## Prismatic honeycombs in H3

Higher-dimensional geometry (previously "Polyshapes").

### Prismatic honeycombs in H3

There are various uniform honeycombs in H3 using pseudohedra, like {7,3,3}, and their derivates with pseudohedral cells. The pseudohedron is inscribed in a pseudosphere and the tiling can be mirrored to the other branch of that pseudosphere as well (on the other side of its central plane).

Mirrored vertices can be connected by straight lines, so {7,3,3}, a honeycomb of {7,3}-pseudohedra, can be transformed into a honeycomb of heptagonal prisms. Is there a prismatic honeycomb of this type that is uniform?

The basic regular honeycombs {x,3,3}, {x,3,4}, {x,3,5}, {x,4,3} and {x,5,3} are standard, but it might work for some derived honeycombs as well.

I think Wendy has been researching non-Wythoffian uniform H3 honeycombs -- is there a page with complete results?
Marek14
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### Re: Prismatic honeycombs in H3

These are the laminate and lamina-apiculates.

In essence, you look for things like xPxQoRo, where P,Q is spherical (ie 3.3, 3,4, 3,5, 4,3 and 4,5 and Q,R is hyperbolic. It works with {4,3,8].

The actual process of laying two figures like {7,3,4} or something, and letting these run parallel, should be within my mental reach, since the height of the prism is a function of the dialation. I'll need to toss the figures around a little, but the results from {3,5,3,3} or {5,3,4,3} did not inspire me.
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### Re: Prismatic honeycombs in H3

I had a quick peek at this.

The issue here is that you have to put the dual polytopes in reciprocal position, so their diameters are equal, so eg

xo3oo3ox ho3oo4oq ho3oo5o(r(3-f))

The tetrahedron gives rise to two solutions. The apiculating triangle is 1:1:q and we see that there are two polygons whose shortchords are in this ratio, 3 , 4 and 4 , U The first leads to 4,3,3 gives 3,4,3 the second gives U,3,3 gives 4,4,3. Note here that to get the 4,4 from the oUx2x prisms, you remove the top and bottom faces, to leave only the square rims.

The case of the cube / octahedron gives verticies at 1,1,1 and sqrt(3),0,0, with a radius of sqrt(3). The cube edge is 2, and the other edge is rss(h-1, 1, 1) = sqrt(4-2r3 + 2 + 2) = sqrt(6-2r3). We get then by casting out common factors with sqrt(4), the square roots of r3 and wr2 (where w = ½r6+¼r2) Since the fourth root of 3 can not arise in a shortchord, we conclude there is no case of p,4,3 or p,3,4 giving a uniform tiling by lamina-apicullation.

The case of the dodecahedron and icosahedron is similar, one would at most be looking at solutions in 15,5,3 or 30,5,3 since the resulting numbers are in the integer set Z15. But i have not done calculations here yet.
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### Re: Prismatic honeycombs in H3

How does this work with other uniform polytopes?

For example, a general {n,3,3} has 15 Wythoffian forms, and all but one (O-n-O-3-O-3-X) contain pseudohedra. Is the laminating dependent here? I.e. if {n,3,3} can be uniformly laminated, can the other forms as well? Or is it independent for all of them?

And of course, then we can think of laminating polytopes with multiple pseudohedra, like bitruncated {7,3,8}...
Marek14
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### Re: Prismatic honeycombs in H3

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
The dream you dream alone is only a dream
the dream we dream together is reality.

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