Bilbirothawroids (D4.3 to D4.9)

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

Bilbirothawroids (D4.3 to D4.9)

Postby quickfur » Tue Feb 04, 2014 4:59 am

I may have discovered a crown jewel candidate(!!!). Well, I'm not 100% sure it exists or is CRF just yet, but I'm posting here for you bright minds out there to verify whether it's possible. :)

Marek's post about the bilunabirotunda (J91) inspired me to investigate possible CRF combinations of this shape. One of J91's interesting characteristics is that it contains a pentagon-triangle-pentagon-triangle combination with exactly the same dihedral angles as the icosidodecahedron o5x3o (probably this is why Johnson named it bilunabirotunda). For convenience of reference, lemme just borrow Wikipedia's J91 image here:

Image

The most interesting thing about this part of J91 is that the angle between the two pentagons around their common vertex is exactly equal to the dihedral angle of the regular dodecahedron, or, equivalently, is the same as in the icosidodecahedron o5x3o. This fact gave me an idea: suppose we take three J91's and paste them together at their pentagonal faces, such that the three J91's share the vertex between two pentagons and two triangles. Since the pentagon-triangle-pentagon-triangle combination is just a section of o5x3o, the angle defect will be positive, which means we can then fold the J91's into 4D so that these pentagonal faces are joined to each other. The dichoral angles around this vertex, then, must be exactly the same as in the rectified 120-cell o5x3o3o. We can therefore insert tetrahedra to fill in the remaining gaps around this vertex, and the dichoral angles that the tetrahedra make around this vertex will be exactly the same as in o5x3o3o.

Now, take a look at this projection of o5x3o3o, showing its equatorial o5x3o cells:

Image

Notice that where three o5x3o's meet, there's the outline of a tetrahedron? So the dichoral angles of the cells around this tetrahedron would match the dichoral angles of the J91's that we have folded together into 4D. Now notice that on the opposite side of the o5x3o's in the above image, we have an identical tetrahedral configuration of cells: so if we ignore the middle portion of the o5x3o, say if we mentally "collapse" the o5x3o's so that they form rhombus-shaped projection images above, then this should represent the configuration of J91's that we have thus assembled (because the dichoral angle around the 5.3.5.3 vertex in the J91 is the same as the dichoral angle of the vertices of the o5x3o). Therefore, it seems reasonable to deduce that if we continue attaching more J91's to our set of three in the same pattern, five J91's should close up into a pentagonal formation. This implies that the J91's edges between two pentagonal faces will be where five J91's will meet (these are the top and bottom edges in the Wikipedia image of J91). Therefore, we should be able to attach 30 J91's to each other and form a closed network with icosahedral symmetry. And since the top and bottom edges of the J91's will have 5 triangles around either end, we can insert pentagonal pyramids to fill up the gaps.

This then produces a pattern of faces with the configuration 3.4.5.4 around each vertex, in icosahedral symmetry, i.e., the faces of a x5o3x. Therefore, we should be able to sandwich our 30 J91's between two x5o3x's to form a closed 4D shape. If my deductions are correct, the result should be CRF, and would be a tetrastratic polychoron with two x5o3x's, 30 J91's, 12 pentagonal pyramids, and 20 tetrahedra.

Can someone verify whether this shape actually exists, and is CRF? :)

P.S. The lace tower of this polychoron should be x5o3x || o5o3A || B5o3o || o5o3A || x5o3x where A and B are as-yet unknown scalars. The height of the tower should be exactly the Golden Ratio (chord of the pentagonal faces of the J91's).
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 04, 2014 6:35 am

student91 wrote:After reading everything carefully, and looking up what a voroni-cell is, I think I understood your post.
Am I right when I say voroni-cells look a bit like the dual cells of s3s4o3o, connected to the center of s3s4o3o to make them 4D-bodies?

One might look at it in that way. But I thought of it (and described it that way too) rather in the sense of a hyperspherical surface representant, i.e. of the surface space only. Thus my Voronoi cells here are 3D only, in fact those are the cells of the dual of sadi.

