Talk:Snub dodecahedron

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Polyhedra

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Is the snub dodecahedron really obtained by rotating the pentagonal faces of the rhombicosidodecahedron, rather than those of some other expanded dodecahedron? According to Mathworld, the snub dodecahedron and the rhombicosidodecahedron with unit edge length have different circumradii: about 2.15 and 2.23 respectively. --Zio illy (talk) 15:43, 6 December 2007 (UTC)[reply]

It's a qualitive description. If you want to rewrite it more clearly, please do. Tom Ruen (talk) 23:27, 6 December 2007 (UTC)[reply]

Ok. I took away the sentence about the rhombicosidodecahedron (as in the snub cube) and re-labeled the corresponding image. IMHO this should provide a good description. --Zio illy (talk) 00:58, 7 December 2007 (UTC)[reply]

I don't understand why you said no rhombicosidodecahedron, since that's what an expanded dodecahedron is, but okay. Tom Ruen (talk) 01:10, 7 December 2007 (UTC)[reply]

Well, unless you require the expansion process to create regular polygons (a regular expansion, I believe), a rhombicosidodecahedron is an expanded dodecahedron and not the converse. Actually, I left the image of the rhombicosidodecahedron, in place of the correct expanded dodecahedron, just for the sake of qualitative description. If you feel like to draw a new one with rectangles in place of squares, you're very welcome. ^^ --Zio illy (talk) 02:27, 7 December 2007 (UTC)[reply]

Ah, I think the issue is a uniform rhombicosidodecahedron has square faces, while a more generalized rhombicosidodecahedron could have rectangle faces. But if you take a uniform rhombicosidodecahedron, delete the squares, but allow the triangle-pentagon connections as rotation points, the figure can smoothly transform until each gap can be filled by two equilateral triangles. The distance of the faces from the center will change as the rotation is performed. Tom Ruen (talk) 02:46, 7 December 2007 (UTC)[reply]

Oh! Right, this way it works. Nice job! --Zio illy (talk) 18:26, 7 December 2007 (UTC)[reply]

Okay, glad you prodded me to write it up better, and I added an animation at rhombicosidodecahedron using Stella (software). For the snub form, I'll have to do my own computing to make a pretty animation - it would be cool to show a icosidodecahedron corner-point rotate to create gaps for a ccw snub, then to the square gaps for a rhombi-form, and then to the cw snub, and then back again to the icosidodecahedron! Someday! Tom Ruen (talk) 22:47, 7 December 2007 (UTC)[reply]

I'm afraid that you've been anticipated... Double sharp (talk) 09:41, 25 May 2012 (UTC)[reply]

This is probably the hardest Archimedean solid to make! (Korthals Altes didn't show a real paper model on his website.) 4 T C 14:26, 29 January 2010 (UTC)[reply]

I've tried to check if the given Cartesian coordinates for the snub dodecahedron were correct or not, and I believe they are not. I may be probably wrong, but the radius for the circumsphere is different for some vertex. All other references to these coordinates in the net redirect us to the Wikipedia. IngGraph (talk) 21:58, 26 April 2014 (UTC)[reply]

Sorry, I was the one who was mistaken.IngGraph (talk) 23:37, 28 April 2014 (UTC)[reply]

The way it reads seems to suggest that we permute the signs of the coordinates and take the even permutations; given that there are only 5 sets specified this seems like it determines only 20 of the 60 vertices. Obviously I'm missing something (some reflection?) - it would be great if we could make that more clear. — Preceding unsigned comment added by 68.134.53.113 (talk) 16:32, 19 February 2017 (UTC)[reply]

