Bilbirothawroids (D4.3 to D4.9)

[Klitzing]

Well, do you know what I consider most interesting part of this polychoron?

It contains no tetrahedra. Tetrahedra are the most basic struts of polychora, like triangles in 3D -- how many Johnson solids lack triangles? None!

Instead, the smallest struts here are square pyramids.

Second interesting feature are those decagonal prisms which appear in towers of 3. They look like elevator shafts.

And then there's the combination of J91 and J92 in a single shape. In short, this is really a great shape!

So, how to call it? Maybe we could call it a CRUSHED or COMPACTED castellated prism. It looks like the middle layer of previous shape was crushed into thinner layer, but the rest of polychoron stayed unharmed.

As for the counts, Stella does that automagically So I can mention that the square pyramids are divided in groups of 120 + 60 + 60 (Stella divides them like this because the second group of 120 is chiral) and the decagonal prism are in two groups as well, 24+12 -- 24 directly adjacent to the x5x3x's and 12 in the middle layer.

As for polygons, there are 1488 in total: 860 triangles, 420 squares, 120 pentagons, 40 hexagons and 48 decagons.

Indeed a great find, Marek!
--- rk

[quickfur]

Well, do you know what I consider most interesting part of this polychoron?

It contains no tetrahedra. Tetrahedra are the most basic struts of polychora, like triangles in 3D -- how many Johnson solids lack triangles? None!

Instead, the smallest struts here are square pyramids.

Second interesting feature are those decagonal prisms which appear in towers of 3. They look like elevator shafts.

And then there's the combination of J91 and J92 in a single shape. In short, this is really a great shape!

So, how to call it? Maybe we could call it a CRUSHED or COMPACTED castellated prism. It looks like the middle layer of previous shape was crushed into thinner layer, but the rest of polychoron stayed unharmed.

As for the counts, Stella does that automagically So I can mention that the square pyramids are divided in groups of 120 + 60 + 60 (Stella divides them like this because the second group of 120 is chiral) and the decagonal prism are in two groups as well, 24+12 -- 24 directly adjacent to the x5x3x's and 12 in the middle layer.

As for polygons, there are 1488 in total: 860 triangles, 420 squares, 120 pentagons, 40 hexagons and 48 decagons.

Indeed a great find, Marek!
--- rk

Now I'm wondering if some kind of Stott expansion is possible, maybe based on o5x3o3x? I'm not sure if the extra pentagonal prisms of o5x3o3x will cause non-CRF edges to appear, but perhaps a partial Stott expansion might work?

[Marek14]

BTW, this won't be a CRF polychoron, but have you ever noticed that the dihedral angle of dodecahedron, icosidodecahedron and tridiminished icosahedron add up to exactly 360?

[quickfur]

BTW, this won't be a CRF polychoron, but have you ever noticed that the dihedral angle of dodecahedron, icosidodecahedron and tridiminished icosahedron add up to exactly 360?

That's what I was getting at in this post. I wonder if that could be the beginnings of a 3D Penrose-style tiling, perhaps?

[wendy]

viewtopic.php?f=25&t=1726&p=20420#p20420

I put a post here about the restrictions for constructing the generalised penrose tiles in N dimensions. Works with octagons and dodecagons, but at the moment, meds are making me somewhat confused.

[student91]

Now you guys are going too fast for me !!those things you've found are extremely awesome (especially the J91/J92-thing).
If I'm right, these things look like J92-thing: xofxF(Vo)Fxfox3xFxoo(xo)ooxFx5xoxFf(oV)fFxox&#xt, where the parentheses mean those things are in the same layer.
and xofxfox3xFxoxFx5xoxFxox&#xt for the J91/J92-thing.

Is the Bilbro-cut also possible on a general o5x3o3o?

I'm also thinking we need to revisit the "castellated prism" name as well. It was appropriate when there was only a single castellated prism, but now that these things are turning out to be rather common, especially here where they are based of various strata of the o5x3o3o, I think we need a more generally-applicable naming scheme, as I'm expecting more CRFs to turn up with similar constructions.

