## 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

quickfur wrote:Haha, looking at the image again, I decided to use my polytope viewer to calculate the angle between the square face of the hex prism and the triangular top of the adjacent cupola. Turns out to be: 110.9051574479 degrees. Is there a CRF 3D cell that contains this angle between a square and a triangle?

Some quick checking on Klitzing's site yields this (from J32, Pentagonal orthocupolarotunda):

between {3} and {4} (across rim): arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157°

Which I assume is the angle in question...
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### Re: axial CRFs

quickfur wrote:Haha, looking at the image again, I decided to use my polytope viewer to calculate the angle between the square face of the hex prism and the triangular top of the adjacent cupola. Turns out to be: 110.9051574479 degrees. Is there a CRF 3D cell that contains this angle between a square and a triangle?

A quick search through my locale copy of incmats files returned:
• Bilbiro.htm - 13.456 bytes - Fr, 21.04.17 at 11:02 - E:\rk\Documents\rk\priv\math\html\Incmats\
2.508 <li>between {3} and {4} (across rim): &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
• cube=ike.htm - 6.800 bytes - Mi, 01.03.17 at 14:54 - E:\rk\Documents\rk\priv\math\html\Incmats\
2.345 <li>at {4} between <a href="cube.htm">cube</a> and <a href="trip.htm">trip</a>: &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
• Pocuro.htm - 5.573 bytes - Mi, 01.03.17 at 14:54 - E:\rk\Documents\rk\priv\math\html\Incmats\
1.806 <li>between {3} and {4} (across rim): &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
• Ri.htm - 4.032 bytes - Mi, 01.03.17 at 14:54 - E:\rk\Documents\rk\priv\math\html\Incmats\
1.751 <li>between {4} and {6} (at pseudo {5/2}): &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
• Thawro.htm - 6.681 bytes - Mi, 01.03.17 at 14:54 - E:\rk\Documents\rk\priv\math\html\Incmats\
1.972 <li>between {4} and {6}: &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
2.125 <li>between {3} and {4} (across rim): &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
• xfoa3fooo5xuFx_zx.htm - 3.851 bytes - Mi, 01.03.17 at 14:54 - E:\rk\Documents\rk\priv\math\html\Incmats\
1.634 <li>between {4} and {6}: &nbsp; arccos(-sqrt[(3-sqrt(5))/6]) = 110.905157&deg;</li>
(For the online ones just replace my local path by https://bendwavy.org/klitzing/incmats/* - and note that this search tool occasionally does not respect casings of filenames...)

--- rk
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### Re: axial CRFs

When I recall correctly, in those day I tried to attach bilbiroes onto the next free lacing squares of the hips.
I don't remember so, whether the acute gap between the next free squares of the tricues and the now neighbouring triangles of those bilbiroes would give place for squippies, as they should...
--- rk
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### Re: axial CRFs

Just dreaming up a different, so non-convex continuation:

Consider the central 3-fold edge facetting sid-0-3-3-3 of sidtid, cf. http://www.polyedergarten.de/polyhedrix/e_sid3pyr1.htm, i.e. oxx3ovo&#xt, where v=1/f and the bottom triangle is empty. It features 3 regular triangles at the top segment, 3 lacing squares at the bottom segment, and 3 diametral pentagons.

That one too could be attached to the next free squares of the hips and the top triangles of the cupolae. It then even provides the pentagons needed for your srid shell. Further the free triangles of the tets, the adjacent free squares of the cupolae and the other pentagons of these added non-convex cells would provide enlarged patches of 8 octagonally shifted further srids.

But then I'm not sure what to place at the opposite square of the hips. And esp. what to do with that hollow face of these new cells...

--- rk
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### Re: axial CRFs

quickfur wrote:A different approach I thought about, was to Stott-expand the cube into a x4o3x, with some adaptations of the cupola cells, the idea being that in a Stott expansion we may be able to avoid the 1/f edges completely. Might be worth exploring.

Not too bad an idea. Because of: (v=1/f) + x = f.
--- rk
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### Re: axial CRFs

quickfur wrote:Actually, no harm posting the image I have so far, maybe it will lead to some new ideas:

P.S. This is the model with the x5o3x completed, so you can see some non-equilateral triangles and non-CRF cells.

