1cs533d-winter-2005

NotesNotes

Assignment 2 going okay?• Make sure you understand what needs to be

done before the weekend Read Guendelman et al, “Nonconvex rigid

bodies with stacking”, SIGGRAPH’03 Mistake last class:

(forgot a transposein calculating torque)

€

ST Fext = δX *T − x i

*T ⎛ ⎝ ⎜

⎞ ⎠ ⎟ f i

i

∑

=f ii

∑x i − X( )

i∑ × f i

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

= Fτ ⎛ ⎝ ⎜ ⎞

⎠ ⎟

2cs533d-winter-2005

Inertia Tensor SimplifiedInertia Tensor Simplified

Reduce expense of calculating I(t):

• Now use xi-X=Rpi and use RTR=

€

I(t) = mi x i − X( )∗T

x i − X( )∗

i∑

= mi x i − X( )T

x i − X( )δ − x i − X( ) x i − X( )T

[ ]i∑

€

I(t) = mi piT RT Rpiδ − Rpi pi

T RT[ ]

i∑

= R mi piT piδ − pi pi

T( )

i∑( )

Ibody

1 2 4 4 4 3 4 4 4 RT

3cs533d-winter-2005

Inertia Tensor Simplified 2Inertia Tensor Simplified 2

So just compute inertia tensor once, for object space configuration

Then I(t)=RIbodyRT

And I(t)-1=R(Ibody)-1RT

• So precompute inverse too In fact, since I is symmetric, know we have an

orthogonal eigenbasis Q Rotate object-space orientation by Q

• Then Ibody is just diagonal!

4cs533d-winter-2005

Degenerate Inertia Degenerate Inertia TensorsTensors

Inertia tensor can always be inverted unless all the points of the object line up (object is a rod)• Or there’s only one point

We don’t care though, since we can’t track rotation around that axis anyways• So diagonalize I, and only invert nonzero elements

5cs533d-winter-2005

Taking the limitTaking the limit

Letting our decomposition of the object into point masses go to infinity:• Instead of sum over particles,

integral over object volume• Instead of particle mass,

density at that point in space

€

mi foo(x i) → ρ(x)x

∫∫∫i

∑ foo(x)dx

6cs533d-winter-2005

Computing Inertia TensorsComputing Inertia Tensors

Do the integrals: Lots of “fun” You may just want to look them up instead

• E.g. Eric Weisstein’s World of Science on the web If not…. align axis perpendicular to planes of

symmetry (of ) in object space• Guarantees some off-diagonal zeros

Example: sphere, uniform density, radius R

€

Ibody = ρ pT pδ − ppT( )dp

p

∫∫∫

€

25 MR2 0 0

0 25 MR2 0

0 0 25 MR2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

7cs533d-winter-2005

Approximating Inertia Approximating Inertia TensorsTensors

For complicated geometry, don’t really need exact answer

Could just take the inertia tensor from a simpler geometric figure (will anyone notice?)

Or numerically approximate integral• If we can afford to spend a lot of time precomputing,

life is simple• Grid approach: sample density…• Monte Carlo approach: random samples

8cs533d-winter-2005

Combining ObjectsCombining Objects

What if object is union of two simpler objects? Integrals are additive

• But DO NOT USE I1(t)+I2(t): World-space formulas (x-X) use the X for the object: X1 and

X2 may be different Simplified Ibody formula based on having centre of mass at

origin

• Let’s work it out from the integral of I(t) Combined mass: M=M1+M2

Centre of mass of combined object:

€

X =ρx

Ω1∪Ω 2∫

ρΩ1∪Ω 2

∫=

M1X1 + M2X2

M

9cs533d-winter-2005

Combined Inertia TensorCombined Inertia Tensor

€

I(t) = ρ x − X( )∗T

x − X( )∗

Ω1∪Ω 2∫

= ρ x − X1 + X1 − X( )∗T

x − X1 + X1 − X( )∗

Ω1∫ + L

Ω 2∫

= ρ x − X1( )∗T

x − X1( )∗

Ω1∫ + ρ X1 − X( )

∗Tx − X1( )

∗

Ω1∫

+ ρ x − X1( )∗T

X1 − X( )∗

Ω1∫ + ρ X1 − X( )

∗TX1 − X( )

∗+ L

Ω 2∫

Ω1∫

= I1(t) + X1 − X( )∗T

ρ x − X1( )∗

Ω1∫

01 2 4 4 3 4 4

+ ρ x − X1( )∗T

Ω1∫

01 2 4 4 3 4 4

X1 − X( )∗

+ M1 X1 − X( )∗T

X1 − X( )∗

+ LΩ 2

∫= I1(t) + M1 X1 − X( )

∗TX1 − X( )

∗+ I2(t) + M2 X2 − X( )

∗TX2 − X( )

∗

10cs533d-winter-2005

Numerical IntegrationNumerical Integration

Recall equations of motion

X and V is just like particle motion Angular components trickier:

R must remain orthogonal, but standard integration will cause it to drift• Can use Gram-Schmidt, but expensive and biased€

ddt V = F

Mddt L = Τ

ddt X = V ω = I(t)−1L

ddt R = ω∗R

11cs533d-winter-2005

Improving on RImproving on R

Instead of 9 numbers for 3 DOF, use a less redundant representation

Euler angles: 3 numbers• But updating with angular velocity is painful

Quaternions: 4 numbers

12cs533d-winter-2005

What are quaternions?What are quaternions?

