Biquadratic Bezier surface point calculation

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

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).