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Dynamic
Simulation
:
Constraint
Kinematics
Objective
The
objective of this module is to show how constraint equations are
used to compute the position, velocity, and acceleration of the
generalized coordinates
.
These equations
are kinematic in nature because they do not consider
the forces required to cause the
motion.
The
kinematic and motion constraints developed in
the previous
module (Module 3)
for the piston

crank mechanism are used to
demonstrate the mathematics.
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Notation
The total set of constraint equations
needed to define a mechanism
includes both kinematic constraints
and drive constraints.
There
are
15
generalized
coordinates
and 15 nonlinear constraint
equations for the piston

crank
assembly used in Module 3.
Since the piston

crank has a mobility
of one,
only one of the
fifteen
equations
will be a
motion constraint
that is an explicit function of time.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 2
0
,
,
t
q
q
t
q
d
k
q
k
i
s the set of kinematic
constraint equations
t
q
d
,
i
s the set of motion
constraint equations
q
i
s the set of generalized
coordinates
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Position
Solving
the constraint equations
will
yield the value of each
generalized
coordinate at a
specific instance of time.
The constraint equations are
non

linear and the Newton

Raphson
method is used as the
solution method.
The Newton

Raphson
method is
iterative and converges when
the constraint equations are
satisfied.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 3
0
,
,
t
q
q
t
q
d
k
Constraint Equations
Newton

Raphson
Equations
t
q
q
t
q
q
q
i
i
,
,
1
1
where
q
t
q
,
i
s the
Jacobian
matrix
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Velocity
The time derivative of the
constraint equations is used
to determine the velocities of
the generalized coordinates.
Since the generalized
coordinates are a function of
time and the constraint
equations are a function of
the generalized coordinates
and time, the chain rule for
partial differentiation must
be used.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 4
0
,
,
t
q
q
t
q
d
k
Constraint Equations
Time Derivative
0
,
t
t
q
q
t
t
q
Velocities
t
q
t
q
1
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Acceleration
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 5
The second time derivative of
the constraint equations is
used to determine the
accelerations of the
generalized coordinates.
Since the generalized
coordinates are a function of
time and the constraint
equations are a function of
the generalized coordinates
and time, the chain rule for
partial differentiation must
be used.
1
st
Time Derivative of Constraint Equations
0
,
t
t
q
q
t
t
q
2
nd
Time Derivative of Constraint Equations
0
2
,
2
2
2
2
2
2
2
t
t
q
q
t
q
t
q
q
q
q
q
t
t
q
Accelerations
2
2
2
1
2
2
2
t
t
q
t
q
q
q
q
q
q
t
q
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Summary of Equations
Newton

Raphson
Equations
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 6
0
,
,
t
q
q
t
q
d
k
Constraint Equations
t
q
q
t
q
q
q
i
i
,
,
1
1
Used to determine the position
(values of the generalized
coordinates) at an instant in time.
Velocities of Generalized Coordinates
t
q
t
q
1
Accelerations of Generalized Coordinates
2
2
2
1
2
2
2
t
t
q
t
q
q
q
q
q
q
t
q
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Jacobian
The
Jacobian
and its inverse is
needed to determine the
position, velocity, and
acceleration of the generalized
coordinates.
Each
i,j
(
row,column
) term in
the
Jacobian
matrix is given by
q
J
Jacobian
matrix
j
i
j
i
q
J
,
i
th
constraint equation
j
th
generalized coordinate
Error messages indicating that
the
Jacobian
is singular are
sometimes encountered when
running multi

body dynamic
programs.
This occurs when there is not a
physically realizable solution.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 7
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Piston

Crank Constraint Equations
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 8
The fifteen constraint
equations developed
for the piston

crank
mechanism in Module
3 are given on the
right.
Note that only the
motion constraint is
an explicit function of
time.
0
)
6
0
)
5
0
)
4
E
cg
E
cg
E
cg
Y
X
0
)
3
0
)
2
0
8
.
156
)
1
A
cg
A
cg
A
cg
Y
X
0
cos
6
.
102
sin
28
)
10
0
sin
6
.
102
cos
28
)
9
C
C
cg
B
B
cg
C
C
cg
B
B
cg
Y
Y
X
X
0
cos
43
sin
3
.
41
)
8
0
sin
43
cos
3
.
41
)
7
D
D
cg
C
C
cg
D
D
cg
C
C
cg
Y
Y
X
X
0
)
12
0
)
11
E
cg
D
cg
E
cg
D
cg
Y
Y
X
X
0
0
1
cos
sin
sin
cos
cos
sin
sin
cos
0
1
1
0
0
1
)
14
B
B
B
B
T
A
A
A
A
T
0
314
)
15
t
D
0
cos
sin
sin
cos
0
1
1
0
0
1
)
13
A
CG
A
CG
B
CG
B
CG
T
A
A
A
A
T
Y
X
Y
X
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Piston

