## Partial Stott expansion of nonconvex figures

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

### Partial Stott expansion of nonconvex figures

Some further thoughts about [D4.11]: I'm pretty sure it is related to the snub 24-cell, which also has demitesseractic symmetry as a subsymmetry. Consider an icosahedron. Its vertices lie on a number of intersecting pentagons -- the faces of the star polyhedron, the great dodecahedron. If you remove two antipodal vertices, you get a pentagonal antiprism. Now suppose you shear this pentagonal antiprism by tearing the two pentagons apart, not radially, but sideways, such that the lacing cells deform into a triangle+square+triangle sequence. This produces two of the bilbiro's pentagonal faces. If we do this to two pairs of parallel pentagonal cross-sections of the icosahedron (or, if you like, the great dodecahedron), then it produces the 4 pentagons of the bilbiro. The convex hull gives you the entire bilbiro. We may call this operation a "distension" of the icosahedron (since it is an "abnormal" shear-expansion of the icosahedron), and say that bilbiro = distended icosahedron.

Now take a snub 24-cell, and pick out 8 tetrahedra not sharing any faces with an icosahedron (i.e., a tetrahedron of the "second kind" as I call it on my website, sharing faces only with other tetrahedra), lying on the vertices of a 16-cell. These tetrahedra do still share edges with 4 icosahedra. Now if we pull these tetrahedra apart radially, while keeping their edge attachments to the icosahedra, this will cause a distension of the 24 icosahedra into 24 J91's. Some of the other tetrahedra remain intact (they become the non-axial tetrahedra), and the remaining gaps can be filled with cuboctahedra and the central octahedron + 4 octahedra formation. Thus, we may call this CRF the distended snub 24-cell. The distension of the icosahedra of the snub 24-cell breaks its 24-cell symmetry, and so the resulting CRF only has the demitesseractic subsymmetry left.

This also suggests that other polychora containing icosahedra may be similarly "distended", and perhaps some of them will also yield CRFs, like this one. Anybody interested to investigate this?

Moderator's note: If splitting any further posts out of this topic, the first post ID needs to be updated in the database, due to an oversight in phpbb's auto updater. ~Keiji
quickfur
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### Temp3

quickfur wrote:Thus, we may call this CRF the distended snub 24-cell.

Can we call it the distended snub demitesseract? Since "snub 24-cell" is a misnomer.

Keiji

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### Re: Johnsonian Polytopes

Fine, call it distended snub-whatever, it doesn't really matter. It's just that old names die hard, even if they're not that accurate (I still frequently see "truncted cuboctahedron" and "truncated icosidodecahedron", which are also misnomers!).
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### Re: Johnsonian Polytopes

quickfur wrote:[...] This also suggests that other polychora containing icosahedra may be similarly "distended", and perhaps some of them will also yield CRFs, like this one. Anybody interested to investigate this?

One instance that occurs to me, is the icosahedral pyramid, which distends into the bilbiro pseudo-pyramid. The line segment apex is a clear exemplification of the distension process, in which the pentagonal cross-sections of the original icosahedron are pulled apart. That the distance of this distension is exactly unit edge length, as indicated by the unit-length apex, is an interesting fact. Here, again, I see some signs of an underlying phi-based 4D tiling: how else would such a seemingly-arbitrary operation along an oblique direction produce a CRF??

Furthermore, this suggests that there is an analogous operation that ostensibly produces the J92 pseudopyramid from the icosahedral pyramid. Some kind of trigonal variation of distension? (tristension? OK OK, just kidding).

This also makes me wonder what happens if the distension process is applied to something other than an icosahedron. Will it produce a CRF? A non-convex fragment of a CRF? Can this process produce new 4D CRFs (or perhaps non-convex fragments that are CRF-able)?
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### Re: Johnsonian Polytopes

In order to more clearly show what I mean by the "distension" of the icosahedron, I decided to make an animation that demonstrates this:

As you can see, the process pulls the icosahedron apart while keeping a pair of triangular faces around an edge, and a pair of pentagonal cross sections intact, pulling them outwards until they become fully separated in the bilunabirotunda. The bilunabirotunda retains 4 of the icosahedron's faces. Four faces are inverted in the process, and a bunch of faces are stretched into coplanar phi-edged triangles, which merge into regular pentagons. A pair of square faces are introduced as the distension splits apart two edges into four. The original top and bottom edges of the icosahedron are merged into a single vertex.

Conversely, you can think of the icosahedron as the result of squeezing the two opposite edges of the bilunabirotunda together, such that its pentagonal faces intersect each other, until their chords coincide. The convex hull then is the icosahedron.

Who knew there's such an interesting relationship between the icosahedron and the bilunabirotunda?
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### Re: Johnsonian Polytopes

To me it seems like some weird stott-contraction, and in fact it is!! when you think of the ike as o5x3(-x), which is just an odd way of writing great dodecahedron (actually it's a bit more similar to the truncated great stellated dodecahedron, but that doesn't really matter, as long as the vertices are the same as those for ike). Now if you do a stott-expansion on this: o5x3(-x) => o5x3o, you make an id directly out of the vertices of an ike. The bilbiro occurs if you do this expansion to just one set of (-x)edges. Now a more regular (and obvious) "distended" polytope is ike || id. This can be gained from o5o3o || o5o3x as follows:
o5o3o || o5o3x => o5o3o || o5x3(-x) => o5o3(o+x) || o5x3(-x+x) => o5o3x || o5x3o.
The bilbiro-pseudopyramid is indeed what you get if you do this expansion only partial.
so actually distension is a more complex example of partial stott-expansion.

btw that GIF is very cool, I would recommend you putting it on the bilbiro-page on your website.
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student91
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### Re: Johnsonian Polytopes

student91 wrote:To me it seems like some weird stott-contraction, and in fact it is!! when you think of the ike as o5x3(-x), which is just an odd way of writing great dodecahedron (actually it's a bit more similar to the truncated great stellated dodecahedron, but that doesn't really matter, as long as the vertices are the same as those for ike). Now if you do a stott-expansion on this: o5x3(-x) => o5x3o, you make an id directly out of the vertices of an ike. The bilbiro occurs if you do this expansion to just one set of (-x)edges. Now a more regular (and obvious) "distended" polytope is ike || id. This can be gained from o5o3o || o5o3x as follows:
o5o3o || o5o3x => o5o3o || o5x3(-x) => o5o3(o+x) || o5x3(-x+x) => o5o3x || o5x3o.
The bilbiro-pseudopyramid is indeed what you get if you do this expansion only partial.
so actually distension is a more complex example of partial stott-expansion.

Whoa. So you're saying, that if you partially expand two sets of (-x) edges, it will produce a thawro??! That's... wow. You just blew my mind.

But how many sets of (-x)-edges are there? Is there something else between a thawro and an id? I'm guessing it won't be CRF, but wow, that's just sooo mind-boggling...

btw that GIF is very cool, I would recommend you putting it on the bilbiro-page on your website.

