Transcript Rigid_Body_Dynamics3..

CS274: Computer Animation and Simulation
Lecture VII
Rigid Body Dynamics
Rigid Bodies
Rigid bodies have both a position and orientation
Rigid bodies assume no object deformation
Rigid body motion is represented by 2 parameters
x(t ) - center of mass
R (t ) - orientation (rotation matrix)
Rigid Bodies
Objects are defined in body space and transformed
by the position and orientation into world space
p(t )  R(t ) p 0  x(t )
Linear Velocity
The change of the center of mass over time
x (t )  v (t )
For a pure translation ( R (t ) constant), all points
move with velocity v (t )
Angular Velocity
The change in orientation over time
ω(t )
Encodes both the axis and speed of the rotation
 direction encodes the axis
 magnitude encodes the speed (rad/s)
But, how are ω(t ) and R (t ) related?
Angular Velocity
ω
For a given vector
r (t )  ω(t )  r (t )
r
The columns of R (t ) represent the transformed axes
 (t )  ω(t )  R (t )
R
.j
.j
e2
R. 2
R.1
e1
e3
R. 3
Angular Velocity
We can represent the cross product with a matrix
0
a  b  a b   ax
 a y
*
az
0
az
Therefore
*

R(t )  ω (t ) R(t )

ax  b
0 
ay
Velocity of a Point
Since a point can be represented at any time by
r(t )  R(t ) r0  x(t )
Total velocity can then be expressed as
 (t ) r  v(t )
r (t )  R
0
Which can be rewritten as
r (t )  ω* R (t ) r0  v(t )
r (t )  ω  (r (t )  x(t ))  v(t )
Force
We can apply forces to the object at any point
F2 (t )
F1 (t )
Total force on an object is simply
F(t )   Fi (t )
No information about where the forces are applied
Torque
Torque describes the “rotational force”
τ i (t )  (ri (t )  x(t ))  Fi (t )
x(t )
F2 (t )
r1 (t )
F1 (t )
Total torque on an object is simply
τ(t )   τ i (t )   (ri (t )  x(t ))  Fi (t )
Tells us about the force distribution over the object
Linear Momentum
Linear momentum of a particle is
pmv
Linear momentum of a rigid body is then
P (t )    r (t ) dV
density
integration over the body
Linear Momentum
Linear momentum can be simplified as follows
P (t )    r (t ) dV
P(t )    v(t )   ω  (r (t )  x(t )) dV
P(t )  M v(t )
Assuming constant mass gives
P (t )  M v (t )
P (t )  F(t )
Angular Momentum
Angular momentum of a rigid body
L(t )  I(t ) ω(t )
inertia tensor
Taking the time derivative
 (t )  τ(t )
L
Angular momentum is conserved for no torque
Inertia Tensor
Describes how mass is distributed in the body
 I xx

I (t)   I yx
 I zx

I xy
I yy
I zy
I xz 

I yz 
I zz 
I xx    ( yˆ 2  zˆ 2 ) dV
I xy     ( xˆyˆ ) dV
xˆ  x  xcen terOfMass
Analogous to mass in linear velocity
Measures the preferred axis of rotation
Expensive to compute this at every time step
Inertia Tensor
Rewrite the tensor as
T
ˆ
ˆ
ˆ
ˆ
I (t )    (r (t )  r (t ) Id  r (t ) r (t ) ) dV
I (t )    (( R (t )r0 )T R (t )r0 Id  R (t )r0 (R (t )r0 )T ) dV
I (t )    (R (t )(r0 r0 Id  r0r0 )R T (t )) dV
T
T
I(t )  R(t ) I body R (t )
T
Integrals can now be precomputed
Rigid Body Equations of Motion
Combining the equations
 x(t ) 
R (t )
d 

dt  P(t ) 


 L(t ) 

 v (t ) 
ω* R (t )


 F (t ) 