I think this polytope is the coolest of the 24-diminished .5.3.3.'s. I mean, it has J83's!! :D

Yep, 8)

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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 04, 2014 1:38 pm

quickfur wrote:I may have discovered a crown jewel candidate(!!!). Well, I'm not 100% sure it exists or is CRF just yet, but I'm posting here for you bright minds out there to verify whether it's possible. :)

Marek's post about the bilunabirotunda (J91) inspired me to investigate possible CRF combinations of this shape.

[...]

... the result should be CRF, and would be a tetrastratic polychoron with two x5o3x's, 30 J91's, 12 pentagonal pyramids, and 20 tetrahedra.

Can someone verify whether this shape actually exists, and is CRF? :)

P.S. The lace tower of this polychoron should be x5o3x || o5o3A || B5o3o || o5o3A || x5o3x where A and B are as-yet unknown scalars. The height of the tower should be exactly the Golden Ratio (chord of the pentagonal faces of the J91's).


Nice find, quickfur! Yes, indeed it is possible.

Your "A" then should be that diagonal of the decagon, which skips 2 vertices, i.e. A=x(10,3) and that value comes out to be F=ff=x+f, i.e. tau^2=2.618 times the unit edge x. - Then for "B". B should be the height of bilbiro (J91, BI-Luna-BI-ROtunda) between the 2 icosidodecahedron-type vertices. Here consider the edge between triangle and square at the bilbiro's lune together with the connecting edges emmanating therefrom to those vertices. This sequence of 3 edges spans a further pentagon. Therefroe B=f, i.e. tau=1.618 times the unit edge x.

Accordingly it could be rewritten as xFoFx3ooooo5xofox&#xt = srid || pseudo F-ike || pseudo f-doe || pseudo F-ike || srid.

Coordinates then are:
  • (tau3/2, 1/2, 1/2, tau/2) & all permutations in all but last coordinate, all changes of sign
  • (tau, tau2/2, tau/2, tau/2) & even permutations in all but last coordinate, all changes of sign
  • (tau2/2, 1+tau/2, 0, tau/2) & even permutations in all but last coordinate, all changes of sign
    (these 3 describe the polar (unscaled) srids)
  • (tau3/2, tau2/2, 0, 1/2) & even permutations in all but last coordinate, all changes of sign
    (this describes the inscribed tropal tau2-scaled ikes)
  • (tau2/2, tau2/2, tau2/2, 0) & all permutations in all but last coordinate, all changes of sign
  • (tau3/2, 0, tau/2, 0) & even permutations in all but last coordinate, all changes of sign
    (these 2 describe the inscribed equatorial tau-scaled doe)
where tau = (1+sqrt(5))/2 = 1.618

The incidence matrix also can be derived as:
Code: Select all
o....3o....5o....     & | 120  *  * |   2   2   1   1  0 |  1  2  1   2   2  2 | 1  1  1  2
.o...3.o...5.o...     & |   * 24  * |   0   0   5   0  1 |  0  0  0   5   0  5 | 0  1  0  5 verf = ox5oo&#f
..o..3..o..5..o..       |   *  * 20 |   0   0   0   6  0 |  0  0  0   0   6  3 | 0  0  2  3 verf = f x3o
------------------------+-----------+--------------------+---------------------+-----------
x.... ..... .....     & |   2  0  0 | 120   *   *   *  * |  1  1  0   0   1  0 | 1  0  1  1
..... ..... x....     & |   2  0  0 |   * 120   *   *  * |  0  1  1   1   0  0 | 1  1  0  1
oo...3oo...5oo...&#x  & |   1  1  0 |   *   * 120   *  * |  0  0  0   2   0  1 | 0  1  0  2
o.o..3o.o..5o.o..&#x  & |   1  0  1 |   *   *   * 120  * |  0  0  0   0   2  1 | 0  0  1  2
.o.o.3.o.o.5.o.o.&#x    |   0  2  0 |   *   *   *   * 12 |  0  0  0   0   0  5 | 0  0  0  5
------------------------+-----------+--------------------+---------------------+-----------
x....3o.... .....     & |   3  0  0 |   3   0   0   0  0 | 40  *  *   *   *  * | 1  0  1  0
x.... ..... x....     & |   4  0  0 |   2   2   0   0  0 |  * 60  *   *   *  * | 1  0  0  1
..... o....5x....     & |   5  0  0 |   0   5   0   0  0 |  *  * 24   *   *  * | 1  1  0  0
..... ..... xo...&#x  & |   2  1  0 |   0   1   2   0  0 |  *  *  * 120   *  * | 0  1  0  1
x.o.. ..... .....&#x  & |   2  0  1 |   1   0   0   2  0 |  *  *  *   * 120  * | 0  0  1  1
ooooo3ooooo5ooooo&#xt   |   2  2  1 |   0   0   2   2  1 |  *  *  *   *   * 60 | 0  0  0  2
------------------------+-----------+--------------------+---------------------+-----------
x....3o....5x....     & |  60  0  0 |  60  60   0   0  0 | 20 30 12   0   0  0 | 2  *  *  * srid
..... oo...5xo...&#x  & |   5  1  0 |   0   5   5   0  0 |  0  0  1   5   0  0 | * 24  *  * peppy
x.o..3o.o.. .....&#x  & |   3  0  1 |   3   0   0   3  0 |  1  0  0   0   3  0 | *  * 40  * tet
xFoFx ..... xofox&#xt   |   8  4  2 |   4   4   8   8  2 |  0  2  0   4   4  4 | *  *  * 30 bilbiro