Good idea!
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### Re: Johnsonian Polytopes

quickfur wrote:In order to more clearly show what I mean by the "distension" of the icosahedron, I decided to make an animation that demonstrates this:

As you can see, the process pulls the icosahedron apart while keeping a pair of triangular faces around an edge, and a pair of pentagonal cross sections intact, pulling them outwards until they become fully separated in the bilunabirotunda. The bilunabirotunda retains 4 of the icosahedron's faces. Four faces are inverted in the process, and a bunch of faces are stretched into coplanar phi-edged triangles, which merge into regular pentagons. A pair of square faces are introduced as the distension splits apart two edges into four. The original top and bottom edges of the icosahedron are merged into a single vertex.

Conversely, you can think of the icosahedron as the result of squeezing the two opposite edges of the bilunabirotunda together, such that its pentagonal faces intersect each other, until their chords coincide. The convex hull then is the icosahedron.

Who knew there's such an interesting relationship between the icosahedron and the bilunabirotunda?

Will an interesting (nonconvex) polyhedron result if you do this with great icosahedron?
Marek14
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### Re: Johnsonian Polytopes

Probably, but currently I don't have the tools to study what that shape might be. (In fact, I had to make the above animation by hand-coding all the edges and faces, since my polytope viewer doesn't support the kind of operation depicted! And furthermore, I had to make some non-convex quadrilateral for the triangles that invert themselves, since that's the only way to get povray to render it properly. I wouldn't know how to even begin to automate this. )
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### Re: Johnsonian Polytopes

Well, the only thing needed would be changing the coordinates -- the structure of icosahedron and great icosahedron is the same.

As for that cell, here's an off file that allows you to view it:
Code: Select all
`OFF# Unnamed# Generated by 'Stella4D', Version 5.3# Author: Robert Webb# Web site: http://www.software3d.com/Stella.php# Licensed to: Marek Ctrnáct# Date: 16:45:55, 13 March 2014# # Copyright (C) Robert Webb, 2001-2013.# This copyright notice and the above# information must be kept intact.6 6 10 4.20584889724015290  6.98142396983825720  2.59935756978407010-4.20584889669585760  6.98142396983825720 -2.59935756974909540 4.20584889667934550 -1.01857603003839770  8.41169779386703630-4.20584889722363990 -5.96284793983797550  6.80520646661570170-4.20584889681414340 -1.01857602983881710 -8.41169779410675740 4.20584889681414340 -5.96284793996132390 -6.805206466410952604 3 2 0 1    255 0 04 5 4 1 0    255 0 03 2 5 0    255 0 03 4 3 1    255 0 03 2 3 5    255 0 03 4 5 3    255 0 0`

Yeah, it might need some work to be reduced to a nice algebraic form
Marek14
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### Re: Johnsonian Polytopes

quickfur wrote:
student91 wrote:To me it seems like some weird stott-contraction, and in fact it is!! when you think of the ike as o5x3(-x), which is just an odd way of writing great dodecahedron (actually it's a bit more similar to the truncated great stellated dodecahedron, but that doesn't really matter, as long as the vertices are the same as those for ike). Now if you do a stott-expansion on this: o5x3(-x) => o5x3o, you make an id directly out of the vertices of an ike. The bilbiro occurs if you do this expansion to just one set of (-x)edges. Now a more regular (and obvious) "distended" polytope is ike || id. This can be gained from o5o3o || o5o3x as follows:
o5o3o || o5o3x => o5o3o || o5x3(-x) => o5o3(o+x) || o5x3(-x+x) => o5o3x || o5x3o.
The bilbiro-pseudopyramid is indeed what you get if you do this expansion only partial.
so actually distension is a more complex example of partial stott-expansion.

Whoa. So you're saying, that if you partially expand two sets of (-x) edges, it will produce a thawro??! That's... wow. You just blew my mind.

I wasn't saying that. In fact, when I wrote my previous post, I had no idea how it should be done with other things than a bilbiro or an id.
I think I have found a way to make a thawro out of an ike as well, it uses 3 edges. I'll try to explain it according to some ugly drawings (It seems I am the only one without a fancy rendering-program)
First I'll show how an x3(-x)-triangle gets deformed into an x3o:
Untitledllll.jpg
The blue lines in this picture are the (-x)-edges, and the black the x-edges. you can see the black lines are staying the same length while the blue edges are shrinking.

now the deformation of an ike into a thawro: first I'll locate the pentagons that will become the pentagons of the thawro:
jojouj.jpg (17.36 KiB) Viewed 4468 times
. (the pentagons are borromean ).
And then finally the deformation itself:
Untitledkjb.jpg (44.88 KiB) Viewed 4468 times
. the left middle triangle (the one with the <=> arrows) should get the (x3(-x) => x3o) deformation of my first picture. The vertices with the <= => arrows should be pulled apart into an edge. The right picture is the back-side. The middle triangle here should be made into a hexagon, by the <= =>-arrows. The edges on the back-side with on both vertices <= =>-arrows should become a square. the colored non-center triangles on the front-side should get deformed the same way that happens at the bilbiro-process.

But how many sets of (-x)-edges are there? Is there something else between a thawro and an id? I'm guessing it won't be CRF, but wow, that's just sooo mind-boggling...
[...]
I don't know how such a meso-thing would look. I don't know really well how partial stott-expansion works on an icosahedral symmetry, so I can't tell you anything about other ways to do it.
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student91
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### Re: Johnsonian Polytopes

Marek14 wrote:Well, the only thing needed would be changing the coordinates -- the structure of icosahedron and great icosahedron is the same.

Hmm. If there's a 1-to-1 mapping between the coordinates of the vertices, then this should be doable within my current povray model. I'm currently using apecs<0, 1, phi> as the icosahedron's coordinates. What would the coordinates of the great icosahedron be? i.e., given some permutation of <0, 1, phi>, what would it map to in the great icosahedron?
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### Re: Johnsonian Polytopes

Well, great icosahedron can have the same set of vertices as icosahedron (of course, its edge length would be longer then). There should be a vertex permutation which would twist all twelve pentagons (when you cut a vertex off) into pentagrams.

Let's see. Let's orient an icosahedron as follows:

Vertex 1 - top
Vertices 2,3,4,5,6 - clockwise around the top.
Vertices 7,8,9,10,11 - around the bottom. Let's say 7 is joined to 6,2, 8 to 2,3, 9 to 3,4, 10 to 4,5 and 11 to 5,6
Vertex 12 - bottom

The permutation will then keep vertices 1 and 12 fixed. The remaining 10 vertices will be swapped like this (there are 5 permutations, I'm only trying one):

2->7->3->9->2
4->11->6->10->4
5->8->5

BTW, I just found out that Stella has a function that tries to modify a polyhedron to make its faces regular -- and it works in 4D as well Well, the results are not so nice, but one day, this may come in handy...
Marek14
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### Re: Johnsonian Polytopes

student91 wrote:[...]
But how many sets of (-x)-edges are there? Is there something else between a thawro and an id? I'm guessing it won't be CRF, but wow, that's just sooo mind-boggling...
[...]
I don't know how such a meso-thing would look. I don't know really well how partial stott-expansion works on an icosahedral symmetry, so I can't tell you anything about other ways to do it.

