M2R IVR, October 12th Mathematical tools 1 - Session 2
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1 Mathematical tools 1 Session 2 Franck HÉTROY M2R IVR, October 12th 2006
2 First session reminder Basic definitions Motivation: interpolate or approximate an ordered list of 2D points P i n Definition: spline curve: C(t) = F i (t)p i Interesting properties: n Normality: t, F i (t) = 1 (affine/barycentric invariance) i=0 i=0 Positivity: t, i, F i (t) 0 (convex hull) Regularity: i, F i has a single max (oscillation regularization) Locality: i, F i has compact support Parametric/geometric continuity
3 Interpolation and approximation 1 Basic definitions 2 Wavelets and multiresolution
4 Decomposing a spline We will look for uniform splines: C(t) = n n 1 F i (t)p i = C i (n.t i) i=0 i=0 with C i : [0, 1] R 3 polynomial in t each C i corresponds to a curve segment, defined for t [ i n, i+1 n ] = [t i, t i+1 ]
5 Interpolation splines Which parametric continuity? C 0 control polygon! C 1 : shared derivatives D i 4 conditions for each curve segment: C i (0) = P i C i (1) = P i+1 C i (0) = D i C i (1) = D i+1 C i = degree 3 polynomial: C i (t) = A + Bt + Ct 2 + Dt 3
6 Cubic Hermite splines (first order) If D i are given, C i (t) is uniquely determined: C i (t) = H 0 (t)p i + H 1 (t)p i+1 + H 2 (t)d i + H 3 (t)d i+1 H 0 (t) = 1 3t 2 + 2t 3 H 1 (t) = 3t 2 2t 3 H 2 (t) = t 2t 2 + t 3 H 3 (t) = t 2 + t 3 Problem: how do we compute D i values?
7 Cubic Hermite splines (second order) First idea: reinforce continuity C 2 continuity conditions: C i (0) = P i C i (1) = P i+1 C i (1) = C i+1 (0) C i (1) = C i+1 (0) if we have m + 1 points P i (i.e. m curve segments), we have 4m unknown values (A, B, C, D for each C i ) and 4(m 1) + 2 equations (only 2 for the last segment) we need 2 other conditions Usually, we set { C 0 (0) = 0 C n 1 = 0
8 Hermite splines (second order) Property For a Hermite spline of order 2, the D i values are given by: D D n = 3(P 1 P 0 ) 3(P 2 P 0 ) (P n P n 2 ) 3(P n P n 1 ) Proof. 4 equations for each C i C i is degree 3. Let C i (t) = a i + b i t + c i t 2 + d i t 3. Solve the previous equations.
9 Cardinal spline Other idea: we want D i parallel to P i 1 P i+1 (no condition on second derivative) conditions: C i (0) = P i C i (1) = P i+1 C i (1) = C i+1 (0) C i (0) = k(p i+1 P i 1 ) (same k for all derivatives) each segment curve depends on 4 points: P i 1, P i, P i+1 and P i+2
10 Cardinal spline Property C i (t) = (t 3 t 2 t 1)M card P i 1 P i P i+1 P i+2 Property M card = with k 2 k 2 + k k 2k 3 + k 3 2k k k 0 k A cardinal spline is normal, regular, local of order 4, and C 1 -continous.
11 Approximation splines Historical background: early 1960s: first numerically-controlled machines in automotive industry need to design (smooth) curves and surfaces starting from a very few number of 3D points (esp. for coachbuilding) Pierre Bézier (Renault, ): theoretical study and tests, conception of a whole system ( UNISURF ) Paul de Faget de Casteljau (Citroën, ): geometric algorithm Robin Forrest (Univ. Cambridge, 1972): link between both, popularization to know more, see Christophe Rabut s webpage: rabut/bezier/
12 Bézier curves Definition (Bézier curve) i, F i (t) = B n i (t) = C i nt i (1 t) n i (Bernstein polynomials). Remember that n + 1 is the number of control points. Property A Bézier curve is normal, positive, regular and C -continuous. Moreover, it goes through first and last points P 0 and P n. Problems: 1 a Bézier curve is not local 2 the higher the number of control points, the higher the degree of polynomials computation time and numerical stability problems
13 de Casteljau s algorithm How can we compute points on a Bézier curve without doing all the computations? Idea: divide each segment P i P i+1, creating a new point Pi 1 = tp i + (1 t)p i+1, then do the same with segments Pi 1 Pi+1 1, etc. Example: t = 0.4
14 Piecewise Bézier curves Definition (Piecewise Bézier curve) If n = 3m + 1, C(t) = m 1 i=0 C i (m.t i) with i m, C i (t) = B 3 0 (t)p 3i + B 3 1 (t)p 3i+1 + B 3 2 (t)p 3i+2 + B 3 3 (t)p 3i+3. That is to say, C i (t) = (1 t) 3 P 3i + 3(1 t) 2 tp 3i+1 + 3(1 t)t 2 P 3i+2 + t 3 P 3i+3. Property A piecewise Bézier curve is local of order 2, normal, positive, regular and C 0 -continuous. It also goes through points P 3i. Problem: only C 0 -continuous Solution: if i m, P 3i+1 = 2P 3i P 3i 1, then C 1 -continuous That means 1/3 of the points cannot be freely chosen If we want C 2 -continuity, the curve must be global
15 B-splines Goal: normal, positive, regular, local and C 2 -continuous splines Property The functions F i are uniquely determined if we assess: normality: F i (t) = 1 i locality of order 4 each F i is made of 4 curve segments, which are cubic polynomials C 2 -continuity between two successive cubic polynomials
16 B-splines Property C i (t) = (t 3 t 2 t 1)M bspline P i 1 P i P i+1 P i+2 M bspline = with Compare to cardinal splines: pros and cons?
