--- title: "Biquadratic Bezier surface point calculation" subject: "Computer Aided Design" chapter: 4 key: "semester_4/computer_aided_design:bezier-surface-numerical" source: https://www.bctengineer.com/q/semester_4/computer_aided_design/bezier-surface-numerical site: BCT Engineering --- # Biquadratic Bezier surface point calculation Computer Aided Design · Chapter 4: Representation of Surfaces and Solids - Asked in: year not given, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/bezier-surface-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-4 ## Question A biquadratic Bezier surface patch has a 3 x 3 control net B(i,j) = (4i, 4j, z_ij), i, j = 0, 1, 2, with the z values z = [[0, 2, 0], [3, 6, 3], [0, 2, 0]] (row i, column j). Write the equation of the patch and calculate the surface point at (u, w) = (0.5, 0.5), and the point at (0.25, 0.5). ## Short answer The biquadratic Bezier surface is a tensor product of quadratic curves. Applying Bernstein blending gives P(0.5,0.5) = (4,4,2.75) and P(0.25,0.5) = (2,4,2.3125). ## Answer #### Equation of the patch For degree 2 in both directions: $$ \mathbf P(u,w)=\sum_{i=0}^{2}\sum_{j=0}^{2}\mathbf B_{ij}\,J_{2,i}(u)\,J_{2,j}(w) $$ with $J_{2,0}=(1-t)^2$, $J_{2,1}=2t(1-t)$, $J_{2,2}=t^2$. #### Control net | $(i,j)$ | j = 0 | j = 1 | j = 2 | |---|---|---|---| | i = 0 | (0,0,0) | (0,4,2) | (0,8,0) | | i = 1 | (4,0,3) | (4,4,6) | (4,8,3) | | i = 2 | (8,0,0) | (8,4,2) | (8,8,0) | #### Point at (0.5, 0.5) Weights for $t=0.5$ in both directions: $0.25,\ 0.5,\ 0.25$. $x$ and $y$ follow from the grid ($x=4i$, $y=4j$ at the net points): $$ x=4\,(0\cdot0.25+1\cdot0.5+2\cdot0.25)=4,\qquad y=4 $$ For $z$, first sum across $j$ for each row ($w$ weights $0.25, 0.5, 0.25$): $$ \begin{aligned} \text{row 0: }&0(0.25)+2(0.5)+0(0.25)=1.0 \\ \text{row 1: }&3(0.25)+6(0.5)+3(0.25)=4.5 \\ \text{row 2: }&1.0 \end{aligned} $$ Then across $i$ ($u$ weights $0.25, 0.5, 0.25$): $$ z=0.25(1.0)+0.5(4.5)+0.25(1.0)=2.75 $$ Surface point $(4,\ 4,\ 2.75)$ (the central net point has $z=6$, so the surface lies well below it, as expected for an approximating surface). #### Point at (0.25, 0.5) $u$ weights for $t=0.25$: $(0.75)^2=0.5625$, $2(0.25)(0.75)=0.375$, $(0.25)^2=0.0625$. The $w$ weights remain $0.25, 0.5, 0.25$, so the row sums are again $1.0, 4.5, 1.0$. $$ \begin{aligned} x&=4(0.375+2\times0.0625)=2.0 \\ y&=4(0.5+2\times0.25)=4.0 \\ z&=0.5625(1.0)+0.375(4.5)+0.0625(1.0)=2.3125 \end{aligned} $$ **Answer: P(0.5, 0.5) = (4, 4, 2.75); P(0.25, 0.5) = (2, 4, 2.3125).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation gives a biquadratic Bezier surface patch? A. P(u,w) = sum over i,j of B_ij J_2,i(u) J_2,j(w), with Bernstein weights (1-t)^2, 2t(1-t), t^2 B. P(u,w) = sum over i,j of B_ij N_i,3(u) N_j,3(w) with knot vectors C. P(u,w) = (1-u)(1-w)P00 + u(1-w)P10 + (1-u)w P01 + uw P11 D. P(u,w) = sum over i of B_i J_3,i(u) only, one parameter Answer: A. A Bezier patch is the tensor product of Bernstein (Bezier blending) functions of degree 2 in u and w. (Key point: Patch equation) 2. What are the quadratic blending weights J2,0, J2,1, J2,2 at t = 0.5? A. 0.5, 0.5, 0.5 B. 0.25, 0.5, 0.25 C. 0.25, 0.25, 0.5 D. 0.5, 0.25, 0.25 Answer: B. (1-t)^2 = 0.25, 2t(1-t) = 0.5, t^2 = 0.25 at t = 0.5. (Key point: Weights at t = 0.5) 3. What is the z value of the surface point at (0.5, 0.5)? A. 6 B. 4.5 C. 2.75 D. 1.0 Answer: C. Row sums across w are 1.0, 4.5, 1.0; then 0.25(1.0)+0.5(4.5)+0.25(1.0) = 2.75. (Key point: z at (0.5, 0.5)) 4. For u = 0.25, the u weights are 0.5625, 0.375, 0.0625. What is P(0.25, 0.5)? A. (2, 4, 2.3125) B. (4, 4, 2.3125) C. (2, 4, 2.75) D. (1, 4, 2.3125) Answer: A. x = 4(0.375 + 2x0.0625) = 2, y = 4, z = 0.5625(1)+0.375(4.5)+0.0625(1) = 2.3125. (Key point: Point at (0.25, 0.5)) ## Related in Computer Aided Design - [Coons patch definition and point calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Acoons-patch-numerical) — Computer Aided Design, chapter 4 - [Cubic Bezier point and tangent calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-numerical) — Computer Aided Design, chapter 3 - [Bezier surface equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-surface-properties) — Computer Aided Design, chapter 4 - [Bezier curve: equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-properties) — Computer Aided Design, chapter 3 - [Convert Hermite data to Bezier cubic](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Ahermite-to-bezier-conversion) — Computer Aided Design, chapter 3 - [Surface of revolution: equation, points, area](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Asurface-of-revolution-numerical) — Computer Aided Design, chapter 4 ## Related in other subjects - [Bezier equation and point at u=0.8](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-equation-point-u-08) — Computer Graphics, chapter 5 - [Column vs row-oriented NoSQL databases](https://www.bctengineer.com/question.md?key=semester_8%2Fbig_data_technologies%3Acolumn-oriented-nosql-row-oriented) — Big Data Technologies, chapter 4 - [Bezier points at u=0.25,0.5,0.75: (10,15)...](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-points-control-10-15) — Computer Graphics, chapter 5, asked 4× - [Bezier points at u=0.25,0.5,0.75: (2,10)...](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-points-control-2-10) — Computer Graphics, chapter 5, asked 2×