Note that here the dihedral angle between the srids and the bilbiros (i.e. at the squares) would be 90 degrees.
Further it should be noted (what also becomes evident from the matrix) that there are no edges connecting the tropal vertices to the equatorial ones!

Btw., as you have 2 srids, the count of the therefrom emanating tets (trigonal pyramids) and peppies (pentagonal ones) gets doubled too (pointing inward from either side). - Those counts you got wrong in your mail. Cf. the above cited bit. - Only the count of the bilbiros remains undoubled, as those connect both opposite basal srids.

Thus, corrected, the cell count here is (as can be read off directly from the Matrix too):
2x srid + 24x peppy + 40x tet + 30x bilbiro.

So, having provided to you even the coordinates, we soon may get some nice pictures of that fellow?   :)

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 3:54 pm

Klitzing wrote:
quickfur wrote:I may have discovered a crown jewel candidate(!!!). [...]
P.S. The lace tower of this polychoron should be x5o3x || o5o3A || B5o3o || o5o3A || x5o3x where A and B are as-yet unknown scalars. The height of the tower should be exactly the Golden Ratio (chord of the pentagonal faces of the J91's).


Nice find, quickfur! Yes, indeed it is possible.

Hooray! Would this be the second CRF crown jewel we found? The first one is your cube||icosahedron.

[...]
Accordingly it could be rewritten as xFoFx3ooooo5xofox&#xt = srid || pseudo F-ike || pseudo f-doe || pseudo F-ike || srid.

Coordinates then are:
[...]

Thanks!! I was dreaming all night about how to calculate the coordinates, but you beat me to it. :XD: Plus, I made several mistakes in my own partial calculations, so it's a good thing I can just reference yours when the numbers don't come out right. :lol:

Note that here the dihedral angle between the srids and the bilbiros (i.e. at the squares) would be 90 degrees.
Further it should be noted (what also becomes evident from the matrix) that there are no edges connecting the tropal vertices to the equatorial ones!

Yes, this much is clear from the construction, since the pentagonal pyramids only reach down to the top/bottom edges of the bilunabirotunda, and the tetrahedra meet at the 5.3.5.3 vertices on the equator.

Btw., as you have 2 srids, the count of the therefrom emanating tets (trigonal pyramids) and peppies (pentagonal ones) gets doubled too (pointing inward from either side). - Those counts you got wrong in your mail. Cf. the above cited bit. - Only the count of the bilbiros remains undoubled, as those connect both opposite basal srids.

Thus, corrected, the cell count here is (as can be read off directly from the Matrix too):
2x srid + 24x peppy + 40x tet + 30x bilbiro.

Ah, you're right, I forgot to count the tetrahedra and pentagonal pyramids on the other side. :oops:

So, having provided to you even the coordinates, we soon may get some nice pictures of that fellow?   :)
[...]

