Johnsonian Polytopes

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

Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 11, 2012 8:30 pm

wintersolstice wrote:[...]Sorry I know how they can be joined I meant how can it be done in a convex way :D (you've just said it is non-convex, so why is it on the the list of CRF polychora?) that's what I was asking :Dbased on what yous aid it should be under a list of shapes that tesselate!!!

Basically i'm asking: "why is it it listed on Convex Regular Faced Polychora page (on the wiki) if it isn't convex?" and based on what you've said: "should it be on a list of shapes that tesselate instead?"

Sorry I should have been more specific!

Oh, that. Well, the 16-cell itself is certainly convex, and its ridges are regular, so it is itself a CRF polychoron. Although implicitly we're looking for non-uniform CRF polychora, the uniform and regular polychora also satisfy the convex regular-faced requirement, so they are included on that page.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 12, 2012 4:10 am

So, there's been a lot of talk about augmented duoprisms but very few pictures. So I decided to render the omni-augmented 5,20-duoprism for your viewing pleasure:

Image Image

The 5,20-duoprism is quite narrow, due to having one very large radius (20-gon) and one comparatively small radius (5-gon with same edge length). Every other pentagonal prism in the 20-membered ring can be augmented with a pentagonal prism pyramid, because its dihedral angle is exactly 18°, and it merges with adjacent cells to form elongated pentagonal pyramids. When every other cell is augmented, the augments merge with the cells between them to form elongated pentagonal bipyramids. I've rendered these in magenta, as you can see on the left.

On the right is a different viewpoint of the 5,20-duoprism, which shows just how narrow it is compared to its major radius. The yellow edges are on the far side, and the black edges on the near side. You can see the closest elongated pentagonal bipyramid to the 4D viewpoint at the center, with its antipodal cell projected within it. This viewpoint shows the square pyramid cells surrounding each "joint" between two elongated pentagonal bipyramids quite clearly.

A quick verification with my polytope viewer confirms that the edge lengths are all equal. (Unfortunately I don't have nice coordinates for it... basically a bunch of hand-calculated constants plugged into sines and cosines. I'm not expecting the algebraic coordinates to be very pretty.)

Have at it!
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Re: Johnsonian Polytopes

Postby wintersolstice » Thu Jan 12, 2012 1:06 pm

quickfur wrote:
wintersolstice wrote:[...]Sorry I know how they can be joined I meant how can it be done in a convex way :D (you've just said it is non-convex, so why is it on the the list of CRF polychora?) that's what I was asking :Dbased on what yous aid it should be under a list of shapes that tesselate!!!

Basically i'm asking: "why is it it listed on Convex Regular Faced Polychora page (on the wiki) if it isn't convex?" and based on what you've said: "should it be on a list of shapes that tesselate instead?"

Sorry I should have been more specific!

Oh, that. Well, the 16-cell itself is certainly convex, and its ridges are regular, so it is itself a CRF polychoron. Although implicitly we're looking for non-uniform CRF polychora, the uniform and regular polychora also satisfy the convex regular-faced requirement, so they are included on that page.


No you've mis-understood again! when I said "why is it it listed on Convex Regular Faced Polychora page" I was refering to the "two 16-cell joined together" not the 16-cell itself
look on the wiki someone said:"The aerochoron has a single augmentation, consisting of two aerochora joined cell-to-cell." on the Convex Regular Faced Polytope page" that's what this was all about
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 12, 2012 6:01 pm

wintersolstice wrote:[...] No you've mis-understood again! when I said "why is it it listed on Convex Regular Faced Polychora page" I was refering to the "two 16-cell joined together" not the 16-cell itself
look on the wiki someone said:"The aerochoron has a single augmentation, consisting of two aerochora joined cell-to-cell." on the Convex Regular Faced Polytope page" that's what this was all about

An aerochoron is a 5-cell. Argh, nevermind. The mistake was that it was supposed to be pyrochoron not aerochoron. My bad.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 25, 2012 5:53 pm

quickfur wrote:[...]So I decided to render the omni-augmented 5,20-duoprism for your viewing pleasure:

