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Preliminary Mathematics of Geometric Modeling (4) Hongxin Zhang and Jieqing Feng 2006-11-30 State Key Lab of CAD&CG Zhejiang University

Preliminary Mathematics of Geometric Modeling (4)

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Page 1: Preliminary Mathematics of Geometric Modeling (4)

Preliminary Mathematics of Geometric Modeling (4)

Hongxin Zhang and Jieqing Feng2006-11-30

State Key Lab of CAD&CGZhejiang University

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Differential Geometry of Surfaces

• Tangent plane and surface normal • First fundamental form I (metric) • Second fundamental form II (curvature)• Principal curvatures • Gaussian and mean curvatures

Explicit surfacesImplicit surfaces

• Euler's theorem

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Principal curvatures

Principal curvature: the extrema of normal curvature

Definition of normal curvature

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Principal curvatures

The normal curvature is determined by

The extrema of the κn is arrived at

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Principal curvatures

Hence

SinceExtreme case

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Principal curvatures

Hence

The extreme values of κn satisfy: Extreme case

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Principal curvatures

The homogeneous linear system of equations for du, dv have a nontrivial solution iff

expanding

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Principal curvatures

Discriminant D of above quadratic equation in κn

• D > 0 : two extreme normal curvatures (max/min)

• D = 0 ⇔ and⇔ const. k, s.t.

Umbilic point: the normal curvature is the same in all directions.

iff

≥ 0

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Principal curvatures

Let

Then

Gaussian (Gauss) curvature

Mean curvature

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Principal curvatures

Solving

We have

κmax and κmin are maximum and minimum principle curvatureThe directions in the tangent plane for which κn takes maximum and minimum values are called principal directions

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Principal curvatures

Computation of principal directions

κn takes either κmax or κmin

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Principal curvatures

• When D=0 or H 2=K, κn is a double root

κmax= κmin=H

The point is an umbilical point, which is locally a part of sphere with radius of curvature

• When K=H=0, the point is an flat/planar point

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Principal directions

removing cubic term about λ

The discriminant is same with that of the principal curvature as follows

D=0

FN=GM, EN=GL and EM=FL

≥ 0

Umbilical point: the principal directions are not defined!

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Principal directions

When discriminant D>0, the principal directions λmax and λmin satisfy

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Principal directions

Recall the orthogonal condition for two tangent vectors

Substitute above principal directions into it

The two principal directions are orthogonal !

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Principal curvatures and directions

Directions at one points on the surface

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An example of line of curvature

• Line of curvature: a curve on a surface whose tangent at each point is in a principal direction

Example of lines of curvature: Solid lines: maximum principal

directionDashed lines: minimum principal

direction

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Lines of curvature

• Surface parametrization: orthogonal net of lines

Lines of curvature The sufficient and necessary condition for the parametric lines to be lines of curvature

Recall that the arc length parameter for curve

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Example: principal curvature

Revolution surface• Meridian curve

x=f(t), z=g(t)rotate along z-axis

• Rotation angle: θ• Parallels curve: circles• Surface equation:

Revolution surface

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Example: principal curvature

Thus

Finally

The meridians and parallels are orthogonal

The meridians and parallels of a surface of revolution are the lines of curvature.

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Gaussian and mean curvatures

• Gaussian curvature: the product of the two principal curvatures

• Mean curvature: the average of the two principal curvatures

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Gaussian and mean curvatures

• Sign of K is same as LN-M2 : EG-F2>0

K>0 : elliptic, κmax and κmin same signK<0 : hyperbolic, κmax and κmin different signK=0 and H≠0 : parabolic, one of κmax and κmin is zeroK=H=0 : flat / planar, κmax= κmin=0

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Computation of Gaussian and mean Curvatures

• Explicit surfaces: z=h(x,y)

• Implicit surfaces: f(x,y,z)=0

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Explicit surfaces

• Explicit surfacez=h(x,y)

can be converted into a parametric formr=(u,v,h(u,v))

where u=x, v=y• Monge form: the parametric form of

explicit surface

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Explicit surfaces

Normal

The first fundamental form coefficients

The second fundamental form coefficients

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Explicit surfaces

Finally

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Example of explicit surface

Hyperbolic paraboloid: z=xy

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Example of explicit surface

Derivatives:

Normal:

First fundamental form coefficients:

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Example of explicit surface

The second fundamental coefficients:

Gaussian and mean curvatures: K<0, hyperbolic point

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Example of explicit surface

L=N=0 and M≠0: The surface intersects its tangent plane at the iso-parametric lines

Principal curvatures:

>0

<0

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Implicit surfaces

• Differential geometry: local geometric properties• The problems of implicit surfaces can be

converted as those of explicit surfaces by using inverse function theorem.

if f(x,y,z)=0 and fz≠0, then z can be expressed as function of x and y

z=h(x,y)It is similar for fx≠0 or fy≠0.

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Implicit surfaces

Computing the partial derivatives z=h(x,y)f(x,y,z)=0 or f(x,y, h(x,y))=0

z=h(x,y)

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Implicit surfaces

Computing the partial derivatives z=h(x,y)f(x,y,z)=0 or f(x,y, h(x,y))=0

Explicitsurface

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Implicit quadric surface

General form

Standard form after transformations

ζ, η, ξ = -1,0,1; δ = 0,1

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Classification of implicit quadrics

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Implicit quadric surface

Gaussian curvature

Mean curvature

Principal curvatures

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Hyperbolic cylinder

When ζ=δ=1, η=-1, ξ=0

we have

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Ellipsoid

When ζ=η=ξ=δ=1

We have

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Ellipsoid

Principal curvatures

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Special ellipsoid: sphere

When a=b=c=R, ellipsoid is a sphere

K=1/R2, H=-1/R

Since H2-K=0 for all points on sphere

We have

A sphere is made of entirely nonflat umbilics

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Elliptic cone

When ζ=η=1, ξ =-1, δ =0

We have

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Euler's theorem

Euler's theoremThe normal curvatures of a surface in an arbitrary direction (in the tangent plane) at point P can be expressed in terms of principal curvatures κ1 and κ2 at point P

where Φ is the angle between the arbitrary direction and the principal direction κ1

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Euler's theorem: simple proof

Assuming the isoparametric curves of the surface are lines of curvature, then

F=M=0Normal curvature is

The principal curvatures are

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Euler's theorem: simple proof

The angle Φ between dv/du and the principle direction corresponding to κ1 (dv1=0, du1) can be evaluated (according to first fundamental form)

Since and

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Download the courses

http://www.cad.zju.edu.cn/home/jqfeng/GM/GM02.zip