If not too much an effort, it would be nice to have layerwise pictures here, like you've done for several of the uniforms on your website.
This then makes the actual outer to be matched cells more obvious, and thus the trials for the next cells get much easier be guessed...

Did someone meanwhile made some progress with that seed?
--- rk
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### Re: axial CRFs

Klitzing wrote:Did someone meanwhile made some progress with that seed?

Went on myself,
Here is the so far extended axially pyritohedral seed:

1) 1 cube embedded centrically at origin in axially pyritohedral symmetry
its vertex coords then are A = ( 1/2, 1/2, 1/2; 0 ) with pyritohedral symmetry applied to the first three.

2) 6 hips to be square adjoined to that cube from (1)
it adds vertex coords B = ( (f+1)/2, 1, 1/2; (f-1)/2 ) for the nearest neighbours
resp. C = ( (2f+1)/2, 1/2, 1/2; f-1 ) for the opposites, each again with pyritohedral symmetry to be applied to the first three.

3) 12 tricues to be hexagon adjoined to the hips of (2)
as well as one lacing square each to be connected to the near lacing square of the neighbouring hip.
Thus all vertices here are already provided, just the single far vertex of the top base triangle,
which then reads D = ( (f+2)/2, 0, (f+1)/2; f-1 ), again with pyritohedral symmetry to be applied to the first three.

4) The near lacing triangles of the tricues of (3) provide around the pyritohedral rotational axes triangular dimples,
which thus can readily be filled by 8 tets. No further vertices arise here.

Thus it occurs that CD = f.
My next try thus was to use
5) 12 bilbiroes to be square attached to the 2 far lacing squares of the hips of (2) each,
resp. triangle attached to the top base of the neighbouring tricues of (3).
One then has lots of new vertices:
(here prime and double prime relate to accordingly applied symmetries)
Code: Select all
` G     H      F     D''      I    E      B'    C' `

which can be calculated as
E = ( (f+1)/2, (f+2)/2, 1; f-1 )
F = ( (f+1)/2, (2f+1)/2, (f+2)/2; (2f-1)/2 )
G = ( 1/2, (f+2)/2, (f+3)/2; (2f-1)/2 )
H = ( 1/2, f+1, (f+2)/2; (3f-2)/2 )
I = ( 0, f+1, 1; (2f-1)/2 )
again all with pyritohedral symmetry to be applied to the first three.

This then allows
6) 24 squippies square attached to far lacing squares of the tricues of (3)
and triangle attached to the overhanging triangles of the bilbiroes of (5)
Again all vertices are already provided:
Code: Select all
`B'  C'  E   B   D'`

7) the open face of the tets of (4) then has to be adjoined to one of 8 octs each,
the neighbouring oct faces then adjoin to the next free squippy triangle.
No new vertices.

8) The B'EFGD'' pentagons of the bilbiroes of (5) adjoin to one of 24 peppies each,
one lacing of which then adjoins to one of the other lacings of the octs of (7).
No new vertices.

9) The opposite triangle of the octs of (7) then adjoin to 8 further octs.
Again no new vertices.

(to be continued ...)
--- rk
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### Re: axial CRFs

Hmm. I'm trying to build the model with bilbiros attached to the cupolae / hex prisms, but I think something is wrong. I can construct most of the vertices of the bilbiros, but the last 2 vertices, i.e., one of the vertices belonging to the pentagon-pentagon edge, my polytope viewer chokes on it, which is an indication that it's either non-convex there, or corealmar with another cell somehow. I'm still trying to narrow down exactly what's going on there... but it seems like there may not be enough room to fit bilbiros there... or it would result in concave regions.
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### Re: axial CRFs

Update: I found the problem. The vertices I mentioned are corealmar with the tetrahedra that share vertices with the starting cube. Which means adding bilbiros would create augmented octahedra (tetrahedron + octahedron blend), but that's unfortunately non-CRF.

I guess we have to try the other alternatives to see what else can fit in that triangle-square configuration with ~110.9° dihedral angle...
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### Re: axial CRFs

Nice, these happen-to-be corealmic tet-oct joins:   -> -> -> ->

For, you already came up with the expanded version idea, replacing the initial cube with sirco.
This then would ask the eight rotational rays of pyritohedral symmetry to be expanded radially in turn,
that is, this corealmic tet-oct join then just would becume a well being allowed tut!!