Cartesian coordinates section question

“This snub dodecahedron has an edge length of approximately 6.0437380841” - could you explain it? What is the exact value? What are the coordinates for the edge length = 2 (or 1)? And: what are the dihedral angles? What is the edge central angle? The midsphere and circumsphere radii? The insphere radius of the dual? The distance centre-to-face of both types? The data are hard to find in the Net (some of the missing info is provided by the German Wiki), and it is almost impossible to find the exact values of some of the above. And perhaps some info on how to obtain the exact values of the above mentioned data? I've found a paper on detailed parametres of the snub cube (and how to compute them from the scratch) but not on the snub dodecahedron... So, help please if you can! 31.11.242.188 (talk) 23:55, 1 January 2015 (UTC)[reply]

Looks like tough to compute exact solutions. But given the edge length for a given coordinates, you can scale the coordinates to match other edge lengths like 1 or 2. Also The Geometrical Foundation of Natural Structure: A Source Book of Design, p 96 has lots of statistics, almost most are decimal based. Tom Ruen (talk) 00:23, 2 January 2015 (UTC)[reply]

Coordinates

There is something wrong with the coordinates. I tried them, and they all lie in the same four octants of the coordinate system.

These coordinates gave the expected result: http://dmccooey.com/polyhedra/RsnubDodecahedron.html

phi = (1 + sqrt(5)) / 2
x = cbrt((phi + sqrt(phi-5/27))/2) + cbrt((phi - sqrt(phi-5/27))/2)

C0  = phi * sqrt(3 - (x^2)) / 2
C1  = x * phi * sqrt(3 - (x^2)) / 2
C2  = phi * sqrt((x - 1 - (1/x)) * phi) / 2
C3  = (x^2) * phi * sqrt(3 - (x^2)) / 2
C4  = x * phi * sqrt((x - 1 - (1/x)) * phi) / 2
C5  = phi * sqrt(1 - x + (phi + 1) / x) / 2
C6  = phi * sqrt(x - phi + 1) / 2
C7  = (x^2) * phi * sqrt((x - 1 - (1/x)) * phi) / 2
C8  = x * phi * sqrt(1 - x + (phi + 1) / x) / 2
C9  = sqrt((x + 2) * phi + 2) / 2
C10 = x * sqrt(x * (phi + 1) - phi) / 2
C11 = sqrt((x^2) * (2 * phi + 1) - phi) / 2
C12 = phi * sqrt((x^2) + x) / 2
C13 = (phi^2) * sqrt(x * (x + phi) + 1) / (2 * x)
C14 = phi * sqrt(x * (x + phi) + 1) / 2

The coordinates of the right snub dodecahedron are the even permutations of the following triples (with ± signs added) with the respective parity of plus signs:

(C2, C1, C14)    even
(C3, C4, C13)    odd
(C0, C8, C12)    even
(C7, C6, C11)    even
(C9, C5, C10)    odd

Probably the current coordinated could be corrected, by being more precise about parity. Maybe wants to do that. Otherwise I will add the above coordinates to the article some time later. Watchduck (quack) 01:49, 13 January 2018 (UTC)[reply]

It looks like they've been there since the original article in 2006. [1] Perhaps it should be checked if those values are the same as what's up now? Tom Ruen (talk) 02:23, 13 January 2018 (UTC)[reply]

The values are different. Those in the article are for edge length 6.043, while the ones above are for edge length 1. At least that's what I calculated:

V0 = (C2, C1, C14) and V2 = (-C2, -C1, C14). V0 and V2 are an edge. So the length is sqrt( (2*V0)^2 + (2*V2)^2 ).

The symbolic result I got from SymPy was long and annoying, but converted to a float it was 1.0.

The other polyhedra I checked on this page also have coordinates for edge length 1. Watchduck (quack) 11:59, 13 January 2018 (UTC)[reply]

Derivation of the snub cube and snub dodecahedron.

Derivation of the snub cube and snub dodecahedron. Regarding one vertex of the icosahedron. Replace one of the five triangles with a square, results in a snub cube. The same is done by replacing one of the five triangles with a pentagon results in a snub dodecahedron. It's that simple, if you regard one vertex of the icosahedron! 87.209.132.7 (talk) 09:16, 4 November 2023 (UTC)[reply]