Lesson 11-C: Body Coordinates Formulation
Velocity Constraints
•
The treatment of the velocity constraints in these notes does not exactly follow the treatment
you find in the textbook. The velocity formulations from these notes will mostly be used
throughout this course.
Velocity Arrays
•
•
•
Velocity constraints are obtained from the time derivative of the position (coordinate)
constraints. For the position constraints
Φ ≡ Φ (q) = 0
Since the velocity array for each body can be expressed in one of the following forms, then
the velocity constraints can also be expressed in three different forms.
⎧⎪ r ⎫⎪
⎧⎪ r ⎫⎪
⎧⎪ r ⎫⎪
v i = ⎨ i ⎬ (a),
h i = ⎨ i ⎬ (b),
q i = ⎨ i ⎬ (c)
⎪⎩ω i ⎪⎭
⎪⎩ω i′ ⎪⎭
⎪⎩p i ⎪⎭
In the following slides we derive the velocity constraints for several common constraint
equations using all three forms of the velocity arrays
Time Derivatives of Vectors
•
•
The time derivative of a vector that is fully attached to a body (fixed magnitude) can be
expressed in any of the following forms:
(a)
s = ω s = −s ω
(b)
s = −s A ω ' = −A s ' ω '
(c)
s = −2 s G p
The time derivative of a vector that locates a point on a body with respect to the global
origin; i.e., r P = r + s P , can be expressed in any of the following forms:
(a)
r P = r + ω s P = r − s P ω
r P = r − s P A ω '
(b)
r = r − 2 s G p
(c)
P
•
P
In this course we choose to work with the global components of the angular velocity vector;
i.e., form (a). The textbook mostly uses form (c) and occasionally form (b). However, the
constraint equations and the equations of motion find simpler forms if we use form (a).
Two Perpendicular Vectors (n1)
•
For this constraint we derive the velocity equation in all three forms that were discussed
previously
–
The time derivative in form (a)
Φ ( n1,1) = a Ti n j + n Tj a i = −a Ti n j ω j − n Tj a i ω i = n Tj a i ω j − n Tj a i ω i = 0
In matrix form:
(a)
⎧ ri ⎫
⎧ ri ⎫
⎪ ⎪
⎪ ⎪
ω
⎪
⎪ω i ⎪
⎪
i
Φ ( n1,1) = ⎡⎣ 0 −n Tj a i 0 n Tj a i ⎤⎦ ⎨ ⎬ = n Tj a i ⎡⎣ 0 −I 0 I ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ r j ⎪
⎪ω ⎪
⎪ω ⎪
⎩ j⎭
⎩ j⎭
The Jacobian is a 1 × 12 matrix.
–
–
•
The time derivative in form (b)
Φ ( n1,1) = a Ti n j + n Tj a i = −a Ti n j A j ω ′j − n Tj a i A i ω ′i = 0
(b)
The time derivative in form (c)
Φ ( n1,1) = a Ti n j + n Tj a i = −2a Ti n j G j p j − 2 n Tj a i G i p i = 0
(c)
For the next set of constraints, we only derive the velocity and acceleration constraints in
form (a). Refer to the textbook for the other forms.
Two Perpendicular Vectors (n2)
•
The time derivative of this constraint is written as
Φ ( n2,1) = d T n j + n Tj d = −d T n j ω j + n Tj (r j − s Pj ω j − ri + s iP ω i ) = 0
In matrix form:
Φ ( n2,1) = ⎡⎣ −n Tj
n Tj s iP
n Tj
⎧ ri ⎫
⎪ ⎪
⎪ω i ⎪
T P
n j (d − s j ) ⎤⎦ ⎨ ⎬ = n Tj ⎡⎣ −I s iP
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
The Jacobian is a 1 × 12 matrix.
⎧ ri ⎫
⎪ ⎪
⎪ω i ⎪
P
I d − s j ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
Two Parallel Vectors (p1)
•
The time derivative can be expressed as
Φ ( p1,2) = a i a j − a j a i = −a i a j ω j + a j a i ω i = 0
In matrix form:
Φ ( p1,2) = ⎡⎣ 0 a j a i
The Jacobian is a 2 × 12 matrix.
⎧ ri ⎫
⎪ ⎪
⎪ω i ⎪
0 − a i a j ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
Two Parallel Vectors (p2)
•
The time derivative can be expressed as
Φ ( p2,2) = d a j − a j d = 0
= − d a j ω j − a j (r j − s Pj ω j − ri + s iP ω i ) = 0
In matrix form:
Φ ( p2,2) = ⎡⎣ − a j
a j s iP
− a j
The Jacobian is a 2 × 12 matrix.
