11cell

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11cell
Hi,
The 11cell or hendecachoron is 11 hemiicosahedra. I have a hemiicosahedra (OFF file below). I was wondering if the 11cell might be attainable (in Stella4D, pressing "u") but it doesn't appear to be. It stops at 10 cells and can't display anything. Is this because the hemiicosahedra is nonorientable?
Roger
OFF
# 6v10f15e
# Generated by 'Stella4D', Version 4.4
# Author: Robert Webb
# Web site: http://www.software3d.com/Stella.php
# Licensed to: Roger Kaufman
# Date: 9:11:49 AM, 29 January 2009
#
# Copyright (C) Robert Webb, 20012008.
# This copyright notice and the above
# information must be kept intact.
6 10 15
0.57735026677154044 0.00000000000000000 0.40824828980302286
0.28867513519687621 0.49999999920719485 0.40824829003990537
0.28867513528105920 0.49999999914905913 0.40824828980179945
0.28867513287350210 0.50000000033526926 0.40824828980302286
0.28867513538130307 0.49999999924622762 0.40824828980302286
0.57735026899467035 0.00000000000000000 0.40824828956736364
3 1 3 0 255 0 0
3 1 0 5 255 255 0
3 1 5 4 255 255 0
3 2 4 1 255 255 0
3 2 1 3 255 255 0
3 5 3 2 255 255 0
3 5 4 3 255 0 0
3 3 4 0 0 0 255
3 2 0 4 255 0 0
3 5 2 0 255 255 0
The 11cell or hendecachoron is 11 hemiicosahedra. I have a hemiicosahedra (OFF file below). I was wondering if the 11cell might be attainable (in Stella4D, pressing "u") but it doesn't appear to be. It stops at 10 cells and can't display anything. Is this because the hemiicosahedra is nonorientable?
Roger
OFF
# 6v10f15e
# Generated by 'Stella4D', Version 4.4
# Author: Robert Webb
# Web site: http://www.software3d.com/Stella.php
# Licensed to: Roger Kaufman
# Date: 9:11:49 AM, 29 January 2009
#
# Copyright (C) Robert Webb, 20012008.
# This copyright notice and the above
# information must be kept intact.
6 10 15
0.57735026677154044 0.00000000000000000 0.40824828980302286
0.28867513519687621 0.49999999920719485 0.40824829003990537
0.28867513528105920 0.49999999914905913 0.40824828980179945
0.28867513287350210 0.50000000033526926 0.40824828980302286
0.28867513538130307 0.49999999924622762 0.40824828980302286
0.57735026899467035 0.00000000000000000 0.40824828956736364
3 1 3 0 255 0 0
3 1 0 5 255 255 0
3 1 5 4 255 255 0
3 2 4 1 255 255 0
3 2 1 3 255 255 0
3 5 3 2 255 255 0
3 5 4 3 255 0 0
3 3 4 0 0 0 255
3 2 0 4 255 0 0
3 5 2 0 255 255 0
Roger Kaufman
http://www.interocitors.com/polyhedra/
http://www.interocitors.com/polyhedra/
 robertw
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Being nonorientable doesn't matter (this is already the case for many uniform polychora, is it not?).
I imagine the problem is that coplanar faces of the vertex figure share an edge. Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse. Actually I'd be surprised if that combo did work.
Rob.
I imagine the problem is that coplanar faces of the vertex figure share an edge. Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse. Actually I'd be surprised if that combo did work.
Rob.

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Carlo's models always were tweaked so that the intersection didn't happen. I can try that.robertw wrote:Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse.
Rob.
I remember some of profesional geometers saying that the 11cell and 57cell can't be represented in the same way as other polytopes. Something to do with hyberbolic geometry; I have little hope of understanding it.
I made the network for the hemidodecahedron. There isn't much promise that this figure can be depicted in anyway to construct the 57cell.
Roger
Roger Kaufman
http://www.interocitors.com/polyhedra/
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 robertw
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I don't really understand the shape you're after. Hitting "u" is for uniform or scaliform polychora. Does the 11cell have all vertex figures the same? I can't imagine that adjusting it to avoid the coplanarity could possibly work. If as you mention that it can only be realised in hyperbolic space, then you won't be able to create it in Stella4D.
Rob.
Rob.

