D4.10 and D4.11

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

Re: D4.10 and D4.11

Postby quickfur » Wed Apr 02, 2014 10:30 pm

Wow, it's hard to get a good render for this CRF. Finally, after lots of trial and error, I managed to get an acceptable image:

Image

The image is kinda cluttered; for best effect you should try stereo viewing to see all elements clearly.

The icosahedra are rendered in red, orange, light orange, and yellow; the cyan cells are cuboctahedra (I added them into the image for reference, since otherwise there are too many stray edges lying around which makes it hard to see the cell shapes).

It's interesting to see the combo of tetrahedra + square pyramids surrounding a triangular prism recurring in various places around the cuboctahedra cells; on the complementary regions of the polytope, we have icosahedra surrounding octahedra in a configuration that looks like fragments of o5o3x3o.

Equally interesting are the 8 clusters of 4 icosahedra surrounding a tetrahedron, a configuration that doesn't occur in any other CRF that I know currently of, not even in the snub 24-cell!

EDIT: Added software models to new wiki page: D4.11.1
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Re: D4.10 and D4.11

Postby student5 » Mon Apr 07, 2014 12:46 am

whoa, D4.10 is awesome :o it has patches of D4.8 and J92 rombochoron aswell :o_o:

take a look at this projection:
Image
the three bilbiro's that share a 3-diminished ike, have a (more or less) equal orientation as D4.8.1
Image
it has, starting from the bottom and a face of the tetrahedron closest to the 4D viewpoint in D4.10
a tridiminished ike (check)
3 bilbiro's surrounding (check)
3 tets filling in the gaps (check displayed green in D4.10)
an ochtahedra on top of the trdim. ike (check)
three octahedra surrounding (check)
a pentagonal cupola (sort of check, it is part of the D4.11's cuboctahedra, which are diminished to show the J92's)
this made me think of three things:
(1) :o_o: these things are awesome :P
(2) this might be a sub-group, as these are probably all be derived from the same operation
(3) this part probably can be expanded in D4.8 aswell, since this patch is expanded both in D.4.11.1 and D4.4.1

P.S. @ quickfur. in your first description, you wrote
Image
The green patches are the 8 axial tetrahedra, lying along the 4 coordinate axes of 4D space. The red, orange, and yellow show 6 of the J91's around the nearest tetrahedron to the 4D viewpoint. The cyan cells are octahedra surrounded by 4 other octahedra: these fill the space where previously the J92's were. These 8 octahedra lie on the vertices of an alternated tesseract. On the vertices of the other alternated tesseract, are 8 cuboctahedra, here outlined in blue edges. These fill the space of where previously there were octahedron + 3 octahedra combos interfacing with J92 cells from the far side -- since the J92 cells are no more, cuboctahedra now fill that space. So here you can see demitesseractic symmetry very clearly.

after some thought regarding the orientation of the bilbiroes, it occured to me that it'd make more sense that the cuboctahedra are where the J92's were, and the cyan cells the octahedra in D4.10, but I can't figure out how this works in D4.11.1.
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Re: D4.10 and D4.11

Postby student5 » Mon Apr 07, 2014 5:44 pm

it seems Klitzing already pointed this out in the modifications of the snub demitesseract thread... :oops:
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Re: D4.10 and D4.11

Postby student91 » Wed Apr 09, 2014 10:04 am

It has been showed that D4.10 and D4.11 have a "parent", D4.11.1, ooxf3foox3oxfo *b3xFxo&#zx. you can get bilbiro's or thawro's by deleting some vertices. Just like how we got bilbiro's and pocuro's in D4.9.0!!

Now D4.9.0, a bilbiro'd rox, has relatives: D4.6 (bilbiro'd srix), and the bilbiro'd rahi.

so I thought, maybe the demicubic-bilbiro'd 600-cell (D4.11.1) has relatives as well, when you take e.g. rox as a start. rox's demicube-representation looks like (if I'm right) all rotatios of the A3x3o*b3o graph(A=2f), and all rotations of the F3o3f*b3x graph, and all rotations of the o3f3x*b3f graph, &#zx. when all in line, you get: AooFxfoxf3xxxooofff3oAofFxxfo*b3ooAxfFfox&#zx.

