Velocity Analysis
Chapter 6
Definition
• Rate of change of position with
respect to time
– Angular
d

dt
– Linear
dR
V
dt
– Position Vector R PA  pe j
– Velocity
R PA
j d
Vpa 
 pje
 pe j
dt
dt
Definition
– Velocity (absolute)
R PA
j d
VPA 
 pje
 pe j
dt
dt
•
The velocity is always in a
direction perpendicular to
the radius of rotation and is
tangent to the path of
motion
VPA  pj cos   j sin    p sin   j cos  
VPA  VP " Absolute"
Definition
– Velocity (difference)
VPA  VP  VA
VP  VA  VPA
" on the same body "
– Relative Velocity
VPA  VP  VA
Graphical Analysis
• Graphical Velocity Analysis
VP  VA  VPA
V  v  r
– Solve for
angular ve locities ; 3 , 4
linear vel ocities; A, B, C
Graphical Analysis
• Graphical Velocity Analysis
VP  VA  VPA
V  v  r
Graphical Analysis
• Example 6-1
– Given θ2, θ3, θ4, ω2 find ω3,
ω4, VA, VB and VC
– Position analysis already
performed
– 1. Start at the end of the
linkage about which you
have the most information.
Calculate the magnitude of
the velocity of point A,
vA   AO2 2
Graphical Analysis
• Example 6-1
– 2. Draw the velocity VA
– 3. Move next to a point
which you have some
information, point B. Draw
the construction line pp
through B perpendicular to
BO4
– 4. Write the velocity
difference equation for point
B vs. A
VB  VA  VBA
Graphical Analysis
• Example 6-1
– 5. Draw construction line qq
through point B and
perpendicular to BA to
represent the direction of VBA
– 6. The vector equation can
be solve graphically by
drawing the following vector
diagram
VB  VA  VBA
Graphical Analysis
• Example 6-1
– 7. The angular velocities of
link 3 and 4 can be
calculated,
4 
vB
BO4
3 
vBA
BA
– 8. Solve for VC
VC  VA  VCA
vCA  c3
Instant Center of
Velocity
• An instant center of velocity is a point, common
to two bodies in plane motion, which point has
the same instantaneous velocity in each body
• The numbers of IC is calculated with;
nn  1
C
2
• Linear graph is a useful way to keep track of
which IC have been found
Instant Center of
Velocity
• Kennedy’s Rule
– Any three bodies in plane motion will have
exactly three instant centers, and they will lie
an the same straight line
Instant Center of
Velocity
Instant Center of
Velocity
• Slider-Crank Linkage
Instant Center of
Velocity
• Slider-Crank Linkage
Instant Center of
Velocity
• Slider-Crank Linkage
• Check Example 6-4:
IC for a CamFollower Mechanism
Velocity Analysis
with IC
• Once the ICs have
been found, they can
be used to do a very
rapid graphical velocity
analysis of the linkage
v A   AO2 2
vA
3 
AI1,3 
vB
4 
BO4 
vB  BI1,3 3
vC  CI1,3 3
Velocity Analysis
with IC
• A rapid graphical
solution for the
magnitudes at B and C
are found from vectors
drawn perpendicular to
that line at the
intersection of the arcs
and line AI1,3 (VB’, VC’)
• Angular Velocity Ratio
– Output angular velocity
divided by the input
angular velocity
4
mV 
2
– Can be derived by
constructing a pair of
effective links
– Effective link pairs is
two lines, mutually
parallel, drawn through
the fixed pivot and
intersecting the coupler
extended
O2 A  O2 Asin
O4 B  O2 Bsin 
VA  O2 A2
VA  VB
4 O2 A O2 A sin
mV 


2 O4 B O4 B sin 
– Now the effective links
are colinear and
intersect the coupler at
the same point, I2,4
4 O2 I 2, 4
mV 

2 O4 I 2, 4
• Mechanical Advantage
– Power in a mechanical system,
 
P  F V  FxVx  FyVy
– For rotating system,
P  T
Tout in

Tin out
Pout
– Mechanical efficiency,  
Pin
– Mechanical Advantage,
Fout  Tout  rin   in  rin   O4 B sin   rin 

