axial CRFs

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

Re: axial CRFs

Postby quickfur » Fri Oct 20, 2017 6:27 pm

Wow, this CRF is quite a pretty little thing! Here's a parallel projection along the axis of the join:

Image

The green tetrahedron is the top cell of this CRF, and the antipodal truncated tetrahedron is outlined in magenta edges. Most importantly, you can see 2 of the 3 pairs of gyrobifastigium (J26) cells sharing edges with the central tetrahedron. The other pair is not shown because it would clutter the image too much.

Here they are, shown separately:

Image

You can easily spot the octahedral cells at the 4 corners of the tetrahedral-like projection envelope, as well as the 4 tetrahedra touching the central one. The triangular face opposite the shared vertices are where these tetrahedra join with the trigonal cupolae that link them to the antipodal truncated tetrahedron. You can also see the 4 trigonal prisms sharing a face with the central tetrahedron, and nestled between the J26 cells.

Here are the coordinates I used (~ denotes vector concatenation; apacs/apecs are Wendy's usual abbreviations):
Code: Select all
# x3o3o:
apecs<1/√2, 1/√2, 1/√2> ~ <-√(5/2)>

# o3x3o:
apacs<0, √2, √2> ~ <0>

# o3x3x:
apecs<1/√2, 1/√2, -3/√2> ~ <√(5/2)>


All in all, a very neatly-assembled CRF, with 1+4=5 tetrahedra, 4 trigonal prisms, 4 octahedra, 6 J26's, 4 trigonal cupolae, and 1 truncated tetrahedron, for a total of 24 cells.
quickfur
Pentonian
 
Posts: 2435
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: axial CRFs

Postby Klitzing » Sun Oct 22, 2017 8:29 pm

quickfur wrote:In other news, I've been searching a long time for a CRF that contains triangular bipyramid cells (other than the obvious prism), but so far haven't found anything. Maybe you have some ideas about some monostratics with tetrahedra that might be able to join up at the right angles to form a bipyramid?

When considering connvex monostratic lace prisms with across symmetries o3oPo, which use lacing tetrahedra, then we just have the following list:
  • pen = pt||tet, relevant dihedral tet-3-tet = arccos(1/4) = 75.522488°
  • octpy = pt||oct, relevant dihedral tet-3-oct = 60°
  • rap = tet||oct, relevant dihedral tet-3-oct = arccos(-1/4) = 104.477512°
  • hex = tet||-tet, relevant dihedral tet-3-tet = 120°
  • tetatut = tet||tut, relevant dihedral tet-3-tut = 60°
  • tetaco = tet||co, relevant dihedral tet-3-co = arccos(1/4) = 75.522488°
  • octacube = oct||cube, relevant dihedral tet-3-oct = arccos[-(3 sqrt(2)-2)/4] = 124.101465°
  • cubaco = cube||co, relevant dihedral tet-3-co = arccos[-(3-sqrt(8))/sqrt(8)] = 93.477707°
  • cubasirco = cube||sirco, relevant dihedral tet-3-sirco = 60°
  • ikepy = pt||ike, relevant dihedral tet-3-ike = arccos(sqrt[7+3 sqrt(5)]/4) = 22.238756°
  • ikadoe = ike||doe, relevant dihedral tet-3-ike = arccos[-sqrt(5/8)] = 142.238756°
  • doaid = doe||id, relevant dihedral tet-3-id = arccos[(3-sqrt(5))/sqrt(32)] = 82.238756°
  • doasrid = doe||srid, relevant dihedral tet-3-srid = arccos(sqrt[7+3 sqrt(5)]/4) = 22.238756°
  • trippy = pt||trip, relevant dihedral tet-3-trip = arccos(sqrt[3/8]) = 52.238756°
thus, at least this list does not leave any such possibilities for according combinations :\
--- rk
Klitzing
Pentonian
 
Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: axial CRFs

Postby Klitzing » Thu Oct 26, 2017 9:01 am

Ha, found further nice ones:

when stacking the segmentochora octasirco and cubasirco, i.e. producing the exterior blend of both (thereby blending out the sirco), then the obtained stack has the description oct||sirco||cube = xxo3ooo4oxx&#xt.

Sure, this one clearly remain a "mere" stacking of segmentochora, i.e. no lacing cells here become corealmic and would combine in turn. But, when actually calculating all those dihedrals across the 3 types of (blended out) sirco faces, then one gets surprised: all 3 independently happen to be 90 degrees exactly!

Furthermore, the around-symmetrical Stott expanded version thereof, i.e. toe||girco||tic = xxo3xxx4oxx&#xt, therefore would share this behaviour.

--- rk
Klitzing
Pentonian
 
Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: axial CRFs

Postby quickfur » Fri Oct 27, 2017 7:08 pm

Interesting. Do all of the lacing cells meet the other lacing cells at 90°? If so, could this CRF be used to produce a tiling of 4-space somehow? Note, interestingly, that one can tile 3-space with x4o3x's, octahedra, and cubes, so if we lay out copies of this CRF such that the equatorial x4o3x's touch each other at the axial cube facets, then we could fill in at least the octahedral gaps with the oct cells of slightly displaced copies of this CRF.

Not 100% sure what would happen to the remaining cubical gaps, though, or what the 4D shape above them would be. But it seems like they ought to be CRF gaps! If I'm not mistaken, they ought to have 2 squippies joined at the square face, and something else to close up the gap.

