Splines II – Interpolating Curvesbased on:

Michael Gleicher: Curves, chapter 15 inFundamentals of Computer Graphics, 3rd ed.

(Shirley & Marschner)Slides by Marc van Kreveld

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Polynomial pieces

n

i

iitt

0

af

m

iii tbt

0

cf

Canonical form of a polynomial of degree n defined with vector coefficients ai

Generalized form of a polynomial defined with vector coefficients ci that are blended by the m polynomials bi(t)

The degree is the max of the degrees of the bi(t)

Using blending polynomials is the way to make splines2

Recall: Basis and constraint matrices• Specifications of a curve give a constraint matrix

p0 = f(0) = a0 + 0 a1 + 02 a2 p1 = f(0.5) = a0 + 0.5 a1 + 0.52 a2 p2 = f(1) = a0 + 1 a1 + 12 a2

• Its inverse B = C–1 is the basis matrix

1110.250.51001

C (quadratic curve)

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Blending functions

• Blending functions (or basis functions) are functions of u and specify how to “mix” the specified constraints (points to pass through, derivatives, …)

• Let u = [ 1 u u2 u3 … un ] be the powers of u• b(u) = u B, a vector whose elements are the blending

functions

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Blending functions• u = [ 1 u u2 u3 … un ]• b(u) = u B so we obtain for the usual quadric example with

three points specified:

24-21-43-001

1CB

2

2

2

21

244231

24-21-43-001

1uuuuuu

uuuCuB

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Blending functions

• We can now blend the control points

i

n

ii ubu pf

0

)()(

)()()(

244231

)(2

1

0

2

2

2

ububub

uuuuuu

b uBu

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Blending functions

i

n

ii ubu pf

0)()(

)()()(

244231

)(2

1

0

2

2

2

ububub

uuuuuu

b uBu

22

12

02 244231 ppp )()()( uuuuuu

We see the contributions of each point depending on u

For fixed u, we linearly interpolate the three points7

Blending functions

22 2uuub )(

20 231 uuub )(

0 10.5

21 44 uuub )(

22

12

02 244231 ppp )()()( uuuuuu

u

1

8

Blending functions

• Note the sum of the contributions

22 210 uuub )(

20 231 uuub )(

21 440 uuub )(

1210 )()()( ububub+

9

Polynomials for interpolation

• Given points p = (p0, p1, … , pn) and increasing para-meter values t = (t0, t1, … , tn), we can make a poly-nomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi

p4

p3

p2p1

p0

p5

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Polynomials for interpolation

• Given points p = (p0, p1, … , pn) and increasing para-meter values t = (t0, t1, … , tn), we can make a poly-nomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi

p4

p3

p2p1

p0

p5

the brown curve has t1’ > t1 and t4’ < t4 11

Polynomials for interpolation

• Method– Set up constraint matrix as before

• Invert to get basis matrix, giving the n+1 basis functions bi(t) and the polynomial f(t) :

• Alternative method (Lagrange form)

i

n

ii tbt pf

0

)()(

n

ijj ji

ji tt

tttb

,

)(0

12

Why not use polynomials to interpolate 5 or more points

• Polynomials of higher degree have– extra wiggles– overshoots

– non-locality: moving the point pn changes the curve even near p0 ; also when we add a point at the end

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Why not use polynomials to interpolate 5 or more points

• This gets worse with higher degree (more points)

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Blending again

• A piecewise-linear curve (polygonal line) can also be defined using blending functions

12

02 uub )(

)(ub0

0 10.5

)(ub1

u

1

0 u 0.5

0.5 u 1

p1

p2

p0

221100 pppf )()()()( ubububu 15

Piecewise cubic polynomials

• Allows position and derivative at each end• Allows C2 continuity• Are in a sense the most smooth curve: minimum

curvature over its length (curvature: second derivative)

curvature 0

curvature low

curvature high16

Cubic polynomials

• f(u) = a0 + a1 u + a2 u2 + a3 u3 in canonical form• Four control points (points or derivatives on curve)

needed• For piecewise cubic curves: n pieces require 4n

control points, but C0 continuity already fixes n – 1 of them

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Cubic polynomials

• The ideal properties:– Each piece is cubic– The curve interpolates the control points– The curve has local control– The curve has C2 continuity

• We can have any three but not all four properties:– Natural cubics do not have local control– Cubic B-splines do not interpolate the control points– Cardinal splines are not C2

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Natural cubics

• The first piece specifies start and end positions, and the first and second derivative at the start

• For each other piece, the position, first and second derivative match with the piece before it only the endpoint can be specified freely

• A curve with n pieces has n+3 control “points” in total (n+1 points and 2 derivative specifications)

p1 p2p0p3

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Natural cubics

• We get (with f(u) = a0 + u a1 + u2 a2 + u3 a3):

p0 = f (0) = a0 + 0 a1 + 02 a2 + 03 a3 p1 = f’(0) = 1 a1 + 20 a2 + 302 a3 p2 = f’’(0) = 2 a2 + 60 a3 p3 = f (1) = a0 + 1 a1 + 12 a2 + 13 a3

1111020000100001

Cconstraint matrix

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Natural cubics

• When you modify your natural cubic, for instance by changing the derivative at the start of the first piece, then the whole curve changes

not local

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Hermite cubics

• Specifies positions of start and end, and the derivatives at these points

• C1 continuous since the start position of a piece must be the same as the end position of the previous piece, and the same is true for the derivatives

• A curve with n pieces has 2n+2 control “points” in total

p2

p0 p3

p1

22

Hermite cubics

• Local control: changing the position or derivative of any point influences only the piece before and the piece after that point

• See slides 34-35 of Curves I lecture

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Cardinal cubics

• Specifies positions only; the derivatives at each point are determined by the points before and after it

• C1 continuous• A curve with n pieces has n+3 control points in total

p2

p0p3

p1

p0 p2p1 p3

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Cardinal cubics

• A tension parameter t [0,1) determines the “strength” of bending

• Each derivative vector pi-1 pi+1 is scaled by (1 – t)/2

p2

p0 p3

p1p2

p0 p3

p1

t = 0 (Catmull-Rom splines) t = 0.525

Cardinal cubics

scaling factor related, not the tension t

???

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Cardinal cubics

• We can set up the constraints:

f(0) = p1

f(1) = p2 f’(0) = ½ (1 – t) (p2 – p0)f’(1) = ½ (1 – t) (p3 – p1)

• Rewrite to get expressions for p0, p1, p2 and p3

• The first and last control points are not interpolated; no derivative would be specified

• Therefore, n pieces require n+3 points

p2

p0 p3

p1

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Summary

• Piecewise cubic curves are popular for modeling• We want C2 continuity, locality, and interpolation;

we can get only two of these with cubics• Natural cubics do not have locality• Hermite cubics are only C1 continuous• Cardinal cubics are only C1 continuous

• The splines that follow next (Bezier and B-spline) will not interpolate but approximate the control points

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