KINEMATIC COEFFICIENTS
Classification of dynamic problems
Forward dynamic
problem
Inverse dynamic
problem
Analysis type
Time-response analysis
Kineto-static analysis
Known
External load exerted on
the system
Displacement, velocity,
and acceleration
Unknown
Displacement, velocity,
and acceleration resulting
from the external loads
Force and torque
necessary for the desired
motion
Newton’s 2nd law Given F and m, solve for a Given a and m, solve for F
Equation type
(Nonlinear) ordinary
differential equations
System of linear algebraic
equations
Solver
Euler’s method or RungeKutta method
Gauss elimination
Purpose
Verify motor selection;
Motor, bearing, and link
transient and steady state selection; shaking
analysis; speed fluctuation force/moment analysis
analysis
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1. Definitions
Kinematic Coefficients
➢ Kinematic coefficient (also known as influence coefficients) is used
for derivatives of motion variables with respect to the input or
reference variables.
➢ Let Si be the input (scalar) variable, and Sk be any other (scalar) variable.
First-Order Kinematic Coefficient:
The first derivative of Sk with respect to Si
Second-Order Kinematic Coefficient:
The second derivative of Sk with respect to Si
dSk
dSi
d 2Sk
dSi2
➢ The significance of the kinematic coefficients lies in the fact that they
are functions of position only. The numerical values are constant for
a fixed position of the input link; They do not depend upon the
velocity or acceleration of the input link.
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Kinematic Coefficients
1. Definitions
➢ Let Si be the input (scalar) variable, and Sk be any other (scalar) variable.
If Sk is an angle k, then
d k / dSi = hk ,
d 2 k / dSi2 = hk
1st-order
2nd-order
If Sk is a vector length rk, then
drk / dSi = fk ,
d 2rk / dSi2 = fk
1st-order
2nd-order
➢ Depending on whether Si is an angle i, or vector length ri, we have
hk = d k / d i (dimensionless)
hk = d 2 k / d i2 (dimensionless)
hk = d k / dri (length)−1
hk = d 2 k / dri 2 (length)−2
fk = drk / dri (dimensionless)
fk = d 2rk / dri 2 (length)−1
fk = drk / d i (length)
fk = d 2rk / d i2 (length)
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Kinematic Coefficients
1. Definitions: Physical Meaning
➢ By the chain rule for differentiation,
dSk dSk dSi
=
dt
dSi dt
or
Sk =
dSk
Si
dSi
or
Sk dSk
=
dSi
Si
1st-order K.C. is the velocity ratio of the output variable's velocity to
the input variable 's velocity.
and
d 2Sk dSk d 2Si d 2Sk  dSi 
=
+
2
2
dt
dSi dt
dSi2  dt 
2
or
dSk
d 2Sk 2
Sk =
Si +
Si
dSi
dSi2
2nd-order K.C. is not the acceleration ratio Sk / Si . It is the rate of
change of the 1-st order K.C. w.r.t. the input variable. Constant 1-st
order K.C. means zero 2nd-order K.C (such as gear trains).
➢ Using kinematic coefficients in velocity and acceleration analysis
 k = hk Si
 k = hk Si + hk Si2
rk = fk Si
rk = fk Si + fkSi2
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Kinematic Coefficients
2. Four-Bar Example
Vector loop equation:
r2 + r3 = r1 + r4
Components scalar equations:
r3 cos 3 − r4 cos 4 = r1 cos1 − r2 cos 2
r1
r3 sin3 − r4 sin 4 = r1 sin1 − r2 sin 2
They can be solved by using Newton-Raphson’s method
when 2 = 100o,  3 = 80o and 4 = 180o
Differentiating the equations with respect to 2:
OA = 4 in, AB = 4 in,
 − r3 sin 3
 r cos 
3
 3
r4 sin 4   h3   r2 sin 2 
=



