hagman wrote: > On 31 Aug., 20:21, Ben Rudiak-Gould <br276delet...@cam.ac.uk> wrote: >> [...] the faces cover equal solid angles with respect >> to the center of gravity. For a center of gravity in the obvious place, I >> get that the six triangle edges should be longer than the other three edges >> by a ratio of sqrt[(15 + 9 sqrt(5)) / 10] ~ 1.87. > > I'd rather use physics and say that a thoroughly thrown die will go > through a lot of bounces and end up in a stable position of potential > energy W with probability proportional to exp(-kT/W). > Since W is proportional to height, the center of gravity should be the > center of a sphere touching all (stable) faces.
Interesting. This (obviously) gives a ratio of sqrt(3) ~ 1.73. I'm not convinced that this is a good model of the final low-energy settling-down part of the toss, but then I don't find my model very convincing either. Now I feel like doing an experiment. (But I'm not going to.)
What I don't get is why anyone would design or use a shape like this in the first place rather than one that's manifestly symmetrical.