### Class-4-Velocity-Analysis1 (1)

```Mechanics of Machines
Class 4
Velocity Analysis
DERIVATIVE OF A ROTATING VECTOR
Derivative of a Rotating Unit Vector
 A unit vector in the θ direction, uθ, is a
vector of unity magnitude and an angle
θ with the x- axis. It is written as:
uθ  cosi  sin j
 If uθ rotates, it angle θ changes with
time, the time derivative of uθ is found
by applying the standard differentiation
rules on the expression of uθ above
duθ
d
d

sin i 
cosj
dt
dt
dt
duθ d
 sin i  cosj 

dt
dt
sinθ j
uθ
θ
cosθ i
Derivative of a Rotating Unit Vector
 Noting that
sin   2  cos
cos   2   sin 
uθ  cosi  sin j
duθ d
 sin i  cosj 

dt
dt
the time derivative of the
vector uθ is
duθ d
 sin i  cosj 

dt
dt
duθ d
cos   2i  sin   2  j 

dt
dt
duθ d

u  2 
dt
dt
sinθ j
uθ
θ
cosθ i
Derivative of a Rotating Unit Vector
 The derivative of a unit vector
whose angle with the x-axis is θ,
(θ changes in time), is a vector
whose angle with the x-axis is
θ+π/2 and whose magnitude is
dθ/dt.
 If we define a vector ω as a
vector in the k direction of
magnitude ω = dθ/dt, then
d
k
dt
duθ d

u 
dt
dt
uθ  cosi  sin j
duθ d

u 
dt
dt
2
duθ /dt
uθ
θ + π/2
θ
ω
2
 ω  uθ
Derivative of a Rotating Vector
 A vector rθ = r uθ , is a general vector of magnitude r
pointing in the θ direction. The time derivative of rθ is
found by applying the normal differentiation rules on
the expression for rθ
rθ  ruθ
drθ
dt
drθ
dt
drθ
dt
drθ
dt




dr
uθ
dt
dr
uθ
dt
dr
uθ
dt
dr
uθ
dt
duθ
r
dt
d
r
u 
dt
 r ω  uθ 
 ω  rθ 
duθ /dt
uθ
θ + π/2
2
θ
Derivative of a Rotating Vector
 Given a vector rθ = ruθ which
rotates relative to the reference
coordinates, the time derivative of
rθ has two components; a
component in the direction of uθ
and a component normal to uθ in
the direction of uθ+π/2
 The magnitude of the component
of drθ /dt in the direction of uθ is
equal to dr/dt; that is the time
derivative of the length of rθ.
 The magnitude of the component
of drθ /dt in the direction of
direction of uθ+π/2 is equal to ωr
ω x rθ
(dr/dt) uθ
θ + π/2 r
θ
θ
VELOCITY ANALYSIS OF FOUR
BAR MECHANISMS
Derivative of the Loop Closure Equation for a
Four Bar Kinematic Chain
 The loop closure equation of a 4-bar
kinematic chain is written as
   
r2  r3  r1  r4
   
r2  r3  r1  r4  0
d    
r2  r3  r1  r4   0
dt



d 

r2u2  r3u3  r1u1  r4u4   0
dt








r22u2  2   r2u2  r33u3  2   r3u3  r11u1  2   r1u1  r44u4  2   r4u4  0

 

 When all the links in the chain are of constant
lengths, the equation above reduces to


r u
2 2
 2  2 


 r u
3 3
3  2 


 r u
1 1
1  2 


 r u
ω2  r2  ω3  r3  ω1  r1  ω4  r4  0
4 4
 4  2 
0
Derivative of the Loop Closure Equation for a
Four Bar Mechanism
 For a four bar mechanism with link 1
fixed we have dθ1/dt = ω1 = 0.


r u
2 2
2  2 


 r u
3 3
3  2 


 r u
4 4
4  2 
 The vector equation above contains
two scalar equations and can be
solved for two unknowns. Knowledge
of dθ2/dt allows the calculation of
dθ3/dt and dθ4/dt .
