Splines II – Interpolating Curves based on: Michael Gleicher: Curves, chapter 15 in Fundamentals of Computer Graphics, 3 rd ed. (Shirley & Marschner) Slides.
Post on 18-Dec-2015
223 Views
Preview:
Transcript
Splines II – Interpolating Curves
based on: Michael Gleicher: Curves, chapter 15 in
Fundamentals of Computer Graphics, 3rd ed.(Shirley & Marschner)
Slides by Marc van Kreveld
1
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
Example
• Canonical form: f(t) = (2,3) + (0,1) t + (5,2) t2 + (-1,-1) t3
• A generalized form could be:f(t) = (7,-3) (t - t3) + (3,1) (t + t2 + t3)
• Or it could be:f(t) = (0,3) (t - t3) + (-4,1) (t - 2t2 + t3) + (0,1) (t + t2) + (-9,2) (2t2 + t3) + (0,1) (7t + t2 + 6t3)
• The canonical form of a cubic always has four coefficients, a generalized form can have any number of coefficients
3
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
111
0.250.51
001
C (quadratic curve)
4
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
5
Blending functions
• u = [ 1 u u2 u3 … un ]• b(u) = u B so we obtain for the usual quadric example
with three points specified:
24-2
1-43-
0011CB
2
2
2
21
2
44
231
24-2
1-43-
001
1
uu
uu
uu
uuuCuB
6
Blending functions
• We can now blend the control points
i
n
ii ubu pf
0
)()(
)(
)(
)(
2
44
231
)(
2
1
0
2
2
2
ub
ub
ub
uu
uu
uu
b uBu
7
Blending functions
i
n
ii ubu pf
0
)()(
)(
)(
)(
2
44
231
)(
2
1
0
2
2
2
ub
ub
ub
uu
uu
uu
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 points8
Blending functions
22 2uuub )(
20 231 uuub )(
0 10.5
21 44 uuub )(
22
12
02 244231 ppp )()()( uuuuuu
u
1
9
Blending functions
• Note the sum of the contributions
22 210 uuub )(
20 231 uuub )(
21 440 uuub )(
1210 )()()( ububub+
10
Polynomials for interpolation
• Given points p = (p0, p1, … , pn) and increasing parameter values t = (t0, t1, … , tn), we can make a polynomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi
p4
p3
p2
p1
p0
p5
11
Polynomials for interpolation
• Given points p = (p0, p1, … , pn) and increasing parameter values t = (t0, t1, … , tn), we can make a polynomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi
p4
p3
p2
p1
p0
p5
the brown curve has t1’ > t1 and t4’ < t4 12
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
13
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
14
Why not use polynomials to interpolate 5 or more points
• This gets worse with higher degree (more points)
15
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 16
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 high17
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
18
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
19
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 p2p0
p3
20
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
1111
0200
0010
0001
Cconstraint matrix
21
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
22
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
23
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 35-36 of Curves I lecture
24
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
25
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 = 0.526
Cardinal cubics
27
Red cardinal spline: very high tension, derivatives at control points ignored, so t = 1 and (1-t)/2 = 0
Orange to purple: lower and lower tension
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 pieces are not yet specified; no derivative is specified at p0 and pn
• Therefore, imagine points p-1, pn+1 as additional first and last points n pieces require n+3 control points
p2
p0 p3
p1
28
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
29
Questions
1. Assume piecewise linear interpolation by blending functions, where the curve passes through points p0, p1, p2, p3, and p4 at u = 0, 0.25, 0.5, 0.75, and 1. Define the five blending functions bi (u) so that
2. Assume that someone defines quadratic blending functions as on slides 5-9, except that b2(u) = u2. Does the resulting curve pass through its defining points. Is the resulting curve translation-invariant?
30
4433221100 )()()()()()( pubpubpubpubpubu f
top related