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In this video Frank Chester claims to have discovered a new, 7-sided solid he calls the "Chestrahedron" See: http://www.youtube.com/watch?v=ql9kh7L91eg It is somewhat interesting because of its symmetry. Is this really a newly discovered form, or is it one of the 34 forms shown in the article? If it is one of the 34 forms, which one is it? Thanks --Lbeaumont (talk) 02:09, 28 July 2013 (UTC)[reply]

It's net is here - [1], looks like faces 4,4,4,3,3,3,3, 12 edges, 7 vertices, same counts as 5 cases. I'm guessing it is closest to this one

, since the 3 quads all meet at a point, so topologically a cube with one vertex removed. Tom Ruen (talk) 03:23, 28 July 2013 (UTC)[reply]

Here's it at Wolfram, [2] SO it also can be seen from a a regular octahedron augmented (or stellated) by a single regular tetrahedron, with some geometric adjustments for equal area. Tom Ruen (talk) 03:36, 28 July 2013 (UTC)[reply]

Here's the canonical form. But the topology isn't the whole point. It may be true, for all I know, that no one before Chester worked out the angles that make the faces equal in area (and the triangles regular). —Tamfang (talk) 03:40, 28 July 2013 (UTC)[reply]

Sure, the topology just shows its form, but the name Chestahedron applies specifically to the equal-area geometry variation. Here's an article [3] about its mystical properties, presumably demonstated in the 2hr video! Tom Ruen (talk) 03:45, 28 July 2013 (UTC)[reply]

I see it might be better to see it from a trapezohedron rather than cube from its symmetry (or an antiprism augmented by a pyramid). Template:Trapezohedra So any of the trapezohedra can have one of its polar vertices removed. Maybe the 3-fold symmetry version is the only one that can be made equal area? Perhaps like the diminished icosahedron this topology could be called a diminished trigonal trapezohedron? Tom Ruen (talk) 04:02, 28 July 2013 (UTC)[reply]

Most intriguing, seems to be a self-dual polyhedron! So the wikipedia article only lists pyramids and elongated pyramid, so diminished trapezohedron may be a new (unrecognized???) class of self-duals!!! Tom Ruen (talk) 04:18, 28 July 2013 (UTC)[reply]

The family is similar to gyroelongated pyramid, a family from Johnson solids, but the pyramid triangles must be merged with the antiprism triangles. Tom Ruen (talk) 04:32, 28 July 2013 (UTC)[reply]

LASTLY, (From Anton's weblink above: http://dmccooey.com/polyhedra/) As self-duals, the family sequence (1+2n faces for C_nv symmetry) can be found here: http://dmccooey.com/polyhedra/SymmetricSelfDuals.html 7F,C_3v:[4] 9,C_4v:[5] 11,C_5v:[6], 13,C_6v:[7], 15,C_7v:[8]. (It shows another (new to me) sequence of self-duals as monotruncated bipyramid, starting with 1+3n faces with C_nv symmetry.) Tom Ruen (talk) 05:11, 28 July 2013 (UTC)[reply]

I added a quick reference to this polyhedron at Diminished trapezohedron#Chestahedron. Tom Ruen (talk) 00:09, 29 July 2013 (UTC)[reply]

Excellent, what symmetry corresponds to the Chestahedron? Is it C4v? If we can decide this perhaps it can be added to your mention of the shape. Should you mention that it is a Heptahedron? Thanks! --Lbeaumont (talk) 12:48, 29 July 2013 (UTC)[reply]

C3v. —Tamfang (talk) 19:07, 29 July 2013 (UTC)[reply]

I wanted to see for myself how difficult it is to specify the shape having equal area faces, as in the Chestahedron. Beginning with the form of the net, See: http://www.frankchester.com/2010/chestahedron-geometry/ I chose to use equilateral triangles with sides of length 2. These have an area of √3. So the problem becomes to specify a kite with area √3 and one side length 2. The formula for the area of a kite is p x q/2 see: Kite (geometry)#Area But there are an infinite number of kites with a side of length 2 and an area of √3. So either I am missing a constraint or there is a family of such solids. Thanks for any help you can provide with this. --Lbeaumont (talk) 16:15, 29 July 2013 (UTC)[reply]

Intuitively I think it's a rhomb, but I can't say why. —Tamfang (talk) 19:07, 29 July 2013 (UTC)[reply]

You have to work from the 3D geometry. There's only a single parameter, the dihedral angles between the base and the triangles. The apex can be computed then simply by assuming planar quad faces. Tom Ruen (talk) 19:38, 29 July 2013 (UTC)[reply]

This page lacks a polyhedron that is known as Heptaeder in German. It has 4 triangles and 3 squares, 12 edges and 6 vertices. It is described in the following sources: Uni Göttingen, Uni Bielefeld
According to the last one it is related to the roman surface. mate2code 23:17, 1 January 2014 (UTC)[reply]

User:Pianostar9

You have mistakenly changed the name of a heptahedron from "The Great Sethahedron" to "Chestahedron". I am therefore going to revert the changes that you made.

The long and short edges of The Great Sethahedron are in golden ratio. This is not the case in the Chestahedron.

The angles in The Great Sethahedron are whole numbers (60º, 132º, and 36º). All of these angles can be constructed using straightedge and compass construction techniques. In contrast, the angles in the Chestahedron are not whole numbers and cannot be constructed with that technique.

The Great Sethahedron is drawn accurately. However, the Chestahedron is incorrectly drawn and that error converts the Chestahedron into a figure that does not have seven sides that have the same area.

Scott Gregory Beach (talk) 07:55, 29 January 2020 (UTC)[reply]

If that's the case, you should respond to the comments on your YouTube video about it, since that's where I received this information, which seemed accurate after looking into it. Also, it might be a good idea to try to contact the author of the Chestahedron about it. Pianostar9 (talk) —Preceding undated comment added 22:05, 29 January 2020 (UTC)[reply]

If you check the source, it has well under 1000 views, and the channel has only 3 subscribers. Pifvyubjwm (talk) Pifvyubjwm (talk) 15:08, 8 August 2022 (UTC)[reply]