I think so too!. Maybe an ad-hoc new naming scheme is usefull, we could make a wiki-page about polystratic CRF's, and index them according to that. What about P for polystratic, then a number for the order of polystratic-ness, so P3. for a tristratic CRF, and then a number, so P3.1 would be my thawrochoron.

Below an attempt is made to explain an idea for a more constructive naming scheme of mine.
The things with such a construction basically take a lace-tower of a Johnson solid, and then place another node to the CD-graph. This node can get different "values" (x, f, F etc.). Most of the time these new values get either x or o. (e.g. castellated prism: xFoFx3ooooo5xofox&#xt, it has ooooo, meaning it closes after the bilbro is completed., another example: oxFx3xfox5xoxx&#xt, my thawro-thing. it has xoxx on the new node, meaning it has oxFx(2)xoxx&#xt and xfox5xoxx&#xt, both can be closed after the thing).
basically what you do when you make such a polytope, you take a Johnson solid %%%%N%%%%&#xt, add a node: %%%%N%%%%P????&#xt. This means the nodes of the N are fixed, and the other node can be changed. This means, if N=3, that you place triangles atop each other, that make the shape %%%%3%%%%&#xt. These triangles are symmetrically placed on the .N.P.-thing, so it's like placing %%%%N%%%%&#xt's symmetrically around a .N.P.
Let's take the ursachora as an example: xfo3oox4ooo&#xt. The triangle gives a trid. ike: xfo3oox&#xt. these triangles are placed on the vertices of an o3o4x, so it's somewhat related to the dual of o3o4x. In the same way the xFoFx3ooooo5xofox&#xt is somewhat related to the dual of o3x5o, because the bilbro's are placed along the verticeds of a o3x5o.
All layers of the trid. ike in xfo3oox4ooo&#xt should be in the same 3-space (i.e. the dichoral angle between xf3oo4??&#xt and fo3ox4??&3xt should be 180 at the f3o). This means that if you choose two ??'s, all other ??'s are set as well. If the other stacks (%%%%P????&#xt and %%%%(2)????&#xt ) give nice things as wel, we have a new polytope. if ???? consists of x's and o's, these stacks are very likely to work out. I think, because of this, we should call things witth only x's and o's on the extra node "simple." if there are other things that x's and o's, it is then considered "complex", and a naming scheme will be difficult to make.
If you have a valid thing %%%%N%%%%P????&#xt, you can also have %%%%N%%%%P(?+l)(?+l)(?+l)(?+l)&#xt, i.e., you can add a constant "l" to all ?'s. Most of the time this only works for ????=oooo=>xxxx. for not-simple polytopes this might also work, but for simple polytopes with ???? not being oooo, it would make it complex.

Clearly for every symmetry there are at most two possible simple polytopes based on a build of a Johnson-solid, and only two if one has only o's, the other one will have only x's. This means we could call the tetahedral ursachoron the simple teddi-based tetahedral polychoron, maybe a shortened form is desired. The castelated prism then is the simple bilbro-based icosahedral polychoron. Things like bilbro's might give problems, as they have multiple lace-tower buildups. Expanded can still be called expanded. This explanation wasn't clear.


Now something I am able to explain: the thawro-ing of the o5x3o3o. To me it seems like a little more complex cut of a o5x3o3o. Compare it with a cut along a dodecahedron of a 600-cell: things get rearranged a bit (tetahedra get replaced by pentagonal pyramids), and if you place the two halves back on each other, the thing isn't CRF anymore. Furthermore it doesn't affect the rest of the polytope. Therefore I am pretty sure the mega-wedge is possible.
The bilbro-ing of the o5x3o3o has some similar properties, and therefore it may be seen as a special cut as well (maybe it's comparable with making a square orthobicupola out of an rhombicuboctahedron). this means both of these operations can be combined with other diminishings, yielding combinatorial explosion

[Klitzing]

Now you guys are going too fast for me !!those things you've found are extremely awesome (especially the J91/J92-thing).
If I'm right, these things look like J92-thing: xofxF(Vo)Fxfox3xFxoo(xo)ooxFx5xoxFf(oV)fFxox&#xt, where the parentheses mean those things are in the same layer.
and xofxfox3xFxoxFx5xoxFxox&#xt for the J91/J92-thing.