Hmm, I forgot to convert it to OFF format for you Stella4D users (hi Marek!). So here it is:

http://teamikaria.com/hddb/dl/HGJHJN82VTV7WC5G8K3V8J0AFF.off

Enjoy!
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Jan 25, 2012 7:12 pm

quickfur wrote:
quickfur wrote:[...]So I decided to render the omni-augmented 5,20-duoprism for your viewing pleasure:

Hmm, I forgot to convert it to OFF format for you Stella4D users (hi Marek!). So here it is:

http://teamikaria.com/hddb/dl/HGJHJN82VTV7WC5G8K3V8J0AFF.off

Enjoy!


Thanks :) Looks nice.
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Re: Johnsonian Polytopes

Postby wintersolstice » Tue Mar 27, 2012 9:10 pm

it mentions on the wiki about the "antiprism rings" (discovered by "Mrrl") and it says that the first is the "square case" well I think there's a triangular case:D this particular polychoron is also a "diminished rectified 5-cell"

the R 5-cell has 10 vertices and it's verf is a 3-prism (and there are 3 octahedra meeting at a vertex) so removing 1 vertex creates 3 4-pryamids and leaves two octahdra (undiminshed) now and octahdron is the same as a 3-antiprism. with one vertex removed it now has nine. (3X3)

so if there are 2 3-antiprisms and 1 3-prism and 9 vertices surely this is the trigonal case? :D (there is a segmentotope "3g || oct" where oct (octahedron) is a 3 antiprism :D )

also if you remove another vertex (it can't be adjacent) you create another segmentochoron: "ortho line || 3p" (as it's called on "orchid palms") the "line || 3p" is a tetrahedron prism posisitioned differently :D

I think but am not sure that the R 5-cell itself might be the "tet || oct" segmentotope

I've also discovered that the "antiprism rings" are the only infinite set of "Johnson Segmentotopes" (all the others are scaliform or less than 4D)

You see that list has all the segmentoptopes up to 4D inclusive (vertex transitive and non-vertex transitive) aswell as including duplicates (like above) so I've been trying to fiter out the "Johnson Polychora" from them hope to get a list up soon for the status of them all :D
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jun 23, 2012 4:21 am

It's been awfully quiet in here... but I just noticed that wintersolstice added four interesting CRFs to the wiki page, one of which is already included in the Klitzing list, but the others are interesting, specially because they are vertex-transitive/facet-transitive, though not uniform. I have a weakness for facet-transitive polytopes... and I'd love to render one of these, if only I had the coordinates.

So this is just a humble request for somebody to post the coordinates, if they happen to have them, or if not, I'll just have to compute them myself (probably programmatically -- I suck at doing algebra by hand). The bi-icositetradiminished 600-cell is especially interesting for me, as it appears to be an evil-twin version of the bitruncated 24-cell (48 diminished icosahedra instead of 48 truncated cubes), and still manages to be vertex-uniform, which is ineffably cool.
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Re: Johnsonian Polytopes

Postby Tamfang » Sat Jun 23, 2012 5:59 am

What interesting wiki page would that be?
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jun 23, 2012 6:35 pm

Tamfang wrote:What interesting wiki page would that be?

This one. See the bottom of the page for the polychora I'm referring to.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jun 25, 2012 4:24 pm

All right, so I've calculated the coordinates of the bi-icositetradiminished 600-cell by hand. It wasn't hard, just very tedious. But anyway, here's a preliminary render:

Image

Due to the non-uniform shape of the cells, it's not so easy to pick them out, so I highlighted the nearest cell to the 4D viewpoint in yellow, and two neighbouring cells in red. These are all tridiminished icosahedra, as is expected. As usual, I culled cells on the far side in order to minimize visual clutter. I'll say more later (gotta run right now), but the cells surrounding the yellow cell are connected to each other in interesting, non-equivalent ways. I'll even post coordinates later: there are 72 of them.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jun 25, 2012 7:39 pm

quickfur wrote:All right, so I've calculated the coordinates of the bi-icositetradiminished 600-cell by hand. [...]

