Equilateral convex polyhedra (new class after 400 years?)
 robertw
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Certainly some polyhedra can't be given equal edge lengths unless they lose their convexity, like many of the duals of the Archimedean solids.
Here's some more convex equilaterals though.
From Stella's near miss library, these two don't need any further variation. I even named them "Equaledged Near Miss", so that's more equilaterals that were known before that paper. These were ones I found, as convex hulls of new Stewart Toroids.
Other near misses in Stella's library can be adjusted for equal edge lengths, such as these near misses discovered by Robert Austin:
Here's some more convex equilaterals though.
From Stella's near miss library, these two don't need any further variation. I even named them "Equaledged Near Miss", so that's more equilaterals that were known before that paper. These were ones I found, as convex hulls of new Stewart Toroids.
Other near misses in Stella's library can be adjusted for equal edge lengths, such as these near misses discovered by Robert Austin:
Yes of course, all those funny groupings of triangles round a vertex for a start.robertw wrote:Certainly some polyhedra can't be given equal edge lengths unless they lose their convexity, like many of the duals of the Archimedean solids.
Something like a heptagonal pyramid requires its equilateral morph to be wound up to give it a star base. That can be done with either a {7/2} or {7/3} base, giving two distinct geometric solutions.
I suspect that those Archimedean duals would also yield two solutions, one with augmentations and the other with excavations.
What could Stella's spring algorithm make of all that?
Cheers,
Guy. Guy's polyhedra pages
Guy. Guy's polyhedra pages

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I haven't been able to make one with the rhombic enneacontahedron truncated on the 5valance vertices. It works on the 3 and 6 valance vertices, but on the 5 it collapses into nonconvex.
Roger
Roger
Roger Kaufman
http://www.interocitors.com/polyhedra/
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There are many that can't work. For example if you augment a cube with cubes it will be nonconvex and is already equilateral.
That models seems familiar. How did you create it?
That models seems familiar. How did you create it?
Roger Kaufman
http://www.interocitors.com/polyhedra/
http://www.interocitors.com/polyhedra/
 robertw
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Or even if you put a pyramid on each side of a cube.polymorphic wrote:There are many that can't work. For example if you augment a cube with cubes it will be nonconvex and is already equilateral.
Weirdly, I started with "6J91 (P4)" which is in Stella under "Stella Library>More Stella Toroids>6J91 (P4)" (it's not a toroid but the shape of a hole in one of the toroids), and took the convex hull, then adjusted for equal edge lengths. But I'm sure there's a more straight forward way.That models seems familiar. How did you create it?

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It can be found by truncating the on 3valance vertices of the truncated octahedron by a factor of 1/2.robertw wrote:Actually, it's a rectified truncated octahedron I suppose.robertw wrote:But I'm sure there's a more straight forward way.
It can also be found by beveling only on the 3valance vertices in conway notation which *would* translate to this operation string. (Beveling itself would not allows the three 'b3').
dk3datO
Conway Notation specification does not specify that beveling can only be done by vertex valances. I think this may have been an oversight! Beveling is actually truncation by 1/2. Therefore since Conway Notation allows for truncation by vertex valance, so should it be with Beveling. I am going to propose this change.
Roger Kaufman
http://www.interocitors.com/polyhedra/
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I experimented with this and the only difference is how much is truncated from the polyhedron before planarization occurs. Normal truncation is only 1/3 of the way. Beveling is 1/2 of the way. So the results of tN and bN are ultimately the same.polymorphic wrote: Conway Notation specification does not specify that beveling can only be done by vertex valances. I think this may have been an oversight! Beveling is actually truncation by 1/2. Therefore since Conway Notation allows for truncation by vertex valance, so should it be with Beveling. I am going to propose this change.
The interesting polyhedron can actually already be generate by ambo of the truncated Octahedron.
conway atO  antiview
Roger
Roger Kaufman
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 robertw
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Are you confusing bevelling with rectification (ambo)? Rectification ("a" operator) is truncating half way so that the neighbouring truncations meet. Bevelling is not just a deeper truncation. It is truncating both vertices and edges. If you limited the set of vertices to be truncated, you'd then also have to decide which edges to truncate.polymorphic wrote:Conway Notation specification does not specify that beveling can only be done by vertex valances. I think this may have been an oversight! Beveling is actually truncation by 1/2. Therefore since Conway Notation allows for truncation by vertex valance, so should it be with Beveling. I am going to propose this change.
Rob.

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Goldberg polyhedron
Rob W. wrote:
Finally, here's a couple of Goldberg polyhedra I made using this new feature, to give them equal edge lengths.
Hi Rob & others:
Great discussion about equilateral Goldberg polyhedra. I've experimented with these on and off in recent years. Here's a SketchUp model (by Taff Goch... from 2010 as I recall), showing the I {3,0} polyhedron.
 Gerry in Quebec
Finally, here's a couple of Goldberg polyhedra I made using this new feature, to give them equal edge lengths.
Hi Rob & others:
Great discussion about equilateral Goldberg polyhedra. I've experimented with these on and off in recent years. Here's a SketchUp model (by Taff Goch... from 2010 as I recall), showing the I {3,0} polyhedron.
 Gerry in Quebec

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Re: Goldberg polyhedron
Let me try posting this again, this time with the link to the 3D warehouse:
https://3dwarehouse.sketchup.com/model. ... 32e8ebf2b1
Hi Rob & others:
Great discussion about equilateral Goldberg polyhedra. I've experimented with these on and off in recent years. Here's a SketchUp model (by Taff Goch... from 2010 as I recall), showing the I {3,0} polyhedron.
 Gerry in Quebec[/quote]
https://3dwarehouse.sketchup.com/model. ... 32e8ebf2b1
Hi Rob & others:
Great discussion about equilateral Goldberg polyhedra. I've experimented with these on and off in recent years. Here's a SketchUp model (by Taff Goch... from 2010 as I recall), showing the I {3,0} polyhedron.
 Gerry in Quebec[/quote]