r22 sin  2  r33 sin 3  r44 sin  4
r22 cos 2  r33 cos3  r44 cos 4
Derivative of the Loop Closure Equation for a
Four Bar Mechanism


r u
2 2
2  2 


 r u
3 3
3  2 


 r u
4 4
4  2 
 Eliminate dθ3/dt by carrying out a dot product with
uθ3 on both sides of the equation
r22 cos 2   2  3   r44 cos 4   2  3 
r  sin      r  sin    
2 2
2
4  4 
3
4 4
4
3
r2 sin  2  3   r2 sin  2  3 
2 
2
r4 sin  4  3 
r4 sin  4  3 
 Alternatively, eliminate dθ4/dt by carrying out a
dot product with uθ4 on both sides of the equation
r22 cos 2   2   4   r33 cos3   2   4   0
r  sin       r  sin    
2 2
2
3  3  
4
3 3
3
4
r2 sin  2   4  
r sin  2   4 
r sin  2   4 
2   2
2  2
2
r3 sin 3   4 
r3 sin 3   4 
r3 sin  4  3 
Angular Velocity Ratio and Mechanical
 The angular velocity ratio mV is defined as the output angular velocity divided
by the input angular velocity. For a four bar mechanism with link 2 as the
input and link 4 as the output this is expressed as
out 4
mV 

in 2
 The efficiency of a four bar linkage is defined as the output power over the
input power,

Pout Toutout

Pin
Tinin
 Assuming 100% efficiency, which is normally approached by four bar
mechanisms, we have
Tout
in
1

 mT 
Tinin out
mV
Angular Velocity Ratio and Mechanical
defined as the ratio between
the output force to the input
force
mA 
Fout Tout rout
r

 mT in
Fin
Tin rin
rout
VELOCITY ANALYSIS OF
SLIDER-CRANK MECHANISMS
Velocity Analysis of a Slider-Crank Mechanism
 Design parameters: r2, r3, r4, θ1.
r3
Position analysis parameters:
r1, θ2, θ3
Velocity analysis parameter.
Find dr1/dt, dθ2/dt , dθ3/dt
   
r2  r3  r1  r4



r22u2  2   r33u3  2   r1u1
 To eliminate ω3 dot product both
sides by uθ3
r22 cos2   2  3   r1 cos1  3 
 r22 sin2  3   r1 cos1  3 
r1 
r2 sin3  2 
2
cos3  1 
r4
r2
rp
r1
 To eliminate dr1/dt dot product
both sides by u(θ1+π/2)
r22 cos2  1    r33 cos3  1 
3  
r2 cos2  1 
2
r3 cos3  1 
Velocity Analysis of an Inverted Slider-Crank
Mechanism
 Given r1 , r2, θ1, θ2 , ω2
Position analysis: Find r3, θ3
Velocity analysis: Find dr3/dt, dθ3/dt
  
r2  r1  r3



r2u2  r1u1  r3u3



r22u2  2   r3u3  r33u3 
2
 To eliminate ω3 and find dr1/dt dot product both
sides by uθ3 r22 cos2   2  3   r3
r22 sin3  2   r3
 To eliminate dr1/dt and find ω3 dot product both
sides by u(θ3+π/2)
r22 cos2  3   r33
r cos3  2 
3  2
2
r3
r2
θ2
r1
r3
θ1
θ3
Velocity Analysis of an Inverted Slider-Crank
Mechanism
 Given r1 , r2, r4 , θ1, θ2 , ω2
Position analysis: Find r3, θ3 , θ4
Velocity analysis: Find dr3/dt, dθ3/dt , dθ4/dt
   
r2  r3  r1  r4




r2u2  r3u 3  r1u1  r4u4



r22u2  2   r3u 3  r33u3 


r

u
2
4 4  4 
with 4  3   2



r22u2  2   r3u 3  r33u3 



r

u
2
4 4 3
2
r3
r2
r1
r4
Velocity Analysis of an Inverted Slider-Crank
Mechanism

r22u2 




r
u

r

u
2
3 3
3 3  3 

Dot bothsidesby u3  2 



r

u
2
4 4 3
r22 cos 2   3   r33  0
r2 cos 2   3 
2
r3

Dot bothsidesby u 3
3  
r22 cos 2   2   3   r3   r44
 r22 sin  2   3   r3   r44
But 4  3  
r2 cos 2   3 
2
r3
 r22 sin  2   3   r3  r4
r2 cos 2   3 
2
r3
 r cos 2   3 

r3   r4 2
 r2 sin  2   3 2
r3


r3
r2
r1
r4
Example
r5
b
a
LoopI




r2u2  a  b u3  r1u1  r4u4
r2
LoopII





r2u2  au3  r5u5  r1u1  su6
s
r4
r1
LoopClosureEquation(1)
a  b u3  r1u1  r4u4  r2u2
Dot bothsidesof the equationby itself
a  b 2  r12  r22  r42  2r1r4 cos4  1   2r1r4 cos2  1   2r2r4 cos2  4 
a  b 2  r12  r22  r42  2r1r4 cos4  2r1r4 cos2  2r2 r4 cos2 cos4  sin 2 sin 4 
A cos 4  B sin  4  C  0
C  At 2  2 Bt  C  A  0  t1,2  B 
Alsofromloopclosurerquation(2)
a  b cos3  r1 cos1  r4 cos4  r2 cos2
2
2
r2
B  2r2 r4 sin 1
3 1,2  tan1  r1 sin 1  r4 cos4  r2 cos2 

 r1 cos1  r4 cos 4  r2 cos 2 
C  r12  r22  r42  2r1r2 cos 2  a  b 
2
t  tan 4 2 
b
a
A  2r1r4  2r2 r4 cos 2