I knew that was coming. :lol: Well, here is one:

Image

This is from the viewpoint <0,0,0,5>, looking towards the origin. The red outline shows where the top x5o3x is, and the green cell with yellow edges is where the bottom x5o3x is. The blue cell shows one of the bilunabirotunda -- they are not very easy to see from this angle, because they're too squished by the projection (they're at a 90° angle to the viewpoint after all, but not completely flattened because I'm using perspective projection rather than parallel projection). But as you can see, the bilunabirotundae have a kind of rhombus-like 3D kite shape that fit into each other in a configuration resembling the faces of the rhombic triacontahedron.

But since we're all more interested in the bilunabirotundae than the boring ol' x5o3x's, here's another viewpoint centered on the blue cell:

Image

This is a "side-view" that shows the blue bilunabirotunda in its full glory. I tweaked the 3D viewpoint a bit so that you can see the outlines of adjacent bilunabirotunda. I also turned on visibility clipping to make the image less cluttered. The tetrahedra should be pretty obvious; the pentagonal pyramids appear a bit skewed here (in 4D they protrude pretty far out, as you can see from the previous image).

Now, the all-important question is, what should we call this polychoron? The only name I could think of is triaconta-bilunabirotunda, but that just doesn't seem to sit very well. :\
Last edited by quickfur on Tue Feb 04, 2014 6:05 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 4:39 pm

Hm, maybe "lunatic srid prism"? :)
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 6:04 pm

Marek14 wrote:Hm, maybe "lunatic srid prism"? :)

:lol: :lol: I like that!

This morning the name bilunabirotundic triacontachoron came to mind, via analogy with the rhombic triacontahedron, since the parallel projection centered on an x5o3x would be an actual rhombic triacontahedron, but this name is not altogether accurate because there are other cells besides the 30 J91's. (And yes, lunatic triacontachoron would be along the same lines. :P)

The "prism" is a good idea, though, 'cos it's sorta like a prism, except more elaborate. So maybe bilunabirotundic rhombicosidodecahedron prism? Kinda wordy, I know, but I do like the Johnson- / Olshevsky-style names. :XD:
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 6:11 pm

Marek14 wrote:Hm, maybe "lunatic srid prism"? :)

Oh, btw, I keep forgetting you want stuff in Stella4D format, so here's the file for this lunatic prism. :P
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 6:29 pm

quickfur wrote:
Marek14 wrote:Hm, maybe "lunatic srid prism"? :)

Oh, btw, I keep forgetting you want stuff in Stella4D format, so here's the file for this lunatic prism. :P


Thanks!

I'm trying to investigate the cuts, but it doesn't seem to have any obvious diminishings.I wonder if the pentagonal pyramids could be augmented -- the height would be relatively small...

I tried to make a VRML file, not sure if it worked...
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 8:16 pm

That thing is just super awesomely cool. :o_o:
I mean, ursachora are pretty cool, and this one is like 20 times cooler.

Speaking of which, it has some similarity with ursachora. Ursachora are based on the xfo3oox-buildup of a trid.ike, and this one is based on the xFoFx(2)xofox-buildup of the bilunabirotunda. Therefore, as the ursachora are possible with other symmetries as well, this one might be too, so e.g. xFoFx3ooooo3xofox&#xt might be possible as well, and maybe expanded ones, so xFoFx3xxxxx5xofox&#xt too. I'm not sure about this, and just extrapolated the ursachora to this one. Of course I couldn't've done this extrapolation without your awesome discovery. It's real awesome :) :D


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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 8:36 pm

Hm, J92 also has a part with o5x3o shape, but the rest of it would probably pose problems...
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 9:00 pm

