Opengl Notes 8

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OpenGL Notes a

Stu Pomerantz

[email protected]

http://www.psc.edu/~smp

December 8, 2004

aMost material is adapted from: OpenGL ARB, et. al, The OpenGL Pro-gramming Guide, Third Ed., Reading: Addison-Wesley, 1999

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2D Parametric Curves

In two dimensions a parametric curve is defined by:

x = x(u)y = y(u)

Where u is the parameter that is free to vary. For example:

x = cos()

y = sin()

x and y will trace out a circle as 0 2 .

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Drawing of the parametric circle:

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Code for the parametric circle:

glBegin(GL_LINES) ;du = PI/32 ;

f o r ( u = 0 ; u < 8 * P I ; u + = d u ) {

x = cos(u) ; y = sin(u) ; z = 0 ;

glVertex3f(x,y,z) ;

x = cos(u+du) ; y = sin(u+du) ; z = 0 ;

glVertex3f(x,y,z) ;

}

glEnd() ;

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In two dimensions a parametric curve is defined by:

x = x(u)y = y(u)

z = z(u)

Where u is the parameter that is free to vary. For example:

x = cos()

y = sin()

z =

x, y and z will trace out a circlular helix as 0 2 .

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Drawing of the parametric helix:

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Code for the parametric helix:

glBegin(GL_LINES) ;du = PI/32 ;

f o r ( u = 0 ; u < 8 * P I ; u + = d u ) {

x = cos(u) ; y = sin(u) ; z = u ;

glVertex3f(x,y,z) ;

x = cos(u+du) ; y = sin(u+du) ; z = u+du ;

glVertex3f(x,y,z) ;

}

glEnd() ;

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3D Parametric Surfaces

In three dimensions a parametric surface is defined by:

x = x(u, v)y = y(u, v)

z = z(u, v)

Where u and v are the parameters that are free to vary. Forexample:

x = cos() sin()

y = sin() sin()z = cos()

x, y and z will trace out a sphere as 0 2 and 0 2 .

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3D Parametric Surfaces

Drawing of the parametric sphere:

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Polynomial Parametric Curves

p(u) =n

k=0

ukc k

where

c k =

cxk

cyk

czk

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For example if n = 3,

p(u) =3

k=0

ukc k

expands to:

p(u) = u3

cx3

cy3

cz3

+ u2

cx2

cy2

cz2

+ u

cx1

cy1

cz1

+

cx0

cy0

cz0

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Continuing,

p(u) = u3

cx3

cy3

cz3

+ u2

cx2

cy2

cz2

+ u

cx1

cy1

cz1

+

cx0

cy0

cz0

expands to

p(

u) =

x(u) = cx3u3 + cx2u

2 + cx1u + cx0

y(u) = cy3u3

+ cy2u2

+ cy1u + cy0z(u) = cz3u

3 + cz2u2 + cz1u + cz0

Notice that 4 coefficient vectors which must be specified.

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Interpolation Revisited

Two points, p0 and p1, can be linearly interpolated like this:

p = a p0 + (1 a) p1

where a [0, 1]

In this context, interpolation is also called blending. The points aremixed together by the a and 1 a which are weights.

The points p0 and p1 can be thought of as control points since they

influence (constrain) the final interpolated value.

This is similar, but not identical to, OpenGLs blending functions.

And, in this case the purpose is not to produce transparent

surfaces but rather to produce a new interpolated point.

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Interpolation using Cubic Polynomials

Recall the cubic polynomial

p(u) =3

k=0

ukc k

Requires 4 vectors of coefficients in order for it to be completely

determined.

Rewrite this equation...

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..using a matrix representation of the coefficient vectors.

p(u) = [x(u) y(u) z(u)] =

u3 u2 u 1

m11 m12 m13 m14

m21 m22 m23 m24

m31 m32 m33 m34

m41 m42 m43 m44

g1x g1y g1z

g2x g2y g2z

g3x g3y g3z

g4x g4y g4z

Or more briefly,

p(u) = U M G

Call M the basis matrix. Call G the Geometry matrix.

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For example,

x(u) = (u3

m11 + u2

m21 + um31 + m41)g1x+(u3m12 + u

2m22 + um32 + m42)g2x+

(u3m13 + u2m23 + um33 + m43)g3x+

(u3

m14 + u2

m24 + um34 + m44)g4x+

This form emphasizes that the curve is a weighted sum of the

elements of G (geometry) matrix.

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The geometry matrix, G, contains 4 points. They are called

control points.

Analogous to the 2 control points in the previous linear

interpolation example.

The meaning of these 4 points is dependant on the basis matrix

M.

The basis matrix M affects

The continuity of the curve.

Which and whether control points are interpolated.

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The Hermite Basis

The Hermite basis is named after the Mathematician. The

construction the Hermite basis matrix is as follows:

Given 4 control points: p1, p2, p3, p4

Constrain the curve so that it interpolates p1 at u = 0 and p4

at u = 1

Let p2 and p3 be the tangent vectors for p1 and p4 respectively.

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The Hermite Basis

Recall,

U = u3 u2 u 1

so,

U =

3u2 2u 1 0

There are 4 equations and 4 unknowns:

p1 = [0 0 0 1] (u = 0)

p4 = [1 1 1 1] (u = 1)

p2 = [0 0 1 0] (u = 0)

p3 = [3 2 1 0] (u = 1)

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The Hermite Basis

Rewriting the 4 equations above as a matrix:

p1

p4

p2

p3

=

0 0 0 1

1 1 1 1

0 0 1 0

3 2 1 0

For these equations to be satisfied M, the Hermitian basis matrix

must be the inverse of this matrix.

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The Hermite Basis

The Hermitian basis matrix:

M =

2 2 1 1

3 3 2 1

0 0 1 0

1 0 0 0

So,

p(u) =(2u3 3u2 + 1)p1 + (2u3 + 3u2)p4 +

(u3 2u2 + u)p3 + (u3 u2)p4

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The Hermite Basis

A picture of the Hermitian basis functions:

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Example Hermite Curves

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Comparison of Curve Types

Curves differ by basis matrix.

Hermite Bezier Non-Uniform B-Spline Catmull-RomA Y Y N Y

B Y N N Y

CC0 C0 C2 C1

A = Interpolates some control points

B = Interpolates all control points

C = Inherent Continuity

If the direction and magnitude of the nth derivative are equal at

the joint point, the curve is called Cn continuous.

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Example NURBS Surface

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