--- rk
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### Re: axial CRFs

Well, before we assume that will work, we should verify whether starting with sirco instead of cube will retain the same hex prism + tri cupola configuration. I'm assuming that we will insert cubes between the hex prisms and tri cupolae, where the 12 non-axial squares of sirco are, and I'm assuming the angles will remain the same (this needs to be checked). But if so, that means the square+triangle gap where we attached J91 will now have a different dihedral angle, since the top triangle of the tri cupola is now 1 edge length separated from the square face of the hex prism, since there is now a cube between the hex prism and the tri cupola. So J91 likely won't fit there anymore.
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### Re: axial CRFs

well these triangles then too get separated, that's right, but what about an accordingly inserted trip?
--- rk
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### Re: axial CRFs

Klitzing wrote:... This then would ask the eight rotational rays of pyritohedral symmetry to be expanded radially in turn, that is, this corealmic tet-oct join then just would become a well being allowed tut!!
--- rk

sadly wrong, expansion here does not apply to the faces, making rhombs into hexagons, rather it will apply to the edges. Thus that tet would become a tricu. And the former corealmic lacing faces of the oct here remain coreamic. Infact you peobably would just get a tut-tricu blend with (again) corealmic triangles, which thus recombine into non-CRF rhombs.
Thus this type of expansion indeed does not help out.
--- rk
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### Re: axial CRFs

quickfur wrote:

It just occured to me an other coincidence here:
Just consider the 9 colored cells (1 yellow cube + 8 red tetrahedra) plus the 12 squares, which are spanned by the tetrahedral lacing edges.
This same partial configuration would be achieved, when instead of hips (= 6-p) and tricues mutually orthogonally attached dips (= 10-p) and pecues would have been attached!

Thus we still not have to abandon this pyritohedral seed fully. We might try to continue it with that ansatz instead!

--- rk
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### Re: axial CRFs

Isn't that just the n=5 case of your original idea? Originally I started with the n=3 case because it appeared to be a simpler case. But perhaps the n=5 case might be easier to close up in a CRF way.

I still have not abandoned the n=3 case yet, actually. Remember that the square-triangle dihedral angle can also fit other things, like pocuro? There might still be a way to get a CRF out of it, even if it would be less symmetric than we set out to find.
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### Re: axial CRFs

quickfur wrote:Isn't that just the n=5 case of your original idea? Originally I started with the n=3 case because it appeared to be a simpler case. But perhaps the n=5 case might be easier to close up in a CRF way.

Indeed, right you are!

You even could make a mixture of both (then leaving axially pyritohedral symmetry, for sure):
take the central cube plus 2 opposing hips (x3x x) plus 2 opposing dips (x5x x). This then is nothing but a finite complex from a mere griddip (x3x5x x)! That is, you'd continue that by stacking atop those grid pseudo facets then further stuff; possibly cupolae each on the hips and dips, thus asking then for the red tets in the original arrangement ...

quickfur wrote:I still have not abandoned the n=3 case yet, actually. Remember that the square-triangle dihedral angle can also fit other things, like pocuro? There might still be a way to get a CRF out of it, even if it would be less symmetric than we set out to find.

I do not like to discourage you, rather the contrary. But I doubt that in view of my list of 110.9 degrees angled cells. For those cells therefrom, which are 3D and use that angle as being found between a triangle and a square, what would be required here, just are: bilbiro, pocuro, and thawro. Moreover those 3 all feature that dihedral within a slightly larger finite complex out of 2 pentagons, being connected by 2 triangles (patch of id = o3x5o), plus a Johnsonian lune (i.e. triangle-square-triangle, patch of srid = x3o5x) being blended to the former at the to be blended out decagonal pseudo facet. The required 110.9 angle then occurs at the triangle-square edge across that blending rim. Thus this larger finite complex seems to be set anyway!

