Constrained rigid body Detect collision simulation Compute - - PowerPoint PPT Presentation
constraints Constrained rigid body Detect collision simulation Compute velocity change for colliding contact Compute contact forces when rigid bodies are in resting contact Bisection Collision detection Determine when the collision occurs
Y(t0) Y(t0 + ∆t) Y(t0 + 1 2∆t) Y(tc)
If two bounding boxes have no overlap, no further comparisons are needed Determining all pairs of boxes that overlap can be done in time O(nlogd-2n+k) for d-dimensional bounding boxes Coherence can substantially improve the performance
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
Sweep and sort algorithm
I1 I2 I3 I4 I5 I6
b3 b5 b1 e5 e1 b6 b2 e3 b4 e6 e4 e2
What is the cost of the sorting process?
What is best sorting algorithm for a nearly sorted list?
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 of simulation or the first time two rigid bodies become close enough to require more than a bounding box test
Defining face
Utilize coherence by caching defining faces or defining edges
Y(t0)
Defining face
Y(t0 + ∆t)
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
vertex/faces edge/edge
4 vertex/face 2 vertex/face 2 edge/edge
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)
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
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
no force no force
∆t
Impulse:
Force Velocity
∆t
Force Velocity
∆t
∆t = 0
Force Velocity
fimp = ∞ fimp∆t = J
In the rigid body world, we want velocity to chance instantaneously
use impulse to change velocity instead of force J = P = M v
If the impulse acts on a point p, then just as a force produces a torque, J produces an impulsive torque
imp = (p - x(t)) J impulsive torque gives rise to a change in angular momentum: imp = L = I(tc) (tc)
A B
ˆ n jˆ n
When two bodies collide, we apply an impulse between them to change their velocity 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
A B
ˆ n
A B
ˆ n ˙ p+
a (tc) − ˙
p+
b (tc)
˙ p−
a (tc) − ˙
p−
b (tc)
jˆ n
Relative velocity before and after the application of the impulse To solve for j, we need one more piece of information If we can solve for j, we then can compute the linear velocity of the rigid body after the collision
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))
Coefficient of restitution: A B
ˆ n V −
r
= 0 = 1 = 0.5
= 0 = 1
Perfect bounce Resting contact
v+
r = −v− r
Now we are ready so solve for the impulse Let’s first 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)
ω+
a (tc) = ω− a (tc) + I−1(tc)(ra × jˆ
n(tc)) ω+
a (tc)
Replace with in ˙ p+
a (tc)
˙ p+
b (tc) = ˙
p−
b + j
ˆ n(tc) Mb + (I−1
b (tc)(rb × ˆ
n(tc))) × rb
p+
a (tc) = ˙
p−
a + j
ˆ n(tc) Ma + (I−1
a (tc)(ra × ˆ
n(tc))) × ra
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
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 J = jˆ n(tc) τimp = ra × J
A: B:
τimp = rb × J J = −jˆ n(tc)
fiˆ n(tc)
In the case of colliding contact, we established the relation between the velocity of the contact points and j, subject to the empirical law For resting contact, we will establish the relation between acceleration of the contact points and fi’s subject to three conditions
di(t) = ˆ ni(t) · (pa(t) − pa(t))
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
Since , we have to keep from decreasing; that is, we have to keep di(tc) = 0 di(tc) ˙ di(tc) ≥ 0 ˙ di(t) = ˙ ˆ ni(t) · (pa(t) − pb(t)) + ˆ ni · ( ˙ pa(t) − ˙ pb(t)) At time tc, pa(tc) = pb(tc) This means that The definition of two bodies in resting contact: di(tc) = ˙ di(tc) = 0 ˙ di(tc) = vr
Since describes the separation distance and describes the separation velocity, measures how two bodies are accelerating towards each other at the contact point ˙ di(tc) di(tc) ¨ di(tc) 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
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
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
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
¨ 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
¨ 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
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
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