MENG 372
Chapter 6
Velocity Analysis
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
1
1
Velocity Analysis
Definitions
Linear Velocity

 dR 
V
R
dt
d 
Angular Velocity  

dt
Velocity of a point

i
RPA  pe



VPA  VP  RPA
i
i

 pe i  pe i 
 
2
Link in pure rotation
Multiplying by i rotates the vector
by 90°
Velocity is perpendicular to
radius of rotation & tangent to
path of motion
2
Velocity Analysis
Vector r can be written as:
rei  r cos  i sin  
Imaginary
r cos 
Multiplying by i gives:
irei  r  sin   i cos 
r
Multiplying by i rotates a vector 90°
r sin 
3

r cos 
r sin 
Real
3
Velocity Analysis
If point A is moving

 
VP  VA  VPA

 VA  pei i 
Graphical solution:
4
4
Graphical Velocity Analysis (3 & 4)
• Given linkage configuration & 2. Find 3 and 4
• We know VA and direction of VB and VBA
(perpendicular to AB)
• Draw vector triangle. V=r.
VBA
VB
VBA Direction
VBA
VA
VB
VB Direction
VBA  3  AB 
5
VB  4  O4 B 
5
Graphical Velocity Analysis (VC)
• After finding 3 and 4, find VC
• VC=VA+VCA
• Recall that 3 was in the opposite direction as 2
VC
Double Scale
VCA
VA
VCA
VC
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6
Instant Center
• A point common to two bodies in plane motion, which
has the same instantaneous velocity in each body.
• In ENGR 214 we found the instant center between
links 1 and 3 (point on link 3 with no velocity)
• Now we also have an instant center between links 2
and 4
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7
Instant Centers
• Kennedy’s rule: any three links will
have three instant centers and
they will lie on a straight line
• The pins are instant centers
• I13 is from links 1,2,3 and 1,3,4
• I24 is from links 1,2,4 and 2,3,4
123 134
8
124 234
I12
I13
I12
I23
I23
I34
I24
I34
I13
I14
I14
I24
I13
Links
IC’s
I24
8
Instant Centers
• I13 has zero velocity since link 1 is
ground
• 3 is the same all over link 3
• Velocity relative to ground=r,
perpendicular to r
• VA2=a2=VA3=p3
p
• From this, 3 must be in the
opposite direction as 2, and
smaller in magnitude since p>a VA2
A
a
9
3
I13
3
VA3
9
Instant Centers
• I24 has the same velocity on link 2
and link 4
• VI2=l22=VI4=l44
• From this, 4 is in the same
direction as 2 and smaller in
magnitude since l4>l2
l4
I24
VI4 V
I2
10
4
l2
10
Instant Centers Practice Problems
A
B
O2
O4
Power=Tinin=Toutout
A
O4
O2
11
B
11
Velocity Analysis of a 4-Bar Linkage
Given 2. Find 3 and 4
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12
Velocity Analysis of a 4-Bar Linkage
• Write the vector loop equation
i 2
ae
i3
 be
i 4
 ce
i1
 de
0
• After solving the position
analysis, take the derivative
 
 
 
aei 2 i2  bei3 i3  cei 4 i4  0
i2aei 2  i3bei3  i4cei 4  0
or




V A  i 2 aei 2
V A  VBA  VB  0

VBA  i 3bei 3

where
VB  i 4cei 4
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13
Velocity Analysis of a 4-Bar Linkage
i2aei 2  i3bei3  i4cei 4  0
• Take knowns to one side:
3bei3  4cei 4   2aei 2
• Take conjugate to get 2nd
equation:
3bei3  4cei 4  2aei 2
i 2 
i 3
i 4  






ae
be

ce
3
2
• Put in matrix form: 



i 2 
i 3
i 4  
 ce
be
  4    2 ae

1
i

i

3   be 3
 ce 4     2aei 2 
• Invert matrix:
    i3
i 2 
i 4  
 ce    2ae 
 4  be
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14
Inverted Crank Slider
Link 3 is a slider link: its effective length, b, changes
Given 2. Find 3 and b
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15
Inverted Crank Slider
• Given 2. Find 3 and b
• Write the vector loop equation:
aei2  bei3  cei4  dei1  0
• After solving the position
analysis, take the derivative:
i2aei 2  bei3  i3bei3  i4cei 4  0
• To get another equation:
 3   4   or 3   4
so
bei3  i3 bei3  cei 4  i 2aei 2

16

16

Inverted Crank Slider

i 3
i 3
i 4
i 2

be  i3 be  ce  i 2ae
• Take conjugate to get second
equation:


bei3  i3 bei3  cei 4  i2aei 2
• Put in matrix form:
 e i 3
 i 3
e
Invert:
17



i bei 3  cei 4
 i bei 3  cei 4
  b   i 2 aei 2 
 
i 2 
  3   i 2 ae 

 b   ei3
    i3
3  e

 i be
i 3
i be
i 4
 ce
i 3

 cei 4




1
 i 2aei 2 

i 2 
 i 2ae 
17
Velocity of any Point on a Linkage
• Write the vector for RP
R p  aei 2  pei  3  3 
• Take the derivative
V p  i 2 ae i 2  i 3 pe i  3  3 
• Similarly
RS  sei 2  2 
VS  i 2 sei  2  2 
RP
RU  uei  4  4 
VU  i 4uei  4  4 
18
18
Offset Crank Slider
Given 2. Find 3 and d
b
c
a
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19