Question

1. (10 points) Find the parametric equation for a Bezier curve with five control points Po, P1, P2, P3, and P4.

Answer 1

HT Solution P.N. I Equation for a Bezier Cusre Standard equation at Bezier Cuę ve 4-Con pol point Plo) = !0°(1-05 po t (? )u(1-uy3.1+) 02 (1-233-2 + ( 3 )(1-0) 3-3 P(4)= (1-0) P6+34(1-0)??,7342 (1-4)P+U3P3 Example: Control point in sequence (30,9,10 (40,0,40 ),(70,0,30), (60,0,20) V=0 U= 0.2 U=0.6 O=0;8, 0-1 Giren data Po i (30,0,10 ) Pi : (40,0,40) Pz : ( 70,0,30) P3 i (60,0,20)

P.N.12. Standard egn is (4-Control points) P(4)= (1-01? Po + 34(1-0)2p, +342(1-4)P+U3P3 + For 2 Coordinate P2 (0) = (1-0) Poz+ 34(1-0)?Piz+ 342 (1-4)P, + uþP32 = [1-34 +34–43] 10+ [34 (+24 +4?)740 + 342 (1-4)30+204? = 10-304 +3042-1043 +1204 - 24042 +12043 +90u? go43 +2043 = you² - 120u² tgou +10 For X Coordinate Rx (4)= (1-4) Pex +34(1–432P, +342 (14) P2 +4 () = [1-34 +342_43] 30+ [34(1-24 +42)]40 +342 (1-4) 70 + 6043 Px (4) = -6043 +604² +304 +30

3 The Can be Calculated some was But hege yzo in in all point hence P(u)y=0 U= 0, 0.2, 0.6 0.8, 100 U = 0 0.2 PY 0.6 0.8 30 1.0 37.9 47.8 56.6 60 Ру O O 0 O Pz lo 23.5 29.9 29.4