That is, whenever you'd use one of these to adjoin at this so far un-matched face pair, then you would ask for the outer triangles of that lune too. And those then require that problematic octahedron atop the red tetrahedra, i.e. require that corealmic pair of cells in the directions of rotational symmetry, which, when blended, featuring rhombic faces!

This is why I doubt that this "pyritohedral seed from 1 cube plus 6 hips" could be continued as CRF.
Allowing for rhombs would one way to go on, thus searching for a still unit-edged convex polychoron,
resp. allowing for non-convex cells (cf. that idea of mine) being used, would be an other.

--- rk
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### Re: axial CRFs

Hmm. What if instead of a single cell to fit into the triangle-square gap, we fit two cells over it? I'm not 100% certain this is possible, but, to use an analogy, it would be similar to the 3D case of being given two edges at 72° dilatral angle, where you could choose to fit a single pentagon, or two triangles.

Of course, whether the result will be CRF depends on what the surroundings, whether the dichoral angles of fitting two cells there will remain convex, or will produce concave gaps with adjacent cells. In this particular case, since the curvature of the starting seed is quite mild, my guess is that our best chances lie in maximizing the angle defect around the square-triangle edge, so that the resulting cell complex won't "protrude" from the rest of the polytope. So perhaps a pair of cupolae... or for simplicity, a square pyramid and tetrahedron.
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### Re: axial CRFs

Not too bad an idea, quickfur.

But then, adding the dihedral angles of tetrahedron and square pyramid, one gets 125.264 degrees. (*)
And the trigonal gyrobicupola (aka cuboctahedron) has the same value. (*)
The square gyrobicupola has a dihedral angle at the rim of 99.736 degrees.
The pentagonal gyrobicupola there even provides 69.095 degrees only.

Thus according pairs of square- or pairs of pentagonal cupolae won't even bridge that 110.905 degrees cavity.

A different possibility would be a pentagonal prism and a pentagonal pyramid, as the elongated pentagonal pyramid features a dihedral angle of 127.377 degrees across the rim.

So your suggested pairings (*) look most promizing.

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### Re: axial CRFs

Cuboctahedron won't work, because the 110.9° angle is rigid (unless we modify the seed cells). You need at least 2 cells to get the degree of freedom needed to account for the angle defect.

Pentagonal prism + pentagonal pyramid looks promising, though, because of the acute angle of the pentagonal pyramid (and also accounting for the current's seed relation to pentagonal polytopes).
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### Re: axial CRFs

quickfur wrote:Cuboctahedron won't work, because the 110.9° angle is rigid (unless we modify the seed cells). You need at least 2 cells to get the degree of freedom needed to account for the angle defect.

Pentagonal prism + pentagonal pyramid looks promising, though, because of the acute angle of the pentagonal pyramid (and also accounting for the current's seed relation to pentagonal polytopes).

I used the Johnsonian corealmic exterior blends just for the measure of the dihedral of both components together. In order to get a wedge that angle thus has to be larger than that of the cavity. (Triangle inequality / relation.) - Thus we have to insert the 2 individual components. I.e. tet+squippy, tricu+(gyrated) tricu (both having same tiny angle defect), peppy+pip (slightly larger angle defect), ...
But squacu+(gyrated) squacu would not work because of negative angle defect.
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### Re: axial CRFs

Ah, I see what you mean. Thanks for clarifying!
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### Re: axial CRFs

I think we mixed up excess and defect here...
shouldn't we get lower than the full circle, in order to bend into the next dimension, when closing like a pyramid?

At least the angular sum around the edge of that {3}-{4}-cavity then would provide
120° (hip) + 125.26° (tricu top-base dihedral) + 70.53° (tet) + 54.74° (squippy bottom dihedral) = 370.53° > 360°
which is an excess rather than a defect!

--- rk
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### Re: axial CRFs

I think there is some confusion here, on the part of both of us.

The requirement of sum of dihedral angles < 360° holds around an edge when we consider a set of cells sharing that edge within a 3D hyperplane. This requirement continues to hold once we fold the cells into 4D, but note that the angle defect in the 3D hyperplane changes as we fold the cells (the idea being that the folding will reduce the defect to 0° so that the cells close up). If we then remove some cells from the folded configuration, the gap opened up by the removed cells will have an angle (in 4D) less than the dihedral angles of the removed cells. Consider for example, removing two adjacent triangles from a pentagonal pyramid. The angle of the gap at the apex will be less than 120°, because it's not measured in the 2D plane orthogonal to the apex-to-base vector of the pyramid, but is measured in a tilted plane.

