Computer Graphics Geometric Modeling - Webcourse...

1

Computer Graphics

Copyright 1996 C. Gotsman & G. Elber

Computer Science Dept. Technion

Geometric Modeling

Geometric

Modeling

Geomertic Modeling I - Center for Graphics and Geometric Computing,

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An Example


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OutlineObjective: Develop methods and algorithms to

mathematically model shapes of real world objects

Categories:

Wire-frame representations

Boundary representations

Primitive based

Freeform based

Meshes & subdivision

Volumetric representations

Freeform based

Voxel Based

Octree Based

Trivariate Functions


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Wire-Frame Representation

Object is represented as as a set of points and edges (a graph)

containing topological information.

Used for fast display in interactive systems.

Can be ambiguous:


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Ambiguity


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Volumetric Representation

Voxel basedSpace subdivided into equal size boxes – each has on/off flag according to if inside/outside object

Scalar values (e.g. density) per voxel are also common

Common in imaging applications (CT,MRI,Ultrasound)

Memory consuming

Extension – Octree (recursive construction).

Advantage: simple and robust Boolean operations, in/out tests, can represent and model the interior of the object.

Disadvantage: memory consuming, non-smooth, time consuming to manipulate.

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Geometric Modeling


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Display – Iso surfaces

Difficulty to display volumetric content.

Instead we can represent the boundary of the volume,

solving for T(u, v, w) = T0

Marching cubes – examine each individual cube (eight

neighboring voxels) and locally decide the shape of the

iso surface there.

Result is a polygonal approximation of the iso surface.

-3

-3

-3

+3-3

+3

-3

+3

-3

+3

-4

+8+8

+8

-8

-8

T0 = 0


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Constructive Solid Geometry

Use set of volumetric primitives

Box, sphere, cylinder, cone, etc…

For constructing complex objects

use Boolean operations

Union

Intersection

Subtraction

Complement


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CSG Trees

Operations performed

recursively

Final object stored as

sequence (tree) of

operations on

primitives

Common in CAD

packages –

mechanical parts fit well

into primitive based

framework

Can be extended with

free-form primitives

C2

B2

S S C3 2 2

S S B2 1 2

C1

B1

S B C1 1 1


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Freeform Representation

Explicit form: z = z(x, y)

Implicit form: f(x, y, z) = 0

Parametric form: S(u, v) = [x(u, v), y(u, v), z(u, v)]

Example – origin-centered sphere of radius R:

],[],2,0[),sin,cossin,coscos(),,(

0

:

:

22

2222

222222

RRRzyx

Rzyx

yxRzyxRz

:Parametric

Implicit

Explicit

Explicit is a special case of

implicit and parametric form


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A representation of choice

What makes a good representation for

solid modeling?

Ease/Intuitive shape manipulation

Class of representable models

Ease of display and evaluation

Ease of query answering such as inside/outside,

subdivision, interference

Robustness and accuracy

Closure under operations such as Booleans.


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Intuitive Curve Editing

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Geometric Modeling


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Splines – Free Form Curves

To allow control of curves need description via set of

basis functions + coefficients

Require geometric meaning of coefficients (base)

Approximate/interpolate set of positions, derivatives, etc..

Typically parametric


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Polynomial Bases

Monomial basis {1, x, x2, x3, …}

Coefficients are geometrically

meaningless

Manipulation is not robust

Number of coefficients =

polynomial rank

We seek coefficients with geometrically intuitive meanings

Polynomials are easy to analyze, derivatives remain polynomial,

etc.

Other polynomial bases (with better geometric intuition):

Lagrange (Interpolation scheme)

Hermite (Interpolation scheme)

Bezier (Approximation scheme)

B-Spline (Approximation scheme) Lagrange


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Mathematical Continuity

C1(t) & C2(t), t [0,1] - parametric curves

Level of continuity of the curves at C1(1) and C2(0) is:

C-1: C1(1) C2(0) (discontinuous)

C0: C1(1) = C2(0) (positional continuity)

Ck, k > 0 : continuous up to kth derivative

Continuity of single curve inside its parameter domain is

similarly defined - for polynomial bases it is C

kjCCjj

0),0()1( )(

2

)(

1


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Geometric Continuity

Mathematical continuity is too strong a requirement

May be relaxed to geometric continuity

Gk, k 0 : Same as Ck

Gk, k =1 : C'1(1) = C'2(0)

Gk, k 0 : There is a reparameterization of C1(t) & C2(t), where

the two are Ck

E.g.