Now I want to expand it, in the same way ex is expanded to get D4.11.1:
Code: Select all
A3x3o*b3o:   F3o3f*b3x:   o3f3x*b3f:

B3x3o*b3o    C3o3f*b3x    x3f3x*b3f
x3x3A*b3o    F3o3x*b3F    o3F3f*b3o       (B=A+x=2f+x, C=F+x=f+2x)
x3x3o*b3A    o3x3F*b3f    F3f3o*b3x

Now the union of this (&#zx) probably has both thawro's and bilbiro's :o_o:
I guess the coordinates can be derived relatively easy from the coordinates of rox, as those coordinates are in demitesseractic symmetry as well, if i'm right.
EDIT: rox' coordinates are unfortunately not in demitesseractic symmetry :(
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Re: D4.10 and D4.11

Postby student91 » Fri Apr 25, 2014 8:19 am

I think I've found coordinates for it, although I'm not sure if they are correct. Here they are:
Code: Select all
all permutations of the following with an even number of changes of signs of:

(0 , 0 , 2, 3-2f)             
(A , A , 2f+2 ,  A)
(-A , A , 2f+2 ,  A)

(F , v , F , -1)
(C , f , C , f)
(B , 1 , B,  2f-1)

(F , -v , 3f+1 , 3f-1)
(f , f , 3f+2 , 3f+2)
(1 ,1 ,2f+1 ,1)

Those should give a polytope with edge-length q
I'm not sure if those coordinates are correct. If they aren't, could you still post some renders, so I can see what's wrong? :)
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Re: D4.10 and D4.11

Postby quickfur » Fri Apr 25, 2014 2:55 pm

The appearance of 3-2*phi, 2*phi-1 and 3*phi-1 seem to be an indication that something went wrong somewhere. Usually you don't see these numbers when you're dealing with edge length 2. Or are you using some other edge length here?

In any case, the result is not CRF; the minimum edge length is 2√2 and there are some long edges that I can't identify (7.404918), might be caused by non-convexity in some areas. Unfortunately, you appear to have some redundant vertices that lie on the hyperplanes of some facets, which caused my polytope viewer to get confused and be unable to identify edges correctly, so it produced nonsensical output. :( Anyway, since you asked, I included it here:

Image

I don't think it's helpful though.
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Re: D4.10 and D4.11

Postby student91 » Sun Apr 27, 2014 8:28 am

quickfur wrote:The apperance 2*phi, 2*phi-1 and 3*phi-1 seem to be an indication that something went wrong somewhere. Usually you don't see these numbers when you're dealing with edge length 2. Or are you using some other edge length here?
Indeed my edge-length is 2sqrt(2). This is because I'm using a demitesseractic orientation (All permutations, even number of changes of signs). I got those coordinates using a formula. I'm not sure if this formula is correct though. You could check this by checking if the first coordinate, together with its permutations, give the first CD-diagram, the second coordinate the second CD-diagram, and so on. If this is not the case, could you tell me what the edge-lengths are?
If this is not the problem, the respective orientations could be. This can be checked by taking a set of coordinates, look if it has a part (let's say the x3x3x*b3.-part) oriented the same way as the other sets. I think if the problem lays here, it can be solved by adding a minus-sign to one coordinate, an then do the permutations.

The weird value becomes F if divided by the edge-length. I guess one part is oriented wrongly, causing the F-edge to be revealed.
Ps. If you don't like the 2sqrt(2) edge-length, you can add a scalar of sqrt(1/2) to the coordinates to make it have edge-lengths of 2.
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Re: D4.10 and D4.11

Postby student91 » Sun Apr 27, 2014 10:16 am

Something more on the formula I used: When you have one vertex of a demitesseractic polytope, you can permute it to get all the vertices of it. This means, if we find coordinates for one vertex, we will be able to get all the other vertices for free. My formula tries to find this vertex that makes the desired polytope when permuted.
I started with an ugly drawing of a cube, saying it has w-coordinate 1, and then addressing coordinates to all vertices. Next I drew the demitesseractic reflective cells in this cube. I tried to find the boundaries of one of these reflective cells by means of formula's. I found x=y, x=-y, x=-z and z=w. Now these give four 3-space-boundaries (wendy probably has a better name for a 3-space, is it chorix?)
special of the area in this boundary is, that when you permute the coordinates the desired way, this area gets copied around like a kaleidoscope-thing. this means tha when the distance from a vertex in this area to a boundary in this area is x, after the permutations this gives a distance of 2x to the vertex over there, so the covex hull has an edge of 2x as well.
The place for the vertex can be found by taking a chorix parallel to boundary 1, with distance a to boundary 1, a chorix parallel to boundary 2, with distance b to boundary 2, a chorix parallel to boundary 3, with distance c to boundary 3 and a chorix parallel to boundary 4, with distance d to boundary 4. The vertex that lies at the intersection of those chorices will get edges of length 2a, 2b, 2c and 2d through boundaries 1, 2, 3 and 4 respectively. these boundaries correspond to the nodes of the CD-diagram. For me, when the CD-diagram is a3c3b*b3d, a corresponds to x=y, b to x=-y, c to x=-z and d to z=w.
These parallel chorices should get formula's, so we can calculate its intersection's coordinates. The formulas I found were x=y-a*sqrt(2), x=-y-b*sqrt(2), z+x=c*sqrt(2) and w=z-d*sqrt(2). Solving this system gave me: x=-sqrt(1/2)[a+b]; y=sqrt(1/2)[a-b]; z=sqrt(1/2)[a+b+2c]; w=sqrt(1/2)[a+b+2(c-d)]
I thought these sqrt(1/2)'s to be ugly, so I jumped to an edge-length of 2sqrt(2) instead of 2. furthermore I omitted the - at x, which is allowed as it only flips the structure, without changing it. This gave me x=a+b; y=a-b; z=a+b+2c; w=a+b+2c-2d.

