quickfur: You shoud check out, eg

http://hddb.teamikaria.com/wiki/Wythoff_polytope where the definition of 'wythoff polytope' is exactly the 'position polytope' of a point in a kaleidoscope, and that you can compute the metric properties (like R_0 etc), directly from the coordinates.

I wrote a spreadsheet based on the theory given on that page, you can fetch Richard's more recent version from his web-page. Basically, you can write, eg o3o5o as a matrix, and then take its inverse.

- Code: Select all
` Dynkin Stott Product`

( 2 -1 0 ) ( 3-f 2 f ) 4-2f 0 0

( -1 2 -f ) * ( 2 4 2f ) = 0 4-2f 0

( 0 -f 2 ) ( f 2f 3 ) 0 0 4-2f

========= ================= ==================

1/ 2 2/(4-2f) 1/(4-2f)

Schläfli function for [3,5] is 4-2f

3 5 5 3

1 f+1 f+1 1 Shortchord^2

1 2 3 4-2f 1 2 3-f 4-2f Schläfli value

The Dynkin matrix is simply the dynkin symbol in matrix form. The main diagonal is all '2', the value d_ij is the cosine between the normals to the plane (ie of the supplement). So (2, 0), (5/2, -v), (3 , -1), (4, -q), (5, -f), (6, -h), (7, -1.801937736) &c.

The Stott-matrix corresponds to S_ij = v_i ̇· v_j, where v_i is an axial vector (like x3o5o or o3x5o, or o3o5x) To preform the dot product over a dynkin graph, you write the vector (eg x3x5o -> (1,1,0). Multiply this by S_ij and then dot it with the second vector, So the norm of (1,1,0) is

S_ij (1,1,0)_i (1,1,0)_j = (5-f, 6, 0)_j (1,1,0)_j gives (11-f). Because there is a factor 2/(4-2f), must be taken to account too.

Let's see. 11 = 111.11 base F. 11-f is 111.11 - f = 10f.ff1 Divide by 2-f = 0.1 gives 10ff.f1 base phi-square.

And that's the radius-square of the x3x5o.

In the spreadsheet, when we calculate the height for the lacing, eg for a polytope xo3oo5ox, the process is to calculate the norm of (1,0,-1), which gives the "drift" across the base symmetry. Because the implied edge-length is 4, the height is H² = 4L² - D², the matrix dot product gives the drift's square.

In short, the notation i designed was specifically to allow vector calculus direct. This is a deep and magical thing.

In fact, i spent a few months 'spotting' stott matrices for groups like o3o3o3o3o4z This involves, for example, calculating the vertex figures of say x3o3ox3o3o4z, and then calculating its diameter. From this, one gets a_ii + a_jj + 2 a_ij. Since we get a_ii and a_jj from the schlafli function, the a_ij come directly.

Schläfli ValueIf you suppose that p, q, r, &c stand for the square of the shortchords of P, Q, R, then eg (ok i don't know the trig, just the values. Most of this work was done when the fashion was a four-function calculator (+, -, * and /). The values are simply memorised.

2->0, 3-> 1, 4-> 2, 5-> 2.61803398875, 6->3, 7-> 3.2467976037, 8-> 3.41421356238, 9-> 3.5320888806, 10-> 3.61803398875, 12 -> 3.73205080757

The diameter of a circum-circle around a triangle 1:1:A is D = 4/(4-A), where all values are squares of lengths. (recall, no sqrt key)

The diameter of a polygon from its 'p' value is then 4/(4-p).

The diameter of a polyhedron-vertex figure for p,q is 4p/(4-q) ie a polygon of this diameter. From this one finds

D(p,q) = 4 / (4- 4p/4-q)) = (4-q) / (4-p-q)

Repeating for a polychoron, *by moving p,q to q,r, and adding in a new p, we get

D(p,q,r) = 4(4-q-r)/(16-4p-4q-r4+pr)

If one has one's eyes open, the denominator is the same for (p,q,r) as it is for (r,q,p).

D(p,q,r,s) = 2. F(q,r,s) / F(p,q,r,s)

Where: F() = 2; F(p) = 4-p ; F(p,q) = 8-2p-2q), F(p,q,r) = 16-4p-4q-4r+pr

and F(p,q) = q.F(p) - 2 F(); F(p,q,r) = r F(p,q)-2F(p), F(p,q,r,s) = s F(p,q,r)- 2 (p,q) &c.

In short, it's kind of like a continued fraction thingie, where one iterates t(n+1) = p t(n) - 2t(n-1).

This also works for the removal of any node. For example, the D for something like oPxQoRo is 2 F(o) F(oRo) / F(oPoQoRo)

= 2 F() F(r) / F(p,q,r).

So you really don't need to derive the matrix to find these figures. It's only when you do something fancier.