Talk:List of uniform polyhedra

message: This is a first pass article, including the complete list of 75 uniform polyhedra, 11 uniform tessellations, and sampling of infinite sets of prism and antiprism.
Tom Ruen 00:01, 7 October 2005 (UTC)
replies: text: I think it would make sense to include Skilling's great disnub dirhombidodecahedron  (Phil. Trans. A 278 (1975), pp 111-135).  In view of the inclusion of tilings in this page and of uniform tilings with hollow tiles in Table 8 of Uniform Polyhedra (Phil. Trans. A 246 (1954), pp 401-450), there is a logical case for covering all the uniform tilings with hollow tiles, apeirogons and zigzags listed by Grünbaum, Miller and Shephard (in The Geometric Vein: The Coxeter Festschrift, pp 17-64), but that's a large collection, without proof of completeness, and if you include them then there's the question of infinite uniform polyhedra in three dimensions where I don't know of a good comprehensive list of the known polyhedra for any definition you might choose.  Joseph Myers 19:42, 8 October 2005 (UTC)
replies: text: Is completeness a requirement for listing a valid class of uniform polyhedra here? If so, it should be stated which such classes are being left out, otherwise it implicitly suggests they are not valid. Cutelyaware (talk) 05:11, 23 July 2015 (UTC)
replies: text:  I rewrote the intro a bit. It probably can be improved more. Tom Ruen (talk) 05:39, 23 July 2015 (UTC)

text: P.S. On "great disnub dirhombidodecahedron", very interesting. Going beyond 2 faces/edge is fun to consider, but I don't think you can call that a polyhedron. Maybe a "Polyhedron network" or something else since it would enclose multiple volumes. Such an ennumeration has a place in geometry, but not here, except as a link if such an article is writen. Tom Ruen 04:24, 15 October 2005 (UTC)
replies: text:  [More recently] I've been rethinking my rejection of this "polyhedron". Initially I rejected it as a true polyhedron for having more than 2 faces per edge. On RETHINKING, technically an even number of faces per edge might be considered as a topological polyhedron with geometrically overlapping edges. Well, I'll have to think further on this! Tom Ruen 20:42, 5 December 2005 (UTC)
replies: text:  It is quite valid for a polyhedron to contain coincident edges that do not interact. You just need to be specific about the topology. That still describes a simple manifold and is a very different situation from the idea of a polyhedron in which at least one edge is explicitly adjacent to more than two faces. Cutelyaware (talk) 05:11, 23 July 2015 (UTC)
replies: text:  Sure, its a matter of how you want to count, but all degeneracies are tricky, and you can understand why some want to exclude them. Tom Ruen (talk) 05:39, 23 July 2015 (UTC)
thread_title:  Start article

message: The "vertex figure" for W79 looks like the polyhedron itself. ?? --Anton Sherwood 06:43, 7 January 2006 (UTC)
thread_title:  Vertex figure error?

message: * I don't fully understand the table, but it looks to me like the 2 7 | 3 entry was missing. I sketched one in. If it belongs, and if there is no image of the "bizarro football" to be found, I can cook one up. --Unzerlegbarkeit (talk) 16:56, 8 August 2008 (UTC)
replies: text: There's infinitely MANY missing from the hyperbolic tilings. There's a newer article Uniform tilings in hyperbolic plane and Uniform tiling with a sampling of the hyperbolic tilings. They were added here to show the relations of vertex figures, but probably should be removed now, since the comparisons are given elsewhere as well. Tom Ruen (talk) 00:02, 10 August 2008 (UTC)
replies: text:  Any list of hyperbolic tilings ought (imho!) to include 433, 542 and 732, because each of these is minimal in the sense that reducing any of the numbers results in a tiling of E2 or S2.  For the same reason I consider such a list sufficient for illustration; or do higher finite numbers bring more interesting properties? —Tamfang (talk) 22:44, 13 August 2008 (UTC)
replies: text: I think we could have *433, *542 and *732 for hyperbolic. Possibly *832, but not higher. Professor M. Fiendish, Esq. 13:44, 6 September 2009 (UTC)
thread_title:  hyperbolica 

message: I ADDED IN THE OCTAGRAMMIC PRISM BUT THEN THE PENTAGRAMMIC ANTIPRISM WAS GONE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASE FIX IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!  — Preceding unsigned comment added by 99.185.0.100 (talk) 21:00, 5 January 2014 (UTC)
replies: text: You were missing a "|-" symbol for a new table row, so the antiprism rows ended up on the far right. Tom Ruen (talk) 22:23, 5 January 2014 (UTC)
thread_title:  OH MY EDITS! 

message: How do you define density for hemipolyhedra? Since there are faces passing through the centre. Double sharp (talk) 13:41, 15 March 2014 (UTC)
thread_title:  density 

message: I corrected some values, substituting the ones from Coxeter's paper. It sorts wrongly though (3, 37, 38, 4...). T_T
In particular, the hemipolyhedra and even-faced nonorientables don't really have a well-defined density, so I replaced them with "--". I think that's what's messing with the sorting.
A few other corrections, based on Coxeter's paper:
* Great truncated cuboctahedron (quitco) – density 1, not 7. We expect 7, as that is the density of the spherical triangle (2 3 4/3), but the vertex figure is an obtuse triangle. A distorted version with an acute vertex figure would indeed have density 7, but as we make the vertex figure's angle grow larger, the eight hexagonal faces pass through the centre and out the other side, and the density now becomes |7 − 8| = 1. Now it has the same density as the triangle (2 3 4) sharing the same great circles. You can see the effect of this in the small triangular holes in the polyhedron.
* Truncated dodecadodecahedron (quitdid) – density 3, not 9. This is essentially the same idea: (2 5/3 5) has density 9, but the vertex figure is obtuse: now we have twelve decagonal faces passing through the centre in the distortion, giving a density of |9 − 12| = 3, the same as (2 5/2 5) which shares the same great circles. Again, you get small triangular holes.
* Snubs – in three cases (great icosahedron, small retrosnub icosicosidodecahedron, great retrosnub icosidodecahedron) the vertex figure is so distorted that it becomes a pentagram/hexagram instead of a pentagon/hexagon. So now we need to subtract the number of snub faces from the density of the Schwarz triangle involved (same idea as the previous two cases), giving |5 − 12| = 7, |22 − 60| = 38, and |23 − 60| = 37. Now the density is the same as some reflex-angled triangles which share the same great circles, respectively: (2/3 3 3) or (2 3 3/4); (2/3 3 5/2), (2 3/4 5/3) or (2 3 5/7); (3 3 5/8) or (3 3/4 5/3).
It's interesting that these cases are exactly the ones where strict application of the Wythoff symbol gives strange results for the vertex figure. I think that means they have to be corrected based on these considerations, using the Schwarz triangles that actually have the same density. In particular, Coxeter writes that for quitco and quitdid, all the faces are retrograde save the aforementioned hexagons and decagons; to my mind that implies true vertex configurations of 4/3.6.8/5 and 4/3.10.10/7. As for the reflex-angled cases, first we'd have to define what Schläfli symbols {n} with n ≤ 1 mean; hopefully they then produce the geometrically correct results! Double sharp (talk) 15:41, 16 March 2015 (UTC)
replies: text: P.S. This is presumably why Stella gives quitco and quitdid negative density values (see above). Double sharp (talk) 16:18, 27 July 2019 (UTC)
thread_title:  Density sorting