## Bilbirothawroids (D4.3 to D4.9)

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

quickfur wrote:
Klitzing wrote:[...]
PS: it just occurs me, that quickfur might have thought of the diminished version of it, chopping off the layers (1) and (2) from the above, i.e. just to consider ..oxFx3..xfox5..xoxx&#xt only.
This then changes the corresponding total cell count into: 1 ti + 30 tets + 20 J92's + 12 peroes + 12 pecues + 60 squippies + 30 trips + 1 grid.
(Here they are again, the peroes or half-ids!)

Even better (more elementary) find!

--- rk

[...]
But anyway, if somebody can compute the coordinates for student91's new CRF, I can run it through my polyview viewer and render some nice images.

Alright, I got impatient today (and needed a break from calculating coordinates for the other constructions under consideration), so I figured out the coordinates for this CRF myself. Thanks to student91's lace tower, it's not that hard to compute these coordinates from the rectified 120-cell o5x3o3o. Since I already have the latter's coordinates, it's just a matter of sorting them by the first coordinate (since viewing from <5,0,0,0> is centered on an icosidodecahedron), and each different value corresponds with a layer in the lace tower. So it wasn't too difficult to identify the layer that needed to be replaced with x5x3x. Turns out that layer is identified by height 1+2*phi (for edge length 2, as is my custom). Replacement with x5x3x wasn't hard either, since I already have the coordinates on my website, so it was just a matter of prepending 1+2*phi to them.

Anyway, enough talk. Let's see some renders!

Here, we have the x5x3o (yellow), which is the cross-section of the o5x3o3o once the bistratic o5x3o cup has been removed from it. Immediately after this point, the 20 J92 cells begin, one of which is shown in magenta above. They fill up the .5.3x elements of the symmetry of the x5x3o. The x5.3. elements are filled up by 12 pentagonal rotundae (yes, there really are pentagonal rotundae ), one of which is shown in green here. These pentagonal rotundae are, in turn, linked to the pentagonal cupola that protrude outwards beyond the limb of the x5x3x, which is outlined in red here. These pentagonal cupola are interfaced to the J92's by triangular prism + 2 square pyramid combos, so again we see how triangular prisms and square pyramids play an important role in interfacing J92's to the pentagonal elements of the 120-cell family polytopes.

So according to my polytope viewer, there are 30 tetrahedra, 60 square pyramids, 30 triangular prisms, 12 pentagonal cupolae, 12 pentagonal rotundae, 1 x5x3x, 1 truncated dodecahedron, and 20 J92's, for a total of 166 cells. There are exactly 300 vertices (nice number!), 840 edges, and 706 polygons (440 triangles, 150 squares, 72 pentagons, 20 hexagons, 24 decagons).

