Spacetime Constraints

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Spacetime Constraints

• Witkin Kass
• Siggraph 1988

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Overview

• Unlike common Newtonian dynamic simulation, the
  due driving force is unknown
• Specify the high-level spacetime constraint and
  let the optimization solve for the position and
  force unknowns by minimizing the energy
  consumption

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Lamp on floor
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Problem Statement (single particle)
Governing Equation (Motion Equation)
Boundary Conditions
f(t)
g
Object Function (Energy Consumption)
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Discretize continuous function
Discretize unknown function x(t) and f(t)
as x1, x2, xi, xn-1, xn f1, f2, fi,
fn-1, fn Our goal is to solve these
discretized 2n values x1?xn satisfies goals
while optimizing f1?fn Next step is to discretize
our motion equation and object equation.
i
1
n
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Difference Formula
h
h
xi - 0.5
xi 0.5
xi - 1
xi
xi 1
Backward
Forward
Central
Central
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Discretized Function
Motion equation
x
x4, f4
x3, f3
x2, f2
Boundary Conditions
x1, f1
t
Object Function
When does R have minimum value?
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Generalize Our Notation
Unknown vector S (S1, S2, Sn)
x
x4, f4
Constraint Functions Ci(S) 0
x3, f3
x2, f2
x1, f1
Minimize Object Function R(S)
t
S (x1, x2, x3, x4, f1, f2, f3, f4)
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Sequential Quadratic Programming (SQP) Step One
Pick a guess S0, evaluate
Most likely
Taylor series expansion of function f(x) at point
a is
Similarly, we have
Set equal to 0
Omit
Sa is the change to S0 that makes derivative
equal to 0
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SQP Step Two
Now we got S1, evaluate our constraints Ci(S1),
if equal to 0, we are done but most likely it
will not evaluate to 0 in the first several steps.
So, lets say Ci(S1) ? 0, lets apply Taylor
series expansion on the constraint function Ci(S)
at point S1
Omit
Set equal to 0
Sb is the change to S0 that makes derivative
equal to 0
Then we will continue with step one and step two
until we got a solution Sn which minimizes our
object function and also satisfies our
constraints.
S0 ? S1 ? S1 ? S2 ? S2 ? ? Sn
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Graphical Explanation of SQP
C(S)
S1
S2
S2
S
S0
S1
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