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Carlos' paper is on line. It shows ways of trying to display the 11cell but using purely 3D constructs. i.e usings 11 vertices and so on. I was bring such models into Stella but I had to double the number of faces (replicated on top of one another) in order for it to be a closed system. This also triples the number of edges. However, since these faces and edges are visually simultaneous, it is still a representation.robertw wrote:I don't really understand the shape you're after.
http://www.cs.berkeley.edu/~sequin/PAPE ... 11Cell.pdf
OK, I didn't know that is what it was specifically for. I had been using it on some of the Johnson solids and would get objects that didn't close yet could still be displayed.robertw wrote:Hitting "u" is for uniform or scaliform polychora. Does the 11cell have all vertex figures the same? I can't imagine that adjusting it to avoid the coplanarity could possibly work.
Mathworld has a short explanation as to why it can't be shown in 4D. The cells are all mobius (I assume this means hemiicosahedrons are mobius).robertw wrote:If as you mention that it can only be realised in hyperbolic space, then you won't be able to create it in Stella4D.
http://mathworld.wolfram.com/11Cell.html
Roger
Roger Kaufman
http://www.interocitors.com/polyhedra/
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 Jabe
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I'm aware of two potential hendecachora with identical cells that could be modeled in 4D, however they aren't the one with the hemiicosahedra  that one requires some strange curved 4space. I have yet to model the two 11celled polychora, but I have done the 7cell, two 9cells, and three distinct 13celled polychora. These polychora belong to a group that Wendy Kreiger calls "steptegums". Below is the net of one of the 13cells.
May the Fourth (dimension) be with you.

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Perhaps those would be nice additions to the Stella library. Are the 7 though 13 (odd) cells unique or are there Ncells at various integer values?Jabe wrote:I'm aware of two potential hendecachora with identical cells that could be modeled in 4D, however they aren't the one with the hemiicosahedra  that one requires some strange curved 4space. I have yet to model the two 11celled polychora, but I have done the 7cell, two 9cells, and three distinct 13celled polychora. These polychora belong to a group that Wendy Kreiger calls "steptegums". Below is the net of one of the 13cells.
I suppose the hemiicosahedron has to be modeled in a curved 4space if all the faces are to be the same. I found something interesting, however, if it means anything significant I am not sure.
The connection system for the hemiicsoahedron can be made in 3D, and I suppose it could be shown to be mobius but doing so is very convoluted. If it means one could hop from face to face and eventually be on the other side of the starting face that would be simple. But from each face there would be two paths to take. The total number of circuits would be fairly high I presume.
Let it be such, though, that it can be represented in 3space with more than one face type, it can also be represented in 2space. Smashed flat, it will still have the 6 vertices, 10 faces, and 15 edges. Those 6 vertices can be placed on the 6 of the 11 vertices of a heptagon. This can be done in Stella.
Using an 11gonal prism, 11.4.4, one one face count off the vertices 1 through 11. Change to No Symmetry, facet out 10 triangles between these vertices.
1 2 5
1 2 8
1 3 6
1 3 8
1 5 6
2 3 5
2 3 6
2 6 8
3 5 8
5 6 8
This closes and creates the flat semiicosahedron. Now change to 11fold Pyramidal symmetry. This repeats the former faceting 11 times around the heptagon. From the faceting window save the faceting diagram as an OFF file (nice trick in Stella!).
Looking at the OFF file in an editor, it has 11 vertices, 55 faces, and 55 edges, the numbers corresponding to the 11cell. Apparently, for every hemiicosahedron 5 of the faces are ajoined to 5 others cancelling them out.
Unfortunately as is, the OFF file can't be brought back into Stella. Being flat is bad enough, one can tweak one of the Zaxis coordinates but that is not enough. The faces need to be doubled to get two faces to an edge, then it can be brought into Stella, then it may become the compound of 10 hemiicosahedra.
Something fun to try if interested. I don't think is the only configuration of hemiicosahedron that can do this. Simulating the 57cell on the 57gon might be possible.
Makes me wonder if an icosahedron could be smashed flat and repeated on some kind of polygon.
Roger
Roger Kaufman
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 Jabe
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These exist for all N greater than 4, for N=5 we get the pentachoron.polymorphic wrote: Perhaps those would be nice additions to the Stella library. Are the 7 though 13 (odd) cells unique or are there Ncells at various integer values?
For N=6 we get the triangle duoprism, for N=8 there are two and one is the tesseract.
May the Fourth (dimension) be with you.

