Formula for the offset surface of an ellipsoid?

Question

If the parametric formula for an ellipsoid is:

$\begin{align} x&=a\cos(\theta)\cos(\varphi)\\ y&=b\cos(\theta)\sin(\varphi)\\ z&=c\sin(\theta)\end{align}\\$

where

$ -\frac \pi 2 \le \theta\le \frac \pi 2 \qquad -\pi\le \varphi\le \pi $

And the parametric formula for an offset curve(!) is:

$x_d(t)= x(t)+\frac{d\; y'(t)}{\sqrt {x'(t)^2+y'(t)^2}}$

$y_d(t)= y(t)-\frac{d\; x'(t)}{\sqrt {x'(t)^2+y'(t)^2}}$

where $d$ is the distance from the curve.

Then what is the formula for the offset surface(!) of an ellipsoid? Thanks.

See:

• https://en.wikipedia.org/wiki/Ellipsoid
• https://en.wikipedia.org/wiki/Parallel_curve

(P.S. what is the derivative of the formula for an ellipsoid? Thanks.)

Answer 1

The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,

• $\vec{x}_d(t) = \vec{x}(t)+d\vec{n}(t)$ with the unit normal $\vec{n}(t)$.

This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)

All that’s left for you to do is to figure out how to compute $\vec{n}$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.