You got them both right.
I could provide their incmats too, if desired.
(I've calculated 'em this weekend. - Be warned, those are rather large, so)

--- rk

[quickfur]

Now you guys are going too fast for me !!those things you've found are extremely awesome (especially the J91/J92-thing).
If I'm right, these things look like J92-thing: xofxF(Vo)Fxfox3xFxoo(xo)ooxFx5xoxFf(oV)fFxox&#xt, where the parentheses mean those things are in the same layer.
and xofxfox3xFxoxFx5xoxFxox&#xt for the J91/J92-thing.

Is the Bilbro-cut also possible on a general o5x3o3o?

I think so!! the reduction of the 30 equatorial o5x3o's is not dependent on the J92's above the J91's to be introduced, since the "special" part of the J92's (the hexagonal faces) are pointing away from the J91's. So it is certainly possible to reduce o5x3o3o into a "flattened" variant with 30 J91's around its equatorial sphere instead of o5x3o's.

[...]
Now something I am able to explain: the thawro-ing of the o5x3o3o. To me it seems like a little more complex cut of a o5x3o3o. Compare it with a cut along a dodecahedron of a 600-cell: things get rearranged a bit (tetahedra get replaced by pentagonal pyramids), and if you place the two halves back on each other, the thing isn't CRF anymore. Furthermore it doesn't affect the rest of the polytope. Therefore I am pretty sure the mega-wedge is possible.
The bilbro-ing of the o5x3o3o has some similar properties, and therefore it may be seen as a special cut as well (maybe it's comparable with making a square orthobicupola out of an rhombicuboctahedron). this means both of these operations can be combined with other diminishings, yielding combinatorial explosion

Yeah, it seems that the J91 and J92 based "crown jewels" are perhaps not that special anymore... or at least, they are special but not as unique as we first thought. There appear to be general construction schemes that yields large numbers of CRFs containing J91's and J92's. But OTOH, this is good news, in the sense that if the general construction principles are known, it can serve as the basis of a more consistent naming scheme for these CRFs than we currently have.

And on that note, I wonder if analogous "bilbro-ing" or "thawro-ing" of o5x3o3x would also be CRF (or CRF-able).

Also, I'm thinking whether a CRF with two orthogonal rings of o5x3o's replaced with J92's is possible. Maybe there's a J92 analogue of the grand antiprism? Would that be the "grand hebesphenorotunda"?

[quickfur]

Now you guys are going too fast for me !!those things you've found are extremely awesome (especially the J91/J92-thing).
If I'm right, these things look like J92-thing: xofxF(Vo)Fxfox3xFxoo(xo)ooxFx5xoxFf(oV)fFxox&#xt, where the parentheses mean those things are in the same layer.
and xofxfox3xFxoxFx5xoxFxox&#xt for the J91/J92-thing.

You got them both right.
I could provide their incmats too, if desired.
(I've calculated 'em this weekend. - Be warned, those are rather large, so)
[...]

Maybe post them on the wiki? Or on your website, then link to them from the wiki? Now that we're finding a lot of new unique shapes, I think we should start applying the one-wiki-page-per-CRF proposal, just so these things don't get lost in the volume of this thread.

[Keiji]

Alright, I've got a working .def and .off file generator on the wiki at :pvtest

For examples, (irregular) pyrochoron should be at :pvtest?coords_hash=0CYRWCRAN6M6PBR2WM56CTAR5S and aerochoron at :pvtest?coords_hash=05K3PZFD87YZMBY59SNNWPCMWC.

