For a fixed dimension n, the total number of CRF n-polytopes with small 2-faces (e.g. 50-sided polygons at the largest) is finite. Any finite set of positive numbers cannot approach 0 . So, if we want to find ditopal angles approaching 0, without increasing the dimension, we must consider polytopes with large 2-faces.

The 3D ones are very familiar: prisms and antiprisms. Given edge length 1, an n-gon prism has height 1, and an n-gon antiprism has height

h = √3/2 √[(2 + 4 cos 180°/n)/(3 + 3 cos 180°/n)]

= √[(1/2 + cos 180°/n)/(1 + cos 180°/n)]

= √[1 - 1/(2 cos 90°/n)²]

which increases with n, approaching √3/2.

- higher-dimensional CRFs 1.png (29.41 KiB) Viewed 26976 times

The 4D ones are also known. Let's consider their "poke sections", perpendicular to an n-gon. The section of an (n,m)-duoprism is just an m-gon, with edge length 1. The section of an antiprismatic prism is a rectangle, with lengths 1,h,1,h. The section of a biantiprismatic ring, or antifastegium, is an isosceles triangle, with lengths 1,h,h.

- higher-dimensional CRFs 2.png (13.76 KiB) Viewed 26976 times

Here I claim (tentatively) that these are the only possibilities for CRF 4-polytopes with large 2-faces.

Now to 5D. Clearly, any CRF 3-polytope with edge length 1 will be a section of a valid CRF 5-polytope; consider the prism product of an n-gon and that polyhedron. Also the polygons shown above, extruded into prisms with height 1, will be valid sections. Many of the CRF polyhedra have variants where some edges have length h; these may or may not be valid sections.

- higher-dimensional CRFs 3.png (114.73 KiB) Viewed 26976 times

How exactly do these correspond to 5-polytopes? Well, each vertex (x,y,z) represents an n-gon with vertices at either

(R cos (2k)180°/n, R sin (2k)180°/n, x, y, z)

or

(R cos (2k+1)180°/n, R sin (2k+1)180°/n, x, y, z)

for various integers k, where R = 1/(2 sin 180°/n) is the circumradius of an n-gon. If two vertices in 3D are connected by an edge with length 1 (depicted blue), then the corresponding n-gons should have the same orientation; both use '2k', or both use '2k+1'. If two vertices in 3D are connected by an edge with length h (depicted orange), then the corresponding n-gons should have opposite orientations; one is the dual of the other.

Is a polyteron made in this way guaranteed to be CRF? It's not clear to me.

...Isn't this construction essentially a "lace hyper-city"?

Anyway, it doesn't look like any of these have angles approaching 0 (though I don't claim to have found all of them, and that pentagonal cupola has a very small angle, from 6.1655° to 10.8123° depending on n).

For 6D, I thought of taking the prism product of an n-gon and a cube, {n}×{4,3} = {n}×{}×{}×{}, and making a segmentotope by aligning vertices (as much as possible) over the centres of the tesseracts: {n}×{4,3} || dual{n}. This continues the dimensional analogy:

{n}||dual{n} = antiprism, with height near √[3/4]

{n}×{}||dual{n} = biantiprismatic ring, with height near √[2/4]

{n}×{4}||dual{n} = polyteron produced by the square pyramid shown above, with height near √[1/4]

But it turns out that {n}×{4,3}||dual{n} has its squared height approaching 0 from the wrong direction: it's negative! Of course that's because the tesseract has circumradius 1, and can't have a CRF pyramid.

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