I was wrong, I found another way to partially stott-expand the ike, this time into a orthocupolarotunda!! ugly drawing:
Sample Picturesjdhg.jpg (22.69 KiB) Viewed 4457 times
. the pentagons are black, red, yellow, blue and green. those should be moved according to the pink arrows, every pentagon has five arows at its vertices that are somewhat parralel. the central vertex should be pulled apart into a pentagon, the 1-further-vertices should be placed atop the ones next to them, and the next vertices should be pulled apart int an edge. the last vertex (not drawn) should be pulled into a pentagon as well,completing the cupola.

maybe the occurence of the orthocupolarotunda's in the bilbiro'd o5x3o3x isn't that random.
I think this is the last such partial stott-contraction of ike
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student91
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### Re: Johnsonian Polytopes

student91 wrote:[...]
I was wrong, I found another way to partially stott-expand the ike, this time into a orthocupolarotunda!! [...]

Wow.

I'm speechless now... and here I thought 4D was mind-boggling... Seriously, this thread has got to be the most awesome thread online. Especially since CRFebruary spiced things up considerably. Ever since then, it has been mind-bogglingly awesome almost every other day.

maybe the occurence of the orthocupolarotunda's in the bilbiro'd o5x3o3x isn't that random.

No kidding!! So there's some kind of hidden connection with the icosahedron going on there... Also, in light of the recent discovery of D4.11, I'm wondering if many of the other icosahedral uniforms can be similarly partial-expanded! They would end up with J91 cells... or if you expand it a different way, you'll get thawroes appearing everywhere. Maybe there's an analogous operation to bilbiroing and thawroing, that produces orthocupolarotundae? So it would be cupolarotunda-ing?

I think this is the last such partial stott-contraction of ike

Oh? Why do you think that?

I looked more carefully at the icosahedron->J91 transformation, and I'm starting to dislike the word "distended", because it doesn't accurately describe what's happening. As you said, this is some kind of complex partial Stott expansion, but it can also be thought of as a kind of "unravelling" or "unfolding". I tried translating "unravel" to Greek and Latin on Google Translate, but didn't find any good candidates for an adjective that might describe such a thing. Any suggestions? Edit: Latin for "unfold" is explico, so maybe "explicated"? As in J91 = explicated icosahedron; D4.11 = explicated snub 24-cell (or snub demitesseract, or whatever)? Edit 2: what about "displeated" (as in, pleating is to double fabric back upon itself and secure it in place, kinda like the pentagons of the icosahedron that intersect each other, and displeating is to pull them apart again).

But in any case, in the icosahedron -> J91 transformation, there are two pairs of pentagonal cross-sections that are sheared apart from each other. In the icosahedron -> J92 transformation, if I understand it correctly, three pentagonal cross-sections are sheared apart. Now in the icosahedron -> pentagonal orthocupolarotunda transformation, if I understand it correctly, 6 pentagonal cross-sections are sheared apart.

Now, the icosahedron has 12 pentagonal cross-sections in total, and the above transformations correspond with the shearing of, respectively, 4, 3, and 6 pentagonal cross-sections, all of which are divisors of 12. This leaves 2 as another possibility. What shape might result from the shearing of 2 pentagonal cross-sections?

Anyway, I'm thinking of making animations for all of these awesome transformations... they will be really cool to watch, if nothing else.

Another direction that I investigated, was the partial Stott expansion of the octahedron (regarded as 3 intersecting squares -- I forget the name of the uniform polyhedron that has these 3 squares as faces). Shearing apart two of the squares produce a funny-looking polyhedron with 2 squares, 2 triangles, and a pair of bisected hexagons as faces. The bisected hexagons join each other at their long edges, and may be regarded as a hexagon folded up along its long diagonal. They are produced by coplanar triangles. This shape may thus be regarded as the octahedral version of the bilunabirotunda, if you allow coplanar faces. Now, if you extend the bisected hexagons into full hexagons, then they cross each other and reverse their orientation, so you can put another 2 squares and 2 triangles around their new edges, and you get a closed self-intersecting polyhedron.

Yet another interesting thing I noticed while thinking about these partial Stott expansions: it seems that the simplex-faced regular polytopes in 3D and 4D have the curious property that they contain their lower-dimensional analogues as cross-sections. For example, the octahedron contains square cross-sections, which can be thought of as 2D crosses, and the icosahedron contains pentagonal cross-sections, which may be regarded as 2D "icosahedra". In 4D, the 16-cell contains octahedral cross-sections, and the 600-cell contains icosahedral cross-sections. Now, these are all obvious, but the thought occurred to me, of what happens if we now apply partial Stott expansion to them, so that these cross-sections and brought out to the surface? What shapes would result, and are they CRF?
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### Re: Johnsonian Polytopes

quickfur wrote:
student91 wrote:[...]
I was wrong, I found another way to partially stott-expand the ike, this time into a orthocupolarotunda!! [...]

Wow.

I'm speechless now... and here I thought 4D was mind-boggling... Seriously, this thread has got to be the most awesome thread online. Especially since CRFebruary spiced things up considerably. Ever since then, it has been mind-bogglingly awesome almost every other day.

maybe the occurence of the orthocupolarotunda's in the bilbiro'd o5x3o3x isn't that random.

No kidding!! So there's some kind of hidden connection with the icosahedron going on there...
I was more thinking in the direction of the stott-expansion x3x3o5o => o5x3o3x, being done only partial, so some ikes become ids, some ikes bilbiro's, and some ikes become orthocupolarotunda's. I have not yet figured the 4D-partial stott-expansions out yet, so I can't investigate it very far. anyway D4.11 clearly is such a partiall stott-expansion as well. that would also explain why there are x3o3x's out there.
Also, in light of the recent discovery of D4.11, I'm wondering if many of the other icosahedral uniforms can be similarly partial-expanded! They would end up with J91 cells... or if you expand it a different way, you'll get thawroes appearing everywhere. Maybe there's an analogous operation to bilbiroing and thawroing, that produces orthocupolarotundae? So it would be cupolarotunda-ing?
that's what I was thinking about as well, has to be investigated
I think this is the last such partial stott-contraction of ike
Oh? Why do you think that?
we've got an edge-first, a vertex-first, a face-first and an origin-first (giving id) complex stott-expansion. Furthermore there aren't any Johnson-solids that seem to be derivable this way. Any other wythoff-representation of ike should be investigated though.

I looked more carefully at the icosahedron->J91 transformation, and I'm starting to dislike the word "distended", because it doesn't accurately describe what's happening. As you said, this is some kind of complex partial Stott expansion, but it can also be thought of as a kind of "unravelling" or "unfolding". I tried translating "unravel" to Greek and Latin on Google Translate, but didn't find any good candidates for an adjective that might describe such a thing. Any suggestions? Edit: Latin for "unfold" is explico, so maybe "explicated"? As in J91 = explicated icosahedron; D4.11 = explicated snub 24-cell (or snub demitesseract, or whatever)? Edit 2: what about "displeated" (as in, pleating is to double fabric back upon itself and secure it in place, kinda like the pentagons of the icosahedron that intersect each other, and displeating is to pull them apart again).

