Constrained rigid body Collision detection Contact point - - PowerPoint PPT Presentation

Constrained rigid body

• Collision detection
• Contact point
• Colliding contact
• Resting contact
• Friction

Collision detection

• Determine whether the collision occurs within

a numerical tolerance

• Determine all pairs of bounding boxes that
• verlap
• Further check contact points between rigid

bodies defined as convex polyhedra

Bisection

Y(t0) Y(t0 + ∆t) Y(t0 + 1 2∆t) Y(tc)

Bounding boxes

• Use bounding boxes to reduce the number of

pairwise collision/contact determinations

• If two bounding boxes have no overlap, no

further comparisons are needed

• Coherence can substantially improve the

performance

Bounding boxes

Sweep and sort algorithm

I1 I2 I3 I4 I5 I6

b3 b5 b1 e5 e1 b6 b2 e3 b4 e6 e4 e2

Bounding boxes

• Each bounding box i can be described as an

interval (bi, ei)

• Find all pairs of interval that intersect
• All the bi and ei are sorted from lowest to

highest in a list

• Maintain an active list which is initially empty
• Whenever bi is encountered, all intervals on the

active list are output as overlapping with i, and i is added to the list

Bounding boxes

Sweep and sort algorithm

I1 I

2

I3 I4 I5 I6

Quiz

• How does sweep and sort extend to 3D cases?
• A: Apply sweep and sort 3 times in x, y, and z

directions, and take the union of overlapped pairs

• B: Apply sweep and sort 3 times in x, y, and z

directions, and take the intersection of

• verlapped pairs

Bounding boxes

• We only need to do a full sort and sweep initially
• What is the cost of the sorting process?
• Subsequent comparisons can be improved by

utilizing coherence

• What is the best sorting algorithm for a nearly

sorted list?

Convex polyhedra

Two polyhedra do not inter-penetrate if and only if a separating plane between them exists

Separating plane

Separating plane

Check all possible edges or faces to find a separating plane at the very first time step or the first time the bounding boxes

• verlap

Y(t0)

Convex polyhedra

Defining face

Utilize coherence by caching defining faces or defining edges

Defining face

Y(t0 + ∆t)

Convex polyhedra

Defining face Defining face

The defining face no longer separate the polyhedra and a new separating plane must be found If no separating plane is found, an inter-penetration has occurred at some earlier time The simulator must back up until the collision time is determined

• Collision detection
• Contact point
• Colliding contact
• Resting contact
• Friction

Contact points

• Once we determine which bodies contact, we

also need to determine the exact contact points

• We consider two types of contacts
• vertex/faces
• edge/edge

Contact points

4 vertex/face 2 vertex/face 2 edge/edge

Contact points

A B A B

ˆ n ˆ n

pa(t) pb(t)

Although pa(t) and pb(t) are coincident at time tc, the velocity of the two points may be different

pa(tc) pb(tc)

Velocity of the contact point

vr = ˆ n(tc) · ( ˙ pa(tc) − ˙ pb(tc))

A B

ˆ n pa pb

˙ pb(tc) = vb(tc) + ωb(tc) × (pb(tc) − xb(tc)) ˙ pa(tc) = va(tc) + ωa(tc) × (pa(tc) − xa(tc))

This quantity gives the component of the relative velocity in the normal direction

Relative normal velocity

vr = 0 vr > 0 vr < 0

A B

˙ pa(tc) − ˙ pb(tc) ˆ n

Separation

A B

˙ pa(tc) − ˙ pb(tc) ˆ n

Resting contact

A B

˙ pa(tc) − ˙ pb(tc)

ˆ n

Colliding contact

• Collision detection
• Contact point
• Colliding contact
• Resting contact
• Friction

Collision process

no force no force

∆t

Impulse:

• fdt = J = M∆v

A soft collision

Force

∆t

Velocity

∆V

A harder collision

Force Velocity

∆t

∆V

∆t = 0

An infinitely hard collision

Force Velocity

fimp = ∞ fimp∆t = J

∆V

Impulse force

• In the rigid body world, we want velocity to

change instantaneously

• use finite impulse to change velocity instead
• f infinite force
• J = ΔP = M Δv

