Toroidal d7
Jul. 10th, 2007 03:52 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
jackThe last post described why a d7 has to have one rotational symmetry, ie. be arranged circularly rather than spherically.
But the maths only applies on the surface of a *sphere*. On the surface of a *torus* it's different:
Look. Seven regions. And look at the symmetries. There's one cycle that slides each block on to the one at the end, repeating after seven times. And there's one nearly perpendicular that slides a block onto a block above it, that repeats after three times. That's not a literal rotation, but it satisfies it morally.
But unfortunately, it's a property of the torus, not the map. You can make a d-anything the same way, without regard to the number. Draw a line that wraps the torus vertically (i) times and horizontally (j) times. (This one has (i,j)=(3,1).) That forms one (or more) strips that wind round the donut. Divide the strip into n equal parts. And lo, n-fold symmetry, together with i-fold symmetry.
Besides, it would be hard to *build* a torus die. Obvious throwing a donut doesn't land at a random point on its surface. I can think of two cheats, make a donut with a slidable surface (like a rubics cube, sort of), so the inner bit can come out to the outer bit. Then it's random. Or switch the central hole for the interior of the band to make another die, and roll both to get two coordinates, which define a random point on it. But you need a ruler or something, there's no way to mark them, let alone divide them into faces.
But the maths only applies on the surface of a *sphere*. On the surface of a *torus* it's different:
Look. Seven regions. And look at the symmetries. There's one cycle that slides each block on to the one at the end, repeating after seven times. And there's one nearly perpendicular that slides a block onto a block above it, that repeats after three times. That's not a literal rotation, but it satisfies it morally.
But unfortunately, it's a property of the torus, not the map. You can make a d-anything the same way, without regard to the number. Draw a line that wraps the torus vertically (i) times and horizontally (j) times. (This one has (i,j)=(3,1).) That forms one (or more) strips that wind round the donut. Divide the strip into n equal parts. And lo, n-fold symmetry, together with i-fold symmetry.
Besides, it would be hard to *build* a torus die. Obvious throwing a donut doesn't land at a random point on its surface. I can think of two cheats, make a donut with a slidable surface (like a rubics cube, sort of), so the inner bit can come out to the outer bit. Then it's random. Or switch the central hole for the interior of the band to make another die, and roll both to get two coordinates, which define a random point on it. But you need a ruler or something, there's no way to mark them, let alone divide them into faces.
no subject
Date: 2007-07-10 03:22 pm (UTC)Now just identify opposite edges of that rhombus to make your torus, and you're done. That's not just a seven-face solid on a torus, it's one in which all the faces are regular hexagons!
no subject
Date: 2007-07-10 03:26 pm (UTC)That could be a constraint. Do you know what other tilings you can draw that torus on? :)
no subject
Date: 2007-07-10 04:47 pm (UTC)Szilassi Polyhedron
Date: 2007-08-07 08:04 am (UTC)It's completely useless as a D7, however
Re: Szilassi Polyhedron
Date: 2007-08-07 02:04 pm (UTC)