So the correct criterion to be applied here is to sum the dihedral angles of the tri-cupola and the hex prism, not the folded angle of 110.9°, because that lies on a tilted hyperplane, so it doesn't actually reflect the actual amount of space available to fit more cells in.
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### Re: axial CRFs

So we have 2 simultanuos inequalities to obey:
a) the sum of the dihedrals of the to be added cells into the 4D gap angle has to be larger than your alread been given value of 110.91 degrees.
b) the flat total angular sum of the dihedrals of the to be added cells plus the already contained cells has to be less than 360 degrees.

And that requirement (b) could be restated into
b*) the sum of the dihedrals of the to be added cells into the 4D gap has to be less than 360 - 120 (hip) - 125.26 (tricu top-base dihedral), i.e. less than 114.74 degrees.

These 2 values are much too close to provide much possibilities!
• Esp. tet+squippy (3-3-4) features an already too large angular sum.
• Tricu+(gyro) tricu (3-6-4) provide the same already too large value.
• Squippy+trip (3-4-4) provide exactly the upper bound. (Thus all cells ought remain within flat 3D space, i.e. giving no way to be bend into 4D!)
• Pecu+pero (4-10-3) provide exactly the lower bound. (This was the already known attempt to use as single to be placed cell, a pocuro.)
Could someone come up with a different, matching pairing?

--- rk
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### Re: axial CRFs

Wow. You're right! There is very little room to maneuvre here.

While this seems like a disappointment in terms of being able to find a CRF solution to your initial seed, I see this as a new step towards tackling the problem of enumerating all 4D CRFs and/or proving that certain constructions cannot exist. In particular, I wonder if this double constraint (sum of all dihedral angles < 360°, and sum of dihedral angles of any subset of cells > angle of gap produced by removing these cells) can significantly narrow down the possible CRF constructions given the set of 3D CRFs, especially in the way of an automated computer search.

In the past, Marek has tried the approach of enumerating all possible vertex figures by enumerating 3D abstract polytopes, but it seems we didn't get very far. I also tried to tackle it from the approach of constraining the dichoral angles around an edge, but the possibilities still seem too numerous. I think somebody (probably Marek?) also proved that once the cells around a vertex are determined in 4D, then the vertex must be rigid, unlike the 3D case where if there are ≥4 polygons around a vertex it has additional degrees of freedom. Now we know an additional constraint that, once we have constructed enough cells to form a rigid cell net, can quickly eliminate impossible cases by a simple computation of two angle sums.

Perhaps it's time to tackle the problem of enumerating all 4D CRFs again?
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### Re: axial CRFs

wendy wrote:In xoxx3oxox5ooxo&#xt, the co-realmic nature of the pentagonal symmetry in ABC can be demonstrated, because the figure oxo5oxo is a singly diminished icosahedron, and showing that the lacing of f gives the same height as the AB+BC sections, is sufficient proof that this face is corealmic. Likewise in the rhomic axis (2) we get BCD gives oxx2oxo = point || square || line. The length of the height is (q+h)/2, the drift at the bottom is 1/2, so the lacing here is (5+2qh)+1 / 4 = (3+qh)/2. Were this used as a lacing, from B to D, we would see it is the same as BC+CD, which is sufficient proof that these faces are corealmic.

The argument of Wendy's 2016 post as such is clearly correct. And its application onto the gyepip part also is right. But her asserting of the autip part has to be pointed out as being wrong. In fact I just realized myself already yesterday, that the corresponding claim in this 2016 post of mine already has been wrong, cf. the there meanwhile being made edit remark.

So what then goes wrong in her above Independent calculation?

The height of autip clearly is (q+h)/2. But when calculating the drift (or shift) we have to compare the vertexwise radius of id (at layer B, which is f) to the edgewise radius of ti (at layer D, which is 3f/2). Thus that drift or shift rather is f/2 instead of the provided value of 1/2!