But in any case, in the icosahedron -> J91 transformation, there are two pairs of pentagonal cross-sections that are sheared apart from each other. In the icosahedron -> J92 transformation, if I understand it correctly, three pentagonal cross-sections are sheared apart. Now in the icosahedron -> pentagonal orthocupolarotunda transformation, if I understand it correctly, 6 pentagonal cross-sections are sheared apart.

Now, the icosahedron has 12 pentagonal cross-sections in total, and the above transformations correspond with the shearing of, respectively, 4, 3, and 6 pentagonal cross-sections, all of which are divisors of 12. This leaves 2 as another possibility. What shape might result from the shearing of 2 pentagonal cross-sections?

Anyway, I'm thinking of making animations for all of these awesome transformations... they will be really cool to watch, if nothing else.
I bet they will, looking forward to seeing them
Another direction that I investigated, was the partial Stott expansion of the octahedron (regarded as 3 intersecting squares -- I forget the name of the uniform polyhedron that has these 3 squares as faces). Shearing apart two of the squares produce a funny-looking polyhedron with 2 squares, 2 triangles, and a pair of bisected hexagons as faces. The bisected hexagons join each other at their long edges, and may be regarded as a hexagon folded up along its long diagonal. They are produced by coplanar triangles. This shape may thus be regarded as the octahedral version of the bilunabirotunda, if you allow coplanar faces. Now, if you extend the bisected hexagons into full hexagons, then they cross each other and reverse their orientation, so you can put another 2 squares and 2 triangles around their new edges, and you get a closed self-intersecting polyhedron.

Yet another interesting thing I noticed while thinking about these partial Stott expansions: it seems that the simplex-faced regular polytopes in 3D and 4D have the curious property that they contain their lower-dimensional analogues as cross-sections. For example, the octahedron contains square cross-sections, which can be thought of as 2D crosses, and the icosahedron contains pentagonal cross-sections, which may be regarded as 2D "icosahedra". In 4D, the 16-cell contains octahedral cross-sections, and the 600-cell contains icosahedral cross-sections. Now, these are all obvious, but the thought occurred to me, of what happens if we now apply partial Stott expansion to them, so that these cross-sections and brought out to the surface? What shapes would result, and are they CRF?
These cross-sections can be visualised with my negative-node things. the octahedron can be seen as x3o4o "=" (-x)3x4o. you have a square (x4o) and a double-winded triangle (-x)3x there. This is expanded by (-x)3x4o => o3x4o, so it's quite logical your figure is part of an o3x4o. The same way, x3o3o5o "=" (-x)3x3o5o, making you see the ike, and x3o3o4o "=" (-x)3x3o4o makes you see the octahedron. I don't know how this is consistently done for things with nodes in the middle yet. Furthermore I don't know yet how partiall stott-expansoins work on those 4D-figures. the outcome should be really interesting though. any help investigating this is welcome
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student91
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### Re: Johnsonian Polytopes

student91 wrote:
quickfur wrote:[...]
Another direction that I investigated, was the partial Stott expansion of the octahedron (regarded as 3 intersecting squares -- I forget the name of the uniform polyhedron that has these 3 squares as faces). Shearing apart two of the squares produce a funny-looking polyhedron with 2 squares, 2 triangles, and a pair of bisected hexagons as faces. The bisected hexagons join each other at their long edges, and may be regarded as a hexagon folded up along its long diagonal. They are produced by coplanar triangles. This shape may thus be regarded as the octahedral version of the bilunabirotunda, if you allow coplanar faces. Now, if you extend the bisected hexagons into full hexagons, then they cross each other and reverse their orientation, so you can put another 2 squares and 2 triangles around their new edges, and you get a closed self-intersecting polyhedron.
[...]
These cross-sections can be visualised with my negative-node things. the octahedron can be seen as x3o4o "=" (-x)3x4o. you have a square (x4o) and a double-winded triangle (-x)3x there. This is expanded by (-x)3x4o => o3x4o, so it's quite logical your figure is part of an o3x4o.

Aha! Now I see the connection. Neat! And on that note, it's rather interesting that the cuboctahedron itself can be considered a kind of "bilunabicupola" -- it consists of two strips of square+2 triangles+square on the top and on the bottom, with two luna (triangle+square+triangle) around the sides. Analogous to J91's two strips of pentagon+2 triangles+pentagon, with two triangle+square+triangle around the sides.

This also suggests a triangular analogue, whereby you have two strips of triangle+2 triangles+triangle, joined by two triangle+square+triangle around the sides, except that there is no room for the triangles of the latter to fit anywhere. One solution is to introduce a strut triangle at either end of the strips, in which case you end up with the tetraaugmented cube (with 4 pairs of coplanar triangles). Another solution is to merge the end triangles of the two luna so that they coincide, in which case there's only room for one strip, and we end up with the (mono)augmented triangular prism.

Now, if we could find a J91 analogue in 4D, perhaps we can obtain something more interesting, because in 4D we have the choice between (at least) tetrahedron, octahedron, cube, dodecahedron, icosahedron, as the "strip" element to fit tetrahedra around. But I currently don't know of any 4D analogue that has a similar structure.

Actually, a thought just occurred to me. You know what a bilunabirotunda is? It's actually two digonal "fat teddies", glued together at the bottom, and it just so happens that the lacing edges between them are unit length! Recall that I described the idea of a "fat teddy" as starting with some (n-1)-polytope, say xAoBo, then stacking it on a phi-scaled version of itself, like in the construction of an ursatope, but inserting extra (n-1)-simplices in between the pentagonal cells -- so the next layer would be fAoBx (instead of the usual fAoBo), then something like oAxBf in the third layer, and then another layer of vertices to make the thing close up in a CRF way. In 3D, the digonal fat teddy is just J62, the metabidiminished icosahedron, and the trigonal fat teddy is J92. (The square fat teddy and above are all hyperbolic, so they won't yield any CRFs, unfortunately.) Now, it just so happens that if you delete the bottom edge from J62 (to get the "pure fat teddy" portion of it), then you can glue two copies of the result together to make the two halves of a J91. The lacing edges between the two halves just happen to be unit length. So you could think of J91 as a "siamese J62".

Now this suggests that in 4D, we may construct an n-gonal fat teddy, and, if we're lucky, one of them will be possible to close up in a CRF way when we glue two copies of them end-to-end, and thus it would serve as a J91 analogue!

Furthermore, the 4D construction of a "trigonal fat teddy" would begin with a tetrahedron, then attaching J63's to its faces, just like in an ursatope, but with extra tetrahedra at the vertices of the tetrahedron. This is how I began the construction of D4.10 and D4.11, in fact. The edges of the central tetrahedron would then have a pentagon+triangle+pentagon+triangle formation, which in the case of D4.10 and D4.11 I continued all the way to make J91's, but technically, I didn't have to. I could've capped them off as J62's instead. That would produce an initial fragment consisting of 1 central tetrahedron, 4 J63's attached to its faces, 6 (partial) J62's attached to its edges, and 4 tetrahedra attached to its vertices. Now consider D4.11: it's just 8 of these initial fragments fused together, such that the partial J62's are joined to each other to make J91's! So D4.11 is a kind of siamese octa-(tetrahedral fat teddy) fusion. But, by analogy with the 3D case where the triangular fat teddy = J92, this suggests also that we may be able to close off a single initial fragment of the tetrahedral fat teddy into a CRF, and then it would serve as a 4D analogue of J92!