a  b sin 3  r1 sin 1  r4 cos4  r2 cos2

r5
B C  A
,  4 1, 2  2 tan1 t1, 2 
AC
2
s
r4
r1
Example
r5
LoopClosureEquation(2)





r5u5  r1u1  su6  r2u2  au3
a
Dot bothsidesof the equationby itself
r2
r5  r12  s 2  r22  a 2  2r1s cos6  1   2r1r2 cos 2  1   2r1a cos3  1 
2
 2r2 s cos6   2   2as cos3  1   2r2 s cos6   2   2r2a cos 2  3 
b
s
r4
r1
Let1  0, 6   2
r5  r12  s 2  r22  a 2  2r1r2 cos 2  2r1a cos3
2
 2r2 s cos6   2   2as cos3  1   2r2 s cos6   2   2r2a cos 2  3 
a  b2  r12  r22  r42  2r1r4 cos4  2r1r4 cos2
C  At
2
 2 Bt  C  A  0  t1, 2
B  B 2  C 2  A2

,  4 1, 2  2 tan1 t1, 2 
AC
r5
Alsofromloopclosurerquation(2)
3 1,2  tan1  r1 sin 1  r4 cos4  r2 cos2 

b
a
a  bcos3  r1 cos1  r4 cos4  r2 cos2
a  bsin 3  r1 sin 1  r4 cos4  r2 cos2

 r1 cos1  r4 cos 4  r2 cos 2 
r2
A  2r1r4  2r2 r4 cos 2
B  2r2 r4 sin 1
C  r12  r22  r42  2r1r2 cos 2  a  b 
2
t  tan 4 2 
s
r4
r1
HW#3
6-17, 6-18 (b+c), 6-21 (b), 6-44.
METHOD OF INSTANT CENTERS
Relative Velocities Between Two Points on a
Rigid Body
 Given any two points A
and B that lie on a rigid
body, let the line AB be
a line passing through A
and B, then the
components of the
velocity of A and the
velocity of B on the line
AB must be equal.
 The velocity of point B
relative to A must be
normal to the line AB
A
B
vA  vB  vB A
A
B
Instant Center Relative to the Ground
 If line AA’ is drawn normal to the
direction of vA, then the velocity of any
point on the rigid body that falls on line
AC must be perpendicular to the line AA’
 In a similar manner, if line BB’ is drawn
perpendicular to the direction of vB, then
the velocity of any point on the rigid
body that falls on this line must be
perpendicular to the line BB’
 If point I is the intersection of lines AA’
and BB’, then the velocity of point I must
be perpendicular to both AA’ and BB’.
This can only happen when the velocity
of point I is zero.
A
B
I
B’
A’
 


v A  v I  k  rA I
 

v A  k  rA I
v A  rA I    v A rA I
vB  rB I
Instant Center Relative to the Ground
 Point I determined as before is
called the instant center of zero
velocity of the rigid body with
respect to the ground. The
direction of the velocity of point A
is normal to the line IA. The
direction of the velocity of line B is
normal to the line IB.
 The direction of the velocity of any
other point C on the same body
must be normal to the line IC.
 The magnitude of the velocity of
any point on the link is
proportional to its distance from
the point I.
A
B
C
I
B’
A’
v A vB v A


IA IB IC
Velocity of a Point on a Link by
Instant Center Method
 Consider two points A and B on a rigid link.
Let vA and vB be the velocities of points A
and B at a given instant. If vA is known in
magnitude and direction and vB in direction
only, then the magnitude of vB may be
determined by the instantaneous centre
method.
 Draw AI and BI perpendiculars to the
directions vA and vB respectively. Let these
lines intersect at I, which is known as
instantaneous centre or virtual centre of
about the centre I at the given instant. The
relations shown may be used to determine
the magnitude of the velocity of point B.
 


v A  v I  k  rA I
 

v A  k  rA I
v A  rA I    v A rA I
vB  rB I
Instant Center Between Two Rigid Bodies
 In the foregoing discussion, the
velocities of points A and B were
assumed to be reference to a
coordinate system attached to the
ground. The resulting center is
called an instant center relative to
the ground.
 The velocities of points A and B
could as well be taken relative to
coordinate system attached to
another body. The resulting point
in this case will have a zero
velocity with respect to that body.
In other words, point I will have
the same velocity for both bodies.
