6.2.5 Example: 4 order basis function (Cubic B-spline case-K= 4 see Figure 6.3 n=6→7 control points m+k+l=ll knots 2,4 七 s,4 Figure 6.3: Cubic B-spline functions From property 1 N04(to)+N1,4(to)=0 (6.19) erefore P(to)=(P1-Po)N.(to) Similarly P(t10)=(P6-P5)N64(to) (6.21)6.2.5 Example: 4 th order basis function (Cubic B-spline case– K = 4 see Figure 6.3) n = 6 → 7 control points n + k + 1 = 11 knots t 1 o t1 t2 t3 t4 t7 t8 t9 t10 t5 t6 N0,4 N1,4 N2,4 N3,4 N4,4 N5,4 N6,4 N0,4 is C -1 N0,4 is C 2 N1,4 is C 2 N5,4 is C 2 N2,4 is C 2 N6,4 is C 0 N1,4 is C 2 N4,4 is C 2 N1,4 is C 0 N2,4 is C 1 N3,4 is C 2 N3,4 is C 2 N4,4 is C 1 N5,4 is C 0 N6,4 is C -1 Figure 6.3: Cubic B-spline functions. From property 1: N˙ 0,4(t0) + N˙ 1,4(t0) = 0 (6.19) Therefore, P˙ (t0) = (P1 − P0)N˙ 1,4(t0) (6.20) Similarly, P˙ (t10) = (P6 − P5)N˙ 6,4(t10) (6.21) 7