0

I am trying to get a half an elliptic torus lie along half a spheroid so that they meet at a curve along the whole surface. I need a general way to find such intersecting surfaces. By elliptic torus, I mean that a circle goes travels through the path of an ellipse.

Here is some Mathematica code for half an elliptic torus:

ParametricPlot3D[{(4 + Cos[$\theta$]) Cos[$\phi$], (2 + Cos[$\theta$]) Sin[$\phi$], Sin[$\theta$]}, {$\theta$, 0, 2 $\pi$}, {$\phi$, 0, $\pi$}]

And then a half a spheroid is right here:

ParametricPlot3D[{2 Cos[$\phi$] Sin[$\theta$], 2 Sin[$\theta$] Sin[$\phi$], Cos[$\theta$]}, {$\phi$, -$\pi$, $\pi$}, {$\theta$, 0, $\pi$/2}]

I am not sure how to mathematically do this. Any reasoning for solution would be much appreciated. I did an approximation using a hemisphere and a torus cut in half so you can understand the orientation. This way you can understand the required orientation.

Embedded half torus in a hemisphere

mathematics2x2life
  • 13,322
  • 1
    I don't think this is possible to do in the way you've stated it, unless the "spheroid" happens actually to be a sphere and the "ellipse" happens actually to be a circle. In order to allow your torus to have a non-circular elliptical centerline, can we allow it to have a non-circular cross-section around the centerline as well? Specifically, I'm thinking of taking a regular circular torus and "stretching" it by a uniform factor in one direction to make it non-circular. – David K Nov 08 '17 at 03:34
  • I was worried about that. I am going to try to work out a proof tomorrow morning. –  Nov 08 '17 at 03:53

1 Answers1

0

Good thing is that it is possible. You have to make sure that the torus

Example:

Spheroid: $ \left\{6 \cos \left(\frac{\left| \phi \right| }{4}\right),4 \cos (\theta ) \sin \left(\frac{\left| \phi \right| }{4}\right),4 \sin (\theta ) \sin \left(\frac{\left| \phi \right| }{4}\right)\right\}$

Elliptical Torus$:\left\{(2 \cos (\theta )+4) \cos \left(\frac{\phi }{4}\right),(2 \cos (\theta )+2) \sin \left(\frac{\phi }{4}\right),2 \sin (\theta )\right\}$

with $0 \leq \theta \leq 2 \pi,0 \leq \phi \leq 2 \pi.$ Giving us the picture: Elliptical Torus

A general formula for when we have a quarter pipe is:

Spheroid: $\left\{(r+3) \cos \left(\frac{\phi }{4}\right),(r+2) \cos (\theta ) \sin \left(\frac{\phi }{4}\right),(r+2) \sin (\theta ) \sin \left(\frac{\phi }{4}\right)\right\}$

Elliptical Torus: $\left\{\cos \left(\frac{\phi }{4}\right) (a+r \cos (\theta )),\sin \left(\frac{\phi }{4}\right) (b+r \cos (\theta )),r \sin (\theta )\right\}$

with $0 \leq \theta \leq 2 \pi,0 \leq \phi \leq 2 \pi.$