by **wendy** » Tue May 24, 2016 11:39 am

What you should look at is the vertex figure of the desired polytope. That's what I did.

Let's suppose you're looking at something like k8tC. Well, first of all, the edges of the tiling must be equal, so you are going to have the vertices of the verf on the same sphere. Every edge of the verf is going to become a polygon in the final structure.

k8tC is the octo-kis truncated cube, in the Conway-Hart notation. This means that we take a truncated cube x4x3o, and put pyramids on the octagons.

The vertex of the tC is given by say, (sqrt(2)+1, sqrt(2)+1, 1) for an edge of 2. Squaring these up and adding, we get 7+4q whose square root is sqrt(7+4q). The second step is to work out the sloping edge, which gives the line between (x,0,0) and (q+1, q+1, 1). We see here that 7+4q is a divisor of 17, and (x-q-1) will turn quite nasty, so that it can never be the case that an apiculated k8tC is going to work as a base for this discussion. This is the fate of most of the examples, which is why I am very picky about what to try.

The tetra-kis ambiate cube k4aC is an example that can work, since it has a good number of the vertices of the octahedral ball.

The CO (cuboctahedron = aC = CO), has nominal vertices (q,q,0) for an edge of 2. The octahedron has a vertex of 2,0,0 which gives a diam2 of 4 as well.

When we apiculate the CO, the edges run from q,q,0 to q,0,q to 2,0,0. The first edge is length 2. The other two edges are (2-q, q, 0) of length. If these edges can be scaled so that everything is a shortchord of a polygon, then it will work.

The sum of squares of the (2-q,q,0) is 6-4q + 2 + 0 = 8-4q. So the Edge-squares stand in the ratio of 8-4q : 4, or 2-q : 1. Multiplying through by 2+q, gives 2 : 2+q for the shortchords square. This is the ratio for the shortchord2 of 4 and 8. So we start with an CO of edge 8, and we can replace the squares of the {8,4} faces with octagon prisms. Likewise, the faces of the triangles get replaced with triangular prisms, and we have a new tiling of triangle and octagon prisms.

Now we turn to o3x5o, and note that the circumradius here is 2f, with a circumdiameter of 4f+4.

D2 of ID = 10.472135954999

D2 of pentagon = 2.894427191

indiam at P = 2.75276384094

elevation at P = 0.48330413655

radial slope2 = 3.12801079

edge2 = 4

ratio = 2 edge2/radial2 = 2.557536