17 Particular cases If P i 1, P i, P i+1 and P i+2 are aligned, then the curve is locally a straight line If P i 1 = P i = P i+1 (triple point), then this point is interpolated by the curve If we want to interpolate first and last points P 0 and P n, we can add extra points P 1 = 6P 0 4P 1 P 2 and P n+1 = 6P n 4P n 1 P n 2
18 B-splines of order k (= degree k 1) Previous were B-splines of order 4/degree 3 (cubic splines) Definition (B-spline of order k, Cox-de Boor recursion formula) F i = B i,k, defined by the following recursive formula: { 1 if ti t < t B i,1 (t) = i+1 0 else Property B i,k (t) = t t i t i+k 1 t i B i,k 1 (t) + t i+k t t i+k t i+1 B i+1,k 1 (t) with i = 0... n, t i [0, 1] and t 0 t1 t n A B-spline of order k is k-local and C k 2 -continuous. Each F i is made of k curve segments, each of them being a polynomial of degree k 1.
19 de Boor s algorithm Generalization of de Casteljau s: construction of points on a B-spline curve What s different: weights used to divide each segment P i P i+1 vary; not all control points are involved, only k + 1 Weights are found with a triangular scheme using Cox-de Boor recursion formula Example Cubic B-spline, 11 control points (n = 10), i, t i = i/10. t = 0.45: we want to compute C(0.45). Cubic 4-local only P 1, P 2, P 3, P 4 involved P 1 a 1 2 P 2 P 3 P 4 1 a 1 2 a a 1 3 a a 1 4 P 1 2 P 1 3 P 1 4 a a 2 3 a 2 4 P 2 3 P 2 4 a a a 3 4 P 3 4
20 de Boor s algorithm Property a j i = t i+k j t t i+k j t i Example Here, a 1 2 = 1 6, a1 3 = 1 2, a1 4 = 5 6, a2 3 = 1 4, a2 4 = 3 4, a3 4 = 1 2.
21 Beta-splines Definition (Beta-spline) C i (t) = (t 3 t 2 t 1)M beta (β 1, β 2 ) P i 1 P i P i+1 P i+2 Generalization of B-splines G k 2 -continuous instead of C k 2 Two parameters β 1 and β 2 : bias and tension, to control slope and curvature
22 N.U.R.B.S. Problem: with these splines we cannot draw some simple curves (e.g. conics, even a circle!) Piecewise polynomials as influence functions are not adequate Example Circle C((0, 0), 1): C(t) = (x(t), y(t)) cannot be represented using polynomials, otherwise x(t) 2 + y(t) 2 = 1 is a polynomial in t C(t) = ( 2t, 1 t2 ) 1+t 2 1+t 2 Idea: rather use rational polynomials
23 N.U.R.B.S. Definition (N.U.R.B.S. of order k) F i (t) = R i,k (t) = w ib i,k (t), n w j B j,k (t) j=1 with w i real numbers (weights, usually 0 what happens if w i = 0?) and B i,k the influence functions of the B-spline of order k having the same control points N.U.R.B.S. stands for Non Uniform Rational B-Spline B-splines are a special case of N.U.R.B.S. : i, w i = 1 R i,k are rational polynomials of degree k 1
24 N.U.R.B.S. Property A N.U.R.B.S. curve of order k is normal, positive, regular, k-local and C k 2 -continuous. Circle: P 4 P 4 P 5 P 6 P 3 P 5 P 3 P 7 P 2 P 1 = P 7 P 6 P 2 P 1 = P 9 P 8 w 2i+1 = 1 w 2i+1 = 1 w 2i = 1 2 w 2i = 2 2
25 Spline surfaces Interpolate or approximate a grid of 3D points P i,j Spline surfaces are constructed by tensor product of spline curves: n m S(t, t ) = F i,j (t, t )P i,j i=0 j=0 If F i,j (t, t ) = F i (t)f j (t ), then isocurves (t = cst or t = cst) are spline curves Same properties as spline curves: normality, regularity, locality, continuity between patches,...
26 See you next week The end!
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