P.S.: Here are the coordinates:
Code: Select all
`# x5x3o<(7+3*sqrt(5))/2, 0, ±1, ±(5+3*sqrt(5))/2><(7+3*sqrt(5))/2, ±1, ±(5+3*sqrt(5))/2, 0><(7+3*sqrt(5))/2, ±(5+3*sqrt(5))/2, 0, ±1><(7+3*sqrt(5))/2, ±1, ±(3+sqrt(5))/2, ±(3+sqrt(5))><(7+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(3+sqrt(5)), ±1><(7+3*sqrt(5))/2, ±(3+sqrt(5)), ±1, ±(3+sqrt(5))/2><(7+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(1+sqrt(5)), ±(2+sqrt(5))><(7+3*sqrt(5))/2, ±(1+sqrt(5)), ±(2+sqrt(5)), ±(3+sqrt(5))/2><(7+3*sqrt(5))/2, ±(2+sqrt(5)), ±(3+sqrt(5))/2, ±(1+sqrt(5))># o5f3x:<(5+3*sqrt(5))/2, 0, ±(7+3*sqrt(5))/2, ±1><(5+3*sqrt(5))/2, ±1, 0, ±(7+3*sqrt(5))/2><(5+3*sqrt(5))/2, ±(7+3*sqrt(5))/2, ±1, 0><(5+3*sqrt(5))/2, ±(1+sqrt(5))/2, ±(2+sqrt(5)), ±(3+sqrt(5))><(5+3*sqrt(5))/2, ±(2+sqrt(5)), ±(3+sqrt(5)), ±(1+sqrt(5))/2><(5+3*sqrt(5))/2, ±(3+sqrt(5)), ±(1+sqrt(5))/2, ±(2+sqrt(5))><(5+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(3+sqrt(5))/2, ±(5+3*sqrt(5))/2><(5+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(5+3*sqrt(5))/2, ±(3+sqrt(5))/2><(5+3*sqrt(5))/2, ±(5+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(3+sqrt(5))/2># x5o3F:<3+sqrt(5), 0, ±(3+sqrt(5)), ±(3+sqrt(5))><3+sqrt(5), ±(3+sqrt(5)), 0, ±(3+sqrt(5))><3+sqrt(5), ±(3+sqrt(5)), ±(3+sqrt(5)), 0><3+sqrt(5), ±1, ±(7+3*sqrt(5))/2, ±(3+sqrt(5))/2><3+sqrt(5), ±(3+sqrt(5))/2, ±1, ±(7+3*sqrt(5))/2><3+sqrt(5), ±(7+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±1><3+sqrt(5), ±(1+sqrt(5))/2, ±(5+3*sqrt(5))/2, ±(2+sqrt(5))><3+sqrt(5), ±(2+sqrt(5)), ±(1+sqrt(5))/2, ±(5+3*sqrt(5))/2><3+sqrt(5), ±(5+3*sqrt(5))/2, ±(2+sqrt(5)), ±(1+sqrt(5))/2># x5x3x (replaces previous o5f3f):<1+2*phi> ~ apacs<1, 1, 4*phi+1><1+2*phi> ~ epacs<1, phi^3, 3+2*phi><1+2*phi> ~ epacs<2, phi^2, phi^4><1+2*phi> ~ epacs<phi^2, 3*phi, 2*phi^2><1+2*phi> ~ epacs<2*phi, 1+3*phi, 2+phi>`

And here's the .off file for Marek.
quickfur
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### Re: Johnsonian Polytopes

Marek14 wrote:... And you're right, octahedron || rhombicuboctahedron is flat.

Hein?

Well, oct || toe = xx3ox4o&#x does come out flat.
But oct || sirco = xx3oo4ox&#x definitely is not flat! It has height 1/2.

In fact, oct || sirco is just the oct-first monostratic cap of spic = x3o4o3x.

--- rk
Klitzing
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### Re: Johnsonian Polytopes

Also, the all-important question: what should we call this new CRF?

Would it qualify as a crown jewel too? I'm inclined to say yes, because it resembles J92 itself in some ways: it contains a piece of the rectified x5o...3o, with the bottom shrunk from a phi-scaled truncated icosahedron to something that contains hexagonal faces (i.e. x5x3x), analogous with J92 being a piece of the rectified dodecahedron with its bottom phi-scaled hexagon shrunk to a hexagon. Perhaps it can be considered as a "true" 4D analogue of the J92?
quickfur
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### Re: Johnsonian Polytopes

Klitzing wrote:
Marek14 wrote:... And you're right, octahedron || rhombicuboctahedron is flat.

Hein?

Well, oct || toe = xx3ox4o&#x does come out flat.
But oct || sirco = xx3oo4ox&#x definitely is not flat! It has height 1/2.

In fact, oct || sirco is just the oct-first monostratic cap of spic = x3o4o3x.

--- rk

You're right, oct || x4o3x is not flat. My bad.
quickfur
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
PS: it just occurs me, that quickfur might have thought of the diminished version of it, chopping off the layers (1) and (2) from the above, i.e. just to consider ..oxFx3..xfox5..xoxx&#xt only.
This then changes the corresponding total cell count into: 1 tid + 30 tets + 20 J92's + 12 peroes + 12 pecues + 60 squippies + 30 trips + 1 grid.
(Here they are again, the peroes or half-ids!)

Even better (more elementary) find!

--- rk

Edit: As pointed out above in bold: it should be tid, not ti, for sure.

quickfur wrote:...
Anyway, enough talk. Let's see some renders!