Thus, when assuming corealmity, the square of the bistratic height between the layers B and D could have been calculated then as [(q+h)/2]^2 - [f/2]^2, i.e. the height itself would come out as sqrt[4+2qh-f]/2 = 1.349161.
OTOH, that here - under assumptions - being calculated height rather equals 1/2 (=BC) + f/2 (=CD) = (f+1)/2 = 1.309017.
Therefore her idea, taken thoroughly, would have shown likewise, that squippy and trip (oriented as digonal cupola) here will not align into an autip! Instead those are tilted slightly.

It shall be pointed out, that xoxx3oxox5ooxo&#xt none the less remains a CRF. It just fails to have the formerly attributed corealmicity of the lower cells. (The corealmicity of the upper cells (gyepip) is trivially fulfilled, as the stack of layers ABC happens to be nothing but the bistratic cap of rox,)

Btw. the dihedral angle between that lacing trip and the sectioning pseudo face (srid, at layer C) already is provided in that mentioned correcting edit: arccos[-(sqrt(5)-1)/sqrt(12)] = 110.905157 degrees. Thus we just have to add the base angle betwwen squippy and srid in that id(=B)||srid(=C) segmentochoron, which equates to 45 degrees exactly. And obviously 45 + 110.905157 = 155.905157 fall short wrt. 180, ain't it?

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### Re: axial CRFs

Having recently disqualified some of the axial CRFs to the effect that those would no longer have (so far stated) corealmic cells (i.e. then uniting therein into single externally blended ones, each) is kind a loss within the corresponding research. But then, even so, yesterday I managed to find a further one, which truely has such corealmic faces and which eluded our Research so far. Even more surprising is its low total number of cells in this context.

In fact, the bistratic stack of tetaco and coatut, when being stacked in the correct mutual orientation, provides the CRF lace tower oxx3oox3xxo&#xt. (The other possible orientation would fail to be convex.)

Here it happens that the equatorial dihedral angle at trip-{4}-trip becomes 180 degrees exactly, thus these pairs unite into singe externally belended ones, each. According to the being used mutual orientation of those trips in each pair that resulting cell then will be a gybef each.

