Ah, this makes a lot more sense, then.
{ ... } is a "true" Schläfli symbol, and "tiling" then means just euclidean polytope, rather than flat 2D fabric.
I for one use here the general term "tesselation" instead, while restricting "tiling" for 2D.
But yours on the other hand is 3D, thus it rather describes a "honeycomb".
In fact you were deeling with {8, oo, 8/3} = x-8-o-oo-o-8/3-o, with {8, oo, 8} = x-8-o-oo-o-8-o, resp. with x-8-o-oo-o-3-o. And you were pointing out, that even so {8, oo} clearly is a hyperbolic (2D) tiling, when taken without multiwrappings, it well can be given a representation within euclidean 3D space, i.e. a polyhedron with regular octagons only, infinitely many at all the vertices. And being an infinite structure, that one would never close again, that is at least not within finite reach. Furthermore you pointed out, that right this non-closing fact leaves a degree of freedom, e.g. the dihedral angle between the octagons. Accordingly you are free to impose an edge figure being a {8/3}, a {8}, resp. a {3} there. Or speaking in dihedral angles again, these would be then 135°, 45°, resp. 120°.
If labeling the octagon vertices alphabetically, and using A and B for the ones at the connecting edge of 2 adjoining octagons, then one might consider the lacing distance between the C Vertices, which provides a different measure for that degree of freedom. Speaking of octagons, one calculates easily that, when the dihedral angle would be given as alpha, that C-C-distance then would be sqrt(1-cos(alpha)).
Esp. we'd get here
- sqrt(1-cos(135°)) = sqrt((2+sqrt(2))/2)
- sqrt(1-cos(45°)) = sqrt((2-sqrt(2))/2)
- sqrt(1-cos(120°)) = sqrt(3/2)
Thus starting with the incidence matrix for the abstract polytope {8, oo} (i.e. independent of its hyperbolic or eucliedean realisation)
- Code: Select all
x-8-o-oo-o (N,M -> oo)
. . . | 8N | M | M
-----------+----+-----+---
x . . | 2 | 4NM | 2
-----------+----+-----+---
x-8-o . | 8 | 8 | NM
one derives accordingly the incidence matrix for that first of your honeycombs:
- Code: Select all
x-8-o-oo-o-8/3-o (N,M,K,L,P -> oo)
. . . . | 2NLP | MK | 4MK | 8M
-----------------+------+-------+-------+-----
x . . . | 2 | NMKLP | 8 | 8
-----------------+------+-------+-------+-----
x-8-o . . | 8 | 8 | NMKLP | 2
-----------------+------+-------+-------+-----
x-8-o-oo-o . | 8L | 4LP | LP | 2NMK
Sure, describing just abstract polytopes, the same matrix likewise describes the second of your honeycombs too, just omit that "/3" part in the leftmost descriptive column.
For the last one you'd then get similarily:
- Code: Select all
x-8-o-oo-o-3-o (N,M,K,L,P -> oo)
. . . . | 8NLP | 2MK | 3MK | 6M
---------------+------+--------+--------+-----
x . . . | 2 | 8NMKLP | 3 | 3
---------------+------+--------+--------+-----
x-8-o . . | 8 | 8 | 3NMKLP | 2
---------------+------+--------+--------+-----
x-8-o-oo-o . | 8L | 4LP | LP | 6NMK
Here note, that this "oo" (infinity symbol) only servs in the sense of a numerator, resp. in the sense of an abstract polytope. But considering the corresponding vertex figure of those "cells" x-8-o-oo-o each, it becomes clear that those surely ask for multiwrappings in the sense of some {n/d} polygram.
To spot onto that we first consider a general flat polygon or polygram. If that one has an vertex angle phi, then the corresponding shortchord would be 2 sin(phi/2). Reverse engineering we derive from our above calculated shortchords CC, that this angle phi can be calculated according to phi = 2 arcsin(CC/2). Accordingly we derive in these three cases
- phi_{oo, 8/3} = 2 arcsin( sqrt((2+sqrt(2))/2) /2) = 81,578941881850578219234172802659°
- phi_{oo, 8} = 2 arcsin( sqrt((2-sqrt(2))/2) /2) = 31,39971480991904220976984000837°
- phi_{oo, 3} = 2 arcsin( sqrt(3/2) /2) = 75,522487814070076121228965200873°
On the other hand we have for some "normal" polygram {n/d} that this vertex angle phi is being calculated as phi = 180°.(1-2d/n). Sure, here we always have q=n/d to be a rational number. But above we supposedly get real numbers r instead. Those would be calculated in these 3 cases according to r = 360°/(180°-phi)
- r_{oo, 8/3} = 3,6577538067903974646494807498297
- r_{oo, 8} = 2,4226063869225328714883422611807
- r_{oo, 3} = 3,4457175756572188802726257456335
I.e. those "oo" rather should be some numbers lim_(n,d -> oo) n/d with limiting values r. And it is that these different real numbers r were refered to by Wendy's A, B, and C.
--- rk