Computer Graphics Geometric Modeling - Arizona Computer … · · 2008-10-23Computer Science -...

Geometric Modeling

Page

Computer GraphicsComputer Graphics

Copyright Gotsman, Elber, Barequet, Karni, Sheffer

Computer Science - Technion

Geometric Geometric ModelingModeling

2

An Example

3 4

5

Outline

� Objective: Develop methods and algorithms to mathematically model shape of real world objects

� Categories:

Wire-frame representations

Boundary representations

Volumetric representations

6

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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7

Volumetric Representation

� Voxel based� Advantages: simple and robust Boolean operations,

in/out tests, can represent and model the interior of the object.

� Disadvantages: memory consuming, non-smooth, difficult to manipulate.

8

Constructive Solid Geometry

� Use set of volumetric primitivesBox, sphere, cylinder, cone, etc…

� For constructing complex objectsuse Boolean operations

UnionIntersectionSubtractionComplement

9

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= +

10

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

ππψπθψψθψθ −∈∈=

=−++

−−−=∪−−+=

RRRzyx

Rzyx

yxRzyxRz

:Parametric

Implicit

Explicit

Explicit is a special case of implicit and parametric form

11

Parametric Curves

� Analogous to trajectory of particle in space.

� Single parameter t∈ [T1,T2] – like “time”.� position = p(t) = (x(t),y(t)),

velocity = v(t) = (x’(t),y’(t))

� Circle:x(t) = cos(t), y(t) = sin(t) t ∈ [0,2π) ||v(t)|| ≡ 1x(t) = cos(2t), y(t) = sin(2t) t ∈ [0,π) ||v(t)|| ≡ 2x(t) = (1-t2)/(1+t2), y(t) = 2t/(1+t2) t ∈ (-∞,+∞)

(x(t),y(t))(x(t),y(t))

v(t) = (xv(t) = (x’’(t),y(t),y’’(t))(t))

(x(t),y(t))(x(t),y(t))

r(t)r(t)

k(t) = 1/r(t)k(t) = 1/r(t)

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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∞

kjCC jj ≤≤= 0),0()1( )(2

)(1

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Geometric Continuity� Mathematical continuity is sometimes too strong� May be relaxed to geometric continuity

(so curves would look the same, but need to change their parameterization) Gk, k ≤ 0 : Same as Ck

Gk, k =1 : C'1(1) =α 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∈[−π/2,0]C2(t)=[cos(t),sin(t)], t∈[0,π/2]C3(t)=[cos(2t),sin(2t)], t∈[0,π/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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Polynomial Bases

� Monomial basis {1 , x, x2, x3, …}Coefficients are geometrically meaninglessManipulation 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)

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Cubic Hermite Basis

�Set of polynomials of degree k is 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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Physical Splines

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Hermite Cubic Basis � 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 cubics

)1()()1()(

)32()(1)32()(2

112

10

201

200

−=−=

−−=+−=

ttthttth

ttthttth

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Hermite Cubic Basis (cont’d)

� Lets solve for h00(t) as an example.� h00(t) = a t3 + b t2 + c t + d

must satisfy the following four

constraints:

� Four linear equations in four unknowns.

.230)1('

,0)0('

,0)1(

,1)0(

00

00

00

00

cbah

ch

dcbah

dh

++====

+++====

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Hermite Cubic Basis (cont’d)

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

i

)()()()()( 111100011000 thTthTthPthPtC +++=

� Then 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

Fit spline independently for x(t) and y(t) to obtain C(t)

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Cubic Splines

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

No derivatives’ specified as input

� Interpolate by n cubic segments (4n DOF):Derive { Ti} i= 0,..,n from C2 continuity constraintsSolve 4n linear equations in 4n unknowns

Interpolation (2 equations):

C continuity constraints ( equations):

C continuity constraints ( equations):

1

2

n

C P C P i n

n

C C i n

n

C C i n

i i i i

i i

i i

( ) ( ) ,..,

( ) ( ) ,..,

( ) ( ) ,..,

' '

'' ''

0 1 1

1

1 0 1 1

1

1 0 1 1

1

1

1

= = =

−

= = −

−

= = −

−

+

+

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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

Periodic end conditionsC1'(0) = Cn'(1), C1''(0) = Cn''(1),

� Question: What parts of C(t) are affected as a result of a change in Pi ?

prescribedprescribed

naturalnatural

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Bezier Curves

� Bezier curve is an approximation of given control points

� Denote by γ (t): t∈[0,1]

� Bezier curve of degree n is defined over n+1control points { Pi} i= 0, n

P1 P3P5

P1

P2

P0P0 P2 P4

)(tγ

)(tγ

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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

nn

ji

ji

ji

ii

=

+−=

===

−−−

γ

γ (1/3)

Select t∈[0,1] value. Then,

t t = 1/3= 1/3

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25 26

Algebraic Form of Bezier Curves

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

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

degree n

Cubic case:

∑∑=

−−−

=n

i

iini

nii tt

ini

nPtBPt

0

n

0=i

)1()!(!