Thus, the cell totals here are:
6 gybefs
4 octs
5 tets
4 tricues
4 trips
1 tut

The incidence matrix of that fellow also could be provided:
Code: Select all
`oxx3oox3xxo&#xt   → both heights = sqrt(5/8) = 0.790569(tet || pseudo co || inv tut)o..3o..3o..     | 4  *  * | 3  3  0  0  0 0  0 | 3  3  6 0 0  0  0  0 0 0 | 1 1 3 3 0 0 0.o.3.o.3.o.     | * 12  * | 0  1  2  2  2 0  0 | 0  2  2 1 1  2  1  2 0 0 | 0 1 2 1 1 1 0..o3..o3..o     | *  * 12 | 0  0  0  0  2 1  2 | 0  0  0 0 0  2  2  1 2 1 | 0 0 1 0 2 1 1----------------+---------+--------------------+--------------------------+--------------... ... x..     | 2  0  0 | 6  *  *  *  * *  * | 2  0  2 0 0  0  0  0 0 0 | 1 0 1 2 0 0 0oo.3oo.3oo.&#x  | 1  1  0 | * 12  *  *  * *  * | 0  2  2 0 0  0  0  0 0 0 | 0 1 2 1 0 0 0.x. ... ...     | 0  2  0 | *  * 12  *  * *  * | 0  1  0 1 0  1  0  0 0 0 | 0 1 1 0 1 0 0... ... .x.     | 0  2  0 | *  *  * 12  * *  * | 0  0  1 0 1  0  0  1 0 0 | 0 0 1 1 0 1 0.oo3.oo3.oo&#x  | 0  1  1 | *  *  *  * 24 *  * | 0  0  0 0 0  1  1  1 0 0 | 0 0 1 0 1 1 0..x ... ...     | 0  0  2 | *  *  *  *  * 6  * | 0  0  0 0 0  2  0  0 2 0 | 0 0 1 0 2 0 1... ..x ...     | 0  0  2 | *  *  *  *  * * 12 | 0  0  0 0 0  0  1  0 1 1 | 0 0 0 0 1 1 1----------------+---------+--------------------+--------------------------+--------------... o..3x..     | 3  0  0 | 3  0  0  0  0 0  0 | 4  *  * * *  *  *  * * * | 1 0 0 1 0 0 0ox. ... ...&#x  | 1  2  0 | 0  2  1  0  0 0  0 | * 12  * * *  *  *  * * * | 0 1 1 0 0 0 0... ... xx.&#x  | 2  2  0 | 1  2  0  1  0 0  0 | *  * 12 * *  *  *  * * * | 0 0 1 1 0 0 0.x.3.o. ...     | 0  3  0 | 0  0  3  0  0 0  0 | *  *  * 4 *  *  *  * * * | 0 1 0 0 1 0 0... .o.3.x.     | 0  3  0 | 0  0  0  3  0 0  0 | *  *  * * 4  *  *  * * * | 0 0 0 1 0 1 0.xx ... ...&#x  | 0  2  2 | 0  0  1  0  2 1  0 | *  *  * * * 12  *  * * * | 0 0 1 0 1 0 0... .ox ...&#x  | 0  1  2 | 0  0  0  0  2 0  1 | *  *  * * *  * 12  * * * | 0 0 0 0 1 1 0... ... .xo&#x  | 0  2  1 | 0  0  0  1  2 0  0 | *  *  * * *  *  * 12 * * | 0 0 1 0 0 1 0..x3..x ...     | 0  0  6 | 0  0  0  0  0 3  3 | *  *  * * *  *  *  * 4 * | 0 0 0 0 1 0 1... ..x3..o     | 0  0  3 | 0  0  0  0  0 0  3 | *  *  * * *  *  *  * * 4 | 0 0 0 0 0 1 1----------------+---------+--------------------+--------------------------+--------------o..3o..3x..     | 4  0  0 | 6  0  0  0  0 0  0 | 4  0  0 0 0  0  0  0 0 0 | 1 * * * * * *  tetox.3oo. ...&#x  | 1  3  0 | 0  3  3  0  0 0  0 | 0  3  0 1 0  0  0  0 0 0 | * 4 * * * * *  tetoxx ... xxo&#xt | 2  4  2 | 1  4  2  2  4 1  0 | 0  2  2 0 0  2  0  2 0 0 | * * 6 * * * *  gybef... oo.3xx.&#x  | 3  3  0 | 3  3  0  3  0 0  0 | 1  0  3 0 1  0  0  0 0 0 | * * * 4 * * *  trip.xx3.ox ...&#x  | 0  3  6 | 0  0  3  0  6 3  3 | 0  0  0 1 0  3  3  0 1 0 | * * * * 4 * *  tricu... .ox3.xo&#x  | 0  3  3 | 0  0  0  3  6 0  3 | 0  0  0 0 1  0  3  3 0 1 | * * * * * 4 *  oct..x3..x3..o     | 0  0 12 | 0  0  0  0  0 6 12 | 0  0  0 0 0  0  0  0 4 4 | * * * * * * 1  tut`

As an aside, this is one of the rare cases of occurance of gybefs within CRFs. In fact so far there are only
Or would anyone be aware of further ones for now?

--- rk
Klitzing
Pentonian

Posts: 1347
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Location: Heidenheim, Germany

### Re: axial CRFs

Wow! Very nice find!!! I would never have thought to look for this combination, because the trigonal prisms are sloping (not perpendicular to the axis of the join), so it's not immediately obvious that the combination would be CRF! Maybe I'll build a model for this later when I get some time.
quickfur
Pentonian

Posts: 2435
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: axial CRFs

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?
quickfur
Pentonian

Posts: 2435
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: axial CRFs

quickfur wrote:Wow! Very nice find!!! I would never have thought to look for this combination, because the trigonal prisms are sloping (not perpendicular to the axis of the join), so it's not immediately obvious that the combination would be CRF! Maybe I'll build a model for this later when I get some time.

Yes indeed they are,
as best can be seen in this lace city display:
Code: Select all
`      x o   o x            -- tet                                             x x  uo ou  x x         -- co                                          o x   x u   u x   x o      -- inv tut                     \                      +-- gybef`

--- rk
Klitzing
Pentonian

Posts: 1347
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

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