Could it be possible that this tiling, if it exists, is some kind of CRF modification of a uniform 4-space tiling? An EKF tiling perhaps? :D
quickfur
Pentonian
 
Posts: 2435
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: axial CRFs

Postby Klitzing » Sat Oct 28, 2017 10:19 am

Wow, great idea of yours, quickfur!
And yes, all three dihedrals are 90 degrees exactly in octasircoacube: that across the triangle, that across the axial squares and that across the squares in rhombical positions of that equatorial (blended out) sirco.

So let's have a look. We need some tetracomb which is an infinite stacking of euclidean layers A = a4x3o4o, B = o4x3o4x, and C = b4o3o4x. Then the part in the column . o3o4o would be exactly our octasircoacube. Here "a" and "b" are to be chosen appropriately such that the lateral extend (distances between according cells) will match in all 3 layers.

So far I managed to calculate "b" to be b=q here. You know, one could provide sidpith = x3o3o4x also as oxxo3oooo4xxxx&#xt. Further we have cope = x o3x4o either oqo4xox xxx&#xt or oo4xx3oo&#x. And then tes = o3o34x being either oo4xx xx&#x or oqo ooo4xxx&#xt. And finally rit = o3o3x4o = oqo4xox3ooo&#xt. Using these tower descriptions of those polychora one then can describe the tetracomb sidpitit = x4o3o3x4o also as infinite stacking :BBC: of the layers B = o4x3o4x and C = q4o3o4x.

Therefore the lower half already is solved. Remains the upper one to be done ...

--- rk
Klitzing
Pentonian
 
Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: axial CRFs

Postby Klitzing » Mon Oct 30, 2017 10:17 am

Klitzing wrote:Wow, great idea of yours, quickfur! [...]

So let's have a look. We need some tetracomb which is an infinite stacking of euclidean layers A = a4x3o4o, B = o4x3o4x, and C = b4o3o4x. Then the part in the column . o3o4o would be exactly our octasircoacube. Here "a" and "b" are to be chosen appropriately such that the lateral extend (distances between according cells) will match in all 3 layers.

So far I managed to calculate "b" to be b=q here. [...]
Therefore the lower half already is solved. Remains the upper one to be done ...

--- rk

In order to get the right distances in the parallel layers, one has to choose furthermore "a" to be a=x. Thus our layers to be considered are A = x4x3o4o = tich, B = o4x3o4x = srich, and C = q4o3o4x (some q,x-variant of chon with x-cubes, prolate x-square prisms of height q, oblate q-square prisms of heigth x, and q-cubes).

I did not see so far, from which uniform tetracomb the segment AB would derive out. In fact, it uses some oct||sirco (the one for our octasircoacube bistratic tower, which is the main point of interest of all this investigation), which happens to be the monostratic cap of spic = x3o4o3x and it uses also elsewhere co||tic, the monostratic cap of srico = x3o4x3o.

But then I desided not to bother about that. Rather we already have all necessary ingrediants for the searched for tetracomb! We just will have to stack those mentioned layers periodically as :ABCB:, i.e. consider the infinte tower of euclidean honeycombs  :xoqo:4:xxox:3:oooo:4:oxxx:&##x.

The to be used polychora in that tetracomb then are:
  • oxo4xxx3ooo ...&#xt = coaticbicu (a mere convex stack of 2 segmentochora) with object frequency 1/22
  • oqo4xox3ooo ...&#xt = rit = o3o3x4o with object frequency 1/22 too
  • xo4xx .. ox&#x = squicuf (the well-known segmentochoron {8}||cube) with object frequency 6/22
  • oqo4xox ... xxx&#xt = cope = o3x4o x with object frequency 3/22
  • xo .. oo4ox&#x = squasc (the well-known segmentochoron line||ortho {4}) with object frequency 6/22
  • oqo ... ooo4xxx&#xt = tes = o3o3o4x with object frequency 3/22
  • ... xxo3ooo4oxx&#xt = octasircoacube (a mere convex stack of 2 segmentochora) with object frequency 2/22

Thus yes, that recently described octasircoacube indeed can be used to tile 4D space (together with some further constituents).

--- rk
Klitzing
Pentonian
 
Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: axial CRFs

Postby Klitzing » Mon Oct 30, 2017 11:18 am

By mere application of around-symmetrical Stott expansion wrt. that axial stacking of vertex layers one can obtain that "other" 4D filling (tetracomb) therefrom, which then incorporates the other recently mentioned special bistratic lace tower toagircoatic = xxo3xxx4oxx&#xt, which likewise has equatorial dihedrals of 90 degrees only.

In fact the respective polychora there would be accordingly:
  • oxo4xxx3xxx ...&#xt = toagircobcu (a mere covex stack of 2 segmentochora), frequency 1/22
  • oqo4xox3xxx ...&#xt = pabdirico (the parabidiminished o3x4o3o), frequency 1/22
  • xo4xx .. ox&#x = squicuf (unchanged), frequency 6/22
  • oqo4xox ... xxx&#xt = cope (unchanged), frequency 3/22
  • xo .. xx4ox&#x = squicuf (further ones here), frequency 6/22 too
  • oqo ... xxx4xxx&#xt = sodip = x4o x4x, frequency 3/22
  • ... xxo3xxx4oxx&#xt = toagircoatic, frequency 2/22

And the vertex layers or crosssections in this periodic stacking :ABCB: here would be A = x4x3x4o (grico), B = o4x3x4x (grico with different orientation / alignment), C = q4o3x4x for sure.

--- rk
Klitzing
Pentonian
 
Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Previous

Return to CRF Polytopes

Who is online

Users browsing this forum: No registered users and 1 guest