Crank
Jacobian
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 9
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
cos
cos
sin
sin
0
0
sin
cos
sin
0
0
0
0
0
0
0
0
0
0
0
0
cos
sin
sin
cos
cos
sin
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
sin
6
.
102
1
0
cos
28
1
0
0
0
0
0
0
0
0
0
0
cos
6
.
102
0
1
sin
28
0
1
0
0
0
0
0
0
sin
43
1
0
cos
3
.
41
1
0
0
0
0
0
0
0
0
0
0
cos
43
0
1
sin
3
.
41
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
2
B
A
B
A
A
B
A
A
A
A
B
cg
A
cg
A
B
cg
A
cg
A
A
B
C
B
D
C
D
C
j
i
Y
Y
X
X
q
J
The
Jacobian
of the constraint equations is given below. Although there are many
terms, there are a lot of zeros and the derivatives are easily computed.
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Velocities
The velocities of the generalized
coordinates are computed from
the equation
Since the
Jacobian
is known, this
equation can be solved if the array
containing the time derivatives of
the constraint equations is found.
Only the motion constraint,
(15),
is an explicit function of time.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page
10
t
q
t
q
1
314
0
0
0
0
0
0
0
0
0
0
0
0
0
0
t
Motion Constraint
Required Array
0
314
)
15
t
D
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Acceleration
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 11
The accelerations can be computed if each term is found.
2
2
2
1
2
2
2
t
t
q
t
q
q
q
q
q
q
t
q
The time
derivative of
the
Jacobian
is zero.
This term is
explained on
the next slide.
Inverse of the
Jacobian
314
0
0
0
0
0
0
0
0
0
0
0
0
0
0
t
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
2
t
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Acceleration Term
q
q
q
q
This term is evaluated by breaking it down
into a series of operations that are easily
done on a computer.
Step 1) Multiply the
Jacobian
by the
velocities. This creates a
column array.
q
q
Step 2) Take the derivative of each row
with respect to each
generalized coordinate. This
operation is similar to finding
the
Jacobian
and results in a
matrix.
q
q
q
Step 3) Multiply the matrix by the
velocities. This results in a
column array.
q
q
q
q
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 12
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Redundant Constraints
The
Jacobian
matrix is an important
quantity and enables the position,
velocity, and acceleration of the
generalized coordinates to be found.
Application of the methods contained
in this module requires that the
Jacobian
have an inverse.
This requires that the determinant of
the
Jacobian
be non

zero or that the
rank be equal to the number of
generalized coordinates.
The rank of a matrix is equal to the
number of independent rows or
columns.
Independent rows or columns can not
be written as a linear combination of
other rows or columns.
If rows or columns of the
Jacobian
are
not independent the
Jacobian
is
singular and the problem does not
have a solution.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 13
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Redundant Constraints: Detection
The Dynamic Simulation environment
within Autodesk Inventor software
assembles the
Jacobian
and
determines its rank as each constraint
is added.
The rank gives the number of
independent constraints.
The difference between the number of
generalized coordinates and the
number of independent constraints is
equal to the degree of mobility.
The difference between the number of
constraints and the number of
independent constraints is equal to
the degree of redundancy.
nq
number of generalized
coordinates
nc
number of constraints
nic
number of independent
constraints
Degree of Mobility
dom
=
nq
–
ni
c
Degree of Redundancy
dor
=
nc

nic
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 14
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Redundant Constraints: Reaction Forces
A redundant constraint occurs when the
motion associated with a DOF is enforced
by too many constraint specifications.
One or
more of the
constraint
specifications can be removed without
affecting the mobility of the system.
The joint reactions can not be
independently determined when
redundant constraints are present.
Although solutions can be obtained they
are based on assumptions by the program
as to which constraints to use.
Different assumptions will yield different
answers.
Joints having friction are
particularly effected by
redundant constraints.
Friction forces are based on
the joint normal forces.
Therefore, the friction forces
are incorrect if the joint
normal forces are incorrect
due to redundant
constraints.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 15
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Redundant Constraint: Example
A simple four bar mechanism will have
redundant constraints if revolute
joints are used at all joints.
The ground link shown in the figure is
fixed.
The revolute joint at 1 prevents the
drive link from rotating about its long
axis and moving normal to the joint
plane.
The revolute joint at 2 prevents the
coupler from rotating about its long
axis and moving normal to the joint
plane.
The revolute joint at 4 prevents the
rocker from rotating about its long axis
and moving normal to the joint plane.
Ground
Drive
Coupler
Rocker
1
2
3
4
A revolute
joint at 3
is redundant
because neither the rocker or
coupler can rotate about their long
axis or move normal to the joint
plane due to the other revolute
joints
. These degrees of freedom
are already restrained.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 16
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Redundant Constraints: Example
A point

line joint must be used at joint
3.
A point

line joint restricts a point
(the
center
point of the
hole at joint
3
on
the
coupler) to remain on a line
(the
centerline
of the
hole at joint
3
on
the
rocker
).
Redundant joints can be confusing and a
detailed analysis of what each joint is
doing is required to figure out how to
remove them.
An example of how to remove redundant
constraints is provided in the next
module: Module 5.
Ground
Drive
Coupler
Rocker
1
2
3
4
Revolute
Revolute
Revolute
Point

Line
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 17
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Module Summary
This module showed how the constraint equations can be used to
find the position, velocity, and acceleration of the generalized
coordinates.
Kinematic relationships were used during the derivation and no
mention of the forces required to impose the motion constraints was
made.
The constraint equations for the piston

crank introduced in the
previous module (Module 3) were used to demonstrate the
mathematical steps.
The Newton

Raphson
method is generally used to solve the
constraint equations.
The
Jacobian
is a key component of the overall solution process and
the rank of the
Jacobian
is used to detect redundant constraints.
Section 4
–
Dynamic Simulation
Module
4
–
Constraint Kinematics
Page 18
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