Yes, I realise it doesn't let you put a name in and just calls the polytope "pvtest" followed by the source hash, I'd like to give the opportunity to name it, but issues arose with that and I've been coding this all day

Also, re incidence matrices, yesterday I added basic integration for the polytope explorer into the wiki, so all the polytopes that were on there now automatically have their imats on the wiki. However, I don't have a way for anyone other than me to add them yet.

[Marek14]

Hm, opening either of those sample links throws error for me.

[Keiji]

That's because you need to log in for it to work - I've just added an explicit check so it now tells you as such instead of erroring

[quickfur]

That's because you need to log in for it to work - I've just added an explicit check so it now tells you as such instead of erroring

I'm logged in, and it still doesn't work. The pvtest link works, but the two sample files don't.

[Keiji]

Ah, now I see the problem.

I've rearranged some of the permission checks, so you can now see already-generated polytopes without having to log in - you need only log in to generate new ones.

[quickfur]

Yay, now it works.

Are you eventually going to support uploading of polyview scripts too?

[Keiji]

I haven't had a good look at the way the scripts work yet, but doesn't the script generally define the projection and camera angle and colors and so on, and then include a .def file?

If so I was thinking of leaving the include out of the script and uploading them separately, so you could combine any .def file with any script.

[Marek14]

OK, tried it for the 4||op and it works nicely

[student91]

I was just thinking on how to bilbro the o5x3o3x, and got something completely unrelated!!
The runcitruncated 600-cell x5o3x3x can have two simultaneous cuts that intersect, giving something that is not CRF, bit it is somewhat interesting. It looks like xxxFxxx5oxuxuxo&#xr.
The o5x3o3x can have a somehow related diminishing, giving xxxxx3oxFLFxo&#xr. (L=2f, I don't know the symol for that).
Those cuts don't have direct applicaions in CRF's, but they're interesting anyhow.
I think the bilbro-ing of the o5x3o3x is most likely to succeed. When looking at your render, I would build it up to this picture, cut the yellow cuboctahedron and the blue icosidodecahedron in half, then apply the bilbro-ing (i.e. cutting the red cuboctahedron in half, copy the whole thing and then place the two halves together at the hexagons of the bisected red cuboctahedra.) . Now where the yellow cuboctahedron and the blue icosidodecahedron are bisected, a x3x5x-pit occurs. this pit shouldn't be CRF, but I think it's easilly CRF-able.

[Klitzing]

I was just thinking on how to bilbro the o5x3o3x, and got something completely unrelated!!
The runcitruncated 600-cell x5o3x3x can have two simultaneous cuts that intersect, giving something that is not CRF, bit it is somewhat interesting. It looks like xxxFxxx5oxuxuxo&#xr.
The o5x3o3x can have a somehow related diminishing, giving xxxxx3oxFLFxo&#xr. (L=2f, I don't know the symol for that).
Those cuts don't have direct applicaions in CRF's, but they're interesting anyhow.

I too already have considered similar cuts. The main problem here is that your flatened Pentagons out of the respective lace cities (just as mine considered rhombs) would use some 144 degrees cutting plane angle. That one might or might not match with a similar partial complex. But that specific angle does not allow for closure within a single circuit. Therefore it would not be CRF.

--- rk

[student91]

I was just thinking on how to bilbro the o5x3o3x, and got something completely unrelated!!
The runcitruncated 600-cell x5o3x3x can have two simultaneous cuts that intersect, giving something that is not CRF, bit it is somewhat interesting. It looks like xxxFxxx5oxuxuxo&#xr.
The o5x3o3x can have a somehow related diminishing, giving xxxxx3oxFLFxo&#xr. (L=2f, I don't know the symol for that).
Those cuts don't have direct applicaions in CRF's, but they're interesting anyhow.
I think the bilbro-ing of the o5x3o3x is most likely to succeed. When looking at your render, I would build it up to this picture, cut the yellow cuboctahedron and the blue icosidodecahedron in half, then apply the bilbro-ing (i.e. cutting the red cuboctahedron in half, copy the whole thing and then place the two halves together at the hexagons of the bisected red cuboctahedra.) . Now where the yellow cuboctahedron and the blue icosidodecahedron are bisected, a x3x5x-pit occurs. this pit shouldn't be CRF, but I think it's easilly CRF-able.

