Normal of a coons patch at a given point

Question

Disclamer: Rendering the Coons patch is part of 3D Graphics homework, but finding the normals at a given point isn't. Just curious.

Here's what I got so far:

It's a Coons patch defined by four Hermite curves. I render the Coons patch by making a mesh of triangles from points calculated on the Coons patch.

The surface could look much smoother than it actually is if I could figure out how to get the normal of the patch at any given point, instead of using the normals of the flat triangles.

Now I know that if I have two tangents of the surface at a given point, their cross product will give me what I look for. I also know that, for an Hermite curve in 2D space, finding the tangent means getting the derivative, which is easy if you see Hermite curve as a product of matrixes and the vector (1, t, t^2, t^3); just use this vector instead (0, 1, 2t 3t^2). The normal is just the perpendicular of the tangent then:

But how do I translate that knowledge into finding the two tangents necessary for finding the normal of the Coons patch at a given point? My shaky programmer maths fail me.

EDIT: Here's how I build my Coons patch. Taken at Wikipedia. The 4 curves are Hermites.

Given four space curves c₀(s), c₁(s), d₀(t), d₁(t) which meet at four corners c₀(0) = d₀(0), c₀(1) = d₁(0), c₁(0) = d₀(1), c₁(1) = d₁(1); linear interpolation can be used to interpolate between c₀ and c₁, that is

$L_c(s,t)=(1-t) c_0(s)+ t c_1(s) $

and between d₀, d₁

$L_d(s,t)=(1-s) d_0(t)+ s d_1(t) $

producing two ruled surfaces defined on the unit square.

The bilinear interpolation on the four corner points is another surface

$ B(s,t) = c_0(0) (1-s)(1-t) + c_0(1) s(1-t) + c_1(0) (1-s)t + c_1(1) s t. $

A bilinearly blended Coons patch is the surface

$L_c(s,t)+L_d(s,t)-B(s,t). $

Answer 1

The term "Coons patch" is used to refer to several different surface types in CAGD, unfortunately. It sounds like you're talking about a standard tensor-product bicubic patch, which can be written in Hermite form: $$ \mathbf{S}(u,v) = [1 \; u \; u^2 \; u^3] \cdot \mathbf{M} \cdot [1 \; v \; v^2 \; v^3]^T $$ where $\mathbf{M}$ is a $4 \times 4$ matrix containing corner points, derivative vectors, and "twist" vectors. There's a good description of these creatures in this book by Mortenson.

You can just differentiate the patch equation from above to get partial derivatives: $$ \frac{\partial\mathbf{S}}{\partial u}(u,v) = [0 \; 1 \; 2u \; 3u^2] \cdot \mathbf{M} \cdot [1 \; v \; v^2 \; v^3]^T $$ $$ \frac{\partial\mathbf{S}}{\partial v}(u,v) = [1 \; u \; u^2 \; u^3] \cdot \mathbf{M} \cdot [0 \; 1 \; 2v \; 3v^2]^T $$ Then, as you mentioned, the normal is just the cross product of the partial derivative vectors $$ \mathbf{N}(u,v) = \frac{\partial\mathbf{S}}{\partial u} \times \frac{\partial\mathbf{S}}{\partial v} $$ You should unitize this normal vector before using it in lighting calculations, of course.

Edit: From the modified question, we now know that the Coons patch is actually a bilinearly blended "boolean sum" one: $$ \mathbf{S}(u,v) = \mathbf{L}_c(u,v) + \mathbf{L}_d(u,v) - \mathbf{B}(u,v) $$ The process described above will work on these kinds of patches, too (or on any parametric surface, actually). We just have to compute partial derivatives, again. These are straightforward: $$ \frac{\partial\mathbf{L}_c}{\partial u} = (1-v)\mathbf{c}'_0(u) + v\mathbf{c}'_1(u) $$ $$ \frac{\partial\mathbf{L}_d}{\partial u} = \mathbf{d}_1(v) - \mathbf{d}_0(v) $$ $$ \frac{\partial\mathbf{B}}{\partial u} = (1-v)\big[\mathbf{c}_0(1) - \mathbf{c}_0(0)\big] + v\big[\mathbf{c}_1(1) - \mathbf{c}_1(0)\big] $$ $$ \frac{\partial\mathbf{S}}{\partial u} = \frac{\partial\mathbf{L}_c}{\partial u} + \frac{\partial\mathbf{L}_d}{\partial u} - \frac{\partial\mathbf{B}}{\partial u} $$ And similarly for the partial derivatives with respect to $v$.