6.5 Interpolation of data points with cubic B-splines Given n data points: Pi,i=1, 2, .. N, and no other derivative data at the boundaries. The problem is to construct a cubic B-spline curve which interpolates(precisely matches) these data points Construction of knot vector Let 1=1+d2 d2+1=P+1-Pl 1.2 d=∑d Remove two knot values u2 and uN-1 from knot vector to obtain proper number of degrees of freedom(instead of having to prescribe boundary conditions, which could be done if needed) u LN=UN==UN knots=n+4+1=(N-4)+4+4=N+4→n=N-1 R(u)=∑RB4() 0<u<1 (6.25) n Require that R(u)=∑RB24(u)=P (6.26 Solve for R(system is banded) There are also other more sophisticated ways to choose knot vector and parameterization attempting to make u proportional to arc length 116.5 Interpolation of data points with cubic B-splines Given N data points: Pi , i = 1, 2, · · · , N, and no other derivative data at the boundaries. The problem is to construct a cubic B-spline curve which interpolates (precisely matches) these data points. Construction of knot vector Let uˆ1 = 0 uˆi+1 = uˆi + di+1 di+1 = |Pi+1 − Pi | i = 1, 2, · · · , N − 1 d = X N i=2 di → ui = uˆi d i = 1, 2, · · · , N − 1 Remove two knot values u2 and uN−1 from knot vector to obtain proper number of degrees of freedom (instead of having to prescribe boundary conditions, which could be done if needed) T = {u1 = u1 = u1 = u1 | {z } 4 times < u3 ≤ u4 ≤ · · · ≤ uN−2 | {z } internal knots=N−4 < uN = uN = uN = uN | {z } 4 times } knots = n + 4 + 1 = (N − 4) + 4 + 4 = N + 4 → n = N − 1 R(u) = Xn i=0 RiBi,4(u) 0 ≤ u ≤ 1 (6.25) n = N − 1 Require that R(uj) = N X−1 i=0 RiBi,4(uj) = Pj j = 1, 2, · · · , N (6.26) Solve for Ri (system is banded). There are also other more sophisticated ways to choose knot vector and parameterization attempting to make u proportional to arc length. 11