The runcitruncated 600-cell diminishing doesn't seem to be unrelated at all!! in fact, it is exactly this cut that can be inserted to close the bilbro'd o5x3o3x up!!.
Think about it this way: the equatorial part of the thing I constructed has a lace tower that looks like this: xxFxx5xuxux3LfxfL&#xt. (there are some vertices skipped in this lace tower, but those vertices are all the vertices that participate in the "pit" EDIT: thus those vertices are only a few of the vertices of the full polytope. in fact, they're exactly the vertices that are part of at least one of the two x5x3x-cuts that made the pits. (one "north", one "south") The below lace city is just one pit, that would occur at the side of a lace city of the complete polytope.). This gives us a lace city that looks like this:

Code: Select all

        x5x

    x5u

F5x

    x5u

        x5x

Again, those are only the vertices that participate in one "pit". The only thing we have to do now, is close up this pit. luckily, the cut-off part of the x5o3x3x is exactly what fits into this, and the non-CRF part of this cut-off becomes part of a bilbro!!! .
the lace city would then become this:

Code: Select all

        x5x

    x5u     x5o

F5x

    x5u     x5o

        x5x

it's beatifully closed. This means a bilbro-ing of the o5x3o3x is indeed possible

[student5]

hi, I am new to this thread and, frankly, not very good at CRF polychora, but an interesting thought crossed my mind;
could you stack J92 on top of the hexagon of a x3x3o, and then fill up some spaces with interesting stuff?

[Marek14]

hi, I am new to this thread and, frankly, not very good at CRF polychora, but an interesting thought crossed my mind;
could you stack J92 on top of the hexagon of a x3x3o, and then fill up some spaces with interesting stuff?

Hmm... interesting. There are two possible orientations, based on whether the triangular faces of x3x3o will be next triangular or square faces of J92.

If you put square faces there, triangular faces of two adjoining J92's can connect. Not sure how to continue or what to put on triangular faces of x3x3o -- triangular cupola looks like a possibility.

What if you use the other orientation? This actually also looks interesting. It should be possible to fit octahedra to triangular faces of x3x3o, triangular prisms between two square faces of J92 and square pyramids or octahedra into the vertices...

Not sure how it would continue, with either shape, but the beginning looks interesting in both cases.

[Klitzing]

sorry student91, cannot follow what you were heading here with that x3o3x5o stuff. In fact the former descriprion was already unclear to me. And then how you would like to fill these parts then in Looks even more unclear to me...

Btw. it is BILunaBIROtunda, and not just a bro'

--- rk

[student91]

I know my second explanation was somewhat unclear, I just didn't have time to type a more clear one.
Basically what I'm trying is "bilbro-ing" (I'm using "to bilbro" as a verb here ) the o5x3o3x. this means I want to change the equatorial id's into bilbro's: These should be changed into bilbro's.
The way I wanted to change those into bilbro's is similar to bilbro-ing the o5x3o3o. To bilbro the o5x3o3o, you delete most of the equatorial vertices, and glue the two halves together. The equatorial id's then become bilbro's.
Similarly, if you want to bilbro the o5x3o3x, you have to delete some of the equatorial vertices. In my first post I tried to explain what vertices should be deleted. I'll explain again what vertices those are, according to the same image: You can see some red cuboctahedra here, that are projected like hexagons. the one in the middle is the clearest. Three edges of that hexagon are connected to pentagonal prisms in this picture. Those edges are exactly the edges of the id's that should be glued "north to south" to make a bilbro. (two pink pentagonal prisms and two triangles of cuboctahedra make a bilbro-depression, two such depressions make a bilbro.) This means, we have to take this hexagon, and place it to the same hexagon on the "south." When this is done, the bilbro-depressions make a bilbro. All the vertices that are in between the north and south hexagons should be deleted. Those vertices are the triangles of the yellow cuboctahedra, the triangles of the red "hexagon" cuboctahedra, and the pentagons of the blue id's. This diminishes the blue id's to pero's, and the red and yellow cuboctahedra to triangular cupola's.

