by **wendy** » Sun May 01, 2016 11:56 am

If you suppose that you are taking say, a tD of edge E1 (or E1, E2), then this is well in the range of calculations, but there are some useful considerations to be made. Been doing this for years.

Horohedra and Bollohedra (aka pseudohedra), are simply polyhedra. It's not really easy to do it for 'uniform polyhedra', as I explained to Richard Klitzing. What you need is a construction, such as the stott-dynkin form. Professor Johnson makes an error when he supposes that the elements of a dynkin symbol are rational numbers: they are real numbers.

The dynkin - stott matrices works in all geometries, including euclidean geometry. But to make this happen, the proper scale to work in is euclidean chords (ie distances along chords), rather than 'straight lines' (ie measures of arc). This makes the curvature issue disappear, and the sizing of things then introduces curvature.

The spreadsheet i wrote for R Klitzing, is pretty much right on the nail on what is needed here, so i shall discuss this.

1. The most useful number of a polygon is its shortchord, p = 2 cos(pi/P). P = 2 acos(p/2). You find the dha = pi(1/2 - 1/P). We normally write this in terms of the square of p though.

2. The diameter2 of a polyhedron (P,Q) is d = 2(4-q)/(8-2p-2q) P,p here is 3,1 4,2, 5,2.618033, 5/2,0.381967 6,3 U,4

The diameter2 of a polygon (P) is 2(2)/(4-q).

3. It follows directly that the d2 of a polychoron is 2(8-2q-2r)/(16-4p-4q-4r+pr) &c.

The denominator is the 'schlafli determinate' of the polytope, and the d2 of a figure 2 * schlafli determinate of vertex figure / sd of polytope, the prism product gives a regular product, so eg o---o---x---o vf = o---o sd=3, and o sd = 2, pt = o---o---o---o = 5, so it's 2 * 3 * 2 / 5 = 12/5 = 2.4.

All of the above work with irrational polytopes, which is why i have W(x) as a general infinite polyhedron. when x=4 you get the horogon, or horosphere, when W(x) > 4 it's a bollogon.

Since one is generally in a tiling, not so interested in the radius of a polytope, but the edge in a sphere whose edge is 4, the point to note here is that E+4/R = 4.

A polyhedron like x5x3o of a given size, would imply some x5o3oWo, where the indiameters over the pentagons/decagons are identical, and this allows one to directly find W, and from this, the dihedral angles and assorted edge lengths associated with both of these figures.

Areas are found by the spherical excess theorm. I don't have a volume formula, though.