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Many apologies ahead of time, I have no idea how to phrase this question, and I'm certainly way out of my element. I'll do my best but please go easy on me.

I wanted to make a polyhedra that was in the shape of a doughnut. For whatever reason, I thought it would be cool if that polyhedra had regular-polygon faces. With some online-research, I found these things called Stewart Toroids that were seemingly what I was looking for. The problem was that they all looked... well... ugly? I think Stewart had some rules for how he made his toroids and maybe that had something to do with why they didn't really look much like a doughnut to me.

Anyway I set about making my own, and I made something that looked more doughnut-like. The driving rule I used to make it was to try and make every vertex (where faces met) as smooth as possible, which I interpreted to mean I needed to minimize the angle change between all the faces. I think for that exact characteristic (least angle change between faces) you can't do better than this shape does (worst angle change is 36 deg)

enter image description here

enter image description here

(sorry, I know those pictures are bad)

So I guess my question is if any of that makes sense.

  • Is it okay thinking one polyhedra could be more "torus" than another?
  • Is there a smart way to measure that, or is this purely a subjective thing?
  • Does the idea of smoothing out a polyhedra actually help or is that really more cosmetic?
  • Is minimizing the angle between adjacent faces the right way to maximize smoothness?

EDIT: It seems I've attracted a number of people also interested in making more such polyhedra, which is fine, I certainly had fun making this thing. But the question is not "can you make these polyhedra", it is "how can we compare them?"

There's nothing stopping me from making a torus with a million little square faces, approximating a torus in the same fashion a bunch of pixels can approximate a circle. The thing I want to know is if math tells us how similar two shapes are, such that there can be an official method for comparing two polyhedral doughnuts.

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    $\begingroup$ Why do yuou say that thes pictures are bad ? They are splendid ! $\endgroup$
    – Jean Marie
    Commented Jan 30, 2019 at 21:30
  • $\begingroup$ you can extend things by adding more rings of squares. My first impression is that each such ring is 10 squares. There are also alternatives that probably work, doubled rings of 20 equilateral triangles. $\endgroup$
    – Will Jagy
    Commented Jan 30, 2019 at 21:36
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    $\begingroup$ @Will Jagy Hi Will ! I didn't notice your comment : I just had the same remark for these "connection rings" with the 20 equilateral triangles that make an antiprism. $\endgroup$
    – Jean Marie
    Commented Jan 30, 2019 at 23:52
  • $\begingroup$ Instead of a polyhedral doughnut, why not a polyhedral pretzel (with more than a single hole...) $\endgroup$
    – Jean Marie
    Commented Jan 30, 2019 at 23:58
  • $\begingroup$ @JeanMarie you might like this o2treehouse.com/honey-sphere which is based on the 72 face Goldberg polyhedron en.wikipedia.org/wiki/Goldberg_polyhedron At home, I have a 42 sided Goldberg solid, made from zometool.com pieces. $\endgroup$
    – Will Jagy
    Commented Jan 31, 2019 at 0:08

3 Answers 3

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I am not sure if this helps.

To construct a torus, you have an area of negative curvature (the inside), and an area of positive curvature (the outside).

The net curvature of a torus is 0.

At each vertex we can sum the angles and subtract from $360$ (or $2\pi$) to find that vertices contribution to the total curvature.

you need a total contribution of $720^\circ$ from the vertices on the outside, and $-720^\circ$ contribution from the vertices on the inside.

If we used only equilateral triangles, we need 12 vertices with 5 triangles at these vertices, and 12 vertices with 7 triangles at the vertex, (and some number of 6 triangles at each vertex).

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  • $\begingroup$ "If we used only equilateral triangles, we need 12 vertices with 5 triangles at these vertices, and 12 vertices with 7 triangles at the vertex, (and some number of 6 triangles at each vertex)." 6 vertices × 4 triangles and 6×8 maybe? $\endgroup$ Commented Jan 30, 2019 at 23:02
  • $\begingroup$ Who says we need these $\pm720^\circ$ in the first place? $\endgroup$ Commented Jan 31, 2019 at 5:08
  • $\begingroup$ There may be something here. My shape, by example, includes negative curvature on the outside, which is balanced by also having more positive curvature. You can take six icosahedrons and connect them using antiprisms (in the exact fashion as I did above), but does end up being more... well "pointy". $\endgroup$ Commented Jan 31, 2019 at 13:21
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Your volume can be described as the assembly of 6 so-called (small) rhombicosidodecahedra (see Figure below) "stitched" together by decagonal-based prismatic shapes

enter image description here

(copied from the site http://www.mi.sanu.ac.rs/vismath/zefiro1/_icosahedral_polyhedra_2007_11_15.htm ; in their fig. 3, one finds a certain generation mode).

I think we have there a way to construct other shapes using other Archimedean polyhedra, the stitching mode being in some cases a little different, for example by using antiprisms (https://en.wikipedia.org/wiki/Antiprism) instead of prismatic shapes.

An article connecting Archimedean solids and in particular rhombicosidodecahedra to quaternions (see its fig. 10) : https://arxiv.org/ftp/arxiv/papers/0908/0908.3272.pdf

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42 sided Goldberg polyhedron made with Zometool parts. For strength, there is a central ball, spokes out to an icosahedron (blue), then many (yellow) spokes out to the polyhedron itself (pentagons in blue, then hexagons with some blue and some red sides).

enter image description here

.............................

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