I'm back with (what I hope is) a new CRF

[New Kid on the 4D analog of a Block]

There's most likely a flaw in this proposed CRF, as to be expected with my "work," but here goes anyway.

A loop of six haps (xo6ox tower) goes through the polychoron, manifesting as columns of three haps in each of the two identical "halves." They make contact with the two hexagons of the "middle boundary." Six peppys connect to each of the "top and bottom" haps with one triangular face, while their pentagons connect to the pentagons of the middle boundary. 36 tets fill in all gaps.
"Central" hap of one half highlighted:

Screen Shot 2019-08-05 at 10.57.04 PM.png

(50.57 KiB) Not downloaded yet

"Top and bottom" haps of one half highlighted:

Screen Shot 2019-08-05 at 10.57.31 PM.png

(43.81 KiB) Not downloaded yet

Like D4.8.x and other known CRFs, it has a near-miss Johnson solid as a middle boundary. This middle boundary is a dodecahedron with two antipodal pentagons replaced with hexagons, and two new pentagons added. (D4.4's middle boundary is like a gyrated, rectified version of this, in case that helps.)
"Top" view:

Screen Shot 2019-08-05 at 10.56.31 PM.png (10 KiB) Viewed 23531 times

Facing one of the pentagons:

Screen Shot 2019-08-05 at 10.56.36 PM.png (9.15 KiB) Viewed 23531 times

I checked and double-checked all the dihedral angles, and didn't see any in excess of 360°. (There were some in the mid-350s, though...) I checked CRF-list.xls, and didn't see any polychora with 6 haps among their cells.
So, what's wrong with this one?

[Klitzing]

At least the pentagonal variant of one of those halves does indeed exist. That one then is nothing but papadoe = pentagonal antiprism atop dodecahedron. Its existance thereby is being seen as a bidiminishing of ikadoe = icosahedron atop dodecahedron, as 2 antipodal tips of the icosahedron get removed at the top level base, while the cutting plane just rasps tangential at the opposing pentagon faces of the bottom level base. And that ikadoe moreover is nothing but a vertex-first subsegment of the 600-cell.

Within lace city display ikadoe is being represented as

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  o5o  o5x      x5o  o5o 
                          
x5o      f5o  o5f      o5x


Here you can see the icosahedron at the top level, showing up its subdimensional layers point atop pentagon atop dual pentagon atop point. Below you see the accordingly oriented dodecahedron level, again with its subdimensional layers pentagon atop f=(1+rt(5))/2 scaled pentagon atop dual f-pentagon atop dual x=unit scaled pentagon.

Accordingly, chopping off those opposite pair of icosahedral tips, you'd get the lace city display of papadoe as

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       o5x      x5o       
                          
x5o      f5o  o5f      o5x


and therefore the pentagonal variant of your idea thus would look like

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       o5x      x5o       
                          
x5o      f5o  o5f      o5x
                          
       o5x      x5o       


i.e. the lace tower pap || doe || pap.

The problem with your hexagonal variant now is that, in order to maintain all faces planar and all unit-edged within that medial layer polyhedron, then those lateral pentagons thereof would no longer remain equiangular, and therefore not regular. But then for your potential CRF you need to errect pyramids on those pentagons. But it is obvious that the lacing edges of such a non-regular pentagonal based pyramid cannot be all of the same size. - Thus your polychoron does exist and is a nice Variation of the above with a different, hexagonal rotational symmetry. But it cannot be made all unit-edged. And therefore fails to be a CRF.

--- rk

[New Kid on the 4D analog of a Block]

Thank you for the clarification.

So I don't make the same mistake again, does the medial layer polyhedron of a CRF (if it has one) need to be unit-edged, regular-faced, and planar-faced? I don't think all these criteria are necessary; D4.8.x has a skew boundary...

And if that doesn't give us an easy way to find potential polyhedra to use as medial layers, is there an easy way to determine whether some lace diagram (for example, the invalid one of the hexagonal variant)

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       o6x      x6o       
                         
x6o      N6o  o6N      o6x
                         
       o6x      x6o

or some polyhedron can be a valid medial layer for a CRF?