Sadly, it doesn't get beyond a tetrahedral fat teddy: we already know the cubical fat teddy is non-CRF because the side cells would have to be square ursahedra, which are non-CRF. Plus, the angle defect for squeezing in tetrahedra at the vertices of the cube would be negative, making it hyperbolic. The octahedral fat teddy is also not possible because it is hyperbolic. So just as in 3D, the CRF fat teddies do not go beyond the simplicial cases. Nevertheless, the n-gonal 4D fat teddies are certainly worth investigation!!

Yet another afterthought: if J92 is the 3D trigonal fat teddy, then we could get "very fat" teddies in 4D by gluing two of them together at their triangular faces (y'know, a kind of 4D n-gonal "very fat" teddy with J92 side cells instead of the usual J63's), and with J62's around the edges. Does this sound familiar??? It should, because it's a fragment of the J92 rhombochoron!!!!! This means that the J92 rhombochoron is nothing other than a siamese triangular very-fat teddy. Which, in light of the above analysis, makes it a candidate for a 4D bilunabirotunda... a fat analogue.

(I better stop now... all of these sudden realizations are going to make my brain explode...)

The same way, x3o3o5o "=" (-x)3x3o5o, making you see the ike, and x3o3o4o "=" (-x)3x3o4o makes you see the octahedron. I don't know how this is consistently done for things with nodes in the middle yet. Furthermore I don't know yet how partiall stott-expansoins work on those 4D-figures. the outcome should be really interesting though. any help investigating this is welcome

Perhaps one way to begin to investigate this, is to simplify the problem by cutting off a cap from the polytope and seeing what partial Stott expansion does to it -- for example, cut off icosahedron||icosidodecahedron from o5o3x3o, then apply partial Stott expansion to it and see what happens.
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### Re: Johnsonian Polytopes

Can't resist yet another afterthought: in 3D, we may also have "very fat" digonal teddies: start with a digon, then attach two hexagons to it, with triangles on either end (of the digon). As it turns out, only another 2 vertices are needed to close this up: it's a truncated tetrahedron. The trigonal 3D fat teddy, unfortunately, is planar, so it doesn't produce a CRF figure (though it does produce the rectified triangular plane tiling). It just so happens that in 4D, the dihedral angles of J92 around its top triangle are just shallow enough to allow a 60° bending when two J92's are joined at their triangles. Inserting J62's around the edges then produce a fragment of the J92 rhombochoron.

This does suggest that it may be possible to close up this "half J92 rhombochoron" in a CRF way, so there would be an actual CRF "very fat" teddy, and the J92 rhombochoron would be the siamese (conjoined) version of it.

Oh, and just in case the analogy isn't clear: the reason we insert J62's around the edges in these fat teddy constructions, is because we're trying to use some analogue of a triangle around the "pentagonal" side elements, but since the side elements are 3D teddies, we need something that's both triangle-like, and has pentagonal faces to join with these side cells. J62 neatly fulfills this purpose, being wedge-like in shape, yet having pentagons on the side for attaching to the 3D teddies. It plays the very important role of being the "middle" element, touching the edges of the starting polytope, between the 3D teddies touching the faces, and the tetrahedra touching the vertices.
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### Re: Johnsonian Polytopes

quickfur wrote:[...]
Anyway, I'm thinking of making animations for all of these awesome transformations... they will be really cool to watch, if nothing else.

that'd be so cool , I thought the Bilbiro-animation was already cool, but these should be awesome

Yet another interesting thing I noticed while thinking about these partial Stott expansions: it seems that the simplex-faced regular polytopes in 3D and 4D have the curious property that they contain their lower-dimensional analogues as cross-sections. For example, the octahedron contains square cross-sections, which can be thought of as 2D crosses, and the icosahedron contains pentagonal cross-sections, which may be regarded as 2D "icosahedra". In 4D, the 16-cell contains octahedral cross-sections, and the 600-cell contains icosahedral cross-sections. Now, these are all obvious, but the thought occurred to me, of what happens if we now apply partial Stott expansion to them, so that these cross-sections and brought out to the surface? What shapes would result, and are they CRF?
[/quote]
to do this, I thought a more thorough understanding of this operation was requiredd, so I investigated it in 3D:
take the edge-centered lace tower of ike and modify it:
Code: Select all
`x2o    x2o    o2(-x)o2f    o2f    f2of2x -> f2(-x) x2fo2f    o2f    f2ox2o    x2o    o2(-x)`

these can be transformed into the bilbiro by adding 1 to the right-hand tower
Code: Select all
`x2x o2oo2F f2xf2x o2Fo2F f2xx2x o2o`

the modified lace towers for vertex and face, so that they become the thawro resp. the orthocup. are (with the help of student91's general note on negative stuff):
Code: Select all
`o5o    x3o    o5x x3xf5(-x) o3f -> f5o o3Fx5o    f3o    x5x f3xo5o    x3(-x) o5x x3o`

the thing I noticed by now, is that this modification process has a certain system in it: you take a row and alter all x's so they become (-x)
(also I hope these towers can be of some help for quickfur's animations)
so now for the octahedron:
Code: Select all
`#no x's in the left tower, so it remains the sameo4o    o4o       x4o o4x #orthobicup and cuboc (gyrobicup)o4x -> q4(-x) -> x4x q4oo4o    o4o       x4o o4x#symetric, so only one modification viewx3o    x3o       x3x #tricupo3x -> x3(-x) -> x3o#left tower againo2x    o2(-x) x2x o2o #elongated square pyramid and a non-CRF thingyq2o -> q2o -> Q2o q2xo2x    o2(-x) x2x o2o`

and the tetrahedron:
Code: Select all
`x3o    x3o      x3x #tricupo3x -> x3(-x) -> x3ox2o    x2o        x2x #squippyo2x -> o2(-x) -> o2o`

well I hope this helped to make the partial stott-expansion easier to understand/elaborate in 4D and I guess I gave quickfur some more stuff to animate
I think it is cool that so much of these things work, a lot of CRFs must be possible to make this way, by having a tower, changing all x's in a row and expanding afterwards, but I'm afraid a lot of things will be like the non-CRF thingy the oct became
February ain't over
student5
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### Re: Johnsonian Polytopes

I know it's a bit of a nagging remark, but if D4.11 is really vertex-transitive, it doesn't fit wintersolstice's original definition anymore...
student5
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### Re: Johnsonian Polytopes

student5 wrote:I know it's a bit of a nagging remark, but if D4.11 is really vertex-transitive, it doesn't fit wintersolstice's original definition anymore...

Huh? Did somebody claim that D4.11 is vertex-transitive? 'cos it's definitely not! It has demitesseractic symmetry, in the sense that if you apply a demitesseractic symmetry operation to it (e.g., rotate by 180° around a coordinate axis), it remains unchanged, but that says nothing about whether its vertices are transitive or not. In this case, they can't be, otherwise it'd have 24-cell symmetry!