⎧ ri ⎫
⎪ ⎪
⎪ω i ⎪
− d a j + a j s Pj ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
Spherical Joint
•
The velocity constraints can be expressed as
Φ (s,3) = r j + s Pj − ri − s iP = r j − s Pj ω j − ri + s iP ω i = 0
In matrix form:
Φ (s,3) = ⎡⎣ −I s iP
The Jacobian is a 3 × 12 matrix.
⎧ ri ⎫
⎪ ⎪
ω
P ⎪ i ⎪
I −s j ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
Spherical-Spherical Joint
•
The velocity constraint can be expressed as
Φ (s−s,1) = d T (r j + s Pj − ri − s iP ) = d T (r j − s Pj ω j − ri + s iP ω i ) = 0
In matrix form:
Φ (s−s,1) = ⎡⎣ −d T
d T s iP
dT
The Jacobian is a 1 × 12 matrix.
⎧ ri ⎫
⎪ ⎪
ω
T P ⎪ i ⎪
−d s j ⎤⎦ ⎨ ⎬ = d T ⎡⎣ −I s iP
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
⎧ ri ⎫
⎪ ⎪
ω
P ⎪ i ⎪
I −s j ⎤⎦ ⎨ ⎬ = 0
⎪ r j ⎪
⎪ω ⎪
⎩ j⎭
Translational (Prismatic) Joint
•
In the Formulation II of the translational joint constraints, the relative orientation of the two
bodies is constrained by the equation Φ (A,3) ≡ A Ti A j − A (c) = 0 . The time derivative of this
equation reveals exactly how the two bodies are constrained to one another:
T A + AT A
= −A T ω A + A T ω A = −A T ω A + A T ω A = 0
(A,3) ≡ A
Φ
i
j
i
j
i
i
j
i
j
j
i
i
j
i
j
j
Pre-multiplying this equation A i and the post-multiplying by A Tj yields:
− ω i + ω j = 0
or
ω j − ωi = 0
General Velocity Constraints -- Type (a)
•
Arrays of velocity for a typical body i and for a multibody system:
⎧⎪ r ⎫⎪
vi = ⎨ i ⎬
⎩⎪ω i ⎭⎪
•
•
⎧ v1 ⎫
⎪ ⎪
v=⎨ ⎬
⎪v ⎪
⎩ n⎭
The velocity constraints for the multibody system are written as
(a)
Φ ≡ D v = 0
Where D is the coefficient matrix in the velocity constraints of Type (a)
We will loosely refer to matrix D as the Jacobian. However, we should note that the correct
⎡ ∂Φ ⎤
definition of the Jacobian yields a matrix as ⎢
⎥ . Here we have defined the arrays of
⎣ ∂q ⎦
coordinates and velocities for body i and for the system as
⎧ q1 ⎫
⎧ q 1 ⎫
⎪ ⎪
⎪ ⎪
⎪⎧ ri ⎪⎫
⎪⎧ ri ⎪⎫
q=⎨ ⎬
q = ⎨ ⎬
qi = ⎨ ⎬
q i = ⎨ ⎬
⎪q ⎪
⎪q ⎪
⎩⎪p i ⎭⎪
⎩⎪p i ⎭⎪
n
⎩ ⎭
⎩ n⎭
The corresponding velocity constraints for the multibody system are written as
∂Φ
Φ ≡
q = 0
(c)
∂q
•
Form (c) of the velocity equations provides the correct description of the Jacobian where
form (a) provides the transformed form of the Jacobian.
•
•
The right-hand-side of the velocity constraints for standard kinematic joints is always zero!
•
Note that whether we use velocity constraints of Type (a), (b), or (c), the position
constraints will be in terms of the Euler parameters!
The Jacobian matrix entries that are listed in Tables 7.1 and 7.2 of the textbook will not be
used in this course. These tables provide the Jacobian entries for Type (c) velocity
constraints. However, you are encouraged to familiarize yourself with these two tables
Example (body coordinates—continued)
•
The velocity constraints for this system are
r1 − s1P ω 1 − r2 + s 2P ω 2 = 0
−b T2 a 1 ω 1 − a1T n 2 ω 2 = 0
r2 − s 2Q ω 2 − r3 + s 3Q ω 3 = 0
We have 7 velocity constraints and 18 velocities (6 per body), therefore the difference is 18
- 7 = 11 that is equal to the number of system’s DoF.
•
The Jacobian matrix based on the velocity constraints is constructed as
⎡I
−s1P
−I
s 2P
0 0⎤
⎢
⎥
D = ⎢ 0 −n T2 a 1 0 −a1T n 2 0 0 ⎥
⎢
⎥
0
I
−s2Q
−I s 3Q ⎥
⎢⎣ 0
⎦
This is a 7 × 18 matrix. The columns correspond to a velocity vector that is formed in the
following order: r1 ω 1 r2 ω 2 r3 ω 3