− r4 cos  4  h4   − r2 cos  2 
MB = 6 in,
M coordinate = (6, 7.878) in
Jacobian matrix, J
 h3 = -1, h4 = 0.232
3 = h3 2
 4 = h42
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Kinematic Coefficients
2. Four-Bar Example
OA = 4 in, AB = 4 in, MB = 6 in,
M coordinate = (6, 7.878) in
r3 ( − sin3 ) h3 − r4 ( − sin 4 ) h4 = −r2 ( − sin 2 )
r3 ( cos  3 ) h3 − r4 ( cos  4 ) h4 = −r2 ( cos  2 )
 h3 = -1, h4 = 0.232
3 = h3 2
 4 = h42
r3 ( − sin 3 ) h3 − r4 ( − sin 4 ) h4 = r2 ( cos  2 ) + r3 ( cos  3 ) h32 − r4 ( cos  4 ) h42
r3 ( cos  3 ) h3 − r4 ( cos  4 ) h4 = r2 ( sin 2 ) + r3 ( sin 3 ) h32 − r4 ( sin 4 ) h42
 h3
&
h4
 k = hk  2 + hk  22
 3
& 4
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3. Disk Cam with an Oscillating Flat Face Follower Kinematic Coefficients
Example
r2 = 0.5 in, r1 = 3 in,
r4 = 1.25 in, R = 1 in,
2 = 120o.
Position analysis: r3 + r4 = r2 − r1
(
)
r3 cos  3 + r4 cos 3 + 90o = r2 cos  2 − r1
(
)
r3 sin3 + r4 sin 3 + 90o = r2 sin 2
  3 = 150o , r3 = 3.031in
Kinematic coefficients:
(
)
r3 ( − sin3 ) h3 + cos3f3 + r4  − sin 3 + 90o  h3 = r2 ( − sin 2 )
r3 ( cos3 ) h3 + sin3f3 + r4 cos 3 + 90o  h3 = r2 ( cos 2 )
(
)

h3 = 0.143, f3 = 0.429 in
 3 = h3 2
&
r3 = f3 2
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3. Disk Cam with an Oscillating Flat Face Follower Kinematic Coefficients
Example
r2 = 0.5 in, r1 = 3 in,
r4 = 1.25 in, R = 1 in,
2 = 120o.
Kinematic coefficients:
r3 ( − sin3 ) − r4 cos3  h3 + cos 3f3 = r2 ( − sin 2 )
r3 ( cos3 ) − r4 sin3  h3 + sin3f3 = r2 ( cos 2 )
 h3 = 0.143, f3 = 0.429  3 = h3 2 &
r3 = f3 2
r3 ( − sin 3 ) − r4 cos  3  h3 + cos  3f3 =
r2 ( − cos  2 ) + 2sin 3f3h3 + r3 ( cos 3 ) − r4 sin3  h32
r3 ( cos  3 ) − r4 sin 3  h3 + sin 3f3 =
− r2 ( sin 2 ) − 2cos 3f3h3 + r3 ( sin3 ) + r4 cos 3  h
2
3
 h3 & f3
 k = hk Si + hk Si2
 3 = h3 2 + h3 22
rk = fk Si + fkSi2
r3 = f3 2 + f3 22
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Kinematic Coefficients
4. Inverted Slider-Crank Example
r2 − r3 − r1 = 0
r2 cos 2 − r3 cos3 = 0
When 2 = 30o,
 r3 = 40.32 cm
r2 sin 2 − r3 sin3 + r1 = 0
3 = 70.894
Differentiating the equations with respect to 2:
r3 sin 3h3 − cos  3f3 = r2 sin 2
−r3 cos  3h3 − sin 3f3 = −r2 cos 2
 h3 = 0.286, f3 = 9.97 cm
OA = r2 = 15.24 cm,
OC = r1 = 30.48 cm
r3 S 3 h3 − C3f3 = −r3 C3 h32 − 2f3 h3 S 3 + r2 C  2
−r3 C3 h3 − S 3f3 = −r3 S 3 h32 + 2f3 h3 C3 + r2 S  2

h3 = 0.10605, f3 = −8.23 cm
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Kinematic Coefficients
4. Inverted Slider-Crank Example
When 2 = 30°,