Edit: oh, I see what you're getting at. Wintersolstice's original definition excluded vertex-transitive polytopes from consideration. But I think he probably intended it to be "non-uniform", rather than non-vertex transitive, because later on he himself added BXD, spidrox, and the expanded snub 24-cell to the CRF polychora project page as examples of "non-uniform scaliform" polychora. Besides, vertex-transitivity in a non-uniform polychoron is a very rare and interesting property, and it wouldn't do to just exclude them!

Ultimately, I think Keiji's approach is best: adopt the inclusive definition that admits any polychoron as CRF, as long as it is (1) convex, (2) does not have coplanar cells, and (3) all 2D ridges are regular polygons. This of course includes the regular and uniform polychora, but since those are already all known, the primary effort now lies in finding CRFs that are not regular nor uniform. And as it turns out, there are CRFs with a high degree of symmetry that fit this definition, and yet they are neither regular nor uniform -- BXD is one such example. D4.11 is another example, and some time ago I found an augmentation of a 24-cell family uniform that has elongated square bipyramid cells, that still retains tesseractic symmetry. Spidrox is particularly notable because in addition to being vertex transitive yet non-uniform, it exhibits swirlprism symmetry, which is unique to 4D. (Well OK, there are probably 6D swirlprisms as well, but 4D is unusual in the sense that here this symmetry is chiral, whereas, if what I read is correct, higher-dimensional equivalents of the Hopf fibration are achiral.) These are of particular interest because of their symmetry. But of course, we usually find some other patterns in the other "irregular" CRFs as well, so they aren't any less interesting, they're just interesting in other ways.
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### Re: Johnsonian Polytopes

Bilbiro

quickfur wrote:

Very good idea! Never thought about it that way.

But still it is not the plain icosahedron, which is the Stott contraction of bilbiro! In your animation you spot surely those lacing edges with varying lengths. Those clearly do not belong to bilbiro. Those are just a side effect of quickfur's generation, which assumes all figures to be convex. Here those come in as additional edges obtained by a further hull process. They do not belong to the Stott contraction / expansion.

Just consider using those 4 pentagons of bilbiro in the whole process of the animation. Then you will have the 2 triangles of the lune, moving together by shrinking the square to a mere edge. This is clearly the effect of Stott contraction. And then there are the 2 triangles at the top (and 2 further ones at the bottom - mostly hided from being seen) which use those crossing sides. These are the triangles x3o --> (x-x)3(o-x) = o3(-x). - Esp. this latter writing Shows that those triangles will be retrograde. And indeed, in that obtained faceting of the icosahedron, those 2 triangles and the 2 intersecting pentagons include a volume with negative density (when calling that at the body center positive).

Thawro

student91 wrote:

The same is true for that second Stott expansion "of ike". The above pic clearly shows ike (from above at the left, from below at the right). And how those pentagonal facets are to be spotted. But those 3 radial edges in the left pic do not belong to the thawro generation. Neither the 2 incident triangles each. Finally, the central triangle in the left pic in fact is that x3(-x), which shall become x3(-x+x) = x3o.

What then is x3(-x)? Right from its number of edges it surely is some hexagon. So consider a regular hexagon: running to the left, then turn 60 degrees up, running still a bit to the right but mostly up, the again 60 degrees to the left side of the momentary direction, etc. Now consider what happens for x3(-x). Let's start with the normal edge. So we run to the right. Then we would like to turn 60 degrees up. But no we run on that new line into the inverse direction. (Kind of having turned 120 degrees down.) Then follows again a normal edge. Thus the next direction will be as in the former regular hexagon to the upper left. Etc. So we finally spooled our hexagon twice around a triangular outline!

That is, that central triangle of the left pic is not really a triangle, it rather is a doubly wound hexagon. The face area thus in fact also is a double cover. Thus is works mostly like 2 completely incident triangles. - But this leads to a different idea. Completely incident elements usually are considered as unusual. One tries to get rid of them. And sure, this is possible here too. Erase that doubly wound hexagon completely, and cross-connect the then the pairs of the then open, completely coincident edges. That is, those will just be identified.

The figure, which then is left from the original icosahedron, would be the following faceting (given in 3 consecutive views: from above, from the side, and from below):

This figure in fact then will be what indeed produces thawro by mere partial Stott expansion. The central triangle of the bottom view expandes into a hexagon. The apparently vertical edges of the side view expand into squares. The top triangles of the side view then are those with the crossing sides in the run of the Expansion: the lower tip becomes a new edge, the top edge degenerates into a single point. And that central gap of the top view (that double wound withdrawn hexagon) just reduces thereby into a point too.

Thus this ike faceting is an axial figure with triangular symmetry. And the Stott expansion here applies with respect to that subsymmetry of the full ike.

--- rk
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### Re: Johnsonian Polytopes

quickfur wrote:Hmm. If there's a 1-to-1 mapping between the coordinates of the vertices, then this should be doable within my current povray model. I'm currently using apecs<0, 1, phi> as the icosahedron's coordinates. What would the coordinates of the great icosahedron be? i.e., given some permutation of <0, 1, phi>, what would it map to in the great icosahedron?

The easiest way to derive the great icosahedron from the (small) icosahedron would be by conjugation: Write the coordinates in terms of sqrt(5), and then just apply sqrt(5) --> -sqrt(5) to every occurance. And this moreover respects all the elemental incidences too! (Not only the vertices.)

When dealing in terms of some constant phi = 1.618 this would translate into transforming all occurances according to phi --> -1/phi = -0.618.

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:Bilbiro

quickfur wrote:

Very good idea! Never thought about it that way.

I never thought about it that way either, until I saw the demitesseractic symmetry of D4.11 and the placement of its tetrahedral cells as strong hints that it is somehow related to the snub 24-cell (snub demitesseract). This prompted me to seek for a possible connection to the snub 24-cell, which then led me to this consideration: if indeed D4.11 is related to the snub 24-cell, then it means that the pentagons of the bilbiroes around the axial tetrahedra must correspond in some way to the icosahedra of the snub 24-cell. That this connection exists, is strongly indicated by the orientation of the tetrahedral cells, which are identical to the orientation of the corresponding tetrahedra in the snub 24-cell: in order for pentagons to connect to their edges in that way, in the configuration 3-5-3-5, must mean that these pentagons must have come from the pentagonal cross-sections of the icosahedra in the snub 24-cell. This suggests that if we were to collapse D4.11 such that these pentagons intersect, then we should get the snub 24-cell again. Or conversely, if we stretch out the pentagonal cross-sections of the ikes in the snub 24-cell, then we should obtain the bilbiroes in D4.11. This then leads directly to the above construction of bilbiro from ike.

But still it is not the plain icosahedron, which is the Stott contraction of bilbiro! In your animation you spot surely those lacing edges with varying lengths. Those clearly do not belong to bilbiro. Those are just a side effect of quickfur's generation, which assumes all figures to be convex. Here those come in as additional edges obtained by a further hull process. They do not belong to the Stott contraction / expansion.