OK, here are the coordinates:

Code: Select all
<0,  1, ±φ,  1/φ>
<0, -1,  φ, ±1/φ>
<0, ±1, -φ, -1/φ>
<0,  1/φ,  1, ±φ>
<0, -1/φ, ±1,  φ>
<0, ±1/φ, -1, -φ>
<0,  φ, ±1/φ, -1>
<0, -φ,  1/φ, ±1>
<0, ±φ, -1/φ,  1>

< 1, 0, -1/φ, ±φ>
<-1, 0, ±1/φ,  φ>
<±1, 0,  1/φ, -φ>
< 1,  1/φ, ±φ, 0>
<-1, ±1/φ,  φ, 0>
<±1, -1/φ, -φ, 0>
< 1,  φ, 0, ±1/φ>
<-1, ±φ, 0,  1/φ>
<±1, -φ, 0, -1/φ>

< 1/φ, 0,  φ, ±1>
<-1/φ, 0, ±φ, -1>
<±1/φ, 0, -φ,  1>
< 1/φ, ±1, 0,  φ>
<-1/φ,  1, 0, ±φ>
<±1/φ, -1, 0, -φ>
< 1/φ, ±φ,  1, 0>
<-1/φ,  φ, ±1, 0>
<±1/φ, -φ, -1, 0>

< φ, 0,  1, ±1/φ>
<-φ, 0, ±1,  1/φ>
<±φ, 0, -1, -1/φ>
< φ, ±1, -1/φ, 0>
<-φ,  1, ±1/φ, 0>
<±φ, -1,  1/φ, 0>
< φ,  1/φ, 0, ±1>
<-φ, ±1/φ, 0, -1>
<±φ, -1/φ, 0,  1>

where φ is the Golden Ratio, (1+√5)/2.

These are basically the coordinates of the snub 24-cell (even permutations and all changes of sign of <φ, 1, 1/φ, 0>), minus an inscribed 24-cell. The 24-cell was found by decomposing it into a 16-cell plus a tesseract:

- First, we choose a pair of antipodal points to be the first axis of the 16-cell;
- Then we take advantage of the fact that the equatorial points of the 16-cell all have zero dot product with first pair of points. So we look for a permutation of <φ, 1, 1/φ, 0> that has zero dot product with the first point; that point with its antipode forms the second axis of the 16-cell.
- Again, the remaining axes of the 16-cell must have zero dot product with the previous two pairs of points, so we go through and find the third pair; the fourth pair is found the same way.

Next, we find the vertices of the tesseract by noting that, since the tesseract must compose with the 16-cell to form a 24-cell, the line segment from the origin to each vertex must pass through the centroid of the corresponding cell in the 16-cell. So we simply add the vectors of each cell in the 16-cell to find the tangent vector, and then scale it so that it is a permutation of <φ, 1, 1/φ, 0>.

- How do we know which vectors to add? It's easy: the 16-cell's vertices are expressed as 4 pairs of vectors of the form ±<A,B,C,D>. Since cells never contain both points of a single axis, and antipodes differ merely in sign, every cell is simply a sum of some combination of coordinate signs of the representative point of each of the 4 pairs. There are exactly 2^4 combinations of signs, so we just enumerate binary numbers from 0000 to 1111, and take 0 to mean + and 1 to mean -, and so we obtain tangent vectors to the 16-cell's facets.

- These sums turn out to be exactly 2*some permutation of <φ, 1, 1/φ, 0>, so the vertices of the tesseract are directly obtained in this way.

So this gives us the vertices of a 24-cell inscribed in the snub 24-cell. To get the coordinates of the bi-icositetradiminished 600-cell, then, we just enumerate the even permutations of <φ, 1, 1/φ, 0>, minus the vertices of this 24-cell.

