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I have a very interesting problem that I can't seems to figure out an answer. I managed to create a 3D polygon that resides on a 3D cone surface: enter image description here

My goal is to somehow project the cone onto a 2D plane but still retains the general shape of the polygon (ex: a rectangular-like shape should not turn into a trapezium-like shape, or at least try to minimise that distortion). Here's the constraint of the problem:

  1. The 2D plane will always be Oxy.
  2. The cone will always be centered around Oz.

Over the last few days, I thought about translating Cartesian coordinate into Cylindrical coordinate and then project the Cylindrical coordinate to Polar Coordinate. However, I very soon realised that by doing so, if the taper angel of the cone is small, a "thick" polygon will turn into a thin line.

Any other suggestions? Thanks :).

Anh Tran
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  • Somehow connected : https://math.stackexchange.com/q/391915 – Jean Marie Jun 15 '20 at 09:38
  • @JeanMarie Hi Jean, that is my initial method. But I find that method suffer from an edge case where the polygon lie on a "transition region" from pi to -pi. It cannot be generalised for arbitrary polygon on the cone. – Anh Tran Jun 15 '20 at 11:26
  • You could map the polyline to an annulus (annular ring?). There will be distortion, but you can minimize it by choosing the scaling so that the mean or median $z$ has no distortion, and distortion increases away from that $z$. If you consider Maazul's answer Jean Marie linked to, it's like you cut out the center circle, then "stretch" what is left into a ring shape. To counteract the angular stretch, you stretch it radially in comparable amount, so the real perceptible distortion is mostly radial. – Example Jun 15 '20 at 13:32
  • Whoops! My previous comment is basically what OP already mentioned trying, albeit with a $z$ stretch factor of say $1/\cos(\theta/2)$ where $\theta$ is the cone aperture angle. Apologies.. – Example Jun 15 '20 at 14:44

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