 be
iq 3
 ce
iq 4
iq 5
 de  f  0
• Separate unknowns and knowns (q5=lq2f)
(same eqn.
iq3
iq5
iq 4
iq 2
be  ce  ae  de  f  Z
as 4bar)
20
Sixbar Linkages
• Watt’s sixbar can be solved as 2 fourbar linkages
• R1R2R3R4, then R5R6R7R8
• R4 and R5 have a constant angle between them
21
Sixbar Linkages
• Stephenson’s sixbar can sometimes be solved as a
fourbar and then a fivebar linkage
• R1R2R3R4, then R4R5R6R7R8
• R3 and R5 have a constant angle between them
• If motor is at O6 you have to solve eqns. simultaneously
22
Position of any Point on a Linkage
• Once the unknown angles have been found it is
easy to find any position on the linkage
• For point S
Rs=sei(q +d )
• For point P
RP=aei q +pei (q +d )
• For point U
RU=d +uei (q +d )
2
2
2
3
4
3
4
23
Using MATLAB (Spring 2007)
24
Transmission Angle
• Extreme value of transmission angle when links 1
and 2 are aligned
2
2
2
2
2
2




b

c

d

a
 b  c  d  a  
1  arccos 

2bc

   arccos 
2





2bc
Extended
Overlapped
25
Toggle Position
• Caused by the colinearity of links 3 and 4.
2
2
2
2

a

d

b

c
bc 
1
 0  q 2toggle  
q 2toggle  cos 

2ad
ad 

• For a non-Grashof linkage, only one of the values
between the () will be between –1 and 1
3
3
4
2
Extended
q2
Overlapped
2
q2
4
26