Impulsive torque

• If the impulse acts on a point p, J produces an

impulsive torque

• τimp = (p - x(t)) × J
• impulsive torque gives rise to a change in angular

momentum: τimp = ΔL

Colliding contact

A B

ˆ n jˆ n

For frictionless bodies, the direction of the impulse will be in the normal direction ˆ

n(tc)

J = jˆ n(tc)

Body A is subject to this impulse J, while body B is subject to an equal but opposite impulse -J If we can solve for j, we then can update the linear momentum of the rigid body after the collision

Colliding contact

A B

ˆ n jˆ n

Similarly, we use impulsive torque to update the angular momentum of the rigid bodies after collision

−rb × jˆ n ra × jˆ n

Colliding contact

A B

ˆ n

When two bodies collide, we apply an impulse between them to change their linear and angular momentum such that the penetration doesn’t happen in the next time step:

v+

r = −v− r

vr < 0

The empirical law

Coefficient of restitution: A B

ˆ n V −

r

= 0 = 1

= 0.5

= 0 = 1

Perfect bounce Resting contact

v+

r = −v− r

Colliding contact

A B

ˆ n

A B

ˆ n ˙ p+

a (tc) − ˙

p+

b (tc)

˙ p−

a (tc) − ˙

p−

b (tc)

jˆ n

Relative normal velocity before and after the application of the impulse before collision after collision

v−

r = ˆ

n(tc) · ( ˙ p−

a (t0) − ˙

p−

b (t0))

v+

r = ˆ

n(tc) · ( ˙ p+

a (t0) − ˙

p+

b (t0))

Compute the impulse

Define the displacements from center of mass of the body

rb = pb(tc) − xb(tc) ra = pa(tc) − xa(tc)

and The pre-impulse velocity of pa: The post-impulse velocity of pa: ˙

p+

a = v+ a (tc) + ω+ a (tc) × ra

˙ p−

a = v− a (tc) + ω− a (tc) × ra

Replace with in

v+

a (tc)

v+

a (tc) = v− a (tc) + jˆ

n(tc) Ma ˙ p+

a (tc)

Replace with in

ω+

a (tc) = ω− a (tc) + I−1(tc)(ra × jˆ

n(tc)) ω+

a (tc)

˙ p+

a (tc)

Compute the impulse

˙ p+

a (tc) = ˙

p−

a + j

ˆ n(tc) Ma + (I−1

a (tc)(ra × ˆ

n(tc))) × ra

• v+

r = ˆ

n(tc) · ( ˙ p+

a (tc) − ˙

p+

b (tc))

ˆ n(tc) · (I−1

b (tc)(rb × ˆ

n(tc))) × rb) = ˆ n(tc) · ( ˙ p−

a (tc) − ˙

p−

b (tc)) + j( 1

Ma + 1 Mb + ˆ n(tc) · (I−1

a (tc)(ra × ˆ

n(tc))) × ra+ = v−

r +j( 1

Ma + 1 Mb +ˆ n(tc)·(I−1

a (tc)(ra׈

n(tc)))×ra+ˆ n(tc)·(I−1

b (tc)(rb׈

n(tc)))×rb) v−

r +j( 1

Ma + 1 Mb +ˆ n(tc)·(I−1

a (tc)(ra׈

n(tc)))×ra+ˆ n(tc)·(I−1

b (tc)(rb׈

n(tc)))×rb) = −v−

r

˙ p+

b (tc) = ˙

p−

b − j

✓ ˆ n(tc) Mb + (I−1

b (tc)(rb × ˆ

n(tc))) × rb ◆

Compute the impulse

j = −(1 + )v−

r 1 Ma + 1 Mb + ˆ

n(tc) · (I−1

a (tc)(ra × ˆ

n(tc))) × ra + ˆ n(tc) · (I−1

b (tc)(rb × ˆ

n(tc))) × rb)

Now we can compute the impulse and impulsive torque for body A and B Finally, apply the change in linear momentum and angular momentum to the current state

L(tc + h) = L(tc) + τimp P(tc + h) = P(tc) + J τimp = ra × J J = jˆ n(tc)

A: B:

τimp = rb × J J = −jˆ n(tc)

• Collision detection
• Contact point
• Colliding contact
• Resting contact
• Friction

Resting contact

• In this case, all n contact points have the zero

relative velocity

• At each contact point there is some force ,

where fi is an unknown scalar

• Our goal is to determine what each fi is and all the

fi’s must be determined at the same time

fiˆ n(tc)