Or... is it pointless to search for such small CRFs when one of you surely would have discovered them already?

[Klitzing]

Just consider what CRF means. It is the abreviation of convex regular-faced (polychoron). Thus all polygonal faces have to be planar, unit-edged, and equi-angular (i.e. regular-faced). Moreover all edges throughout the polychoron have to be the same size. Within an intermedial layer however you also might find pseudo faces, i.e. faces which only occur in that section, but else are contained within some through-going polyhedral cell. Those clearly neither have to be equiangular nor need to be completely planar (then not even being pseudo faces, just edge circuits). Even so, those all have to be unit-edged too.

--- rk

[New Kid on the 4D analog of a Block]

I tried replicating the three-mibdi edge of D4.16 with bilunabirotundae since they have the same 5-5 dihedral. It's a simple enough idea that someone's probably tried it already, but I couldn't find anything similar in CRF-list.xls or my previous ideas.

Screen Shot 2022-04-07 at 5.25.58 PM.png

8 tets (pink) join the bilbiros at their "id" triangles, 8 more tets (orange) join them at their "mibdi" triangles. 6 bilbiros in white, 4 tricus in blue (unless they combine into 2 coes?)

(37.04 KiB) Not downloaded yet

Screen Shot 2022-04-07 at 5.26.27 PM.png

The dyad at the top of the lace is visible in the center of the left figure (just the second picture), and the point at the bottom of the lace can be seen in the center of the right figure (easier to see in the first picture).

(38.87 KiB) Not downloaded yet

I'm pretty sure it has this lace, but in it I can only find one bilunabirotunda at a time. (Switching the first two columns reveals the other bilunabirotunda in the lace.) So I suspect that something's not right. Maybe I'm trying to fold some facets in two opposite directions again.

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o3o2x
o3x2f
o3f2o
x3x2f
F3o2x
x3o2f
o3o2o

I also had a couple of other ideas inspired by this one. One is to change the "3" in the lace to 4 or 5 (but since that doesn't seem to work for D4.16, I doubt it'd work for this), and the other one has a lot more cells crammed into the "right" (with regards to the pictures) half of the figure (but again, I doubt that'll work if the basic idea here doesn't work).

[Klitzing]

I tried replicating the three-mibdi edge of D4.16 with bilunabirotundae since they have the same 5-5 dihedral. It's a simple enough idea that someone's probably tried it already, but I couldn't find anything similar in CRF-list.xls or my previous ideas.

8 tets (pink) join the bilbiros at their "id" triangles, 8 more tets (orange) join them at their "mibdi" triangles. 6 bilbiros in white, 4 tricus in blue (unless they combine into 2 coes?)

I'm pretty sure it has this lace, but in it I can only find one bilunabirotunda at a time. (Switching the first two columns reveals the other bilunabirotunda in the lace.) So I suspect that something's not right. Maybe I'm trying to fold some facets in two opposite directions again.
[...]

Interesting idea of yours.

What you tryed is

• having 4 vertical unit edges (where 3 bilbiroes will have to meet), those being at a pairwise distance of F each (long diagonal of bilbiro). Thus those alone would be described by an "x o3o3F" layer.
• Next there are the various squares of all those 6 bilbiroes, which come at a distance of f within each bilbiro, thence those are described by a "f x3o3x" layer (which thus shows as well, that it would be indeed 2 coes instead of the 4 blue tricues).
• Finally there are the id-like vertices of those bilbiroes. Again those are 4 such, situated likewise at a distance of f (across each individual bilbiro). Thence we here have some "o f3o3o" layer.

This then indeed would show the bilbiroes as "xfo oxf ... Fxo&#zxt" and the orange 8 tets (pointing at those first vertical edges) as "... ox.3oo. ...&#x" and the 8 pink tets (pointing on the last id-like vertices) as "... ... .oo3.xo&#x".

However, in order to work like that, the respective lacings between these 3 layers ought be x (cf. the final bits &#x, &#zxt) in each transitional case. But when typing in those layers, the respective radii would vary and esp. the heights between consecutive layers each come out to be imaginary (when unit lacings would be assumed), that is, they would be required to be larger than unity instead! And thereby one derives not even the same value between the first and second versus second and third layer pairs.

