CRFs containing J12

[quickfur]

Recently I've been trying -- so far unsuccessfully -- to construct a CRF containing J12 (trigonal bipyramid) as cells. The only one I could find is the trivial J12 prism; all other constructions don't seem to work out.

Does anyone know of a 4D CRF that has a 90° dichoral angle between a tetrahedron and some other cell? If so, we might be able to glue two of them together at that other cell to make J12 from the tetrahedra. I can't think of any, though, except the tetrahedral prism, which just gives the J12 prism.

Interestingly enough, if we glue two tetrahedral prisms in a gyrated orientation, we get that segmentotope that's a 4D analogue of the gyrobifastigium (I was trying to find the post with my renders of it but the search function in phpBB is rather crippled ). I believe it was something like triangular_prism||digon or something to that effect.

Another route is to augment the J12 prism, but I think the dichoral angles of the triangular prisms are too big to fit a triangular prism pyramid augment without becoming non-convex.

Any other ideas?

[Keiji]

Interestingly enough, if we glue two tetrahedral prisms in a gyrated orientation, we get that segmentotope that's a 4D analogue of the gyrobifastigium (I was trying to find the post with my renders of it but the search function in phpBB is rather crippled ). I believe it was something like triangular_prism||digon or something to that effect.

Do you mean the gyrated octahedral prism?

[quickfur]

Ah yes, that's the one. Thanks!

EDIT: Ah but wait, that doesn't produce the J12 prism at all. I must've confused the two. This is a different kind of gyrobifastigium, made from gluing two tetrahedral prisms together in gyro orientation.

[Klitzing]

Recently I've been trying -- so far unsuccessfully -- to construct a CRF containing J12 (trigonal bipyramid) as cells. The only one I could find is the trivial J12 prism; all other constructions don't seem to work out.

Does anyone know of a 4D CRF that has a 90° dichoral angle between a tetrahedron and some other cell? If so, we might be able to glue two of them together at that other cell to make J12 from the tetrahedra. I can't think of any, though, except the tetrahedral prism, which just gives the J12 prism.

[...]

Any other ideas?

I've just scanned all my dihedral angle values contained within my incmats. There is indeed so far no such angle at a {3} with coincident tet, which has a value of 90 degrees (except of the trivial one, you mentioned).

The only case, where tridpy (J12) occurs so far, is ditoh, a euclidean honeycomb.

--- rk

[quickfur]

Hmm. Does this mean that the J12 prism is the only CRF that contains J12 as cells? (Besides the honeycomb, that is.)

If so, that may serve as an example that may give us a way to determine which CRF constructions are impossible. There are a number of topological possibilities I've considered, involving J12, but the edge lengths and dichoral angles don't work out to be CRF. It appears that there's some kind of restriction on CRF possibilities imposed by the equatorial dihedral angles of J12, that make it either very hard, or impossible, to build a CRF out of it. This is quite surprising to me, since the tetrahedron itself appears almost in every CRF in many different situations, yet J12 hardly appears anywhere!

If it's possible to make a CRF out of J12 (other than the prism), I suspect it will have to be a CVP 3 polytope.

[Klitzing]

Hmm. Does this mean that the J12 prism is the only CRF that contains J12 as cells? (Besides the honeycomb, that is.)

No, not at all. It just tells, that we haven't found one so far.
But that again tells, that it would not be so easy to get one, I fear.

--- rk

[Polyhedron Dude]

No, not at all. It just tells, that we haven't found one so far.
But that again tells, that it would not be so easy to get one, I fear.

--- rk

I know this all too well - I was once trying to find uniform polychora with gidrids (Miller's monster) for cells - but all I could find was the prism of it. Although I did get lucky with sirsid and seside and found rapsady.

[wendy]

J12 sounds like oxo3ooo&#t. I'll play around with it to see if i can find anything interesting, although don't hold one's breath.

[Keiji]

Hmm...

Obviously J12 pyramid is not CRF since J12 is not orbiform.

But, how about these possibilities:

J12 || digon
J12 || J14 (Elongated triangular bipyramid)
J12 || J27 (Triangular orthobicupola)
J12 || J35 (Elongated triangular orthobicupola)

[Klitzing]

Keiji, the main problem here always is, to establish that the lacing cells shall not get sheered.

E.g. in your first case, line || J12 we have: J12 = Ao ox3oo&#x with some A. In fact A/2 = height of ox3oo&#x = rt(2/3).
Then your intend would be BAo oox3ooo&#x with some B, which you assume to respect B=x.

So consider the coordinates:

Code: Select all

x3o (J12-equator):
1/2, -1/rt12, 0, 0
-1/2, -1/rt12, 0, 0
0, 1/rt3, 0, 0

A (J12-poles):
0, 0, rt2/rt3, 0
0, 0, -rt2/rt3, 0

B (ends of the line atop):
0, 0, B/2, h
0, 0, -B/2, h

Then any of the B-points has to unit-laced to any of the equator points as well as to the corresponding A-point. Thus:

Code: Select all

B <-> x3o
1 = 1/4 + 1/12 + B^2/4 + h^2
-> h^2 = 2/3 - B^2/4

B <-> A
1 = (rt2/rt3 - B/2)^2 + h^2
= 2/3 - B.rt2/rt3 + B^2/4 + h^2
h^2 = 1/3 + B.rt2/rt3 - B^2/4

equating h^2 =>
1/3 + B.rt2/rt3 = 2/3
B.rt2/rt3 = 1/3
B.rt2 = 1/rt3
B = 1/rt6

That is, B = 1/rt6 is definitely not the same as x = 1!

--- rk

[quickfur]

[...]
But, how about these possibilities:

J12 || digon
J12 || J14 (Elongated triangular bipyramid)
J12 || J27 (Triangular orthobicupola)
J12 || J35 (Elongated triangular orthobicupola)

The challenge, as Klitzing said, is that while it's relatively easy to build these topological constructions, it's not so easy to be sure that (1) there are no concave parts, and (2) tetragons are actually squares, rather than rhombuses.

I've considered J12 || digon before. The trouble here is that the centroids of two conjoined tetrahedra of unit edge (which correspond with the orthogonal projection of the digon's endpoints) are not a unit edge apart, so lacing edges will not be of equal length. So some of the lacing triangles will not be equilateral.

For J12 || J14, the main trouble is that the tetragons of the lacing triangular prisms get sheared into rhombuses. You can see an analogous issue in 3D if you start with two coplanar triangles and try to lace it to an elongated digonal bipyramid (i.e., coplanar sequence of triangle-square-triangle). The tetragons lacing the exposed sides of the triangles to the elongated digonal bipyramid will be sheared. A similar problem occurs with J12 || J27.

I haven't checked J12 || J35 yet, but given that it's derived from J27 and J14 via various Stott expansions, it will share many of the same dihedral angles, so will most likely exhibit the same shearing problem.