FM 04 Parametrization

8/8/2019 FM 04 Parametrization

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Surface Parametrization

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R Grosso


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Content

Motivation Some History Differential Geometry Background Conformal and Harmonic mappings The Spring Model Some remarks: closed surfaces Deformation Analysis

Most Isometric ParametrizationS

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Motivation

Literature Siggraph Course Notes: Mesh Parametrization: Theory and

Practice, 2007. Kai Hormann, Bruno Levy and Alla Sheffer

Eurographics 2008 Full-Day Tutorial Geometric Modeling Basedon Polygonal Meshes

Surface Parameterization: a Tutorial and Survey. Michael S.Floater and Kai Hormann

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Motivation

Texture mapping

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Motivation

Surface Fitting

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Parameter domain


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Motivation

Mesh completion, surface reconstruction

Mesh completion, Kraevoy and Sheffer 2005

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Motivation

Remeshing

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parameter domain


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Motivation

Mesh editing

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Motivation

Detail Mapping

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Siggraph Course Notes 2007: Mesh Parameterization: Theory and Practice


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Motivation

Complex topology

segmentation and texture atlases for Isocharts, Zhou et al. 2004

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Some History

Notes from Carlos A. Furuti,www.progonos.com/furuti/index.html Azimuthal orthographic

projection

mentioned by Hipparchus200 B.C. but probably older

do not preserve angles andareas

Copyright 2008 C.A.Furu:

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Some History

Azimuthal stereographic projection attributed to Hipparchus,

190 -120 B.C.

name given by dAiguillon1613

conformal: preserves anglesbut does not preserve areas

maps circles to circles


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Some History

Mercators cylindrical conformal projection Flemish cartographer, 1512 -1594 used navigation from

the 16th century until present

any straight line between twopoints is a loxodrome (line ofconstant course on the sphere)

Mercator loxodrome bearsthe same angle from all

meridians.Copyright 2008 C.A.Furu:

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Some History

Lamberts azimuthal equal-area projection Johann H. Lambert 1772 preserves areas strong distortion in the

boundary of the worldwidemap


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Differential Geometry

First fundamental form or metric tensorg = xx, , = 1,2

ds2 = g du du quadratic form

det(g ) = g > 0

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(g) = (If ) = g11 g12

g21

g22

= E FF G


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Differentiable ManifoldTopological manifold

Locally homeomorphic to euclidean space V, V open chart pair (V , ), : VR3Atlas: set {(V , )} which covers with transition maps

if (RnRn) is Ck for all transition maps then the atlas {(V ,)} is Ck. In this case we say that the pair {, {(V , )} isCkdifferentiable manifold

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u2

u1

u2

u1


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Parametrization

Notation: S surface, (u1,u2) coordinates S* second surface fmap, f: SS*A parametrization ofS* is given by * = fA pointp in S* has the same coordinates as in S.

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u2

u1

S

*

S*

f


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Isometric Mapping

Definitionf: SS* is an isometric or length-preserving map if the length of

any arc on S* is the same as its pre-image on SWe call f an isometry.

TheoremA map f: SS* is an isometry iff the coefficient of the firstfundamental form are the same, i.e. I = I*

The surface S and S* are called isometric.

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Conformal Mapping

Definitionf: SS* is a conformal or angle-preserving map if the angle at the

intersection of two curves on S* is the same as the angle ofintersection of the corresponding curves on S.

Theorem f: SS* is conformal iff the first fundamental form areproportional, i.e. I=(u1,u2)I* ; 0 scalar

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Note: Jf: Jacobian off If=JT J (2x2 Matrix) Singular values ofJ (3x2 Matrix) are the square roots of the

eigenvalues ofgSingular values ofJf: i, Eigenvalues ofg: i i2 = i

Then

f is isometric

f is conformal

f is equiareal

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Strategy for computing a surface parametrization Ideal case: isometric mapping, but not possible in general.