Therefore it seems to me that your brilliant idea sadly would not work out after all.

--- rk

[New Kid on the 4D analog of a Block]

All right, this time I have proper CRFs, but the question is are they known already? Since they're only segmentotera, it's hard for me to believe that nobody thought of them before, but on the other hand, I think they would be on Incmats if someone had used or mentioned them, and they aren't there.

I've been affectionately referring to the general category of this find as "mutantifastegia." What this entails is a lace isosceles triangle

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  A 
B   B

where "A atop B" is a segmentotope with height > 1/2. Hypothetically, A and B can be "exchanged" to form a different mutantifastegium, but this may not always work.
The polygonal pucofastegia and cupofastegia fit this definition. Nonconvex versions of these might even use a cupoliprism instead of a proper prism. Wherever segmentotopes can be sliced out of Archimedeans, these can likely be made. They might look similar to the "pseudopyramids" in 4D, rising to a sub-facet-dimensional apex that isn't quite as small as a point. None of that is what I'm concerned with here.
Some polytera that fit this description are already listed on the always-useful Segmentotopes page of Incmats. They often occur as slices of Archimedean polytera, for example, Opeatut, Tetacope, and Copatut are all derived from spix. But if we made a mutantifastegium based on H₃-symmetric polyhedra instead of small A₃ or B₃ ones, it clearly wouldn't come from a slice of any Archimedean polyteron, it might even be novel.

Mutantifastegia.zip

Contains OFF files of ten potentially new segmentotera, my scrawled notes on them, and a bonus segmentoteron of the same kind that I didn't realize was already known. I would have uploaded the OFF files on their own, but the forum doesn't allow that.

(40.81 KiB) Downloaded 347 times

About half of my attempts failed, and at first I didn't know why. I gave it a bit of thought, and while I haven't arrived at any solid conclusions, it probably has to do with the relative size of the two polytopes that make up the used segmentotope. For example, while Cubaope works and can be found on Incmats, its "opposite" (oct atop tes) doesn't work. And as for the other mutantifastegia that use segmentochora of the same height as octacube (≈0.67610), both of the Toatic possibilities fail while both of the Coasirco ones work.

It seems likely that the polytera I found aren't very notable, probably because of a lack of relations to Archimedean polytera, which would explain their absence from Incmats. Has someone already considered polytopes of this form?

[Klitzing]

It well might be that not all of your mutantifastegia have their own incmats file already.
However, your mutantifastegia are just a special case of the much more general already well defined trigonics:

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   A   
       
B    C


where A,B,C all are various decorations of a common symmetry group symbol, and all pairs, i.e. A||B, A||C, and B||C, would do exist as segmentotopes.
Yours then is just the special case of B=C thereof. And, for sure, the case A=B=C is still more well-known, as it then is nothing but the duoprism {3} x A.

--- rk

[New Kid on the 4D analog of a Block]

Ah, thank you! As it turns out, that mostly answers another question I had, proposing "lace scalene triangles," which now I know are well-understood. I didn't look very deeply into it, only came up with xxo4xxx3xox&#x and xxo4oox3oxo&#x, and I doubt those could be made CRF. I'm glad we were able to hit two flies with one swatter there.

[Klitzing]

In fact there is, even more general, the lace simplex for any number of "layers".

Thus the classical Wythoffian CD symbol is nothing but a one-layered lace simplices (with 0D position space),
the well-known lace prism is nothing but the two-layered lace simplices (with 1D position space),
those just mentioned trigonics are the three-layered lace simplices (with 2D position space),
etc.

--- rk

[New Kid on the 4D analog of a Block]

I have heard of lace simplices, but I haven't given them enough thought. They are important for determining the verfs of polytopes from their CD diagrams. When I did this for the Archimedean polypeta, I'm not sure I fully understood the concept of lace simplices.
It's good to see that the 6D lace simplices are now available on Incmats, thank you very much for making that information available.