!)( = )(γ

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Algebraic Form of Bezier Curves

�

why?

� Curve is linear combination of basis functions

� Curve is affine combination of control points

∑ ∈∀=n

0=i

]1,0[,1)( ttBni

∑=

−−−

n

i

iini tt

ini

nPt

0

)1()!(!

! = )(γ

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Properties of Bezier Curves� γ (t) is polynomial of degree n� γ (t) is contained in the convex hull CH(P0,…,Pn)� γ (0) = P0 and γ (1) = Pn

� γ '(0) = n(P1−P0) and γ '(1) = n(Pn−Pn-1)� γ (t) is affine invariant and variation diminishing� γ (t) is intuitive to control via Pi and it follows the general

shape of the control polygon� γ '(t) is a Bezier curve of one degree less� Questions:

What is the shape of Bezier curves whose control points lie on one line?How can one connect two Bezier curves with C0 continuity? C1

? C2 ?

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Drawbacks of Bezier Curves� Degree corresponds to number of control points

Global support: change in one control point affects the entire curveFor 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.

For example, a 90 degrees arc as a rational Bezier curve:

.2where

)()()()()()(

)(

21

20

222

211

200

2222

2111

2000

=

++++=

w

ww

tBwtBwtBw

tBPwtBPwtBPw tCP0

=(0,1)

P1=(1,1)

P2 =(1,0)

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

across the domains with 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 called knots, { τi } .

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

� The knots are called uniform knots if τi - τi-1 = c,constant. WLOG, assume c = 1.

∑−

=

=1

0

)()(n

iii tNPtC

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31

Uniform Cubic B-Spline Curves� Definition (uniform knot sequence, τi - τi-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

iii

t

t

t

t

tr

tr

tr

tr

r

rr

rrr

r

tN

nttNPt

ττ

ττ

ττ

ττ

τ

τ

τ

τ

γ

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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 a convex combination of at most four control points

1)(1

0

3 =∑−

=

n

ii tN

[ )

.)(

)()(

Then, .,Let

3

3

0

30

3

1

00

3

430

∑

∑

−=

−

==

=

=

∈

τ

τ

γ

ττ

iii

n

iiitt

tNP

tNPt

t

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� B-Splines do not interpolate control pointsin 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-1

3(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 CurvesControl point Pi

affects γ (t) only fort∈(τi, τi+4)

B-spline

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Properties of B-Spline Curves�

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

degree 3, defined over t∈[3, n)�

� γ(t) is affine invariant and variation diminishing� γ(t) follows the general shape of the control polygon and

it is intuitive and ease to control its shape� Questions:

What is γ (τi) equal to?What is γ '(τi) equal to?What is the continuity of γ (t) ? Prove !

U4

03),..,()(

−

=+∈

n

iii PPCHtγ

),3[,)()(1

0

3 nttNPtn

iii ∈= ∑

−

=

γ

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From Curves to Surfaces

� A curve is expressed as inner product of coefficients Pi and basis functions

� Treat surface as a curves of curves. Also known as tensor product surfaces

� Assume Pi is not constant, but are functions of a second, new parameter v:

∑=

=n

iii uBPuC

0

)()(

∑=

=m

jjiji vBQvP

0

)()(

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From Curves to Surfaces (cont’d)Then

or

∑∑

∑ ∑

∑

= =

= =

=

=

=

=

n

ii

m

jjij

n

ii

m

jjij

n

iii

uBvBQ

uBvBQ

uBPuC

0 0

0 0

0

)()(

)()(

)()(

∑∑= =

=n

ii

m

jjij uBvBQvuS

0 0

)()(),( BezPatch

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Surface Constructors

� Construction of the geometry is a first stage in any image synthesis process

� Use a set of high level, simple and intuitive, surface

constructors: Bilinear patchRuled surfaceBoolean sumSurface of RevolutionExtrusion surfaceSurface 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 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 +−+−+−−=

40

PP1010

PP1111

PP0000

PP1010

PP0101

xx

yy

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� Given two curves a(t) and b(t), the corresponding ruled surface between them is:

The corresponding points on a(u) and b(u) are connected by straight lines

� Questions: When is a ruled surface a bilinear patch?When is a bilinear patch a ruled surface?

Ruled Surfaces

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

a(u)b(u)

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Surface of Revolution� Rotate a, usually planar,

curve around an axis

� Consider curve β(t) = (βx(t), 0, βz(t))

and let Z be the axis of revolution. Then,

).(),(

),sin()(),(

),cos()(),(

uvuz

vuvuy

vuvux

z

x

x

βββ

===

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Extrusion

� Extrusion of a, usually planar, curve along a

linear segment.

� Consider curve β(t) andvector

� Then

.)(),( VvuvuSr

+= β

Vr

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Sweep Surface� Rigid motion of one (cross section) curve along another

(axis) curve:

� The cross section may change as it is swept

Question: Is an extrusion a special case of a sweep?a surface of revolution?

Computer Graphics Geometric Modeling - Arizona Computer … · · 2008-10-23Computer Science -...

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