I think so too. J92 has C3v-symmetry, so it should be viewed as xfox3oxFx&#xt. (J91 has D2h-symmetry, so that one can be seen as either xfofx(2)oxfxo&#xt, ofxfo(2)oxFxo&#xt or xFoFx(2)xofox&#xt. the last one does work, the other ones haven't been tested) If this one is extended to xfox3oxFx3oooo&#xt, we have oooo(2)xfox&#x, not giving a facet, xfox3ox&#xt, giving J92, and oxFx3oooo&#xt, giving a non-CRF. EDIT:this view isn't correct yet, as it would say the new thing doesn't exist as well.
Last edited by student91 on Tue Feb 04, 2014 9:05 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 9:04 pm

student91 wrote:I think so too. J92 has C3v-symmetry, so it should be viewed as xfox3oxFx&#xt. (J91 has D2h-symmetry, so that one can be seen as either xfofx(2)oxfxo&#xt, ofxfo(2)oxFxo&#xt or xFoFx(2)xofox&#xt. the last one does work, the other ones haven't been tested) If this one is extended to xfox3oxFx3oooo&#xt, we have oooo(2)xfox&#x, not giving a facet, xfox3ox&#xt, giving J92, and oxFx3oooo&#xt, giving a non-CRF.


For the third possibility, could two be glued together to restore CRF properties?
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 9:08 pm

Marek14 wrote:
quickfur wrote:
Marek14 wrote:Hm, maybe "lunatic srid prism"? :)

Oh, btw, I keep forgetting you want stuff in Stella4D format, so here's the file for this lunatic prism. :P


Thanks!

I'm trying to investigate the cuts, but it doesn't seem to have any obvious diminishings.

As befits a true crown jewel. ;)

I wonder if the pentagonal pyramids could be augmented -- the height would be relatively small...

That's an interesting idea... I'm not sure, it doesn't look like it would remain convex, but you never know with these things until you try it. Maybe some time later today I can try to add a point above the pentagonal pyramids to see if the result still comes out CRF. If it does, then since the pentagonal pyramids are well-separated from each other, it would have an omni-augmentation as well, which would be even more lunatic. :lol:

I tried to make a VRML file, not sure if it worked...

Hmm, I don't have a VRML player at work, and probably the one on my PC doesn't work anymore (it's been a loooong time since I've used/updated it), so I'll have to try this another time.

student91 wrote:That thing is just super awesomely cool. :o_o:
I mean, ursachora are pretty cool, and this one is like 20 times cooler.

Ooh, I forgot about the ursachora... I think they may qualify as crown jewels too, and Wendy discovered the first one. So the "lunatic prism" should count as the third crown jewel we found? Note that although the original icosahedral ursachoron is a diminishing of the 600-cell, the tetrahedral and octahedral variants with their expanded siblings are not diminishings of any uniforms, so they would qualify as crown jewels too.

Speaking of which, it has some similarity with ursachora. Ursachora are based on the xfo3oox-buildup of a trid.ike, and this one is based on the xFoFx(2)xofox-buildup of the bilunabirotunda. Therefore, as the ursachora are possible with other symmetries as well, this one might be too, so e.g. xFoFx3ooooo3xofox&#xt might be possible as well, and maybe expanded ones, so xFoFx3xxxxx5xofox&#xt too. I'm not sure about this, and just extrapolated the ursachora to this one. Of course I couldn't've done this extrapolation without your awesome discovery. It's real awesome :) :D
[...]

Thanks!

And yes, I think it should be possible to apply Stott expansion to the "lunatic prism" to get a new CRF. You'd get x5x3x's for the top and bottom cells, and in between J91's separated by pentagonal prisms, then the remaining gaps can get filled up by triangular cupolae and pentagonal cupolae (which are nothing but Stott-expanded trigonal/pentagonal pyramids). So it should be a CRF with two x5x3x's, 30 J91's, 60 pentagonal prisms, 40 triangular cupolae, and 24 pentagonal cupolae.

I'm not so sure about the possibility with other symmetries, because a degree-3 edge is rigid, and we have 3 J91's around an edge. So it may not close up if we have only 4 J91's, I think. (In fact, it was this rigidity that initially led me to consider icosahedral symmetry, because I was originally thinking of tetrahedra or square pyramids for those J91 triangles around the top/bottom edges, but I wasn't sure if the angles will match up properly. At least with icosahedral symmetry there's precedent for a 5-fold symmetry around that edge, which allows CRF pentagonal pyramids to fill the gap.)

Marek14 wrote:Hm, J92 also has a part with o5x3o shape, but the rest of it would probably pose problems...

