Timeline for Odds on rolling a rhombicosidodecahedron
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2018 at 23:16 | comment | added | user44143 | Note circumradius/inradius = 1.05 for this solid (mathworld.wolfram.com/SmallRhombicosidodecahedron.html) vs 2.55 for the die in the linked arxiv article | |
| Apr 29, 2018 at 18:32 | vote | accept | TwoScoopsOfHot | ||
| Apr 29, 2018 at 17:30 | comment | added | Joseph O'Rourke | See Bill Thurston's musings in the posting to which Carlo linked. E.g., "If the projected image of a die along a certain axis is almost round, then at low energy levels it should roll more easily about those axes than about axes where the projection is bumpy, other things being equal. This suggests larger components of the phase space for these kinds of rolls, when the phase space becomes disconnected." | |
| Apr 29, 2018 at 17:22 | comment | added | user44143 | @TimothyChow, it's a poor fit for an oblong solid, but this solid is close to spherical. | |
| Apr 29, 2018 at 16:43 | answer | added | user44143 | timeline score: 9 | |
| Apr 28, 2018 at 14:42 | comment | added | TwoScoopsOfHot | "Theoretical answer", I'll try to explain. For example, on a 6-sided die (d6), the theoretical odds for each side are 1 in 6. That can be effected by people weighting the die, or rolling in a certain manner, or even by the die having some edges slightly scuffed, etc. I'm not concerned about any of those factors. So, if I had asked the question about a d6, I would be expcting the answer of 1 in 6 for each side. | |
| Apr 27, 2018 at 23:38 | comment | added | Timothy Chow | @MattF. : Your "spherical approximation" sounds like Simpson's model, which, as the paper cited by Carlo Beenakker explains in detail, is a poor fit to experimental data. It might be okay for an "adhesive surface" if by that you mean a surface that brings the die's motion to an abrupt halt the instant it first touches the surface, but this is not what most people think of as "rolling" a die. | |
| Apr 27, 2018 at 20:15 | comment | added | Timothy Chow | I agree with Carlo Beenakker; it is unclear what is meant by a "theoretical answer." As soon as you talk about "rolling," you must at minimum bring physics into the picture, and then lots of complicated considerations immediately arise. See for example this paper: researchgate.net/publication/… | |
| Apr 27, 2018 at 14:45 | comment | added | TwoScoopsOfHot | Not sure what you mean by "spherical approximation". But I'll take a stab at it.... If I take the orthographic projection and measure the surface area for each face type, the ratios there would give the answer? If so, would that not be the same as the ratio of their areas without the projection? | |
| Apr 27, 2018 at 14:33 | comment | added | user44143 | The spherical approximation should do for “a theoretical answer” or for the limit of a die thrown high above an adhesive surface. | |
| Apr 26, 2018 at 21:57 | comment | added | Carlo Beenakker | I am doubtful that simple geometry will resolve this issue; here is an article from the physics literature that addresses some of the complicating factors. | |
| Apr 26, 2018 at 21:22 | comment | added | user44143 | This can be calculated by finding the solid angles, e.g. with the tetrahedral formulas at en.wikipedia.org/wiki/Solid_angle#Tetrahedron, using the Cartesian coordinates in the article linked in the question. | |
| Apr 26, 2018 at 21:14 | history | edited | user44143 | CC BY-SA 3.0 |
added 70 characters in body; edited tags
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| Apr 26, 2018 at 21:04 | comment | added | Carlo Beenakker | related: mathoverflow.net/questions/46684/… | |
| Apr 26, 2018 at 20:03 | review | First posts | |||
| Apr 26, 2018 at 21:26 | |||||
| Apr 26, 2018 at 20:03 | history | asked | TwoScoopsOfHot | CC BY-SA 3.0 |