C1(t)=[cos(t),sin(t)], t[/2,0]

C2(t)=[cos(t),sin(t)], t[0,/2]

C3(t)=[cos(2t),sin(2t)], t[0,/4]

C1(t) & C2(t) are C1 (& G1) continuous

C1(t) & C3(t) are G1 continuous (not C1)C1

C2, C3


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More on Continuity Consider a set of points {Pi}i=0,n

A piecewise constant interpolant will be C-1

A piecewise linear interpolant will be C0

Polyhedra are usually piecewise linear approximation

of freeform polynomials

Polyhedra are C0 continuous

Higher order polynomials posses better continuity


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Cubic Hermite Basis

The set of polynomials of degree k is a linear vector

space of degree k+1

The canonical, monomial basis for polynomials is

{1, x, x2, x3,…}

Define geometrically-oriented basis for cubic

polynomials

hi,j(t): i, j = 0,1, t[0,1]

Has to

satisfy:

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Geometric Modeling


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Hermite Cubic Basis (cont’d)

The four cubics which satisfy these conditions are

Obtained by solving four linear equations in four

unknowns for each basis function

Prove: Hermite cubic

polynomials are linearly

independent and form a

basis for cubic

polynomials

)1()()1()(

)32()(1)32()(

2

11

2

10

2

01

2

00

ttthttth

ttthttth


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Lets solve for h00(t) as an example.

h00(t) = a x3 + b x2 + c x + d

must satisfy the following four

constraints:

Four equations and four unknown to solve for.

.230)1('

,0)0('

,0)1(

,1)0(

00

00

00

00

cbah

ch

dcbah

dh


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Let C(t) be a cubic polynomial defined as the linear

combination:

What are C(0), C(1), C’(0), C’(1) ?

Pi

)()()()()( 111100011000 thTthTthPthPtC

Hermite

C(0) = P0, C(1) = P1, C’(0) = T0, C’(1) = T1

To generate a curve through P0 & P1 with slopes T0 &

T1 use

)()()()()( 111100011000 thTthTthPthPtC


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Parametric Splines

So far we solve for a single explicit function C(t)

Cannot represent closed curves such as a circle

Solution: let Pi = (xi, yi) and Ti = (tix, ti

y) be vector

functions

Solve for each individual

axis independently

Here is an example of

four Hermite cubics that

approximate a circle


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Cubic Splines

Standard spline input – set of points {Pi}i=0, n

No derivatives’ specified as input

Can we prescribe derivatives values in order to achieve C2?

Interpolate by n cubic segments (4n DOF):

Derive {Ti}i=0 , n from C2 continuity constraints

Solve 4n equations

1,..,1,)0()1(

:equations) 1( sconstraint continuityC

1,..,1,)0()1(

:equations) 1( sconstraint continuityC

,..,1,)1(,)0(

:equations) (2ion Interpolat

''

1

''

2

'

1

'

1

1

niCC

n

niCC

n

niPCPC

n

ii

ii

iiii


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Cubic Splines Have two degrees of freedom left (to reach 4n DOF)

Options

Natural end conditions: C1''(0) = 0, Cn''(1) = 0

Complete end conditions: C1'(0) = 0, Cn'(1) = 0

Prescribed end conditions (derivatives available at the ends):

C1'(0) = T0, Cn'(1) = Tn

Quadratic end conditions

C1''(0) = C1''(1), Cn''(0) = Cn''(1),

Periodic end conditions

C1'(0) = Cn'(1), C1''(0) = Cn''(1),

Question: What parts of C(t) are affected as a result of

a change in Pi ?

prescribed

natural

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Geometric Modeling


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The assumption t [0,1] (uniform parameterization)

is arbitrary

Implicitly implies same curve length for each segment

Not natural if points are not equally spaced

One alternative - chord-length parameterization:

Question: Can chord-length parameterization be

used for cubic Hermite splines?

Parameterization

].,0[ :segmentth ' For the

)()( Denote 2

1

2

1

i

y

i

y

i

x

i

x

ii

dti

PPPPd


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Parameterization

Question: What do we need arc/chord length curves for?

