quickfur wrote:Thus, we may call this CRF the distended snub 24-cell.
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?
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.
btw that GIF is very cool, I would recommend you putting it on the bilbiro-page on your website.
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?
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.
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-4.20584889681414340 -1.01857602983881710 -8.41169779410675740
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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 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.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...
[...]
Marek14 wrote:Well, the only thing needed would be changing the coordinates -- the structure of icosahedron and great icosahedron is the same.
student91 wrote:[...]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.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...
[...]
student91 wrote:[...]
I was wrong, I found another way to partially stott-expand the ike, this time into a orthocupolarotunda!! [...]
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
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.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...
that's what I was thinking about as well, has to be investigatedAlso, 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 bet they will, looking forward to seeing themI think this is the last such partial stott-contraction of ikewe'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.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.
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 welcomeAnother 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?
student91 wrote: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.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.
[...]
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
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.
[/quote]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?
x2o x2o o2(-x)
o2f o2f f2o
f2x -> f2(-x) x2f
o2f o2f f2o
x2o x2o o2(-x)
x2x o2o
o2F f2x
f2x o2F
o2F f2x
x2x o2o
o5o x3o o5x x3x
f5(-x) o3f -> f5o o3F
x5o f3o x5x f3x
o5o x3(-x) o5x x3o
#no x's in the left tower, so it remains the same
o4o o4o x4o o4x #orthobicup and cuboc (gyrobicup)
o4x -> q4(-x) -> x4x q4o
o4o o4o x4o o4x
#symetric, so only one modification view
x3o x3o x3x #tricup
o3x -> x3(-x) -> x3o
#left tower again
o2x o2(-x) x2x o2o #elongated square pyramid and a non-CRF thingy
q2o -> q2o -> Q2o q2x
o2x o2(-x) x2x o2o
x3o x3o x3x #tricup
o3x -> x3(-x) -> x3o
x2o x2o x2x #squippy
o2x -> o2(-x) -> o2o
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...
quickfur wrote:
student91 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?
Klitzing wrote:Bilbiroquickfur 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.
[...]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.
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
x3o3o
o3o3f
o3f3o
f3o3x
o3x3f
f3x3o
etc.
x3x3o o3x3o
o3x3f x3o3f
o3F3o x3f3o
f3x3x F3o3x
x3o3F x3x3f
F3o3x F3x3o
x3x3f o3x3f
o3F3o x3f3o
f3x3o F3x3o
o3x3x x3o3x
x3x3o //x3o3f somehow disappears?
o3F3o
f3x3x
x3o3F
o3f3x
f3o3x
o3x3o
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
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
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