Now there is a non-CRF part in the polytope derived by the explained construction. That is, where the id's and the yellow cuboctahedra are diminished, a deppresion occurs. This depression looks like a part of a x5x3x. In fact, it is the xxF5xux&#xt-part. This part can be recognised as follows: the bottom of the blue id looks like x5x. Then the diagonals of the yellow cuboctahedra, together with some edges of the to-be bilbro's make a x5u. Finally the edges of the red "hexagon" cuboctahedron together with a diagonal of the to-be bilbro makes a F5x. The hexagons, (that means the F5x3x's, only the F5x-part is used) are placed atop each other, and that's how I derived my lace city:

Code: Select all

        x5x

    x5u

F5x

    x5u

        x5x

Now as I said, this can be made CRF by inserting a part of a x5o3x3x. I don't think that will cause problems .
I hope it's clear now, I know I'm not always that clear .

A thawro-ing of o5x3o3x doesn't seem to work out nicely, it would be capped of with a u5x3x, and that's not CRF, so it's not a nice cut. However, it might be CRF-able in another way. Anyway it's not as general-applicable as the other things we've found recently.

[Klitzing]

Hmmm,

you diminish the red coes to tricues. The hexagons of which then will be connected of the opposite partial cap, that is already would be dyaically closed. (In fact you either could consider those here as pairs of red tricues, or alternatively as one single new triangular orthobicupola = tobcu!) The introduction of the bilbiroes onto the the pink pips faces also is clear. - It just remains, how to fill the dimples of halved blue ids (then peroes, the decagon of which has to be reconnected somehow) and the halved yellow coes (then further tricues, the hexagons of which have to be connected somehow). - Also the visible squares of the deeper coes then are to be reconnected to something, and likewise the lacing squares and triangles of the lunes of the new bilbiroes remain to be connected.

Consider the yellow coes first. Those show up now their diametral hexagons. One edge of which is connected to the red hexagons (or diametrals of the tobcues). Thus at this edge 2 yellow hexagons will connect (from either sandwiching cap). The incident angles moreover are filled by the incident bilbiro's lunes, i.e. by triangles each. Thus those yellow replacements would become tuts. Therefrom it now follows that the visible red squares of the deeper coes (of either cap) and the square of the lune of the incident bilbiro would be fitted by an hexagonal prism (hip). The hexagons of which then connect to the remaining hexagons of the newly introduced tuts.

This finally shows that the remainder, i.e. the replacement at the blue id withdrawn part would ask for the there created decagons to be adjacent to pecues: the squares of which then connect to further squares of the new hips, and the triangles would connect to the triangles of the new tuts. And between the 2 pentagons of the pecues of euther side there will be incident those edges of the new tuts connecting the just dihedrally closed triangles. Between each pair of those lacing edges there will be the still remaining square face of the new hips. This outlines thus a pip!

That is, any of the 12 blue dimples (after halving the blue ids to peroes and the yellow coes to tricues) gets filled by a ring of 5 tuts and 5 hips, around an axis out of 2 pecues and 1 pip. - You further might like to do the above mentioned replacement of the 20 to tricues diminished red coes (of either cap) into 20 tobcues instead. - The remainder of the 2 srix (= x3o3x5o) caps so would remain as before. - Just that you might could apply usual (non-intersecting) diminishings to each cap as well...

And yes, those introduced tuts, hips, pecues (as parts of srids), and pips are the cells of prahi (= x3x3o5x). That is, your nose led you into the right direction here. - Even so, I still cannot see, how you aimed to get those flattened pentagonal parts of the prahi's lace city into those dimples... (Maybe this is exactly what I was describing here in detail, but I don't see that so far.)