Whoa, you guys are fast... I'm still not done with J91 yet. ;) I still have a few other ideas up my sleeve with J91, that might lead to more CRF crown jewels. So I'm going to stick with J91 for a little bit longer. :P

OTOH, J92 does look like you should be able to stick four of them around a tetrahedron and get something analogous to one of the 120-cell uniforms out of it. Not sure how the rest of the shape will close up, but that's a start!
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 9:18 pm

Marek14 wrote:For the third possibility, could two be glued together to restore CRF properties?

I think you didn't understand what I ment, probably because my way of thinking has an error. I'll try to eplain my thoughts anyway. xfox3oxFx3oooo&#xt is a stack of sets of vertices. viewed form the first 3, it has xfo3oxFx&#xt. this occurs in the stack of vertices, and gives the J92. The second 3 gives oxFx3oooo&#xt, which also occurs in the stack of vertices. This stack is not CRF, although maybe the x's connect, and skip the F, making a x3o2x. I'm not sure, but I still think it's impossible

quickfur wrote:OTOH, J92 does look like you should be able to stick four of them around a tetrahedron and get something analogous to one of the 120-cell uniforms out of it. Not sure how the rest of the shape will close up, but that's a start

I already checked if it'd be possible to place it around a octahedron resp. icosahedron. the angle defect at the octahedron exceeds 360, so that one is clearly impossible.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 9:32 pm

Well, you can't place J92's around octahedron or icosahedron since that would be a "shard" of o5x3o4o or o5x3o5o, which are hyperbolic.

Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.
Last edited by Marek14 on Tue Feb 04, 2014 9:33 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby Keiji » Tue Feb 04, 2014 9:32 pm

quickfur wrote:Now, the all-important question is, what should we call this polychoron? The only name I could think of is triaconta-bilunabirotunda, but that just doesn't seem to sit very well. :\


Now this is a task for Keiji :D

It's very similar to a birotunda, having 5 layers of vertices just like our 3D birotundae.

However, it doesn't have a contour for the center layer, rather it has a kind of castellated edge like the dodecahedron does.

The ends are rhodoperihedra.

This leads me to call it the castellated rhodoperihedral birotunda - or for short, a carper! :D

(If the stauro- form exists, that would be a casper, and if the pyro- form exists, that would be a capper. But I have a feeling those wouldn't exist for the same reason that you don't get trigonal or square rotundae in 3D.)
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 9:37 pm

quickfur wrote:[...]
I'm not so sure about the possibility with other symmetries, because a degree-3 edge is rigid, and we have 3 J91's around an edge. So it may not close up if we have only 4 J91's, I think. (In fact, it was this rigidity that initially led me to consider icosahedral symmetry, because I was originally thinking of tetrahedra or square pyramids for those J91 triangles around the top/bottom edges, but I wasn't sure if the angles will match up properly. At least with icosahedral symmetry there's precedent for a 5-fold symmetry around that edge, which allows CRF pentagonal pyramids to fill the gap.)

Remember that a symmetrical edge is also rigid, and it's dihedral angle is calculated even easier. This means that your discovered polytope has rigid edges as well.
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 9:40 pm

Keiji wrote:[...]
(If the stauro- form exists, that would be a casper, and if the pyro- form exists, that would be a capper. But I have a feeling those wouldn't exist for the same reason that you don't get trigonal or square rotundae in 3D.)

You got a point there, let me check.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 9:47 pm

Keiji wrote:
quickfur wrote:Now, the all-important question is, what should we call this polychoron? The only name I could think of is triaconta-bilunabirotunda, but that just doesn't seem to sit very well. :\


Now this is a task for Keiji :D

It's very similar to a birotunda, having 5 layers of vertices just like our 3D birotundae.

The problem is that it's not a rotunda (in the sense of being dome-shaped) at all! There is nothing rotunda-like about it; it's two parallel x5o3x's connected by a bunch of lacing cells, so it's more like a chicken sandwich than a bun (i.e. birotunda :P).

However, it doesn't have a contour for the center layer, rather it has a kind of castellated edge like the dodecahedron does.

I like the adjective castellated. So maybe we can call it a castellated rhombicosidodecahedral prism?