r3 = 40.32 cm
3 = 70.894
Link 2
X g 2 = r4 cos  2 = 6.60 cm
Yg 2 = r4 sin 2 = 3.81cm
Link 3
X g 3 = r2 cos  2 = 13.20 cm
Yg 3 = r2 sin 2 = 7.62 cm
Link 4
X g 4 = r5 cos  3 = 7.48 cm
OA = 15.24 cm, OC = 30.48 cm,
OG2 = 7.62 cm, CG4 = 22.86 cm
Yg 4 = − r1 + r5 sin3 = −8.88 cm
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Kinematic Coefficients
4. Inverted Slider-Crank Example
fg 2 x = dX g 2 / d 2
fg 4 x = dX g 4 / d 2
= − r4 sin 2 = −3.81 cm = − r5h3 sin 3 = −6.17 cm
fg 2 y = dYg 2 / d 2
fg 4 y = dYg 4 / d 2
= r4 cos  2 = 6.60 cm
= r5h3 cos  3 = 2.14 cm
fg 3 x = dX g 3 / d 2
= − r2 sin 2 = −7.62 cm
fg 3 y = dYg 3 / d 2
= r2 cos  2 = 13.20 cm
➢ Velocity components of mass centers
Vgix = fgix2
Vgiy = fgiy 2
2
2
Vgi = Vgix
+ Vgiy
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Kinematic Coefficients
4. Inverted Slider-Crank Example
fg2 x = d 2 X g 2 / d 22
fg4 x = d 2 X g 4 / d 22
= − r4 cos  2 = −6.60 cm
= − r5h3 sin3 − r5h32 cos 3
fg2 y = d 2Yg 2 / d 22
= −2.90 cm
= − r4 sin 2 = −3.81 cm
fg4 y = d 2Yg 4 / d 22
fg3 x = d 2 X g 3 / d 22
= r5h3 cos  3 − r5h32 sin3
= − r2 cos  2 = −13.20 cm = −0.970 cm
fg3 y = d 2Yg 3 / d 22
= − r2 sin 2 = −7.62 cm
➢ Acceleration components of mass centers
2
2
 22 Agiy = fgiy 2 + fgiy
 22 Agi = Agix
Agix = fgix 2 + fgix
+ Agiy
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Kinematic Coefficients
5. Finding Dead Positions Using Kinematic Coefficients
➢ When the determinant of the Jacobian matrix goes to zero, all the
kinematic coefficients go to infinity, which means that the input
velocities become zero. This means the input stops moving and the
mechanism is locked.
| J |=
− r3 sin 3
r4 sin 4
r3 cos  3
− r4 cos  4
= − r3r4 sin( 4 − 3 )
➢ Dead positions correspond to when IJI = 0.
That is, when (θ4 - θ3) = 0° or 180°.
➢ Geometrically, this means that dead positions of the
four bar mechanism occur when link 3 (the coupler)
and link 4 (the output) are in-line. The transmission
angle of the four-bar mechanism is 0° or 180°.
Moment cannot be transmitted to link 4.
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Kinematic Coefficients
5. Finding Dead Positions Using Kinematic Coefficients
| J |=
r3 sin 3
− cos 3
− r3 cos  3
− sin 3
= − r3 = 0
➢ So dead positions of the inverted crank-slider
with θ2 as the input occur when the joint
variable, which is the vector magnitude r3,
goes to zero. The inverted crank-slider cannot
realize this dead position.
➢ If we change r2 to realize this dead position,
then the crank is no longer capable of
continuous rotation. The input would become
a rocker.
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Kinematic Coefficients
6. Finding Limit Positions Using Kinematic Coefficients
➢ When a motion variable of a mechanism reaches
a limit position, its instant velocity must be zero.
➢ For a motion variable to be in a limit position, the
associated first-order kinematic coefficient must
be zero.
 r3 sin 3
 − r cos 
3
 3
− cos  3   h3   r2 sin 2 
=



− sin 3   f3   − r2 cos  2 
J
➢ Using Cramer’s rule yields
r2 sin 2 − cos  3
− r cos  2 − sin3
− r cos( 3 −  2 ) r2
h3 = 2
= 2
= cos( 3 −  2 )
|J|
− r3
r3
cos( 3 −  2 ) = 0  ( 3 −  2 ) = 90
➢ Geometrically, this means that vectors r2 and r3 have to be perpendicular.16
Kinematic Coefficients
6. Finding Limit Positions Using Kinematic Coefficients
 − r3 sin 3
 r cos 
3
 3
r4 sin 4   h3   r2 sin 2 
=



− r4 cos  4  h4   − r2 cos  2 
J
➢ Using Cramer’s rule yields
r2 sin 2
h3 =
r4 sin 4
− r2 cos  2 − r4 cos  4 r2 sin( 2 −  4 )
=
|J|
r3 sin( 4 −  3 )
➢ h3 is zero when sin( 2 −  4 ) = 0  ( 2 −  4 ) = 0 ,180
➢ This corresponds to when r2 and r4 are parallel.
The angular velocity of link 3 is zero.
➢ Similarly, h4 is zero when sin( 3 −  2 ) = 0  ( 3 −  2 ) = 0 ,180
➢ This corresponds to when r2 and r3 are in-line. The angular velocity
of link 4 is zero.
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