Actually, I didn't use my polytope generator for this animation. Those edges were added by me. Or rather, I started with the icosahedron, together with all of its edges, and left them in as I proceeded with the Stott expansion to bilbiro. Leaving those edges in actually had another side-effect: in order to produce the new edges in bilbiro where the retrograde triangles are, I needed to double the base vertices -- since otherwise it would require some ugly hacks in the povray code to make the animation work. Here, I had two options: in the mid-sequence of the transformation, I could either have trapeziums in place of those triangles, or the crossed tetragon in the final animation. In order to maintain a closed polyhedron during the transformation, I actually needed to use these trapeziums as faces (you can actually spot them if you look carefully at how the light source illuminates the area left of the front and top retrograde triangle as it transforms -- I didn't put in edges for those trapeziums because they were distracting). However, since those edges were left in there, it became obvious to me that I needed the edges to show the crossed tetragon instead of the actual outline of the trapezium faces, otherwise it would fail to convey the source of the bilbiro's pentagonal faces as pentagonal cross-sections of ike. So that's how those crossed tetragons came about -- and they do reflect what's happening with the actual Stott expansion.

[...]Thawro
[...]Thus this ike faceting is an axial figure with triangular symmetry. And the Stott expansion here applies with respect to that subsymmetry of the full ike.

This is all very interesting. So now I'm wondering what kind of figures we might obtain, if we were to perform this kind of partial Stott expansion to, say, the 600-cell? I suppose we would have to consider some subsymmetry, perhaps derived from (parts of) the regular star polychora? The far vaster numbers of surtopes involved do make it hard to visualize, though!
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### Re: Johnsonian Polytopes

Klitzing wrote:
quickfur wrote:Hmm. If there's a 1-to-1 mapping between the coordinates of the vertices, then this should be doable within my current povray model. I'm currently using apecs<0, 1, phi> as the icosahedron's coordinates. What would the coordinates of the great icosahedron be? i.e., given some permutation of <0, 1, phi>, what would it map to in the great icosahedron?

The easiest way to derive the great icosahedron from the (small) icosahedron would be by conjugation: Write the coordinates in terms of sqrt(5), and then just apply sqrt(5) --> -sqrt(5) to every occurance. And this moreover respects all the elemental incidences too! (Not only the vertices.)

When dealing in terms of some constant phi = 1.618 this would translate into transforming all occurances according to phi --> -1/phi = -0.618.

--- rk

Thanks! That indeed produced a great icosahedron, and all element incidences are correct. Moreover, it produced a most curious animation:

It's not very obvious from this viewpoint, but this Stott-expanded great icosahedron has two square faces (they look like rectangles but they're actually square). This "unravelled" great icosahedron also has a weird double-spike configuration protruding from the square faces. Between the two squares, some of the "arms" of the original star converge on a pair of new peaks of order 6. Together, these spikes and peaks form a pair of pouch-like formations around the center of the polyhedron. All in all, a rather strange non-convex polyhedron!

One of the intermediate stages look like the spikes converge to a single compound spike; this appears to be at the point when what will become square faces are golden rectangles. I'm not 100% sure I have all the faces right; it's not obvious what the faces are supposed to be in this odd non-convex shape! Its convex hull is a partial dodecahedron, a parabiaugmented cube with the roof-shaped augments that produces a dodecahedron when 6 of them are applied in pyritohedral symmetry. Here, however, 4 of the augments are missing. I've no idea what the inner peaks between the two square faces are supposed to correspond with.

Here's another version of the animation where the "extra edges" Klitzing mentioned earlier are removed from the end-product:

This makes the shape of the peaks and spikes harder to see, though.
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Here is what i know of ursulatopes.

Ursulatopes are a kind of partially truncated fostrums, literally the prototype is xu&#ft. If the truncation happens at 1/F of the height from the large base, and the large base is rectified, the style of the figure is xfo&xt, a pentagon. The base figure is rectified.

The ursulation of a polygon gives xfoPoop&#t, where p is the shortchord of P (eg x=3, q=4, f=5, h=6, u=infinity. The largest value of P which leads to mostly unit-edge ursulate is P=10. A polytope with this size is x3o3o5o, so xfo3oox3ooo5ooo&#xt is in fact flat.

All of the parameters of the ursulates are given by the a single property: since i use the square of the diameter as the size-property, the discussion shall be with this.

The ursulates have four kinds of face: the top, the base, the lacing faces (eg the pentagons), and the valences (a valance here is something that covers the legs of furniture, or the wheels where these are open.

The base is a rectate specifically of the top face of edge 2. This produces in general, a figure for which the D2 is (in L2 form:) b = 4t-4 So, eg 3.618*4-4 = 10.944. ie an ursulated x3o5o gives a base o3x5o, which is a decagon-round.

The valance faces slope inwards on the teddi xfo3oox&#t, but on xfo10oop&#xt, they are clearly facing upwards. One can find out the condition where the valence faces become vertical, by equating fPo with oPp. This means that the apex of the valence-faces are directly over the base (they're in essence op&#t pyramids. The value gives a base diameter of t=5.236067=2F. For a top base larger than this, the figure is "vertically convex": you can stand it on a face, and all the falls from any point perpendicular to the base is inside the figure. This means we can use this as a capping for a pyramid extension, or for back to back constructions.

One can place a peaking pyramid on the top of the ursulate, in the form of oxfo3ooox&#xt. This is allowed as long as the diameter2 of the base is less than 4.

So we get
• 1.3333 < t< 4 top face can be raised to a peak
• 4.000 <= t < 5.236 can be neither top-peaked or used as a pyramid cap.
• 5.236< t <10.472 The figure is vertically convex, and might be used as pyramid-capping.

While this might be applied to conventional simple-figures + truncates (eg t = x3oXo.... b=o3xXo....), it also applies to the rectates, t=..o3x3o.., b=..x3o3x..

One can derive a formula for the diameter of a truncate, directly from the dynkin symbol, where D is the diameter square of the 'regular figure' (including the likes of 2_21 &c).

D' = (n+1)² D - 4(n)², where n is the order of rectate, When one sets D to the diameter of a simplex D=2n/(n+1), then the n'th rectate corresponds to a point.

One puts D'=10.472, and finds D for various values of N, ie D = (D'+4n²)/(n+1). For the cross polytope, and simplexes in high dimensions, n is limited to 4, but for lower dimensions, eg up to 22D, all values of N work with the simplex.

So, apart from 2_21 = /4B with d=2.66666, one has 1/3B with d=6.666666, both work.

Stott-expansions

It is permissible to make any node, not being used in either the top or base figures, to cause all nodes in the tower to be marked. This creates a new kind of face along the edges running from the vertices of the sections xPo to fPo.

One can comprise the stott-expanded figure of an ursulate by eg T = t+S, B=b+S. For example, one might have oxfo3ooox5xxxx&#xt, can be constructed by way of the stott-sum u = oxfo3ooox5oooo&xt and the stott-expansion consist S=oooo3oooo5xxxx&xt. The element s is S with the the nodes involved in the ursulation, removed, ie s = .3.5x = x.

The top and bottom bases are T=t+S, B=b+S. The lacing-faces which have pentagons and teddies, are expansions of the face-elements as well (ie the lacing-faces are ursulates of the faces of t). So these expand to T'=t'+S and B'=b'+S. The valence faces start as pyramids, from point || vf of t, becomes expanded by s to s || s+vf. New faces follow the former edges from the vertices of t to the valences apex, gives now a {5}*s pyramid.