The ends are rhodoperihedra.

This leads me to call it the castellated rhodoperihedral birotunda - or for short, a carper! :D

(If the stauro- form exists, that would be a casper, and if the pyro- form exists, that would be a capper. But I have a feeling those wouldn't exist for the same reason that you don't get trigonal or square rotundae in 3D.)

I really don't see what's rotundic about it; I think prism would be a much better word here. After all, it's just a prism of x5o3x with slightly more elaborate side cells (the J91's are at 90° to the x5o3x's, btw!). I like "castellated", though. So "castellated x5o3x prism" would be my choice of name. It's a lot saner than "lunatic prism", that's for sure. :lol:

But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.
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Re: Johnsonian Polytopes

Postby Keiji » Tue Feb 04, 2014 9:55 pm

quickfur wrote:the J91's are at 90° to the x5o3x's, btw!


Ooh, now I see what's going on. Yes, that's not rotundaeic at all. Castellated rhodoperihedral prism it is!
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 9:57 pm

quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.


Could something more be inserted, if you'd elongated the pyramids or something in that vein? Probably not...
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 10:00 pm

quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.

Note that the ursachora have similar edges (3 tridiminished icosahedra placed face-to-face, and a pyramid filling the gap). This one also works out.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 10:15 pm

student91 wrote:
quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.

Note that the ursachora have similar edges (3 tridiminished icosahedra placed face-to-face, and a pyramid filling the gap). This one also works out.

The difference there is that the number of tridiminished icosahedra around an edge is flexible. Here, the number of J91's around the 5.3.5.3 vertex can only be 3, because otherwise it becomes hyperbolic (the angle of the two pentagons is too wide to fit more than 3 J91's around that vertex). So the dichoral angles between the J91's is fixed, which means that if you continue the pattern of 3 J91's around the 5.3.5.3 vertices, it will eventually form an icosahedral framework of J91's, which means the top/bottom edges will have 5 J91's surrounding it.

Now, it may be possible to have other numbers of J91's around that edge, but it means that you cannot also have the repeating pattern of 3 J91's around the 5.3.5.3 vertices anymore, but you'll need some other arrangement of cells to close the shape up.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 10:20 pm

Marek14 wrote:Well, you can't place J92's around octahedron or icosahedron since that would be a "shard" of o5x3o4o or o5x3o5o, which are hyperbolic.

Quite so.

Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.

What's wrong with putting 4 J92's around a tetrahedron? It would appear that we should be able to at least insert 4 more tetrahedra into the result (i.e. following the pattern of o5x3o3o), which then leaves only the outer surface of hexagons, squares, and triangles, to be closed up. Seems like it should be possible? Or am I missing something obvious?
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 10:29 pm

quickfur wrote:
Marek14 wrote:Well, you can't place J92's around octahedron or icosahedron since that would be a "shard" of o5x3o4o or o5x3o5o, which are hyperbolic.

Quite so.

Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.

What's wrong with putting 4 J92's around a tetrahedron? It would appear that we should be able to at least insert 4 more tetrahedra into the result (i.e. following the pattern of o5x3o3o), which then leaves only the outer surface of hexagons, squares, and triangles, to be closed up. Seems like it should be possible? Or am I missing something obvious?


Well, that was something which was already suggested. My suggestion was meant to be "in addition to", not "in place of" the previous one.
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 10:35 pm

quickfur wrote:
student91 wrote:
quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.

Note that the ursachora have similar edges (3 tridiminished icosahedra placed face-to-face, and a pyramid filling the gap). This one also works out.

The difference there is that the number of tridiminished icosahedra around an edge is flexible. Here, the number of J91's around the 5.3.5.3 vertex can only be 3, because otherwise it becomes hyperbolic (the angle of the two pentagons is too wide to fit more than 3 J91's around that vertex). So the dichoral angles between the J91's is fixed, which means that if you continue the pattern of 3 J91's around the 5.3.5.3 vertices, it will eventually form an icosahedral framework of J91's, which means the top/bottom edges will have 5 J91's surrounding it.

