Transcript L14_RigidBody

Rigid Body Dynamics
simulate basic physics of an object
subject to forces
Keyframing can be tedious - especially to get ‘realism’
Simulate physics by
- programming equations of motions
- setting initial conditions
Animator gives up control
Animator gets ‘realistic’ motion automatically.
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Siimulation Update Cycle
Object properties
Position, orientation
Linear and angular velocity
Linear and angular momentum
mass
Update object properties
Calculate forces
Calculate accelerations
Using mass, momenta
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Object under Forces
Forces
Gravity
Wind
Springs
Collision avoidance
Soft constraints
Calculate acceleration due to forces
Calculate update to object velocity & position
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Object under Linear Force
Given: force, f(t), gives rise to acceleration, a(t), or x(ti )
Use a(t) to update current state of point mass: x(t), v(t)
x(ti 1 )  x(ti )  x (ti )t
x (ti 1 )  x (ti )  x(ti )t
( x (ti )  x (ti 1 ))
x(ti 1 )  x(ti ) 
t
2
1
x(ti 1 )  x(ti )  x (ti )t  x(ti )t 2
2
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Equations of Motion
Linear force
Angular force: torque
Linear momentum
Angular momentum
Conserved in a closed system
Resistance to linear force: mass
Resistance to angular force: inertia tensor.
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Rotational Movement
Represent orientation with rotation matrix: R(t)
Represent angular velocity with vector: w(t)
- direction is axis of rotation
- magnitude is speed of rotation
Angular velocity insensitive to distance from center or rotation
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Rotational Movement
w(t) - angular velocity
w(t)
a  b  r(t)
r(t) - position of a relative to b
b - point on axis of rotation

r (t)
q - angle r(t) makes with axis of rotation
q
b
r(t )  w (t )  r (t )
r(t )  w (t ) r (t ) sin( q )
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Rotational Movement
R(t) - rotation matrix representing the orientation of a rigid body
Columns are vectors to positions a unit length on principle axis
R(t)  R1(t) R2 (t) R3 (t)

Change in rotation matrix can be computed by
taking cross product of w(t) with each column
R (t )  w (t )  R1 (t ) w (t )  R2 (t ) w (t )  R3 (t )
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Rotational Movement
R (t )  w (t )  R1 (t ) w (t )  R2 (t ) w (t )  R3 (t )
Define the following for notational convenience to
define cross product as a matrix operation:
 w y  Az  w z  Ay 


w  A   w z  Ax   w x  Az 
 w x  Ay  w y  Ax 


 0

  wz
 w y

 wz
0
wx
w y   Ax 
 
 w x   Ay   w * A
0   Az 
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Velocity of a Point
q - original position of point in rigid body
x(t) - position of rigid body
v(t) - velocity of rigid body
q(t) – time-dependent position of point in rigid
body
q(t )  R(t )q  x(t )
q (t )  R (t )q  x (t )
*

R(t )  w (t ) R(t )
q (t )  w (t ) R(t )q  v(t )

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Center of Mass
Object’s “position” is the position of its center of mass
Integration of differential mass times position in object
Approximate by summing over representational particles in object
M   mi
m q (t)

x(t) 
i i
M
Mass is the object’s resistance to a change in velocity
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Force and Torque
F  ma
a  F /m
Force on center of mass results
in linear acceleration
F
F (t )   f i (t )
 i (t )  (q(t )  x(t ))  f i (t )
 (t )   i (t )
Force off-center of mass is torque and
creates angular acceleration - what is
angular mass?
F
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But first…Linear Momentum
p  mv
Mass times velocity
P(t )   mi vi (t )
P(t )  Mv (t )
Sum over all particles of object
Use total mass for linear momentum
P (t )  Mv(t )  F (t )
Change in (linear)
momentum is equal
to force
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Angular Momentum -
measure of rotating
mass weighted by its distance from the axis of rotation
Relative position of particle
Relative velocity of particle
mass
L(t )   ((q(t )  x(t ))  mi (q (t )  v(t ))
  ( R(t )q  mi (w(t )  (q(t )  x(t ))))
  (mi ( R(t )q  (w(t )  R(t )q)))
Factor out mass
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Angular Momentum
L(t)  (mi (R(t)q  (w(t)  R(t)q)))
Define Inertia Tensor, I(t), that captures distribution of mass

L(t )  I (t )w (t )
L (t )   (t )
Similar to the linear case, the change in
Angular Momentum is equal to torque
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Inertia Tensor -
object’s resistance to a
change in rotational velocity
Ixx

Iobject  Ixy

Ixz
Ixx 

Ixz 

Iyz 
Izz 

Ixy
Iyy
Iyz
 (q)(q
y
2
2
 qz )dxdydz
I(t)  R(t)Iobject R(t)
T
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Inertia Tensor
Ixx 
 (q)(q
y
 Ixx   mi (y i  zi )
2
2
2
2
 qz )dxdydz
Ixy   m i x i y i
Iyy   mi (x i  zi )
Ixz   m i x i zi
Izz   mi (x i  y i )
Iyz   m i y i zi
2
2
2
2
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Inertia Tensor for a Cuboid
M (b2 + c2)
0
0
0
M (a2 + c2)
0
0
0
M (a2 + b2)
1
12
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Finally, the Equations - State vector
S (t) =
x (t)
Update with velocity
R (t)
Update with angular velocity
P (t)
Update with force
L (t)
Update with torque
Given: S(t), M, Iobject
I(t)  R(t)Iobject R(t)
*

R(t )  w (t ) R(t )
P (t )  Mv(t )  F (t )
w(t)  I(t) L(t)
L (t )   (t )
v(t)  P(t) / M
T
1
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
Update State Vector
x(t)  v(t) 
   *

d
d R(t) w (t) R(t)
S(t) 

dt
dt P(t)  F (t) 
  

L(t)   (t) 
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Update Orientation - options
R(t) += w (t) * R(t) and re-orthonormalize
Extract axis-angle from w (t) and rotate columns of R(t)
Use quaternions to rotate quaternion version of R(t)
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QuickTime™ and a
Video decompressor
are needed to see this picture.
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Now, what about collisions?
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