So if ü represents the ursulated bottom face node, we get then o3ü5x has 30 pentagonal prisms following the edges, and a further 12 prisms following the pyramids from the pentagons of x3o5x to f3o5x.

The parameters evaluated above do not affect the stott expansions, so the total diameter of t can be much larger than 10.944.

[b]Multiple ursulates[b]

Not only can one have ursulates top-and-tailing a pyramid (especially if the top diameter2 is > 5.236), but when expansion is taken to account, one can have different ursulates at each end. A simple case might be

o3x3o3o3oBo is vertical-convex, so it might be used as an an apex, eg

ox3xo3ox3oo3ooBoo (omitting medial layers)

We see that the bottom is given by x3o3xo3oBo. This can be stott-expanded by x3o3xo3xBo (or the last node by 'x'). But this same base can arise from a different figure, eg

oox3xfo3oox3ooo3xxxBooo&xt and xxx3ooo3oox3xfo3oox&Booo both have the same base, but are the ursulates of 1/3B and 3/1B.

So, we can produce an assymetric (like the gyrated and gyrated-extended bicupolae), like

ooxxx 3 xfooo 3 ooxoo 3 ooofx 3 xxxoo B ooooo.

What is happening here is the top is an ursulate of node 2 into 1 and 3, the total expanded along 5 (and probably 6). The vertex-layers 3,4,5 are another ursulate from node 4 into 3 and 5, the lot expanded by 1. So the middle layer is to have nodes 1,3,5 marked, which can be expanded into a prism.

One can use a similar process on the simplex-as-simplex faces. For example, the larger simplex-rectates give rise to vertical-convex ursulates, which might be used as face adornments on the simplex and cross polytopes having these faces.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
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### Re: Johnsonian Polytopes

hi
while researching the partial strott-expansion of ex, I've come across the following:
I wanted to make some tetraheddron-oriented CRF, so I started with the following lace tower of ex:
Code: Select all
`x3o3oo3o3fo3f3of3o3xo3x3ff3x3oetc.`

and, because ex is symetrical there were two expansions possible:
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`x3x3o o3x3oo3x3f x3o3fo3F3o x3f3of3x3x F3o3xx3o3F x3x3fF3o3x F3x3ox3x3f o3x3fo3F3o x3f3of3x3o F3x3oo3x3x x3o3x`

now I don't know if these are CRF, but the interesting thing is, that they fit together at x3x3f||F3o3x and then become D4.8.4
Code: Select all
`x3x3o //x3o3f somehow disappears?o3F3of3x3xx3o3Fo3f3xf3o3xo3x3o`
student5
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### Re: Johnsonian Polytopes

Last night, I discovered a surprising connection between the 16-cell, the 5-cell, and the pentagonal polychora.

As we all know(?), the 16-cell cannot be augmented in a CRF way, because it becomes non-convex. However, the last time we considered this, we just wrote it off and never thought about it again. Last night, though, I was experimenting with different things that might possibly produce a CRF with trigonal bipyramid cells, and it just so happened that I sat down and calculated the coordinates of an augmented 16-cell (i.e., a 16-cell with a 5-cell attached to one of its facets). I took the 16-cell of edge length 2, as is customary, which gives the coordinates as apacs<0,0,0,√2>. If we attach a 5-cell to one of its hexadecants, say the one with all positive coordinates, the apex of the 5-cell should have coordinates <x,x,x,x> for some value of x. Since the 5-cell has equal edge lengths, it's just a matter of solving for x such that the distance between ap<0,0,0,√2> and <x,x,x,x> is unit length (i.e., distance = 2 since I'm using edge length = 2). All of this is pretty obvious and routine work, of course.

The surprise comes with the value of x: it turns out, that x = phi/√2. If you haven't fallen out of your chair yet, read that again. The value of x came out to be exactly phi/√2. I fell out of my chair, because ... where on earth did the golden ratio pop out of?!?!?! OK, maybe this is not surprising to you. Well, what about this: I then ran these coordinates through my polytope generator, which runs a convex hull on these vertices, and I got a (non-CRF) polychoron out of it that has phi-scaled edges lacing the apex of the 5-cell to the non-adjacent vertices of the 16-cell. This means that it should be possible to "complete the pentagon" around these phi-edges, and get a CRF out of it. Have you fallen out of your chair yet? So I wasn't dreaming when I suspected some kind of special hidden thing going on between 4D and the golden ratio... this is now showing up as a hitherto unsuspected connection between a 5-cell, a 16-cell, and some kind of pentagonal polychoron (perhaps a 600-cell?) that could possibly be a 4D tessellation.

Or maybe this is all already known, and I'm just making a fool of myself? In any case, I was very, very surprised by the way I stumbled across this strange coincidence. It was very uncanny. Now I'm going to try to see if I can "complete the pentagon" in my 16-cell + 5-cell complex, to see what comes out of it...
quickfur
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### Re: Johnsonian Polytopes

Here's the animation of an icosahedron transforming into a J92 via partial Stott expansion:

This time I took Klitzing's advice and got rid of the edges that aren't part of the Stott expansion. There are 3 such edges, which you can see lacing the top part of the icosahedron to the bottom. They just vanish when the transformation begins, and they're not present in the J92.

I also assigned a distinct color to each group of edges moved by the Stott expansion, so that you can see what's being expanded more easily. Well, actually, the colors were left from when I was trying to figure out some wrong edges that I had put in by mistake. This animation was a lot more complex than I first expected (and far more complicated than it looks at the end!), and it wasn't easy figuring out exactly which edges should connect which vertices. There was a lot of splitting and merging going on that made it really confusing to keep track of. The colors were my attempt to group things that were moving in the same direction together, so that it's easier to see which edges were wrong. So you guys are lucky, that my mistakes turned out to be a nice visual aid in the final animation.
quickfur
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### Re: Johnsonian Polytopes

hello again

I've been trying to stott-expand ex, and arrived at the following:

in this lace city of ex, you can see a hexagonal ring of ikes
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`                o3o                                                           x3o   o3f f3o   o3x                                                                               o3o o3f   f3x       x3f   f3o o3o                                       f3o       o3F   F3o       o3f                                                                         o3x   x3f F3o   f3f   o3F f3x   x3o                                                                         f3o       o3F   F3o       o3f                                       o3o o3f   f3x       x3f   f3o o3o                                                                               x3o   o3f f3o   o3x                                                           o3o                `

together with the o3o at each "vertex" of this lace city, when expanded, they form the thawro pseudo-pyramid, this process, of adding x to each left part of the 3 can be further applied to ex, although I still doubt how x3f or F3o should be converted/expanded (either into o3F or u3f and (F+1)3o or v3x) but here's the lace city so far (I'm experiencing issues while trying to visualize these)
Code: Select all
`                  x3o                                                           o3x   x3f F3o   x3x                                                                               x3o x3f   F3x      o3F    F3o x3o                                       F3o       x3F  A3B        x3f                                                                         x3x  o3F A3B   F3f   x3F f3x   o3x                                                                         F3o       x3F  A3B        x3f                                       x3o x3f   F3x      o3F    F3o x3o                                                                               o3x   x3f F3o   x3x                                                           x3o      `
student5
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