Spacetime Constraints

Witkin & Kass

Siggraph 1988

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”

2

3

Lamp on floor

4

Problem Statement (single particle)

0 mgfxm

dttfRT

0

2|)(|

bTxax )(,)0(

Governing Equation (Motion Equation):

Object Function (Energy Consumption):

Boundary Conditions:

gf(t)

5

Discretize continuous function

1n

i

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… x1xn satisfies goals while optimizing f1fn

Next step is to discretize our motion equation and object equation.

6

Difference Formula

h

xx

h

xx

h

xxx iiiiiii 2

1111

xi - 1 xi xi + 1

xi - 0.5 xi + 0.5

h h

211

11

5.05.0 2

h

xxx

hhxx

hxx

h

xxx iii

iiii

iii

Backward Forward Central

Central 7

Discretized Function

3,2,0)2(

0

112

imgfxxxh

m

mgfxm

iiii

bxbTx

axax

4

1

)(

)0(t

x

x1, f1

x2, f2

x3, f3x4, f4

Motion equation:

Boundary Conditions:

24

1

0

2

||

|)(|

ii

T

fR

dttfR

Object Function: When does R have minimum value?

0000

0

4321

f

R

f

R

f

R

f

R

f

R

i

8

Generalize Our Notation

0S

R

3,2,0)2( 112 imgfxxx

h

miiii

bxax 41

t

x

x1, f1

x2, f2

x3, f3

x4, f4

Unknown vector:

S = (S1, S2, …Sn)

Constraint Functions:

Ci(S) = 0

Minimize Object Function R(S):

0000

0000

4321

4321

f

R

f

R

f

R

f

R

x

R

x

R

x

R

x

R S = (x1, x2, x3, x4, f1, f2, f3, f4)

j

iij S

CJ

jiij SS

RH

2

9

Sequential Quadratic Programming (SQP) Step One

S

R

0S

R

))(())((!2

1))(()()( 22''' axOaxafaxafafxf

Pick a guess S0, evaluate

Taylor series expansion of function f(x) at point a is:

))(()( 2002

2

00

SSOSSS

R

S

R

S

R

SSSS

'100

00 ))((0

SSSSorSSS

SSSHS

Rij

SS

Most likely

Similarly, we have:

Set equal to 0Omit

S is the change to S0 that makes derivative equal to 0

10

SQP Step Two

))(()()()( 2'1

'1

'1

'1

SSOSSS

CSCSC

SS

SSSSSSorSSS

SSSJSC ij

0'1

'1

'1

'1

'1 ))(()(

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, let’s say Ci(S1’) ≠ 0, let’s apply Taylor series expansion on

the constraint function Ci(S) at point S1’ :

1001 SSsoSSSS

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.

Set equal to 0 Omit

S0 S1’ S1 S2

’ S2 … Sn

S is the change to S0 that makes derivative equal to 0

11

Graphical Explanation of SQP

S0 S1’

S1S2’ S2

C(S)

S

R

S

12

13

Homework

Spacetime Particle (2D version)

15

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… x1xn satisfies goals while optimizing f1fn

1n

i

Discretize, (xi,fi) as variables, minimize sum of fi.fiFormulate as constrained optimization.