(I have this nagging feeling that there's a shorter way to obtain these coordinates, but this way is relatively straightforward and in any case, is how I found the coordinates. I did have another idea of how to find these coordinates, by taking advantage of the correspondence of the snub 24-cell's icosahedra with the cells of a tesseract + a 16-cell along with their orientations, which form a sort of non-parallel orientation assignment of the tesseract's cells (the 4D analog of bisecting the cube's faces such that no two adjacent faces are bisected by the same plane), but eventually it turned out to involve exactly the same calculations as I describe above, and is more complicated to deal with to boot. So if you have a better way to do this, I'm all ears. :) )
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Re: Johnsonian Polytopes

Postby wendy » Tue Jun 26, 2012 8:27 am

You could use the method that Coxeter used to get them, which is by using directed edges of s3s4o3o.

You start off with a 24-choron, divided into three sets of 16-choron. All of these are all-changes of sign.

A 1,1,0,0 ; 0,0,1,1 cf 3d 1,0,0
B 1,0,1,0 ; 0,1,0,1 cf 3d 0,1,0
C 1,0,0,1 ; 0,1,1,0 cf 3d 0,0,1

The vertices of A+f.B, B+f.C and C+fA give in 4D the snub 24-choron, and in 3d, an icosahedron. Each of these gives 4 vertices, corresponding to 32 on full reflection, or 96 for all three.

You then add to the 4d figure, a 24-choron of size , viz o3o4o3f. The notional vertices of this gives 2f,0,0,0 and f,f,f,f all permutations, all changes of sign.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jun 26, 2012 3:03 pm

wendy wrote:You could use the method that Coxeter used to get them, which is by using directed edges of s3s4o3o.

You start off with a 24-choron, divided into three sets of 16-choron. All of these are all-changes of sign.

A 1,1,0,0 ; 0,0,1,1 cf 3d 1,0,0
B 1,0,1,0 ; 0,1,0,1 cf 3d 0,1,0
C 1,0,0,1 ; 0,1,1,0 cf 3d 0,0,1

The vertices of A+f.B, B+f.C and C+fA give in 4D the snub 24-choron, and in 3d, an icosahedron. Each of these gives 4 vertices, corresponding to 32 on full reflection, or 96 for all three.

You then add to the 4d figure, a 24-choron of size , viz o3o4o3f. The notional vertices of this gives 2f,0,0,0 and f,f,f,f all permutations, all changes of sign.

What's f? The golden ratio?
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Re: Johnsonian Polytopes

Postby wendy » Wed Jun 27, 2012 7:19 am

f is the golden ratio 1.61803398875, and ff is its square 2.61803398875. It's basically phi and phi square.

To get the 0.618, 1, 1.618, 0, divide through by phi (which gives v, x, f, o).

What is being done here is that you can colour the vertices of a 343 by its three inscribed 334 (here, A,B,C). The edges can get a direction A -> B, B -> C and C-> A. You can effect this division of 1:f by adding A + fB etc.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jun 27, 2012 3:35 pm

wendy wrote:f is the golden ratio 1.61803398875, and ff is its square 2.61803398875. It's basically phi and phi square.

To get the 0.618, 1, 1.618, 0, divide through by phi (which gives v, x, f, o).

What is being done here is that you can colour the vertices of a 343 by its three inscribed 334 (here, A,B,C). The edges can get a direction A -> B, B -> C and C-> A. You can effect this division of 1:f by adding A + fB etc.

Yes, I know that. But doesn't this just give you a regular 600-cell? I was looking for the coordinates of the bi-icositetradiminished 600-cell.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jun 29, 2012 5:21 pm

This is a continuation to the gyration thread, but I decided to post here in order to liven up this thread a little. :mrgreen:

The following CRF was described by wintersolstice as a gyrated version of the cube cupola:

Image

Basically, take a cube cupola, cut off a square orthobicupolic ring, and replace it with a square gyrobicupolic ring. The rhombicuboctahedron base of the cupola gets truncated to an elongated square cupola (due to the shorter height of the square antiprism in the gyrocupolic ring, we do not get a pseudo-rhombicuboctahedron: the square cupola cell of the gyro ring is not coplanar with the truncated base). The result has 1 elongated square cupola (the remnant of the original "base"), 6 cubes, 8 triangular prisms, 4 tetrahedra, 1 square antiprism, 8 square pyramids, and 1 square cupola, for a total of 29 cells. All edge lengths are equal, so it is CRF.