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And the number of configurations increases as N grows larger I would presume. Is there a formula for that?Jabe wrote:These exist for all N greater than 4, for N=5 we get the pentachoron.
For N=6 we get the triangle duoprism, for N=8 there are two and one is the tesseract.
Roger
Roger Kaufman
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 Jabe
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A step tegum requires two numbers  N and M  where N is the number of cells, and M represents the "step". I'll use a 13 sided one for an example.polymorphic wrote:And the number of configurations increases as N grows larger I would presume. Is there a formula for that?
Roger
Consider the 13gon, we can label its vertices from 0 to 12  now consider the 13gon duoprism, we can label its vertices with two numbers  both going from 0 to 12. To find the vertices of the 13  3 step prism (where 3 is the step), we start with vertex 0,0  then 1,3  2,6  3,9  4,12  5,2 (2 is 15 mod 13)  6,5  7,8  8,11  9,1  10,4  11,7  12,10. The step prism is the convex hull of those points, where the step tegum is the dual.
So the question is "what M's give us distinct step tegums"  when M = 1, we only get a polygon  so it never works. When M > N/2 it looks just like a NN/2 step tegum (in otherwords, it might be a step 10 if we count the vertices clockwise, but going counterclockwise it would be a step 3  so the only M's to check are those between 2 and N/2. N/2 gives us a duotegum (dual of duoprism)  so no need to count it. So for 13, M can be 2,3,4,5, or 6  but there is one more trick  a 13gon duoprism is the cross product of 13gon #1 by 13gon #2  if we change the order to 13gon #2 by 13gon #1, the step may also change  it turns out that 13  2 = 13  6, what is step 2 in one orientation is step 6 in the other. Here is a trick to see which steps are equal: Let M and P be two different steps, if MP (M times P) is one off from a multiple of N, then NM = NP. 2x6 = 12 which is one off from 13, 3x4 = 12 which is one off from 13, the only number left is 5  5x5 = 25 is one off from 26 a multiple of 13. The 135 has twice the symmetry also. We can call P the alterstep of M
Lets check the 17 sided step tegums. We need to check steps 28
2x8 = 16 = 171
3x6 = 18 = 17+1
4x4 = 16 = 171
5x7 = 35 = 34+1
So there are 4 17sided step tegums 172, 173, 174 (double symmetry), and 175
One more thing, if N and M are not relatively prime  then M will not pair up with another step  since MP is never one off from a multiple of N. So there are three types of step tegums: M,N not relatively prime (i.e. 104), M,N relatively prime and M<>P (P is the alterstep), and M,N relatively prime with M=P (has double symmetry).
Another example N = 20: need to check steps 2 through 9
20  2: not relatively prime
20  3: alter step = 7, 3x7=21
20  4: not relatively prime
20  5: not relatively prime
20  6: not relatively prime
20  8: not relatively prime
20  9: alter step = 9, 9x9=81  double symmetry (this is actually the decagon duoprism)
it turns out that the highest possible step of an even N is a duoprism.
May the Fourth (dimension) be with you.

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Since it is systematic I wonder if these could be generated. If the connectivity could be systemized it could.
You mentioned that there are two 8cells. Try to look something like that up on google! If there exists information on an 8cell it is inevitably the tesseract.
Looking at what you've written about even N, it might have to be a duoprism. But the 4duprism (generated in stella) or the prism built on the cube, is still looks like a tesseract.
Roger
You mentioned that there are two 8cells. Try to look something like that up on google! If there exists information on an 8cell it is inevitably the tesseract.
Looking at what you've written about even N, it might have to be a duoprism. But the 4duprism (generated in stella) or the prism built on the cube, is still looks like a tesseract.
Roger
Roger Kaufman
http://www.interocitors.com/polyhedra/
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Not trying to be a lamer here, and I really tried to see if this was out there some where. In any search I would ultimately land on your 4D dice page!Jabe wrote:the 8  2 looks totally different.
I'm convinced you may be the only person in the world who has seen it.
Roger
Roger Kaufman
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How do you do this?!
It looks like there is no other way to make these than by hand. It is amazing to me that such an oddball looking outer cell can contain copies of itself inside.
Do you have the stel file for this? I had thought I might try to construct what you had just by looking but I'd have to guess at too much. But you must have to do this by hand, so I am wondering how you do it.
Roger
It looks like there is no other way to make these than by hand. It is amazing to me that such an oddball looking outer cell can contain copies of itself inside.
Do you have the stel file for this? I had thought I might try to construct what you had just by looking but I'd have to guess at too much. But you must have to do this by hand, so I am wondering how you do it.
Roger
Roger Kaufman
http://www.interocitors.com/polyhedra/
http://www.interocitors.com/polyhedra/