Talk:Snub cube

Why's it also called a snub cuboctahedron? Professor M. Fiendish 04:37, 23 August 2009 (UTC)

With an alternative construction, you take a cuboctahedron, cut all the edges, and alternate vertices, you can twist the 6 square and 8 triangular faces until gaps remain that can be filled by pairs of triangles, like shown in these images (swapping blue and yellow triangle colors in the second image): Tom Ruen (talk) 04:46, 23 August 2009 (UTC)
 * Cuboctahedron.png Snub hexahedron.png

Coordinates
I'm having trouble getting
 * $$C_3=\sqrt{\frac{1}{3}+\frac{1}{12c_3}+\frac{c_4}{12}}\approx1.14261$$
 * $$c_3=\sqrt[3]{199+3\sqrt{33}}$$
 * $$c_4=\sqrt[3]{199-3\sqrt{33}}$$

to come out to 1.14261... The numerical values seem correct (they give a snub cube with unit length edges), so I suspect there may be a typo in the formula somewhere. Can anybody spot it? - IP71.127.175.197, 15:29, April 25, 2011


 * The same goes for edge length &alpha;.


 * $$\alpha = \sqrt{\frac{4}{3}-\frac{8\sqrt[3]{4}}{3\beta}+\frac{4\beta}{3}}\approx1.60972$$
 * $$\beta = \sqrt[3]{13+3\sqrt{33}}$$
 * doesn't give &alpha; = 1.60972.


 * I found another formula which does give the correct &alpha;:


 * $$\alpha = \sqrt{\frac{4}{3}-\frac{32}{6\sqrt[3]{2}\beta}+\frac{6\sqrt[3]{2}\beta}{9}}$$


 * I don't know if this is the official foruma (I'm sure it isn't). I got it by solving for &alpha; in &alpha;6-4&alpha;4+16&alpha;2-32=0, extracting out common factors equal to &beta; and simplifying the rest by hand. I'm putting this in the article.


 * I can't find an easy replacement for C3, so I'm just going to leave that for the moment and use 1/&xi;/&alpha; as the value. --Zom-B (talk) 17:52, 23 May 2011 (UTC)

Names for the chiral versions


Does someone know if there are common names to distinguish the chiral versions of the snubs and their duals? (I ask this question here, but also mean it for Pentagonal icositetrahedron, Snub dodecahedron and Pentagonal hexecontahedron.)

Currently the names of the green files use "clockwise" and "counter-clockwise" in a way, where duals have the same direction:

On http://dmccooey.com/polyhedra I found a convention that uses "left" and "right" in a way that duals have opposite directions:

Whether to use "red-based" or "yellow-based" names is a random choice, and I expect that both conventions exist. The more important question is if duals have the same or opposite directions. Watchduck (quack) 20:40, 26 July 2018 (UTC)

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 * Rhombicdodecahedron.jpg