- Asked 4/25
- 2079 Baisakh · 5 marks
Bezier points at u=0.25,0.5,0.75: (10,15)...
Short answer
With control points (10,15),(15,20),(20,35),(25,10), the cubic Bezier yields P(0.25)=(13.75,19.84), P(0.5)=(17.5,23.75), P(0.75)=(21.25,22.03). These are weighted sums using the Bernstein blending functions, and the curve passes through the first and last control points.
Find the coordinates of Bezier curve at u = 0.25, 0.5 and 0.75 with respect to the control points (10, 15), (15,20), (20, 35), (25, 10) using Bezier function.
Control points P0(10,15), P1(15,20), P2(20,35), P3(25,10). Cubic Bezier:
P(u) = (1−u)³P0 + 3u(1−u)²P1 + 3u²(1−u)P2 + u³P3
| u | B0 | B1 | B2 | B3 |
|---|---|---|---|---|
| 0.25 | 0.4219 | 0.4219 | 0.1406 | 0.0156 |
| 0.50 | 0.1250 | 0.3750 | 0.3750 | 0.1250 |
| 0.75 | 0.0156 | 0.1406 | 0.4219 | 0.4219 |
u = 0.25 x = 0.4219·10 + 0.4219·15 + 0.1406·20 + 0.0156·25 = 4.219 + 6.328 + 2.813 + 0.391 = 13.75 y = 0.4219·15 + 0.4219·20 + 0.1406·35 + 0.0156·10 = 6.328 + 8.438 + 4.922 + 0.156 = 19.84
u = 0.5 x = 0.125·10 + 0.375·15 + 0.375·20 + 0.125·25 = 1.25 + 5.625 + 7.5 + 3.125 = 17.5 y = 0.125·15 + 0.375·20 + 0.375·35 + 0.125·10 = 1.875 + 7.5 + 13.125 + 1.25 = 23.75
u = 0.75 x = 0.0156·10 + 0.1406·15 + 0.4219·20 + 0.4219·25 = 0.156 + 2.109 + 8.438 + 10.547 = 21.25 y = 0.0156·15 + 0.1406·20 + 0.4219·35 + 0.4219·10 = 0.234 + 2.813 + 14.766 + 4.219 = 22.03
Answer
| u | Point |
|---|---|
| 0.25 | (13.75, 19.84) |
| 0.50 | (17.50, 23.75) |
| 0.75 | (21.25, 22.03) |
✓ Answer checked by Claude Opus 5.5 (high effort)
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