--- rk

[student91]

Hmmm,

you diminish the red coes to tricues. The hexagons of which then will be connected of the opposite partial cap, that is already would be dyaically closed. (In fact you either could consider those here as pairs of red tricues, or alternatively as one single new triangular orthobicupola = tobcu!)

no, the tricues will have a dichoral angle of 164.4775..., so they won't make tobcu

[...] That is, your nose led you into the right direction here. - Even so, I still cannot see, how you aimed to get those flattened pentagonal parts of the prahi's lace city into those dimples (Maybe this is exactly what I was describing here in detail, but I don't see that so far.)

--- rk

Indeed you described exactly what I did. . If you'd cut two x5x3x's out of the o5x3o3x before the bilbiro-ing, you get a lace-city that looks like this:

Code: Select all

x5x

    x5u

        F5x

            U5x

        F5x

    x5u

x5x

Left to this is where the vertices inside the x5x3x's are deleted. Right of this is where the rest of the polytope is. I didn't draw that. You can see this are two x5x3x-cuts, at an angle of 144 to each other. If we bilbiro the o5x3o3x, the U5x gets deleted, and the F5x's get placed atop each other. The angle between the towers stays the same. This means it looks like this:

Code: Select all

x5x

    x5u

        F5x

    x5u

x5x

Now this is exactly the outline of the x5o3x3x-cut (that one also has an angle of 144):

Code: Select all

        x5x

    x5u     x5o

F5x

    x5u     x5o

        x5x

This means we can put this outline on top of the outline we have in the lace-city above (i.e. in the bilbiro'd o5x3o3x). Furthermore, because the x5o3x3x has a bigger circumradius than o5x3o3x, everything stays CRF.
It wasn't really my nose that led me there, I searched for something that could be inserted, but didn't see it could be at first.

[quickfur]

I know my second explanation was somewhat unclear, I just didn't have time to type a more clear one.
Basically what I'm trying is "bilbro-ing" (I'm using "to bilbro" as a verb here ) the o5x3o3x.

Hooray for bilbro-ing! (Or bilbiroing, as Klitzing would have it.)

this means I want to change the equatorial id's into bilbro's: These should be changed into bilbro's.
The way I wanted to change those into bilbro's is similar to bilbro-ing the o5x3o3o. To bilbro the o5x3o3o, you delete most of the equatorial vertices, and glue the two halves together. The equatorial id's then become bilbro's.
Similarly, if you want to bilbro the o5x3o3x, you have to delete some of the equatorial vertices.

Note that there are multiple ways to "bilbiro" the o5x3o3x according to the above image. If you place the J91 (bilbiro) such that its pentagonal faces and the triangles between them fall on the outline of pentagons and triangles in each o5x3o in the image, you'll get what you're trying to describe here. But if you place the J91's sideways, such that they form rhombus-shaped projections, then you can collapse each o5x3o seen above sideways into rhombuses (i.e., J91's), and the result would be the Stott-expanded castellated prism!

So then, this makes me wonder if there are actually many other ways to "squash" the o5x3o's into J91's that we just haven't considered yet, that could lead to novel CRFs. For example, might it be possible to "bilbiro" a o5x3o3x (or, for that matter, o5x3o3o), such that the J91's lie in non-parallel hyperplanes in skew orientations? If this is possible (granted, very big "if"), it could become a whole new class of crown-jewel CRFs that feature skewed Johnson solid cells. That would be truly cool!