Now, it may be possible to have other numbers of J91's around that edge, but it means that you cannot also have the repeating pattern of 3 J91's around the 5.3.5.3 vertices anymore, but you'll need some other arrangement of cells to close the shape up.


Ooh, you ment that edge!!, After a quick reflection I see what you mean, and I understand why it's impossible :( . I apologise for being so stubborn. the way I concluded It was impossible is by looking at the edge with the tetahedron and two bilunabirotunda's. This edge fixes the bichoral angle for the pentagon, and thus 5 is the only possibility :sweatdrop: .
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 04, 2014 10:39 pm

student91 wrote:
quickfur wrote:
student91 wrote:
quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.

Note that the ursachora have similar edges (3 tridiminished icosahedra placed face-to-face, and a pyramid filling the gap). This one also works out.

The difference there is that the number of tridiminished icosahedra around an edge is flexible. Here, the number of J91's around the 5.3.5.3 vertex can only be 3, because otherwise it becomes hyperbolic (the angle of the two pentagons is too wide to fit more than 3 J91's around that vertex). So the dichoral angles between the J91's is fixed, which means that if you continue the pattern of 3 J91's around the 5.3.5.3 vertices, it will eventually form an icosahedral framework of J91's, which means the top/bottom edges will have 5 J91's surrounding it.

Now, it may be possible to have other numbers of J91's around that edge, but it means that you cannot also have the repeating pattern of 3 J91's around the 5.3.5.3 vertices anymore, but you'll need some other arrangement of cells to close the shape up.


Ooh, you ment that edge!!, After a quick reflection I see what you mean, and I understand why it's impossible :( . I apologise for being so stubborn. the way I concluded It was impossible is by looking at the edge with the tetahedron and two bilunabirotunda's. This edge fixes the bichoral angle for the pentagon, and thus 5 is the only possibility :sweatdrop: .

It's partly my fault, I was thinking of the 120-cell's edges, which the 5.3.5.3 vertex is symmetry-equivalent to, but in this case it's actually a vertex, not an edge. :oops: So the confusion started from me.

And don't worry about being stubborn, sometimes it takes stubbornness to try cell combinations that nobody has explored far enough before, and you may get lucky and discover something new. ;)
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 11:01 pm

Marek14 wrote:
quickfur wrote:
Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.

What's wrong with putting 4 J92's around a tetrahedron? It would appear that we should be able to at least insert 4 more tetrahedra into the result (i.e. following the pattern of o5x3o3o), which then leaves only the outer surface of hexagons, squares, and triangles, to be closed up. Seems like it should be possible? Or am I missing something obvious?


Well, that was something which was already suggested. My suggestion was meant to be "in addition to", not "in place of" the previous one.

it seems that your polytope is an expansion of mine, so xfox3oxFx3oooo&#xt => xfox3oxFx3xxxx&#xt.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 11:06 pm

student91 wrote:
Marek14 wrote:
quickfur wrote:
Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.

What's wrong with putting 4 J92's around a tetrahedron? It would appear that we should be able to at least insert 4 more tetrahedra into the result (i.e. following the pattern of o5x3o3o), which then leaves only the outer surface of hexagons, squares, and triangles, to be closed up. Seems like it should be possible? Or am I missing something obvious?


Well, that was something which was already suggested. My suggestion was meant to be "in addition to", not "in place of" the previous one.

it seems that your polytope is an expansion of mine, so xfox3oxFx3oooo&#xt => xfox3oxFx3xxxx&#xt.


Yes, looks like it.
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 11:29 pm

A further investigation of xfox3oxFx3oooo&#xt gave the following:
the first part works out well.
the second part is, according to xfox3oxFx3oooo&#xt, build up by placing a trigonal prism at a triangle of the ....3ox..3oo..&#xt-tetahedron, and then filled up with square pyramids.
The problem here is that [dichoral angle of trigonal prism]+[dichoral angle of square pyramid]=144.7356103<159.0948426=[dichoral angle of J92 at 4.3]. This means it is not possible with a trigonal prism and a square pyramid. you might want to insert a polytope in beween somewhere, it's dihedral angle should then be between 14.3592323 and 56.169471. Of course you could also use different polytopes. fact is that a lot of what is already build is rigid.
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