Resting contact

• To solve for fi, we need to consider
• three relations between acceleration of the

contact points and the contact forces

• all the fi that are in contact at the same time

Non-penetration

A B

ˆ n

pa(t) pb(t)

di(t) = 0

B

ˆ n pb(t)

A

pa(t)

di(t) < 0

A B

ˆ n pa(t) pb(t)

di(t) > 0

This is what we want to avoid di(t) = ˆ ni(t) · (pa(t) − pb(t))

Non-penetration

˙ di(t) = ˙ ˆ ni(t) · (pa(t) − pb(t)) + ˆ ni · ( ˙ pa(t) − ˙ pb(t)) At time tc, pa(tc) = pb(tc) The definition of two bodies in resting contact: di(tc) = ˙ di(tc) = 0 This means that ˙ di(tc) = vr

Non-penetration

If , the bodies have an acceleration towards each other and the penetration will occur

¨ di(tc) < 0

Therefore, the first condition is

¨ di(tc) ≥ 0 di(tc) = ˙ di(tc) = 0

When

Repulsive force

The contact forces can push bodies apart, but can never act like “glue” and hold bodies together Therefore, each contact force must act outward

fi ≥ 0

Workless force

The contact force at the a contact point becomes zero if the bodies begin to separate If contact is breaking , then fi should be zero

¨ di(tc) > 0

If fi is not zero, then the contact is not breaking

¨ di(tc) = 0

What is the equation that satisfies these two cases?

fi ¨ di(tc) = 0

Resting contact conditions

A B

ˆ n f ˆ n

Condition 1: Non-penetration Condition 2: Repulsive force Condition 3: Workless force

di(t) = ˆ ni(t) · (pa(t) − pb(t))

¨ di(tc) ≥ 0 fi ≥ 0

where

fi ¨ di(tc) = 0

Compute contact forces

¨ di(tc) = ai1f1 + ai2f2 + · · · + ainfn + bi

Step 1: compute A and b

   ¨ d1(tc) . . . ¨ dn(tc)    = A    f1 . . . fn    + b

Step 2: solve f using quadratic programming

Compute contact forces

¨ di(tc) = ˆ ni(tc) · (¨ pa(tc) − ¨ pb(tc)) + 2 ˙ ˆ ni(tc) · ( ˙ pa(tc) − ˙ pb(tc))

Factor out the terms that depend on fj and assign them to aij Assign the rest of terms to bi

= ai1f1 + ai2f2 + . . . + ainfn+bi

See details in Baraff and Witkin’s course notes

Quadratic programming

Solve for Subject to

f1, f2, · · · , fn fi ≥ 0

A    f1 . . . fn    + b ≥ 0

(ai1f1 + · · · + ainfn + bi)fi = 0

i = 1 · · · n i = 1 · · · n

• Collision detection
• Contact point
• Colliding contact
• Resting contact
• Friction

Contact friction

• Dry friction occurs when surfaces in contact

are free of any contamination

• Wet friction occurs when the surfaces are

separated by a thin film of lubricants

Dry contact friction

• Coulomb’s law
• if sliding, the kinetic friction is
• if static (|vT|= 0) then stay static as long as

Ffriction = −µk|Fnormal| vT |vT | |Ffriction| ≤ µs|Fnormal|

Coulomb friction cone

static friction dynamic friction

Quiz

• Given µ for static friction, can you compute the angle
• f the friction cone θ?

|Ffriction| ≤ µs|Fnormal| θ

Quiz

• True or False
• Glass has bigger mu value than rubber.
• The larger the mu is, the wider the angle of the

friction cone is.

• Suppose I am pushing a furniture on the floor.

The faster I push horizontally, the more I damage the floor.

Collision friction

• Collision takes place over a very small time

interval with very large forces

• Assume forces don’t vary significantly over that

interval, then we can replace forces in friction laws with impulses

Dry collision friction

• Assume
• Static case is , when
• Sliding case is

µs = µk v+

T = 0

|v−

T | =≤ µ|∆vN|

v+

T = v− T − µ|∆vN| v− T

|v−

T |

Summary

• What’s the physical meaning of the rotation

matrix?

• How is the rotation matrix related to angular

velocity?

• How do you compute inertia tensor efficiently?
• Why do we compute impulse instead of force in

the case of colliding contact?