In the above render, I disabled vis clipping so that you can see all cells (the viewpoint is from underneath the elongated square cupola). The red cube is the top of the original cube cupola. The square antiprism is shown in green.

The coordinates of this pretty little CRF are:

# The elongated square cupola
<±1, ±1, ±(1+√2), 0>
<±1, ±(1+√2), ±1, 0>
<-(1+√2), ±1, ±1, 0>

# The rotated square
<1+√(√8-1), ±√2, 0, √2-1>
<1+√(√8-1), 0, ±√2, √2-1>

# The red cube at the top
<±1, ±1, ±1, √2>

Note that this CRF is not in Klitzing's list, because its vertices lie on more than two parallel hyperplanes.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jun 29, 2012 5:47 pm

The astute reader who looked at the previous CRF (which I will tentatively name the gyro cube cupola) will realize that it's possible to cut off another square orthobicupolic ring from the opposite end of the shape, and replace it with a second square gyrobicupolic ring. The result looks like this:

Image

Here I've colored both square antiprism cells in green, so that they're easier to pick out.

The cells are: 5 cubes, 2 square antiprisms, 16 square pyramids, 4 triangular prisms, 2 square cupolae, and 1 octagonal prism (30 cells in total).

Again, all edge lengths are equal, so this polychoron is CRF. I propose the name "bigyro cube cupola". It may be thought of as the linear stacking: square gyrobicupolic ring + square cupola prism + square gyrobicupolic ring.

Here are its coordinates:

# Octagonal prism
<±1, ±1, ±(1+√2), 0>
<±1, ±(1+√2), ±1, 0>

# Two rotated squares
<±(1+√(√8-1)), ±√2, 0, √2-1>
<±(1+√(√8-1)), 0, ±√2, √2-1>

# The red cube
<±1, ±1, ±1, √2>
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Re: Johnsonian Polytopes

Postby wintersolstice » Sun Jul 01, 2012 1:10 pm

here's what I call "the 29" :D

segmentotopes whose bases are either Platonic/Archimedean (excluding vertex transitive cases)

Key:

here are the catagories, underneath which are the Platonic solid/Platonic solid pairs that are Johnson cases, underneath that is why the others are invalid. Next to the name is a bracket showing the pair of bases for the segmentotope. the "X" isn't part of the name it just means you change it for the name of the Platonic Solids for it to name the shapes in that catagory.

meanings:

snubdis means truncate then alternate (based on snubdis 24-cell)

Birectate = Dual in 3D

Bitruncate = Truncate of the dual in 3D

Platonic = original Platonic solid

other notes:

I know the cupolas and antiprisms are already known (they're here just for completeness :D )

Cupoliprism was not my invention (based on the non-uniform scaliform polychoron name :
"Truncated tetrahedral cupoliprism"

the forward slashes mean that either platonic solid (or hypertruncate) can be used in the pair and it would be the same

the other names I did invent :D

the list:

X Cupola (Platonic||Cantellate)

1) Tetrahedron
2) Cube
3) Octahedron
4) Dodecahedron

Note: the icosahedron cupola is not CRF

X Antiprisms (Platonic||Birectate)

5) Octahedron/cube
6) Dodecahedron/icosahedron

Note: the tetrahedron antiprism is vertex transitive( 16-cell)

X Anticupola (Platonic||Rectate)

7) Cube
8 ) Octahedron
9) Dodecahedron
10) Icosahedron

Note: the tetrahedron anticupola is vertex transitive (rectified 5-cell)

Partially-base truncated X cupola (Truncate||Cantellate)

11) Tetrahedron
12) Cube
13) Octahedron
14) Dodecahedron
15) Icosahedron

Partially-base rectified X cupola ((Rectate||Cantellate)

16) Octahedron/cube
17) Dodecahedron/icosahedron

Note: the “Partially-base rectified tetrahedron cupola” = “octahedron anticupola”

Partially-base truncated X anticupola (Truncate|| Rectate)

18) Tetrahedron
19) Cube
20) Octahedron
21) Icosahedron

Note: the “Partially-base truncated dodecahedron anti cupola” is not CRF

Completely-base truncated X anticupola (Truncate||Cantitruncate)