  

  


mA 
 
Fin  Tin  rout   out  rout   O2 A sin  rout 
Centrodes
• The path, or locus, created by a IC at
successive positions
Centrodes
Velocity of Slip
• Used when there is a sliding joint between
two links and neither one is the ground
– Example 6-5, 6-6
Vslip42  VA4 slip  VA2 slip
4  3 
VA4
AO4
Velocity of Slip
4  3 
VA3
AO3
Analysis Solution
• Position Analysis (revisited)
  

R2  R3  R4  R1  0
ae
j 2
 be
j3
 ce
j 4
 de
j1
0
3
  E  E 2  4 DF 

 2 arctan 


2D


4
  B  B 2  4 AC 

 2 arctan 


2
A


1, 2
1, 2
Analysis Solution
• Velocity Analysis
ae j2  be j3  ce j4  de j1  0


d
ae j 2  be j3  ce j 4  de j1  0
dt
d
d
d
jae j 2 2  jbe j3 3  jce j 4 4  0
dt
dt
dt
ja2e
j 2
 jb3e
j3
 jc4e
j 4
0
Analysis Solution
• Velocity Analysis
VA  VBA  VB  0
VA  ja2 e j 2
VBA  jb3e j3
VB  jc4 e j 4
Euler identity
real part
imaginary part
Analysis Solution
• Velocity Analysis
a2 sin  4   2 
3 
b sin 3   4 
a2 sin  2  3 
4 
c sin  4  3 
VA  ja2 cos  2  j sin  2   a2  sin  2  j cos  2 
VBA  jb3 cos  3  j sin  3   b3  sin  3  j cos  3 
VB  jc4 cos  4  j sin  4   c4  sin  4  j cos  4 
Analysis Solution
– Slider-Crank
   
R2  R3  R4  R1  0
ae j2  be j3  ce j4  de j1  0
ja2e j2  jb3e j3  d  0
VA  VAB  VB  0
VAB  VBA
VB  VA  VBA
Analysis Solution
– Slider - Crank
3 
a sin  2
2
b sin  3
d  a2 sin  2  b3 sin 3
VB  VA  VBA
VA  a2  sin  2  j cos  2 
VAB  b3  sin  3  j cos  3 
VBA  VAB
Review - Inverted Slider - Crank
Geared Fivebar
    
R2  R3  R4  R5  R1  0
ae j2  be j3  ce j4  de j5  fe j1  0
ja2e
j 2
 jb3e
5   2  
j3
 jc4e
5   2
j 4
 jd5e
j5
0
Geared Fivebar
3  
2 sin  4 a2 sin  2   4   d5 sin  4  5 
bcos3  2 4   cos 3 
a2 sin  2  b3 sin 3  d5 sin 5
4 
c sin  4
VA  a2  sin  2  j cos  2 
VBA  b3  sin  3  j cos  3 
VC  d5  sin  5  j cos  5 
VB  VA  VBA
Velocity of Any Point
• Once the angular velocities of all the links are
found it is easy to define and calculate the
velocity of any point on any link for any input
position of the linkage
Velocity of Any Point
• To find the velocity of points S & U
R S 02  R S  se j 2  2   scos 2   2   j sin 2   2 
VS  jse j 2  2 2  s2  sin2   2   j cos2   2 
RU 04  ue j 4  4   ucos 4   4   j sin 4   4 
VU  jse j 4  4 4  u4  sin4   4   j cos 4   4 
Velocity of Any Point
• To find the velocity of point P
R PA  pe j 3  3   pcos3   3   j sin 3   3 
R P  R A  R PA
VPA  jpe j 3  3 3  p3  sin 3   3   j cos3   3 
VP  VA  VPA