A
B
I
B’
A’
Instant Center Between Two Rigid Bodies
 A instant center between two
bodies is a:
 A point on both bodies
 A point at which the two
bodies have no relative
velocity.
 A point about which one
body may be considered
to rotate around the other
body at a given instant.
A
B
I
B’
A’
Instant Center Between Two Rigid Bodies
connected to one
another by a revolute
joint, the center of the
connecting joint is an
instant center for the two
 When two links are not
connected, an instant
center between the two
can be determined if the
known.
A
A’
Instant Center Between Two Rigid Bodies
 The number of instant centers in a constrained
kinematic chain is equal to the number of
 The number of pairs of links or the number of
instantaneous centers is the number of
combinations of L links taken two at a time.
Mathematically, number of instant centers is,
LL  1
N
2
Types of Instantaneous Centers
 The instant centers for a mechanism
are of the following three types :
1. Primary (Permanent) instant
centers. They can be fixed or
moving
2. Secondary Instant centers (Not
permanent)
C
n n  1
2
Instant Centers of a Four Bar Mechanism
D
 Consider a four bar mechanism ABCD as
shown. The number of instant centers (N) in a
four bar mechanism is given by
N
44  1 12

6
2
2
 The instant centers I12 and I14 are fixed instant
centers as they remain in the same place for
all configurations of the mechanism. The
instant centers I23 and I34 are permanent
instant centers as they move when the
mechanism moves, but the joints are of
permanent nature. The instantaneous centres
I13 and I24 are neither fixed nor permanent as
they vary with the configuration of the
mechanism.
3
2
C
A
4
1 (Ground)
B
n n  1
C
2
Location of Instant Centers
 When the two links are connected by a pin
joint (or pivot joint), the instant center lies
on the center of the pin as. Such an instant
center is of permanent nature. If one of the
links is fixed, the instant center will be of
fixed type.
 When the two links have a pure rolling
contact (without slipping), the
instantaneous centre lies on their point of
contact, as this point will have the same
 When the two links have a sliding contact,
the instant center lies at the center of
curvature of the path of contact. This
points lies at the common normal at the
point of contact.
Kennedy (or Three Centers in Line) Theorem
 Kennedy’s theorem states that any
three bodies in plane motion will
have exactly three instant centers,
and the three centers will lie on
the same straight line.
 Note that this rule does not
require that the three bodies be
connected in any way. We can use
this rule, in conjunction with the
linear graph, to find the remaining
lCs which are not obvious from
inspection.
Example: Instant Centers of a Four Bar
Mechanism
 Draw a circle with all links numbered around
the circumference
 Locate as many ICs as possible by inspection.
All pin joints will be permanent ICs . Connect
the links numbered on the circle to create a
linear graph and record those found.
 Identify a link combination for which the IC
has not been found, and draw a dotted line
connecting those two link numbers. Identify
two triangles on the graph which each
contains the dotted line and whose two other
sides are solid lines representing the ICs the
already found. Use Kennedy’s theorem to
locate the needed IC.
Example: Instant Centers of a Four Bar
Mechanism
Example: Instant Centers of a Slider-Crank
Mechanism
Example: Instant Centers of a Slider-Crank
Mechanism
Example: Instant Centers of a Cam and
Follower: Common Normal Method
Example: Instant Centers of a Cam and
Velocity Analysis with Instant Centers
 Once the instant centers (ICs)
of a linkage with respect to
found, they can be used for a
very rapid graphic analysis for
 Note that some of the ICs may
be very far removed from the
and 4 are nearly parallel,
their extended lines will
intersect at a point far away
and not be practically
available for velocity analysis.
Velocity Analysis with Instant Centers
 From the definition of the instant center, both
links sharing the instant center will have
identical velocity at that point.
 Instant center I13 involves the coupler (link 3)
which is in general plane motion, and the
ground link which is stationary. All points on
the ground link have zero velocity in the global
coordinate system, which is embedded in link 1.
Therefore, I13 must have zero velocity at this
instant, and it can be considered to be an
is in pure rotation with respect to link 1.
 A moment later, I13 will move to a new location
center.
Velocity Analysis with Instant Centers
 If ω2 is known for the mechanism shown,
the magnitude of the velocity of point A can
be computed as vA = ω2 O2A Its direction
and sense can be determined by inspection.
 Note that point A is also instant center and
it has the same velocity as part of link 2 and
pivoting about I13 at this instant, the angular
velocity ω3 can be found by ω3 = vA /AI13 .
Once ω3 is known, the magnitude of vB can
also be found from vB = ω3 AI13
 Once vB is known, ω4 can also be found from
ω4 = vB /BO4 = vB /BO4 . Finally, vC (or the
velocity of any other point on the coupler)
can be found from vC = ω3 CI13
3 
vA
AI13
vB  3BI13
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