22) Tetrahedron
23) Cube
24) Octahedron
25) Dodecahedron

Note: the “Completely-base truncated icosahedron anticupola” is not CRF

Truncated X cupoliprisms (Truncate||Bitruncate)

26) Octahedron/cube
27) Dodecahedron/icosahedrons

Note: “Truncated tetrahedron cupoliprism” is vertex transitive

Partially-base snubdis X antiprism (Snubdis||Birectate)

28) Octahedron

Note: only the octahedron can be “snubdis”ed

Semi cupola (Platonic|| Truncate)

29) Tetrahedron

Note: this shape is unique, the others in this group are not CRF
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jul 03, 2012 8:14 pm

@wintersolstice: will take a look at your constructions later; been busy along other lines.

Anyway, I've been trying to find more CRFs, and here's a report of failure (so that other people won't have to waste time going down blind alleys):

I had an idea that I might be able to produce CRFs by taking an n-gonal prism pyramid (which is CRF), and augmenting one of its side cells (an n-gonal pyramid) with an n-gonal pyramid pyramid. I tried this for n=4 and n=3, and unfortunately, it turns out that independent of the value of n, the tetrahedral cells introduced by the augment will always be coplanar with the surrounding n-gonal pyramids, and so they will merge into a non-CRF, sheared triangular prism (triangular prism with rhombic faces instead of squares).

So no CRFs can be produced along these lines. :-( Oh well, gotta search in other directions now...
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Re: Johnsonian Polytopes

Postby wintersolstice » Tue Jul 03, 2012 9:51 pm

btw 28 is cube to icosahedron I struggled with naming that shape
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jul 05, 2012 5:48 pm

wintersolstice wrote:btw 28 is cube to icosahedron I struggled with naming that shape

Regardless of naming difficulties, it's a pretty one!

Image

The cells are clearly visible here: 1 cube, 6 triangular prisms, 12 square pyramids, 8 tetrahedra, and 1 icosahedron. I especially like how the square pyramids interface the triangular prisms in different orientations. More and more, I'm realizing square pyramids are very versatile shapes for connecting squares to triangles in various ways!
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jul 05, 2012 5:51 pm

Oh, and I keep forgetting about Marek (and other Stella4D users), so here's the OFF file:

Code: Select all
4OFF
# Automatically generated by def2off from data/crf/C100_H001.def
# Polytope: cube__icosahedron
20 74 66 28

# Vertices
0.00000000000000000000 -1.00000000000000000000 -1.61803398874989490253 0.00000000000000000000
0.00000000000000000000 -1.00000000000000000000 1.61803398874989490253 0.00000000000000000000
0.00000000000000000000 1.00000000000000000000 -1.61803398874989490253 0.00000000000000000000
0.00000000000000000000 1.00000000000000000000 1.61803398874989490253 0.00000000000000000000
-1.00000000000000000000 -1.61803398874989490253 0.00000000000000000000 0.00000000000000000000
-1.00000000000000000000 1.61803398874989490253 0.00000000000000000000 0.00000000000000000000
1.00000000000000000000 -1.61803398874989490253 0.00000000000000000000 0.00000000000000000000
1.00000000000000000000 1.61803398874989490253 0.00000000000000000000 0.00000000000000000000
-1.61803398874989490253 0.00000000000000000000 -1.00000000000000000000 0.00000000000000000000
1.61803398874989490253 0.00000000000000000000 -1.00000000000000000000 0.00000000000000000000
-1.61803398874989490253 0.00000000000000000000 1.00000000000000000000 0.00000000000000000000
1.61803398874989490253 0.00000000000000000000 1.00000000000000000000 0.00000000000000000000
1.00000000000000000000 1.00000000000000000000 1.00000000000000000000 1.61803398874989490253
1.00000000000000000000 1.00000000000000000000 -1.00000000000000000000 1.61803398874989490253
1.00000000000000000000 -1.00000000000000000000 1.00000000000000000000 1.61803398874989490253
1.00000000000000000000 -1.00000000000000000000 -1.00000000000000000000 1.61803398874989490253
-1.00000000000000000000 1.00000000000000000000 1.00000000000000000000 1.61803398874989490253
-1.00000000000000000000 1.00000000000000000000 -1.00000000000000000000 1.61803398874989490253
-1.00000000000000000000 -1.00000000000000000000 1.00000000000000000000 1.61803398874989490253
-1.00000000000000000000 -1.00000000000000000000 -1.00000000000000000000 1.61803398874989490253