Therefore, attempt one of the following mappings:

1. Conformal: no distortion in angles2. Equiareal: no distortion in areas3. Minimize some combination of both

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Conformal Map

Conformal map complex transformation w=f(z) w,z Cthat preserves angles fis analytic z=x+iy; w=u+iv It follows u=0; v=0;

i.e. u(x,y) and v(x,y) are harmonic

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Conformal Map

Minimal surfaces: Plateaus Problem (x,y) displacements on domain

Potential energy

Minimize surface area with boundary conditions =0 on. The Euler-Lagrange equationsxx(1+y

2)+yy(1+x2)-2xyxy=0 (nonlinear!)

Linearized problem: minimize Dirichlets energy

=0 (=0) in !

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Conformal Map

Result harmonic map can be viewed as minimizing the Dirichlets

energy over the surface S.

Note: In this case the domain is planar, i.e. R2 For S R3 use the generalized Laplace equation: use operators

defined on S: s=divsgrads

Result Isometric conformal harmonic

Remark: Not every harmonic map is conformal.

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Harmonic Map

Harmonic maps: discretization ST piecewise linear surface T={ti} triangulation ofST

Parametrization problem: Given ST Find polygonal domain S*R2 Piecewise linear map f: STS* f is uniquely determined by the values offat the vertices in T

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Harmonic Map

1. Set boundary mapping

2. Compute piecewise linear f: STS* which minimizes

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vi

f(vi)


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Harmonic Map

Note: Quadratic minimization problem For a triangle t=[v1v2v3]

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Harmonic Map

Lest squares minimization:The normal equations are: Let V be the set of all vertices and VI the set of all interior vertices of

T.

LetNi be the set of the indices of the neighbors ofvi VI

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vj

vi


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Harmonic Map

Remark:The matrix is symmetric and positive definiteThe bandwidth of the matrix is given by the max. valence of

viVI

at the boundary for viVI

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boundary edges


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Harmonic Map

Minimize the Dirichlet energy functional: Notation fi=f(vi).

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ED min solve ED = 0,

where the gradient is taken with respect tofi,v iVI


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Harmonic Map

The linear system:

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Harmonic Map

Remark: Harmonic map IfS* is convex fis one-to-one

Discrete harmonic map S* is convex and the weights wij are positive fis one-to-one orientation of triangles is preserved t=[v

1,v

2,v

3] f(t)=[f(v

1) f(v

2) f(v

3)]

This can be seen as follows:

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Convex Combination

Convex combination Set of points {pi} i=1,,N. Convex combination

Ifp is a convex combination p is in the convex hull of{pi}.

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vi

ij>0 ij


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Convex Combination

The discrete harmonic map ij>0 wij>0 wijij>0 fi is a convex combination of the fjTherefore fi is in the convex hull offj the orientation of thetriangles in S does not change in S* and the map is one-to-one,

i.e. there is no flip-over

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flip-over, weight < 0


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Convex Combination

The weights are negative if:

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quadrilateral opposite angles sum <


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General discrete formulationThe one dimensional case

At the boundary f(p0) = x0, f(p5) = x5.

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p0

p1 p2

p3p4

p5

x0 x1 x2 x3 x4 x5


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The linear system

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Aij =

wikkN( i) , i = j

wij , jN(i)

0 , otherwise

bi = wijx jjN(i)jB

A x = b


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Procedure Step 1: Parametrization of the boundary: Compute aparametrization for the boundary points

Step 2: Set the weights wij (use some heuristic model) Step 3: Solve linear system for the inner points!

Analysis

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p0

p1 p2

p3 x0 x1 x2 x3

x0 = f(p0)x3 = f(p3)

boundary, fix

f


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The system of equations

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wij =1

pi pjp

(the spring model)

x1 x0

x2 x1=

p1 p0

p2 p1

p

p = 0 uniform

p =1 chordal

p =1

2centripedal

Example: choose

it follows


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The general two dimensional case Step 1: Parametrization of the boundaryAsumption: the boundary is a closed closed polygon Map the boundary using e.g. the spring model Set the weight, choose p=0,1 or (uniform, chordal,

centripetal) Problem: Form/geometry of boundary curveThe boundary should be mapped to a convex polygon!

Choosing the geometry of the boundaryi. Circle

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convex


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Geometryii. n-polygon

iii. Choose best fitting plane (least squares) project (orthogonally) boundary vertices

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berfaltung


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The inner vertices Requirements for a good mapi. Identity property:

Preserves topology For planar meshes, the resulting map in the identity

ii. Regularity with respect to convolutions (berfaltung) (one-to-one!)

