Rigid Body Dynamics
CSE169: Computer Animation
Instructor: Steve Rotenberg
UCSD, Winter 2005
Cross Product
i
j
k
a  b  ax
ay
az
bx
by
bz

a  b  a y bz  a z by
a z bx  a x bz
a x by  a y bx

Cross Product

a  b  a y bz  a z by
a z bx  a x bz
c  ab
c x  0  bx  a z by  a y bz
c y  a z bx  0  by  a x bz
c z  a z bx  a x by  0  bz
a x by  a y bx

Cross Product
c x  0  bx  a z by  a y bz
c y  a z bx  0  by  a x bz
c z   a z bx  a x by  0  bz
cx   0
c    a
 y  z
 c z   a y
 az
0
ax
a y  bx 
  
 a x   by 
0  bz 
Cross Product
cx   0
c    a
 y  z
 c z   a y
a  b  aˆ  b
 0

aˆ   a z
 a y

 az
0
ax
 az
0
ax
a y  bx 
  
 a x   by 
0  bz 
ay 

 ax 
0 
Derivative of a Rotating Vector


Let’s say that vector r is rotating around the
origin, maintaining a fixed distance
At any instant, it has an angular velocity of ω
dr
 ωr
dt
r
ω r
ω
Derivative of Rotating Matrix



If matrix A is a rigid 3x3 matrix rotating with
angular velocity ω
This implies that the a, b, and c axes must be
rotating around ω
The derivatives of each axis are ωxa, ωxb, and
ωxc, and so the derivative of the entire matrix is:
dA
 ω A
dt
ˆ A
ω
Product Rule

The product rule defines the derivative of
products
d ab  da
db

ba
dt
dt
dt
d abc  da
db
dc

bc  a c  ab
dt
dt
dt
dt
Product Rule

It can be extended to vector and matrix products
as well
d a  b  da
db

b  a 
dt
dt
dt
d a  b  da
db

b  a
dt
dt
dt
d A  B  dA
dB

B  A
dt
dt
dt
Dynamics of Particles
Kinematics of a Particle
x
dx
v
dt
2
dv d x
a
 2
dt dt
position
veloc ity
accelerati on
Mass, Momentum, and Force
m
mass
p  mv
momentum
dp
f
 ma
dt
force
Moment of Momentum

The moment of momentum is a vector
L  rp



Also known as angular momentum (the two
terms mean basically the same thing, but are
used in slightly different situations)
Angular momentum has parallel properties with
linear momentum
In particular, like the linear momentum, angular
momentum is conserved in a mechanical system
Moment of Momentum

L is the same for all three of these particles
p
•
p
•
r3
L  rp
r2
p
r1
•
Moment of Momentum

L is different for all of these particles
p
•
p
•
L  rp
r2
p
r3
•
r1
Moment of Force (Torque)

The moment of force (or torque) about a
point is the rate of change of the moment
of momentum about that point
dL
τ
dt
Moment of Force (Torque)
L  rp
dL dr
dp
τ
 p  r
dt dt
dt
τ  vp  rf
τ  v  mv   r  f
τ  rf
Rotational Inertia


L=rxp is a general expression for the
moment of momentum of a particle
In a case where we have a particle
rotating around the origin while keeping a
fixed distance, we can re-express the
moment of momentum in terms of it’s
angular velocity ω
Rotational Inertia
L  rp
L  r  mv   mr  v
L  mr  ω  r   mr  r  ω 
L  mrˆ  rˆ  ω
L  I ω
I  mrˆ  rˆ
Rotational Inertia
I  mrˆ  rˆ
 0

I  m  rz
 ry

 rz
0
rx
 r  r

I  m  rx ry
 rx rz

2
y
2
z
ry   0
 
 rx    rz
0   ry
rx ry
ry rz
0
rx
2
z
ry 

 rx 
0 


ry rz 
2
2
 rx  ry 
rx rz
r r
2
x
 rz
Rotational Inertia

m r  r

I    mrx ry
  mrx rz

2
y
L  I ω
2
z

 mrx ry

m r r
2
x
2
z
 mry rz

 mrx rz 

 mry rz 
2
2 
m rx  ry 


Rotational Inertia


The rotational inertia matrix I is a 3x3 matrix that
is essentially the rotational equivalent of mass
It relates the angular momentum of a system to
its angular velocity by the equation
L  I ω

This is similar to how mass relates linear
momentum to linear velocity, but rotation adds
additional complexity
p  mv
Systems of Particles
n
mtotal   mi
tota l mass of all particles
i 1
x cm
mx


m
i
i
position of center of mass
i
p cm   p i   mi v i
to tal momentum
Velocity of Center of Mass
v cm
v cm
v cm
dx cm d


dt
dt
m x
m
dx i
 mi dt


 mi
p cm

mtotal
p cm  mtotalv cm
i
i
i
m v
m
i
i
i
Force on a Particle


The change in momentum of the center of mass
is equal to the sum of all of the forces on the
individual particles
This means that the resulting change in the total
momentum is independent of the location of the
applied force
p cm   p i
dp cm d  p i
dp i


 f i
dt
dt
dt
Systems of Particles

The total moment of momentum around
the center of mass is:
L cm   ri  p i
L cm   x i  x cm  p i
Torque in a System of Particles
L cm   ri  p i
τ cm
τ cm
τ cm
dL cm d  ri  p i


dt
dt
d ri  p i 


dt
  ri  f i 
Systems of Particles


We can see that a system of particles behaves a lot like
a particle itself
It has a mass, position (center of mass), momentum,
velocity, acceleration, and it responds to forces
fcm   fi