# Ridges
3 0 2 8
3 0 2 9
4 0 2 13 15
4 0 2 17 19
3 0 4 6
3 0 4 8
3 0 4 19
3 0 6 9
3 0 6 15
3 0 8 19
3 0 9 15
3 0 15 19
3 1 3 10
3 1 3 11
4 1 3 12 14
4 1 3 16 18
3 1 4 6
3 1 4 10
3 1 4 18
3 1 6 11
3 1 6 14
3 1 10 18
3 1 11 14
3 1 14 18
3 2 5 7
3 2 5 8
3 2 5 17
3 2 7 9
3 2 7 13
3 2 8 17
3 2 9 13
3 2 13 17
3 3 5 7
3 3 5 10
3 3 5 16
3 3 7 11
3 3 7 12
3 3 10 16
3 3 11 12
3 3 12 16
4 4 6 14 18
4 4 6 15 19
3 4 8 10
3 4 8 19
3 4 10 18
3 4 18 19
4 5 7 12 16
4 5 7 13 17
3 5 8 10
3 5 8 17
3 5 10 16
3 5 16 17
3 6 9 11
3 6 9 15
3 6 11 14
3 6 14 15
3 7 9 11
3 7 9 13
3 7 11 12
3 7 12 13
4 8 10 16 17
4 8 10 18 19
3 8 17 19
4 9 11 12 13
4 9 11 14 15
3 9 13 15
3 10 16 18
3 11 12 14
4 12 13 15 14
4 12 13 17 16
4 12 14 18 16
4 13 15 19 17
4 14 15 19 18
4 16 17 19 18

# Cells
20 0 1 4 5 7 12 13 16 17 19 24 25 27 32 33 35 42 48 52 56 58 127 179
5 0 3 9 29 62 4 127 163
5 1 2 10 30 65 4 127 163
5 2 3 11 31 71 91 127 163
5 4 6 8 11 41 4 127 163
4 5 6 9 43 30 127 43
4 7 8 10 53 30 127 43
5 12 15 21 37 66 4 127 163
5 13 14 22 38 67 4 127 163
5 14 15 23 39 70 91 127 163
5 16 18 20 23 40 4 127 163
4 17 18 21 44 30 127 43
4 19 20 22 54 30 127 43
5 24 26 28 31 47 4 127 163
4 25 26 29 49 30 127 43
4 27 28 30 57 30 127 43
5 32 34 36 39 46 4 127 163
4 33 34 37 50 30 127 43
4 35 36 38 58 30 127 43
5 40 41 45 55 72 91 127 163
5 42 43 44 45 61 4 127 163
5 46 47 51 59 69 91 127 163
5 48 49 50 51 60 4 127 163
5 52 53 54 55 64 4 127 163
5 56 57 58 59 63 4 127 163
5 60 61 62 66 73 91 127 163
5 63 64 65 67 68 91 127 163
6 68 69 70 71 72 73 253 127 249
# The end.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Jul 05, 2012 6:05 pm

Looks really nice :) How many CRF ways are there to glue two of these together?
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jul 05, 2012 6:48 pm

Marek14 wrote:Looks really nice :) How many CRF ways are there to glue two of these together?

Hmm... Let's see... there are 60 ways (60=order of icosahedral group without reflections) to orient the icosahedral cells relative to each other, and the cube has two distinct orientations depending on how the triangular prisms line up, making a total of 120. But the cube also has its own symmetries, which mean we have to divide by 24 (order of cubic symmetry group), which gives 5.

