Mesh Deformation Based on Discrete Differential Geometry Reporter: Zhongping Ji 4.4.2007

Mesh Deformation Based on

Discrete Differential Geometry

Reporter: Zhongping Ji

4.4.2007

Leonhard Euler 1707.4.15-1783.9.18

Papers

(A) Linear Rotation-invariant Coordinates for Meshes [SIG 2005]

• Yaron Lipman Olga Sorkine David Levin Daniel Cohen-Or

• Tel Aviv University

(B) Volume and Shape Preservation via Moving Frame Manipulation [TOG 2007]

• Yaron Lipman Daniel Cohen-Or Gal Ran David Levin

• Tel Aviv University

About authors

A Ph.D. student at The School of Mathematical Sciences, Tel-Aviv University.His supervisors are Prof. David Levin and Prof. Danny Cohen-Or.

http://www.math.tau.ac.il/~lipmanya/

Mesh deformation

( A ) Rotation-Invariant

A rotation-invariant representation for triangular mesh based on two discrete forms;

The first form locally discribes the tangential component of the surface;

The second form locally discribes the normal component of the surface.

Classical differential geometry

The first and second fundamental forms;Given two fundamental forms I and II, a

nd they hold some compatibility conditions, Gauss-Codazzi-Mainardi equations;

Given an initial right-hand frame, then solve a first-order linear PDE.

Discrete Forms

Notations

First form

Coefficients:

Parameterization:

First form:

The length and the signed angl

e

Second form

Coefficients:

Second form:

Height function of a vertex above the tange

nt plane

LemmaGiven the discrete form coefficients and

the orientation bits at vertex i, the 1-ring neighborhood of i is defined up to a rigid transformation.

Fixing the position of vertex i, and fixing the direction of one edge emanating from vertex i and the normal, uniquely defines all the rest of the 1-ring vertex positions.

Proof

Discrete surface equationsDefine a discrete frame at vertex i,

Discrete surface equations:

(1)

Theorem 1

The coefficients of the discrete surface equations can be expressed by the discrete forms.

Theorem 2

Given an initial discrete frame at an arbitrary vertex i0, and given that the first and second discrete forms with the orientation bits are taken from an existing mesh, there exists a unique solution to the set of discrete surface equations.

Theorem 3

• Given the discrete frame at each vertex, there exists a unique embedding of the vertices of the mesh in R3 up to a translation, such that the resulting mesh has the given discrete forms and discrete frames.

Proof

More constraints can be added to this equation.

Geometry difference equations:

Mesh representation

The first discrete form is represented by the lengths of the projected edges and angles between adjacent projected edges.

The second discrete form represents the orthogonal component.

This sums to 3di scalars, and the scalars are rigid motion invariant.

To reconstruct the mesh given its discrete form coefficients, an initial discrete frame and a position of one vertex in space.

• Construct the discrete surface equations;• Add the initial discrete frame as an additional equation;• Solve the augmented linear system in the least-squares se

nse to obtain the discrete frames;• Construct the geometry difference equations;• Add the given initial vertex position as an additional equ

ation;• Solve the augmented linear system in the least-squares se

nse to get the positions of the vertices.

Mesh editing

Two constraints and two sparse linear systems;

• Transformation constraints on the local frames and translation constraints on the vertex positions;

• Discrete surface equations and geometry difference equations.

Examples of basic editing operations

Results

Compatible constraints

Shape interpolation

Shape interpolation

Future work

Directly manipulate the local frames;Improve the co-existence of constraints

on the local frames and the vertex positions;

Use this representation for detail editing, such as detail enhancement.

( B ) Moving Frame

Isometric shape-preserving Aiming at minimizing distortion of the shape.

Local volume preservation Control the local volume by scaling the movin

g frames.

Motivation

Look for a deformation which maintains the first fundamental form intact (isometry), and minimizes changes in the second fundamental form.

Based on the Stokes’ theorem, using a carefully designed differential form which establishes a connetion between local volum and surface properties.

Shape-preserving

Isometric deformation Defining a rigid motion-invariant geometri

c distance(GD) between two isometric surfaces;

Given a surface and a set of rotational constraints, look for an isometric deformation minimizing the GD to the original surface under the constraints.

Shape operator Orthonormal frame: 1 2 3( ( ), ( ), ( ))e p e p e p

, ,p pHr r r T M

11 12

21 22

h hH

h h

1 1 2 2{ | ( ) ( )}pT M v v e p e p

Shape operator:

Second fundamental form:

Tangent plane:

Geometric distance2

, ( )EM f F

dist p H H

, ,( , ( )) M f M fMDist M f M Dist dist d

Point to point:

Surface to surface:

meaning: measures to what extent the surface M and deformed surface are rigid motion of each other.

Geometric deformation problem: Given a surface M, the goal is to deform it into another surface, subject to some prescribed constraints such that the geometric distance between them is minimal.

Solve the problem

: (3)R M SO

2

,

1

2M f M FDist R d

Dirichlet-type integral:

An energy functional on the rotation’s field.

2 21/ 2

FFH H R

The minimizer of the intergral, subject to constraints, is a harmonic function defined on a Lie group.

