The equation of a surface created by the extrusion of a 2D closed curve along a path

Question

How do I obtain the equation of a surface created by the extrusion of a circle (or ellipse) created on the XY plane along a parabola or a parametric curve which lies on the YZ plane. The goal is to calculate the normal vector at each point of such surface so I can apply a reflection condition. Any literature on the subject will be highly appreciated.

Answer 1

Let's parameterize the circle or ellipse, so that points of the circle/ellipse are given by $(x(t), y(t), 0)$ for $t$ in some interval. For a circle, something like

$x(t) = r \cos(t)$

$y(t) = r \sin(t)$

for $0 \le t \le \pi$ works pretty well. For an axis-oriented ellipse, you can use

$x(t) = A\cos(t)$

$y(t) = B\sin(t)$

instead (for radii $1/A$ and $1/B$ on the $x$- and $y$-axes, respectively).

Let's suppose that the extrusion curve is $\gamma(t) = (0, b(t), c(t))$. And for later use, define $perp( [0, u, v] ) = ([0, -v, u])$. (That gives, for a vector in the $yz$-plane, another vector perpendicular to it, and also in the $yz$-plane).

With those things in hand, here's a parametric surface whose image is the extruded shape:

$S(u, v) = \gamma(v) + x(u) * perp(\gamma'(v)) + y(u) * [0, 0, 1]$

If $\gamma$ has tight curvature (e.g., if its radius of curvature is less than $r$, the radius of the circle being extruded in the simple case), then the resulting surface will have singularities.

To find the normal vector at the point $S(u, v)$, you can simply compute

$\frac{\partial S}{\partial u} (u, v) \times \frac{\partial S}{\partial u} (u, v)$,

which you'll probably want to normalize when you're done. Let's do that computation:

$\frac{\partial S}{\partial u} (u, v) = x'(u) * perp(\gamma'(v)) + y'(u) * [0, 0, 1]$

while

$\frac{\partial S}{\partial v} (u, v) = \gamma'(v) + x(u) * perp(\gamma''(v))$

You can probably write out the cross product and simplify a few terms, but if I were coding it, I'd just leave it in this form.

-John