OK, I've no idea if I counted it correctly. 5 seems kinda low...?
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Jul 05, 2012 7:50 pm

The way I see the shape in Stella, there are two shapes that share faces with the icosahedron: tetrahedra and square pyramids. Tetrahedra form two groups of four, which have opposite chirality when you look at them from the icosahedron. Twelve square pyramids form six pairs around six edges where the triangular prisms are.

The important thing is that the whole icosahedron has only a tetrahedral symmetry.

So, we have a direct join -- tetrahedra on tetrahedra, opposite chirality.
If we join a tetrahedron with a tetrahedron of same chirality, the same kind of join will repeat on the antipodal tetrahedra, and otherwise we get joins of tetrahedra and square pyramids and square pyramids with square pyramids.

And now, is it possible to join the icosahedra in a way that NEVER pairs two tetrahedra? As it turns out, it's not. If you take the icosahedron as "outside" and look at the faces, each triangle has two other triangles adjoining at each corner (not counting those that share a whole edge). Tetrahedra have all three wedges of the type (pyramid-tetrahedron) or all three of the type (tetrahedron-pyramid), but pyramids have three different wedges: (pyramid-pyramid), (pyramid-tetrahedron) and (tetrahedron-pyramid). And at least one of those will always line with a wedge of tetrahedron when you join two icosahedra.

So, basically, it turns out there are only TWO different ways to connect the icosahedra: direct and gyrated or what do you want to call it, since there are only two ways to have two tetrahedra joined together, and you must always join at least one pair of them.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jul 05, 2012 8:14 pm

You're right, I made a mistake about the possible relative orientations of the two cubes. The orientations of the triangular prisms on the cube's faces form a kind of "parity", where each face has two possible orientations. However, one of these orientations is not allowed because the two joining icosahedra would have an edge of one intersecting the other at 90° -- i.e., the bases don't line up correctly. So my count was wrong.
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Re: Johnsonian Polytopes

Postby wintersolstice » Thu Jul 05, 2012 9:47 pm

when I can, I'll re-enumerate the 48 gyrations and dimishes of the 29:D for now, the remaining of the 100 segmentotopes (there are 129 Johnson segmentochora not including the antiprismatic rings)

12 are fragments (there are 6 more but there pyramids)

the 9 bicupolic rings (already known)

square bipyramidal bicupolic ring (4pyr||4cup) if that makes sense!
pentagonal cupolarotunda ring (only gyro form is CRF) (Rotunda||5g)
augmented pentagonal cupolarotunda ring (5pyr||rotunda)

12 pyramids (already known)

25 prims of orbiform* Johnsons soilds (already known)

three oddballs:

diminished triangular antiprismatic ring/bidimished rectified 5-cell (Note the triangular antiprismatic ring is the dimished rectified 5-cell) (perp line||3-prism

Gyrated octahedral prism (3-prism||refl ortho 3-prism)

and the unamed "3g||tridiminished icosahedron"

with the 48 and the 29 it becomes 129.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jul 06, 2012 7:57 pm

And while I'm at it, here are more .off files for Marek's viewing pleasure:

Bi-icositetradiminished 600-cell

Gyro cube cupola

Bi-gyro cube cupola
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jul 10, 2012 2:17 pm

I just realized this morning that a 24-cell can be cut into two segmentochora, both are octahedron||cuboctahedron (in Klitzing's notation). Each segmentochoron consists of an octahedron top face, 8 octahedra surrounding that, 6 square pyramids (half an octahedron), and a cuboctahedral base.

If I'm reading this correctly, this means that you can assemble an octahedron||cuboctahedron with a cube||cuboctahedron, to get a CRF with 9 octahedra, 6 square pyramids, 6 square antiprisms, and 1 cube. Half of it is a half-24-cell, and the other half contains square antiprisms. :)

Edit: This 24-cell hybrid can also be elongated by inserting a cuboctahedral prism between its two halves; the square pyramids (I believe) will be coplanar with the cubes, so they will become elongated square pyramids. So the result will have 9 octahedra, 6 elongated square pyramids, 8 triangular prisms, 6 square antiprisms, and 1 cube. Would this be the first non-trivial CRF that contains elongated square pyramid cells? :)
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