Shape-preserving method

Compute a rotation field R such that the enerty is minimized subject to constraints;

Apply rotations over the moving frames, and reconstruct the isometry f.

Parametrization of SO(3)

Dim(SO(n)) = n(n-1)/2

Two parametrization: • orthogonal parametrization; • conformal parametrization.

Orthogonal Parametrization1 2 3 9: ( , , ) [0, ) [0,2 ) [0,2 ) (3)orth SO R

1 2 3

1 2 31 2 3

1 2

1

sin( / 2)sin( )sin( )

sin( / 2)sin( ) ( )( , , )

sin( / 2)cos( )

cos( / 2)

cos

2 2

2 2

2 2

1 2 2 2 2 2 2

2 2 1 2 2 2 2

2 2 2 2 1 2 2

xy z xy wz xz wy

yxy wz x z yz wx

zxz wy yz wx x y

w

Solution 21

2 M FR d

2 2 21 1 2 2 2 32 4sin( / 2)( sin ( ) )M

d

21

MMin d

Conformal Parametrization1 2 3 3 9: ( , , ) (3)conf R SO R

1

2

2

2

1 2 33

2

2

2

4

4

4

4( , , )

4

4

4

4

2 2

2 2

2 2

1 2 2 2 2 2 2

2 2 1 2 2 2 2

2 2 2 2 1 2 2

xy z xy wz xz wy

yxy wz x z yz wx

zxz wy yz wx x y

w

Solution

21

2 M FR d

2

2 2

164

(4 )Md

1 2 3( , , )

Piecewise-Linear Case (I)

1 1(cot( ) cot( ))( ) 0,j

r rj j j r

r N

j V

1 1

23 1 1

1, coti i j j jiT T i i iT T j

Q Q Q

21

Md

Piecewise-Linear Case (II)

( , , 1) ( , , 1) 2 3

2(cot( ) cot( ) )( )

(4 )ii j

j

ljr r l l

j j r r j j r r j r TT NTr N

j

W W Q

1 2 3 1 2 3

2 2( , , ) ( ( ) ( ) ( )), ( ) 1/(4 )i i i i i iW w w w w

1 2 3

2

( , , )2 2

1 1

6(4 ) i

i

i i i TMT T

d W Q

Iterative scheme

An example

Volume preservation

Motivation, Strokes’ theorem:

Designing a two-form:

D Dd

( )D

volume D d

is a volumetric form and reflects the local volume underlying the point p on surface M.d

Notations

Requirements:

1 2u h w w

1 2 3 1 2 3d w w w dx dx dx (1)

(2)

Illustration:

Derive two-form Let (b1,b2,b3) denote the standard basis in R3,

1 2 3 1 2 3( , , ) ( , , )e e e b b b M

1 2 3 1 2 3( , , ) ( , , )w w w dx dx dx M

1 2 1 3 2 3dx dx dx dx dx dx

3 2 11 x x x

3 3,3, 1( )i j i jM m R

3 2 11 2 3 1 2 3( )x x xdx dx dx d dx dx dx

3 31 ( )h e h e

1 2 1 3 2 3

11 1 21 2 31 3 12 1 22 2 32 3( ) ( )

dx dx dx dx dx dx

h m dx m dx m dx m dx m dx m dx

12 13 23hM hM hM

12 1 2 13 1 3 23 2 3( )h M dx dx M dx dx M dx dx

1 2

1 2

i i

ijj j

m mM

m m

1 2h w w

Calculate scaling factor

( )D V M V M

Volume D hd hd hd

1 2d w w

1/ 2( / )h h

( )V M

Volume D hd

d d

Scale the moving frame, scaling factor:

( )V M

Volume D hd

Calculate h(t)• Defining the depth function

1 2( ) min{ ( ), ( ) , ( ) }p L p r r

1 2

3

1 21 2

0 0

1( ) 0, 0,{ , } {1,2}

1 10. 0

j ij

k k

e t k k i ja t

k ka t a t

[0, ( )]t p

1 2

2

3 21 2 1 2

1 21 2

0

/ 2( ) 0, 0,{ , } {1,2}

/ 3 ( ) / 20, 0

( )( )

jj i

j

t k k

t a th t k k i j

a t

t a a t a a tk k

a t a t

3 31 ( )h e h e

• Solve the ODE

Implementation Select the set of Constraints: static and handle vertice

s;Create and factorize the Laplace-Beltrami matrix of

ROI;Calculate the local depth field L of the mesh;Calculate the rotation field R: MSO(3) and apply to

the moving frames at each vertex;Reconstruct the mesh in a least-squares sense;Calculate the scaling factor for volume correction;Scale the moving frames and reconstruct the mesh.

Results

Without volume correction

Results

Without volume correction

Results

With volume correction

Results

Comparison

Results

Comparison

Results

Irregular mesh

Relative change in volume

Future work

It cannot preserve simultaneously the surface area and volume of an object. An exciting avenue for future research is to investigate orher complete local surface descriptors, whose preservation yields a shape preservation of both surface and volume.

Thank you!