So this means that for a CD-diagram a3c3b*b3d, the coordinates should be given for edge-length 2sqrt(2) by (a+b;a-b;a+b+2c;a+b+2c-2d), with all permutations and an even number of changes of signs. I have to admit that when w<0, I flipped it's sign, which may have caused errors. Furthermore I don't trust these formulas to be completely correct.
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Re: D4.10 and D4.11

Postby student91 » Sun Apr 27, 2014 10:41 am

I recalculated the coordinates. most of them are still the same, although some of them have changes
Code: Select all

(0 , 0 , 2, -2f)       <--       
(A , A , 2f+2 ,  A)
(-A , A , 2f+2 ,  A)

(F , v , F , -1)
(C , -f , C , -f)       <--
(B , 1 , 2F,  2F)       <--

(F , -v , 3f+1 , 3f-1)
(f , f , 3f+2 , 3f+2)
(1 , -1 , 2f+1 , -1)       <--

Those should give a polytope with edge-length q

Ps. if this is nonsense again, could you colour edges with different edge-lengths differently in your renders :)
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Re: D4.10 and D4.11

Postby quickfur » Sat May 03, 2014 2:56 pm

Really sorry for the laaaate response, I have been really busy the past week and didn't get around to trying out your new coordinates until now. Unfortunately, it's still not CRF, I'm still getting a min edge length of 2√2 and a max edge length of 7.404... (unidentified). The long edge is probably a sign that somewhere in your construction there's a non-convex patch.

Unfortunately, my polytope viewer currently doesn't have a good way of coloring edges by length (I really should add that, though: it will be indispensible once I start dealing with non-CRFs like the 4D Catalans). The non-closed edges are still present. :( I think I'm just going to compute the coordinates myself, based on your lace city. But I might not get around to it until tomorrow.
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Re: D4.10 and D4.11

Postby student91 » Sat May 03, 2014 6:23 pm

I, quite elaborately said in my previous post that my edge-length is 2sqrt(2). This makes the "unidentified" edge become F. I guess my coordinates are still wrong then. Maybe the expansion doesn't work after all though. Could you tell me what the firstlast triplet of coordinates looks like, does it make bilbiros?
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Re: D4.10 and D4.11

Postby quickfur » Fri May 16, 2014 2:39 am

student91 wrote:I, quite elaborately said in my previous post that my edge-length is 2sqrt(2). This makes the "unidentified" edge become F. I guess my coordinates are still wrong then. Maybe the expansion doesn't work after all though. Could you tell me what the firstlast triplet of coordinates looks like, does it make bilbiros?

Really sorry for the super late reply, I've had a totally insane week, busy both at work and at home, and haven't had a chance to do anything 4D-related except answer a few posts.

It seems that it's these last triplet of coordinates that are causing grief. I think you have some redundant coplanar vertices here, which are confusing my polytope viewer, so it can't find the edges correctly. The output is the same as before, with those strange isolated edges (and the other edges which are supposed to be there aren't showing up -- 57 edges got deleted because they are zero-length, which is a sign of program confusion caused by redundant coplanar vertices).

Either that, or I completely misread your coordinates. Here is what I used:
Code: Select all
apecs<0, 0, 2, -2*phi>
apacs<2*phi, 2*phi, 2*phi+2, 2*phi>

apecs<phi^2, 1/phi, phi^2, -1>
apecs<phi+2, -phi, phi+2, -phi>
apecs<2*phi+1, 1, 2*phi^2, 2*phi^2>

apecs<phi^2, -1/phi, 3*phi+1, 3*phi-1>
apecs<phi, phi, 3*phi+2, 3*phi+2>
apecs<1, -1, 2*phi+1, -1>
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Re: D4.10 temp

Postby Klitzing » Sun May 17, 2015 10:22 pm

Klitzing wrote:
quickfur wrote:Image
Image


Did you already make up a good name for that one?
Or should we ask HedronDude to set up one?