Uniform

Chord- length

[0,1]

[0,1]

[0,8]

[0,1]

[0,3]

[0,1]

[0,1]

[0,4]

Lagrange


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Bezier CurvesThe Bezier curve is an approximation

of a given control polygon (a polyline)

Denote by (t): t[0,1]

Bezier curve of degree n is defined over n+1

control points,

{Pi}i=0, n

Have two

formulations

Constructive

Algebraic

P1 P3P5

P1

P2

P0P0 P2P4

)(t

)(t


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De Casteljau Construction

);(:)(

);()()1(:)(

do to:=For

do to1:For

;:)( do to0:For

][

]1[]1[

1

][

]0[

tPt

ttPtPttP

nji

nj

PtPni

n

n

j

i

j

i

j

i

ii

(1/3)

Select t[0,1] value. Then,


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Algebraic Form of Bezier Curves


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Bezier curve for set of control points {Pi}i=0, n :

where {Bin(t)}i=0, n Bernstein basis of polynomial

of degree n

Here is a

Cubic case

example:

n

i

iin

i

n

ii ttini

nPtBPt

0

n

0=i

)1()!(!

!)( = )(


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why?

Curve is linear combination of basis functions

Curve is affine combination of control points

n

0=i

]1,0[,1)( ttBn

i

n

i

iin

i ttini

nPt

0

)1()!(!

! = )(

Bezier


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Properties of Bezier Curves (t) is polynomial of degree n

(t) CH(P0,…,Pn)

(0) = P0 and (1) = Pn

'(0) = n(P1P0) and '(1) = n(PnPn-1)

(t) is affine invariant and variation diminishing

(t) is intuitive to control via Pi and it follows the general shape of the control polygon

'(t) is a Bezier curve of one degree less

Questions:What is the shape of a Bezier curve whose control points lie on a line?

How can one connect two Bezier curves with c0 continuity? c1 ? c2 ?


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Bezier Curve Editing


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Some Trivia

Bezier curves introduced nearly

simultaneously by Pierre Bezier at

Renault and Paul De Casteljau at

Citroen (late 60s)

Attributed to Bezier as he was the

first to make the results public

Bezier


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Degree corresponds to number of control points

Global support: change in one control point affects the entire

curve

For large sets of points – curve deviates far from the points

Cannot represent conics exactly. Most noticeably

circles

Can be resolved by introducing a more powerful representation

of rational curves.

As an example, a 90 degrees arc as a rational Bezier curve:

.2where

)()()(

)()()()(

2

1

20

2

22

2

11

2

00

2

222

2

111

2

000

w

ww

tBwtBwtBw

tBPwtBPwtBPw tCP0

=(0,1)

P1=(1,1)

P2 =(1,0)

Drawbacks of Bezier Curves

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B-Spline Curves Idea: Generate basis where functions are continuous

across the domains while presenting local support

For each parameter value only a finite set of basis functions is non zero.

The parametric domain is subdivided into sections at

parameter values denoted knots, {i }.

The B-spline functions are then defined over the knots.

The knots are denoted uniform knots if i - i-1 = c,

constant. W.l.o.g.,

assume c = 1.

1

0

)()(n

i

ii tNPtC


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Uniform Cubic B-Spline Curves Definition (uniform knot sequence, i - i-1 = 1):

),[

),[

),[

),[

6/)1(

6/)463(

6/)1333(

6/

)(

),3[,)()(

43

32

21

1

3

2

1

3

23

23

3

3

1

0

3

ii

ii

ii

ii

i

i

i

i

i

n

i

ii

t

t

t

t

tr

tr

tr

tr

r

rr

rrr

r

tN

nttNPt


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Uniform Cubic B-Spline Curves For any t [3, n]:

(prove it!)

For any t [3, n] at most four basis functions are non

zero

Any point on a cubic B-Spline is an affine combination

of at most four control points

1)(1

0

3

n

i

i tN

.)(

)()(

Then, .,Let

3

3

0

3

0

3

1

0

0

3

430

i

ii

n

i

iitt

tNP

tNPt

t


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In general, B-Splines do not interpolate control points

in particular, the uniform cubic B-spline curves do not interpolate

the end points of the curve.

Why is the end points’ interpolation important?

Two ways are common to force endpoint interpolation:

Let P0 = P1 = P2 (same for other end)

Add a new control point (same for other end) P-1 = 2P0 – P1 and a

new basis function N-13(t).

Question:

What is the shape of the curve at the end points if the first

method is used?

What is the derivative vector of the curve at the end points if the

first method is used?