Instead of R, use q=(s,x,y,z) with |q|=1• Can think of q as a “super complex number”

s+xi+yj+zk• i2=j2=k2=-1, ij=-ji=k, jk=-kj=i, ki=-ik=j

• Quaternions don’t commute! q1q2≠q2q1 in general Represents “half” a rotation:

• s=cos(/2)• |x,y,z|2=sin2(/2)• Axis of rotation is (x,y,z)

Conjugate (inverse for unit norm) is

€

q = (s,−x,−y,−z)

13cs533d-winter-2005

Rotating with quaternionsRotating with quaternions

Instead of Rp, calculate Composing a rotation of ∆t to advance a time step:

For small ∆t approximate:

From this get the differential equation:

€

q(0, p)q

€

qn +1 = cosΔtω

2, sinΔt

ω

2

ω

ω

⎛

⎝ ⎜

⎞

⎠ ⎟qn

€

qn +1 = 1, Δtω

2

⎛

⎝ ⎜

⎞

⎠ ⎟qn = qn + Δt

(0,ω)

2qn

€

˙ q = 12 (0,ω)q

14cs533d-winter-2005

Integrating RotationIntegrating Rotation

Can update like Symplectic Euler, but need to renormalize q after each step

For reasonable accuracy, limit time step according to rate of rotation• Don’t try for more than a quarter turn per time

step, say• Stability is not an issue due to renormalization

For more accurate methods, see S. R. Buss, “Accurate and efficient simulation of rigid body rotations”, JCP 2000

15cs533d-winter-2005

Converting q to RConverting q to R

Clearly superior to use quaternions for storing and updating orientation

But, slightly faster to transform points with rotation matrix

If you need to transform a lot of points (collision detection…) may want to convert q into R

Basic idea: columns of R are rotated axes R(1,0,0)T, R(0,1,0)T, and R(0,0,1)T

Do the rotation with q instead.• Can simplify and optimize for the zeros - look it up

16cs533d-winter-2005

GravityGravity

Force on a point is mig Net force:

Net torque:

€

F = mig = Mgi

∑

€

τ = x i − X( ) × migi

∑

= mix i

i

∑ ⎛

⎝ ⎜

⎞

⎠ ⎟− MX

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟× g

= 0

17cs533d-winter-2005

Collision ImpulsesCollision Impulses

Can use same collision detection as deformable objects• Since geometry is fixed, may be cheaper• E.g. can use level set approximation to geometry

But applying collision impulses is more complicated than for simple particles• Need to take into account angular motion too

Use same principle though for the colliding points• What is the impulse that causes their relative velocity

to change as desired?

18cs533d-winter-2005

Frictionless impulseFrictionless impulse

Object velocities at point:• vi=i(x-Xi)+Vi

Relative velocity v=v1-v2

• Normal component vn=v•n Want post-collision relative normal velocity to be

vnafter=-vn

Apply an impulse j=jnn in the normal direction to achieve this

€

Viafter = Vi + M i

−1 ji

Liafter = Li + x − X i( ) × ji

ωiafter = ωi + Ii(t)

−1 x − X i( ) × ji

ji = (−1)i+1 jnn

19cs533d-winter-2005

Computing frictionless Computing frictionless impulseimpulse

€

K i =1

M i

δ + x − X i( )∗T

Ii−1 x − X i( )

∗

€

j =− 1+ ε( )vn

nT K1 + K2( )nn

20cs533d-winter-2005

Computing frictionComputing friction

Static friction valid only in “friction cone”

Approach:• Calculate static friction impulse (whatever it

takes to make relative velocity zero)• Check if it’s in the friction cone• If so, we’re done• If not, try again with sliding

€

jT ≤ μ jn

21cs533d-winter-2005

Computing static frictionComputing static friction

€

v after = −εvnn

€

j = K1 + K2( )−1

−v −εvnn( )

22cs533d-winter-2005

Sliding frictionSliding friction

If computed static friction impulse fails friction cone test

We’ll assume sliding direction stays constant during impact: tangential impulse just in the initial relative velocity direction• Not true in some situations…

23cs533d-winter-2005

Computing sliding frictionComputing sliding friction

€

T =v − vnn

v − vnn

€

j = jnn − μjnT

€

jn =− 1+ ε( )vn

nT K1 + K2( )(n − μT)

24cs533d-winter-2005

Rigid Collision AlgorithmsRigid Collision Algorithms

Use the same collision response algorithm as with particles• Identify colliding points as perhaps the deepest

penetrating points, or the first points to collide• Make sure they are colliding, not separating!

Problem: multiple contact points• Fixing one at a time can cause rattling.• Can fix by being more gentle in resolving contacts -

negative coefficient of restitution Problem: multiple collisions (stacks)

• Fixing one penetration causes others• Solve either by resolving simultaneously

or enforcing order of resolution