Btw., here is the incmats of that figure (in its full hexadecachoral symmetry):
Code: Select all
32  *  * |  3  3  0  0   0  0 |  3  6  3  0  0  0  0  0  0  0 | 1  3  3  1 0  0 0
 * 96  * |  0  1  2  2   2  0 |  0  2  2  1  1  2  1  2  0  0 | 0  1  2  1 1  1 0
 *  * 48 |  0  0  0  0   4  4 |  0  2  0  0  0  0  1  2  2  2 | 0  2  1  0 0  2 1
---------+--------------------+-------------------------------+------------------
 2  0  0 | 48  *  *  *   *  * |  2  2  0  0  0  0  0  0  0  0 | 1  2  1  0 0  0 0
 1  1  0 |  * 96  *  *   *  * |  0  2  2  0  0  0  0  0  0  0 | 0  1  2  1 0  0 0
 0  2  0 |  *  * 96  *   *  * |  0  0  1  1  0  1  0  0  0  0 | 0  0  1  1 1  0 0
 0  2  0 |  *  *  * 96   *  * |  0  0  0  0  1  1  0  1  0  0 | 0  0  1  0 1  1 0
 0  1  1 |  *  *  *  * 192  * |  0  1  0  0  0  0  1  1  0  0 | 0  1  1  0 0  1 0
 0  0  2 |  *  *  *  *   * 96 |  0  0  0  0  0  0  1  0  1  1 | 0  1  0  0 0  1 1
---------+--------------------+-------------------------------+------------------
 3  0  0 |  3  0  0  0   0  0 | 32  *  *  *  *  *  *  *  *  * | 1  1  0  0 0  0 0
 2  2  1 |  1  2  0  0   2  0 |  * 96  *  *  *  *  *  *  *  * | 0  1  1  0 0  0 0
 1  2  0 |  0  2  1  0   0  0 |  *  * 96  *  *  *  *  *  *  * | 0  0  1  1 0  0 0
 0  3  0 |  0  0  3  0   0  0 |  *  *  * 32  *  *  *  *  *  * | 0  0  0  1 1  0 0
 0  3  0 |  0  0  0  3   0  0 |  *  *  *  * 32  *  *  *  *  * | 0  0  0  0 1  1 0
 0  4  0 |  0  0  2  2   0  0 |  *  *  *  *  * 48  *  *  *  * | 0  0  1  0 1  0 0
 0  1  2 |  0  0  0  0   2  1 |  *  *  *  *  *  * 96  *  *  * | 0  1  0  0 0  1 0
 0  2  1 |  0  0  0  1   2  0 |  *  *  *  *  *  *  * 96  *  * | 0  0  1  0 0  1 0
 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  * 32  * | 0  1  0  0 0  0 1
 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  *  * 32 | 0  0  0  0 0  1 1
---------+--------------------+-------------------------------+------------------
 4  0  0 |  6  0  0  0   0  0 |  4  0  0  0  0  0  0  0  0  0 | 8  *  *  * *  * *  tet
 3  3  3 |  3  3  0  0   6  3 |  1  3  0  0  0  0  3  0  1  0 | * 32  *  * *  * *  teddi
 4  8  2 |  2  8  4  4   8  0 |  0  4  4  0  0  2  0  4  0  0 | *  * 24  * *  * *  bilbiro
 1  3  0 |  0  3  3  0   0  0 |  0  0  3  1  0  0  0  0  0  0 | *  *  * 32 *  * *  tet
 0 12  0 |  0  0 12 12   0  0 |  0  0  0  4  4  6  0  0  0  0 | *  *  *  * 8  * *  co
 0  3  3 |  0  0  0  3   6  3 |  0  0  0  0  1  0  3  3  0  1 | *  *  *  * * 32 *  oct
 0  0  6 |  0  0  0  0   0 12 |  0  0  0  0  0  0  0  0  4  4 | *  *  *  * *  * 8  oct


--- rk


Hi Quickfur, student91, et al.!

Well, this finding of Quickfur from Mar 12, 2014, which meanwhile was called "pretasto" (short for "pentagonorhombic trisnub trisoctachoron"), and in that above once more cited post already got its fully demitessic incidence matrix, appeared today also as a true EKF (expanded kaleido-faceting)!

In fact, start with sadi (snub 24-cell), and describe that in demitessic subsymmetry. This then reads fox3ooo3xfo *b3oxf&#zx . Next apply an edge flip onto the first node of the third layer. The resulting kaleido-faceting then reads fo(-x)3oox3xfo *b3oxf&#zx . And finally apply a partial Stott expansion wrt. the first node in all layers simultanuously. This now results in Fxo3oox3xfo *b3oxf&#zx . This figure then is nothing but what once was described by Quickfur!

You even could spot in that description the bilbiroes (J91) as Fxo ... xfo oxf&#zx , or e.g. the teddies (J63) as ... oox3xfo ...&#xt .

--- rk
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Re: D4.10 and D4.11

Postby quickfur » Mon May 18, 2015 4:44 am

Wow! So there is a pattern to the madness after all? :) Sorry I can't say more at the moment, been really busy with other things.
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