Boundary Conditions for

Cubic B-Spline Curves


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Local Control of B-spline Curves

Control point Pi

affects (t) only for

t(i, i+4)


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Intuitive B-spline Curve Editing

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Properties of B-Spline Curves

For n control points, (t) is a piecewise polynomial of

degree 3, defined over t[3, n)

and for specific t = t0?

(t) is affine invariant and variation diminishing

(t) follows the general shape of the control polygon and it is intuitive and ease to control its shape

Questions:What is (i) equal to?

What is '(i) equal to?

What is the continuity of (t) ? Prove it!

4

0

3),..,()(

n

i

ii PPCHt

B-spline

),3[,)()(1

0

3 nttNPtn

i

ii


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From Curves to Surfaces

A curve is expressed as inner product of coefficients

Pi and basis functions

One way of extending curves into surfaces is to treat

surface as a curves of curves. This approach is also

known as tensor product surfaces

Assume Pi is not constant, but are functions of a

second, new parameter v:

n

i

ii uBPuC0

)()(

m

j

jiji vBQvP0

)()(


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From Curves to Surfaces (cont’d)Then

or

Question: What is the difference between u and v?

n

i

i

m

j

jij

n

i

i

m

j

jij

n

i

ii

uBvBQ

uBvBQ

uBPuC

0 0

0 0

0

)()(

)()(

)()(

n

i

i

m

j

jij uBvBQvuS0 0

)()(),( BezPatch


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Intuitive Surface Editing


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Surface constructors

Construction the geometry is a first stage in any

synthetic image synthesis process

There exist a set of high level, simple and intuitive,

surface constructors for novice users:

Bilinear patch

Ruled surface

Boolean sum

Surface of Revolution

Extrusion surface

Surface from curves (skinning)

Swept surface


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Bilinear Patches

Bilinear interpolation of 4 3D points - 2D analog of 1D

linear interpolation between 2 points in the plane

Given P00, P01, P10, P11 the parametric Bilinear surface

for u,v[0,1] is:

Questions:

What does an isoparametric curve of a Bilinear patch look

like ?

Can you represent the Bilinear patch as a Bezier surface?

When is a Bilinear patch planar?

11100100 )1()1()1)(1(),( uvPPvuvPuPvuvuP

..\..\Applets\applets\bspline\GermanApplet.html

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Ruled Surfaces

Given two curves a(t) and b(t), the corresponding ruled

surface between them is constructed by

The corresponding points on a(t) and b(t) are connected by

straight lines

Questions:

When is a ruled surface a bilinear patch?

When is a bilinear patch a ruled surface?

S(u, v) = v a(u) + (1-v) b(u)

a(u)b(u)

11100100 )1()1()1)(1(),( uvPPvuvPuPvuvuP


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Boolean Sum

Given four connected curves i i=1,2,3,4, Boolean sum

S(u, v) fills the interior. Let,

Then

)()1()(),(

)()1()(),(

)1(

)1()1)(1(),(

312

201

1110

0100

vuvuvuS

uvuvvuS

uvPPvu

vPuPvuvuP

),(),(),(),( 21 vuPvuSvuSvuS


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Boolean Sum (cont’d)

S(u,v) interpolates the four

i along its boundaries.

For example, consider the

u = 0 boundary:

Question: Must i be coplanar?

)(

)1()()1(

)1()(1)(0)0()1()0(

),0(),0(),0(

),0(

3

010030001

01003120

21

v

vPPvvPvvP

vPPvvvvv

vPvSvS

vS

)()1()(),()()1()(),(

)1()1()1)(1(),(

312201

11100100

vuvuvuSuvuvvuS

uvPPvuvPuPvuvuP


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Surface of Revolution Rotate a, usually planar,

curve around an axis

Consider curve

(t) = (x(t), 0, z(t))

and let Z be the axis of

revolution. Then,

Question: sin and cos are not rational. Can we represent

surfaces of revolution as rational?

).(),(

),sin()(),(

),cos()(),(

uvuz

vuvuy

vuvux

z

x

x


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Extrusion Surface

Extrusion of a, usually

planar, curve along a

linear segment.

Consider curve (t) and

vector

Then

.)(),( VvuvuS

V


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Surface from Curves (Skinning)

An intuitive surface fitting (approximation or

interpolation) scheme

Fit a surface through a set of prescribed curves:

Question: Must the curves be planar? Closed?

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Sweep Surface A general surface construction scheme

Rigid motion of one (cross section) curve along another (axis) curve:

The cross section can change as it is swept along

Question: Is an Extrusion as special case of sweep?

Surface of revolution?

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