Mesh-Based Inverse Kinematics - Computer...

Mesh-Based Inverse Kinematics

R. W. Sumner, M. Zwicker, C. Gotsman, J. Popovic

CS468, Wed Nov 9th 2005

SIGGRAPH 2005

The problem

Mesh-Based Inverse Kinematics — R. W. Sumner, M. Zwicker, C. Gotsman, J. Popovic 1

Mesh-Based Inverse Kinematics — R. W. Sumner, M. Zwicker, C. Gotsman, J. Popovic

General approach

2

Learn from experience...

3

As-rigid-as-possible shape interpolation [ACL00]

Outline

4Mesh-Based Inverse Kinematics — R. W. Sumner, M. Zwicker, C. Gotsman, J. Popovic

• Related work

• Overview of the method

• The method step by step

• Numerics

• Experiments


Related workSubdivision

[ZSS97, KCVS98, KBS00, GSS99]


Skeleton-based kinematics

Related work

[JT05, ASKTR05]


Relation to mesh editing

Inverse kinematics

coating transfer mesh transplanting

8

Intrinsic mesh editing methodsRelated work


Laplacian (or differential) coordinates [SLCARS04]

Overview of the method


Let M0, · · · ,Mk be the input meshes (M0 is the reference mesh).

1. map each transformation T i (i ≥ 1) to a feature vector fi ∈ R9m,where m is the number of triangles of the base mesh

2. define the space Σ of admissible transformations as the span ofthe {fi}i≥1

3. given a set of specified vertex positions v1, · · · , vl, find a featurevector f close to Σ that maps each constrained vertex vj to a pointclose to its specified position vj

9

Feature vectors

Given i ≥ 1, fi encodes the deformation gradients of the piecewise affinemap from M0 to Mi


For the jth triangle of M0, let φj be the affine transformation that mapsthe triangle onto its image in Mi:

∀p ∈ R3, φj := Tj p + tj

linear (3x3 matrix)

translation vector

• only the gradient of φj is encoded: Dφj(p) = Tj

• to make Tj unique, a 4th vertex is added to the jth triangle, alongthe normal direction

v1

v2v3

v4

v1

v2

v3

v4

φj

from [SP04]

Feature vectors




∀p ∈ R3, φj := Tj p + tj

linear (3x3 matrix)

translation vector



Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1

tj = v4 − Tj v4

Feature vectors




∀p ∈ R3, φj := Tj p + tj

linear (3x3 matrix)

translation vector



Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1

tj = v4 − Tj v4

Feature vectors


11

Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1


linear in v1, v2, v3, v4

(3× 3)

Feature vectors


11

Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1



(3× 3)

fij :=

TjxT

TjyT

TjzT

= Gj

v1x...

v4x

v1y...

v4y

v1z...

v4z

Tj :=

TjxTjy

Tjz

Gj :=

GjxGjy

Gjz

∈ R9

∈ R12

(9× 12)

Feature vectors


11

Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1



(3× 3)

fij :=

TjxT

TjyT

TjzT

= Gj

v1x...

v4x

v1y...

v4y

v1z...

v4z

∈ R9

∈ R12

(9× 12)

fi :=

f i1...

f im

= G v

G :=

Gx

Gy

Gz

Feature vectors


11

Tj = [v1 − v4, v2 − v4, v3 − v4] · [v1 − v4, v2 − v4, v3 − v4]−1



(3× 3)

fij :=

TjxT

TjyT

TjzT

= Gj

v1x...

v4x

v1y...

v4y

v1z...

v4z

∈ R9

∈ R12

(9× 12)

fi :=

f i1...

f im

= G v∈ R9m

(9m× 3(n + m))

∈ R3(n+m)

G :=

Gx

Gy

Gz

(3m× (n + m))

Feature vectors

12

f = G v∈ R9m

∈ R3(n+m)


Inverse transformation

(9m× 3(n + m))

”v = G−1 f ”

Feature vectors

12

f = G v∈ R9m

∈ R3(n+m)



(9m× 3(n + m))

”v = G−1 f ”

9m > 3(n + m) ⇒ G is not a square matrix

By fixing one vertex, one reduces the dim. to zero

To a feature vector f ∈ R9m corresponds ≤ 1tranformed mesh (up to translation)

Feature vectors

12

f = G v∈ R9m

∈ R3(n+m)



(9m× 3(n + m))

”v = G−1 f ”

9m > 3(n + m) ⇒ G is not a square matrix

By fixing one vertex, one reduces the dim. to zero

least squares problem:

v∗ = arg minv‖G v− f‖ most ”plausible” positions

To a feature vector f ∈ R9m corresponds ≤ 1tranformed mesh (up to translation)

If too many constraints (or imprecisions), theset of solutions can be empty

Linear feature space

Σ := Span(f1, · · · , fk)

To every vector w of coordinates in the basis(f1, · · · , fk

)corresponds

a feature vector fw =∑

i wi fi ∈ R9m

Once one or more vertex positions are set, compute

v∗,w∗ = arg minv,w

∥∥∥G v− (f1, · · · , fk) w∥∥∥



Σ := Span(f1, · · · , fk)

To every vector w of coordinates in the basis(f1, · · · , fk

)corresponds

a feature vector fw =∑

i wi fi ∈ R9m

Once one or more vertex positions are set, compute


∥∥∥G v− (f1, · · · , fk) w∥∥∥

To penalize solutions that are far from the fi, take the mean f of the fi

as the origin: fw = f +∑k

i=1 wi fi

and compute:


∥∥∥G v− f− (f1, · · · , fk) w

∥∥∥+ α ‖w‖






f1

f2

f( 13 , 2

3 )

Nonlinear feature space


For each triangle j of M0, separate the rotational component from thescale and shear components of Tj by polar decomposition: Tj := Rj Sj

Interpolate the Sij linearly, but the T i

j by composition:

”T ij (w) :=

k∏i=1

Rij

wi ·k∑

i=1

wi Sij ” R2

θ = R2θ





j by composition:

”T ij (w) :=

k∏i=1

Rij

wi ·k∑

i=1

wi Sij ” R2

θ = R2θ

For non-integer wi, use the exponential map:

T ij (w) := exp

(k∑

i=1

wi log Rij

)·

k∑i=1

wi Sij





j by composition:

”T ij (w) :=

k∏i=1

Rij

wi ·k∑

i=1

wi Sij ” R2

θ = R2θ

(∗)non-linear

To retrieve v∗,w∗, solve the following least squares problem:


‖G v− fw‖

For non-integer wi, use the exponential map:

T ij (w) := exp

(k∑

i=1

wi log Rij

)·

k∑i=1

wi Sij



linearinterpolation

exponential map

scale/shear

rotational

Numerics


• To solve (∗), use Gauss-Newton algorithm

• To accelerate the resolution of the linear least squares problem, useCholesky decomposition

see details in the paper

reduces (∗) to solving a linear least squares problem at eachiteration of the algorithm (∼ 6 iterations in practice)

Experimental results


Pentium IV @ 3.4 GHz2GB RAM

Cholesky

one iteration

Conclusion


• MeshIK is an easy-to-use mesh manipulation tool, based on an in-tuitive interface

• the algorithm adapts to each model by learning from example meshes

• the method is effective in practice and provides interactive timings

• What about meshes of genus > 0?

• Try other feature spaces, to enhance the feature interpolation

• Try other numerical techniques, to enhance speed

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