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Step 2: Set the weights wija) The spring model

wij= || pi pj ||-p, withp=0 (uniform),p=1 (chordal) orp=1/2 (centripetal) p=0 andp=1/2 do not satisfy (i), but satisfy (ii)

b) Harmonic parametrization Solution of Laplace (elliptic) equation

Su=0; Sv=0

Weak formulation

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b)

Harmonic parametrization cont.

Weak formulation min(ED);

Use Dirichlet boundary conditions wij = cotij + cotij

One-to-one only if

ij+ij < (wij > 0) it satisfy (i) but not (ii)!

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vj

vi


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c)

Shape preserving

Remark: convex combination regular triangulation The method:

Step 1: Flattening

Step 2: Compile weights

Step 3: Solve linear system

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c)

Shape preserving

Flattening by exponential mp: consider P and corresponding trianglefan

Compute mapping ofP and corresponding triangle fan to a plane

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Q4

Q6Q1

Q2

Q3

Q5

Pq

3q4

q6

q5q1

q2p 1

x

y


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c)

Shape preserving: map of triangle fan

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P p =0

0

Q1 q

1= Q

1 P

1

0

Qi qi = Qi Pcosi

sini

i = 2,K,N QN = Q1

i=

=2

i

; =2

ii=2

N

; i=(Q

i,P,Q

i+1)


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c)

Shape preserving, cont.

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P p

Qi

qii Qi-1

i qi


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Shape preserving, example Step 1: flattening the 1-neighborhood ofvi (using theexponential map)

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vi

v`i

v j1 v j2

v j3v j4


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Shape preserving, example Step 2: computing the barycentric coords.

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v'i = 12v' j1 +2

2v' j2 +4

2v ' j4

vì

v j1 v j2

v j3v j4

vì vì vì


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shape preserving, example cont. use the average of the barycentric coords as weights

Result: no deformations, no overlaps! Shape preserving satisfy requirements i) and ii) for a good map!

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v`i

v j1 v j2

v j3v j4


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Closed Surfaces1) Surfaces with genus 0, i.e. spherei. Find map to the unit sphereii. A harmonic map is conformal, Gu. And Yan, 2002iii. Many methods are nonlinear

2) Surfaces with arbitrary genusi. Segment surface into open disk like meshesii. Use network for open surfacesiii. Problem: smooth transition across patch boundaries

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MIPS

Most Isometric ParametrizationS Fact: except for developable surfaces, such as planes or

cylinders, any parametrization will deform the shape of thetriangles

Problem statement: find a way to measure this deformation

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f


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MIPS

Statement:

the map between two triangles is linear: fj: Tj tj. call the map fjatomic linear map.

Idea find a functional that measures the distortion (or deformation) of

the atomic linear map minimize a functional over all these atomic maps E(fj).

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R3 R2

fj


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MIPS

The deformation functional should be invariant under

translation orthogonal transformations, e.g. rotations uniform scaling

The functional should penalize triangle collaps

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MIPS

The functional

consider atomic linear map

ignore in the following the translation partb.

Singular value decomposition

2-norm condition

this number gives information on deformation and collaps A value of 1 corresponds to an isometry

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g:R2R

2, xa Ax + b

UTAV = =

1

0

0 2

2(A) = A

2A

1

2=

1

2


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MIPS

Problem

the computation of the two norm condition number is difficult Solution

use the Frobenius norm condition instead.

The global functional takes the form

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F(A) = A F A1

F

=

trace(ATA)

det(A)

= F( fj)

j


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MIPS

The global deformation functional is non linear

solution through an iterative method, e.g. Gau-Seidel likeiteration

the parametrization of the boundary is determined during theiteration and must not be specified before.

needs an efficient technique to compute the Frobeniouscondition number for each atomic linear map.

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Final remarks

linear methods, e.g. the spring model, are easy to implementand fast

the harmonic energy can produce overlaps, e.g. if some of theweights are negative

non linear methods, e.g. MIPS or more realistic spring models,produce better results. Implementation is difficult and requiremuch more computation time.

Most methods require to specify first the parametrization of theboundary, except MIPS

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FM 04 Parametrization