We can also define it’s angular momentum and relate a
change in system angular momentum to a force applied
to an individual particle
τ cm   ri  fi 
Internal Forces



If forces are generated within the particle system
(say from gravity, or springs connecting
particles) they must obey Newton’s Third Law
(every action has an equal and opposite
reaction)
This means that internal forces will balance out
and have no net effect on the total momentum of
the system
As those opposite forces act along the same line
of action, the torques on the center of mass
cancel out as well
Kinematics of Rigid Bodies
Kinematics of a Rigid Body

For the center of mass of the rigid body:
x cm
dx cm
v cm 
dt
2
dv cm d x cm
a cm 

2
dt
dt
Kinematics of a Rigid Body

For the orientation of the rigid body:
A
3x3 orientatio n matrix
ω
angular ve locity
dω
ω
dt
angular accelerati on
Offset Position


Let’s say we have a point on a rigid body
If r is the world space offset of the point relative
to the center of mass of the rigid body, then the
position x of the point in world space is:
x  x cm  r
Offset Velocity

The velocity of the offset point is just the
derivative of its position
x  x cm  r
dx dx cm dr
v


dt
dt
dt
v  v cm  ω  r
Offset Acceleration

The offset acceleration is the derivative of the
offset velocity
v  v cm  ω  r
dv dv cm dω
dr
a


r  ω
dt
dt
dt
dt
a  a cm  ω  r  ω  ω  r 
Kinematics of an Offset Point

The kinematic equations for an fixed point
on a rigid body are:
x  x cm  r
v  v cm  ω  r
a  a cm  ω  r  ω  ω  r 
Dynamics of Rigid Bodies
Rigid Bodies




We treat a rigid body as a system of particles, where the
distance between any two particles is fixed
We will assume that internal forces are generated to
hold the relative positions fixed. These internal forces
are all balanced out with Newton’s third law, so that they
all cancel out and have no effect on the total momentum
or angular momentum
The rigid body can actually have an infinite number of
particles, spread out over a finite volume
Instead of mass being concentrated at discrete points,
we will consider the density as being variable over the
volume
Rigid Body Mass

With a system of particles, we defined the total
mass as:
n
m   mi
i 1

For a rigid body, we will define it as the integral
of the density ρ over some volumetric domain Ω
m   d

Rigid Body Center of Mass

The center of mass is:
x cm
xd


 d
Rigid Body Rotational Inertia


  ry2  rz2 d

I     rx ry d

   rx rz d
 I xx

I   I xy
 I xz

I xy
I yy
I yz
I xz 

I yz 
I zz 
  rx rz d 

2
2

r

r
 x z d   ry rz d 
2
2
  ry rz d

r

r
 x y d
  rx ry d




Diagonalization of Rotational Inertial
 I xx

I   I xy
 I xz

I xy
I yy
I yz
I  A  I0  A
T
I xz 

I yz 
I zz 
where
I x

I0   0
 0
0
Iy
0
0

0
I z 
Derivative of Rotational Inertial
dI
dt
dI
dt
dI
dt
dI
dt
dI
dt

d A  I0  A

dt
T
  dA  I
dt
 dA 

0  A  A  I0  
 dt 
T
 ω  A  I 0  AT  A  I 0  ω  A 
T
ˆ  A
 ω  I  A  I 0  ω
T


ˆ T  ωI  I ω
ˆT
 ω  I  A  I 0  AT  ω
ˆ
 ω I  I ω
T
Derivative of Angular Momentum
L  I ω
dL dI
dω
τ
 ω  I 
dt dt
dt
ˆ  ω  I  ω
τ  ω  I  I  ω
ˆ ω  I  ω
τ  ωI ω  I ω
τ  ωI ω  I  ω
Newton-Euler Equations
f  ma
τ  ωI ω  I  ω
Applied Forces & Torques
f cg   f i
τ cg   ri  f i 
1
a f
m
1
ω  I  τ  ω  I  ω 
Properties of Rigid Bodies
m
I
x
v
A
ω
a
ω
p  mv
L  I ω
f  ma
τ  r f  ω I ω  I  ω
Rigid Body Simulation
RigidBody {
void Update(float time);
void ApplyForce(Vector3 &f,Vector3 &pos);
private:
// constants
float Mass;
Vector3 RotInertia; // Ix, Iy, & Iz from diagonal inertia
// variables
Matrix34 Mtx;
// contains position & orientation
Vector3 Momentum,AngMomentum;
// accumulators
Vector3 Force,Torque;
};
Rigid Body Simulation
RigidBody::ApplyForce(Vector3 &f,Vector3 &pos) {
Force += f;
Torque += (pos-Mtx.d) x f
}
Rigid Body Simulation
RigidBody::Update(float time) {
// Update position
Momentum += Force * time;
Mtx.d += (Momentum/Mass) * time;
// Mtx.d = position
// Update orientation
AngMomentum += Torque * time;
Matrix33 I = Mtx·I0·MtxT
// A·I0·AT
Vector3 ω = I-1·L
float angle = |ω| * time;
// magnitude of ω
Vector3 axis = ω;
axis.Normalize();
Mtx.RotateUnitAxis(axis,angle);
}