In my first post I tried to explain what vertices should be deleted. I'll explain again what vertices those are, according to the same image: You can see some red cuboctahedra here, that are projected like hexagons. the one in the middle is the clearest. Three edges of that hexagon are connected to pentagonal prisms in this picture. Those edges are exactly the edges of the id's that should be glued "north to south" to make a bilbro. (two pink pentagonal prisms and two triangles of cuboctahedra make a bilbro-depression, two such depressions make a bilbro.) This means, we have to take this hexagon, and place it to the same hexagon on the "south." When this is done, the bilbro-depressions make a bilbro. All the vertices that are in between the north and south hexagons should be deleted. Those vertices are the triangles of the yellow cuboctahedra, the triangles of the red "hexagon" cuboctahedra, and the pentagons of the blue id's. This diminishes the blue id's to pero's, and the red and yellow cuboctahedra to triangular cupola's.

I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...

[Marek14]

Hm, I'm not getting any output from pvtest when trying this set of vertices:

Code: Select all

1.00000000000000000000 1.00000000000000000000 1.28718850581116524947 0.00000000000000000000
1.00000000000000000000 -1.00000000000000000000 1.28718850581116524947 0.00000000000000000000
-1.00000000000000000000 1.00000000000000000000 1.28718850581116524947 0.00000000000000000000
-1.00000000000000000000 -1.00000000000000000000 1.28718850581116524947 0.00000000000000000000
1.41421356237309514547 0.00000000000000000000 0.910179721124454682609 0.910179721124454682609
-1.41421356237309514547 0.00000000000000000000 0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 1.41421356237309514547 0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 -1.41421356237309514547 0.910179721124454682609 0.910179721124454682609
1.00000000000000000000 1.00000000000000000000 0.00000000000000000000 1.28718850581116524947
1.00000000000000000000 -1.00000000000000000000 0.00000000000000000000 1.28718850581116524947
-1.00000000000000000000 1.00000000000000000000 0.00000000000000000000 1.28718850581116524947
-1.00000000000000000000 -1.00000000000000000000 0.00000000000000000000 1.287188505811165249470
1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 0.910179721124454682609
-1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 1.41421356237309514547 -0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 -1.41421356237309514547 -0.910179721124454682609 0.910179721124454682609
1.00000000000000000000 1.00000000000000000000 -1.28718850581116524947 0.00000000000000000000
1.00000000000000000000 -1.00000000000000000000 -1.28718850581116524947 0.00000000000000000000
-1.00000000000000000000 1.00000000000000000000 -1.28718850581116524947 0.00000000000000000000
-1.00000000000000000000 -1.00000000000000000000 -1.28718850581116524947 0.00000000000000000000
1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 -0.910179721124454682609
-1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 -0.910179721124454682609
0.00000000000000000000 1.41421356237309514547 -0.910179721124454682609 -0.910179721124454682609
0.00000000000000000000 -1.41421356237309514547 -0.910179721124454682609 -0.910179721124454682609
1.00000000000000000000 1.00000000000000000000 0.00000000000000000000 -1.28718850581116524947
1.00000000000000000000 -1.00000000000000000000 0.00000000000000000000 -1.28718850581116524947
-1.00000000000000000000 1.00000000000000000000 0.00000000000000000000 -1.28718850581116524947
-1.00000000000000000000 -1.00000000000000000000 0.00000000000000000000 -1.287188505811165249470
1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 0.910179721124454682609
-1.41421356237309514547 0.00000000000000000000 -0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 1.41421356237309514547 -0.910179721124454682609 0.910179721124454682609
0.00000000000000000000 -1.41421356237309514547 -0.910179721124454682609 0.910179721124454682609


It should be a ring of 8 square antiprisms. I tried four and six, but didn't see anything that could be added to them to form a CRF...

[Keiji]

Found the problem in my logs:

ZGTDM4RXYD2K889SJ5ZGAV028Q: failed: Error: Command failed: Caught exception: Face 39 not closed!

At some point I'll need to make it report these errors back to the user...

[Marek14]

Why is face 39 not closed...?

BTW, ring of 4 antiprisms had side cells as two types of non-CRF tetrahedra and sing of 6 had tetrahedra and octahedra, both non-CRF. I'm trying to see whether the ring will eventually allow a kind of snub -- two orthogonal rings of square antiprisms with an additional layer of vertices or cells between them.