 τ (t ) 
Discretize these continuous equations and integrate
Using Quaternions
Use quaternions to represent orientation
 cos( / 2) 
q(t )  

sin(  / 2) axis
The update rule is then
1
q (t )  ω(t ) q(t )
2
Rigid Body Equations of Motion
Using quaternions gives
 x(t ) 
 q(t ) 
d 

dt  P (t ) 


L(t )
 v (t ) 
 1 ω q(t )

 2
 F (t ) 


 τ (t ) 
Discretize these continuous equations and integrate
Collisions and Contact
So far, no interaction between rigid bodies
Collision detection –
determining if, when and where a collision occurs
Collision response –
calculating the state (velocity, …) after the collision
Collisions and Contact
What should we do when there is a collision?
x(t0 )
x(t1 )
x(t2 )
x(t3 )
Rolling Back the Simulation
Restart the simulation at the time of the collision
x(t0 )
x(t1 )
x(t2 )
x(tc )
x(t3 )
Collision time can be found by bisection, etc.
Collision Detection
Exploit coherency through witnessing
Two convex objects are
non-penetrating iff there exists a
separating plane between them
separating plane
First find a separating plane and
see if it is still valid after the next
simulation step
Speed up with bounding boxes, grids, hierarchies, etc.
Collision Detection
Conditions for collision
p a (t )  v a (t )  ωa (t )  (p a (t )  x a (t ))
A
p a  p b
N
A
N
B
N  (p a (t )  p b (t ))  0
separating
A
p a  p b
N
B
N  (p a (t )  p b (t ))  0
contact
B
p a  p b
N  (p a (t )  p b (t ))  0
colliding
Collision
Soft Body Collision
Force is applied to prevent interpenetration
Collision
Soft Body Collision
Apply forces and change the velocity
Collision
Harder Collision
Higher force over a shorter time
Collision
Rigid Body Collision
Impulsive force produces a discontinuous velocity
Impulse
We need to change velocity instantaneously
Infinite force in an infinitesimal time
J  F t
An impulse changes the velocity as
J
v 
M
or
P  J
Impulse
An impulse also creates an impulsive torque
τ impulse  (p(t )  x(t ))  J
The impulsive torque changes the angular velocity
1
ω  I (t ) τ impulse
or
L  τ impulse
Impulse
For a frictionless collision
J jN
A Ja
N
B
Jb
But how do we calculate j ?
Impulse
For a frictionless collision
N  (p a (t )  p b (t ))   (N  (p a (t )  p b (t )))
Given this equation and knowing how j affects the
linear and angular velocities of the two bodies,
we can solve for j
.̂
.̂
.̂
.̂
Resting Contact
Bodies are neither colliding nor separating
We want a force strong enough to resist
penetration but only enough to maintain contact
Resting Contact
We want to prevent interpenetration
d (t )  N  (p a (t )  pb (t ))  0
Since d (tc )  0 we should keep it from decreasing
  (p (t )  p (t ))  N  (p (t )  p (t ))  0
d (t )  N
a
b
a
b
Since d (tc )  0 we should keep it from decreasing
  (p (t )  p (t ))  0
a (tc )  p
b (tc ))  2 N
d(tc )  N  (p
a c
b c
Describes the objects’ acceleration towards one another
Resting Contact
Contact forces only act in the normal direction
Fc  f N(tc )
Contact forces should
 avoid interpenetration
d(tc )  0
 be repulsive
f 0
 become zero if the bodies begin to separate
f  d(tc )  0
workless force
Resting Contact
The relative accelerations can be written in terms
of all of the contact forces
di (tc )  a0 f 0  ...  an f n  bi
So we can simply solve a Quadratic Program to
find the solution to all the constraints
Simulation Algorithm
Algorithm with collisions and contact
current state
compute new state
next state
detect collisions and backtrack
collision state
compute and apply impulses
post-collision state
compute and apply constraint forces