-2

I was wondering how to formulate the equation of an ellipse which is rotated about an elliptical orbit. Here's the example:

Say I have ellipse in the x-y plane:

$$ \frac{x^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$ where $a>b$.

Suppose I want to rotate this ellipse about the x-axis, but instead of a typical circular revolution (which would give a torus), I would like to do this revolution in an elliptical fashion such that the center of the ellipse follows the equation

$$\frac{y^2}{c^2}+\frac{z^2}{d^2}=1$$

I've been scratching my head for hours but can't quite seem to get a function representing this 3D shape in the form $f(x,y,z)$.

This shape should represent something like a single chain link: chain link

Would appreciate any help out there ! Thanks !

Ivan Neretin
  • 12,835
RGG
  • 19

1 Answers1

0

You can establish the parametric equation of a suitable surface by writing the parametric equation of the directrix ellipse and computing its Frenet frame.

Then writing the parametric equation of the generatix ellipse, and plugging it in the above Frenet frame (normal plane), you will obtain the desired equation.

Care that if the radius of curvature at the apex of the directrix is smaller than the axis of the generatrix, you will get a self-intersecting surface.

  • Hello, Thank you for getting back so soon. Would then the following equations suffice (i simply adjusted the parametric of such elliptical torus very slightly):

    \begin{align} x(u,v) &= \big(c + a \cos(v)\big)\cos(u), \ y(u,v) &= \big(d + a \cos(v)\big)\sin(u), \ z(u,v) &= b \sin(v). \end{align}

    Would this lead to a self-intersecting surface? If so, how do I check? Thank you!

     – RGG Oct 12 '17 at 02:25
  • @RGG: did you read my comments ? I can't answer this. And did you read my answer (which you seem to neglect) ? –  Oct 12 '17 at 06:54