Here, we have the x5x3o (yellow), which is the cross-section of the o5x3o3o once the bistratic o5x3o cup has been removed from it. Immediately after this point, the 20 J92 cells begin, one of which is shown in magenta above. They fill up the .5.3x elements of the symmetry of the x5x3o. The x5.3. elements are filled up by 12 pentagonal rotundae (yes, there really are pentagonal rotundae ), one of which is shown in green here. These pentagonal rotundae are, in turn, linked to the pentagonal cupola that protrude outwards beyond the limb of the x5x3x, which is outlined in red here. These pentagonal cupola are interfaced to the J92's by triangular prism + 2 square pyramid combos, so again we see how triangular prisms and square pyramids play an important role in interfacing J92's to the pentagonal elements of the 120-cell family polytopes.

So according to my polytope viewer, there are 30 tetrahedra, 60 square pyramids, 30 triangular prisms, 12 pentagonal cupolae, 12 pentagonal rotundae, 1 x5x3x, 1 truncated dodecahedron, and 20 J92's, for a total of 166 cells. There are exactly 300 vertices (nice number!), 840 edges, and 706 polygons (440 triangles, 150 squares, 72 pentagons, 20 hexagons, 24 decagons).

...

Cell counts just as have been stated before (up to my initial typo).

And here now the complete incmats:
Code: Select all
`oxFx3xfox5xoxx&#xt   → height(1,2) = height(3,4) = 1/2                       height(2,3) = (sqrt(5)-1)/4 = 0.309017(tid || pseudo (x,f)-ti || pseudo (F,x)-srid || grid)o...3o...5o...     | 60  *  *   * |  2  1   2  0   0   0  0   0  0  0  0 |  1  2  1  2  2  0  0  0   0  0  0  0  0  0  0 | 1  1  1  2  0  0  0 0.o..3.o..5.o..     |  * 60  *   * |  0  0   2  1   2   2  0   0  0  0  0 |  0  0  2  2  1  2  1  1   2  0  0  0  0  0  0 | 0  2  1  1  1  1  0 0..o.3..o.5..o.     |  *  * 60   * |  0  0   0  0   2   0  2   2  0  0  0 |  0  0  0  1  0  0  2  0   2  1  1  2  0  0  0 | 0  1  0  1  0  2  1 0...o3...o5...o     |  *  *  * 120 |  0  0   0  0   0   1  0   1  1  1  1 |  0  0  0  0  0  1  0  1   1  0  1  1  1  1  1 | 0  1  0  0  1  1  1 1-------------------+--------------+--------------------------------------+-----------------------------------------------+----------------------.... x... ....     |  2  0  0   0 | 60  *   *  *   *   *  *   *  *  *  * |  1  1  0  1  0  0  0  0   0  0  0  0  0  0  0 | 1  1  0  1  0  0  0 0.... .... x...     |  2  0  0   0 |  * 30   *  *   *   *  *   *  *  *  * |  0  2  0  0  2  0  0  0   0  0  0  0  0  0  0 | 1  0  1  2  0  0  0 0oo..3oo..5oo..&#x  |  1  1  0   0 |  *  * 120  *   *   *  *   *  *  *  * |  0  0  1  1  1  0  0  0   0  0  0  0  0  0  0 | 0  1  1  1  0  0  0 0.x.. .... ....     |  0  2  0   0 |  *  *   * 30   *   *  *   *  *  *  * |  0  0  2  0  0  2  0  0   0  0  0  0  0  0  0 | 0  2  1  0  1  0  0 0.oo.3.oo.5.oo.&#x  |  0  1  1   0 |  *  *   *  * 120   *  *   *  *  *  * |  0  0  0  1  0  0  1  0   1  0  0  0  0  0  0 | 0  1  0  1  0  1  0 0.o.o3.o.o5.o.o&#x  |  0  1  0   1 |  *  *   *  *   * 120  *   *  *  *  * |  0  0  0  0  0  1  0  1   1  0  0  0  0  0  0 | 0  1  0  0  1  1  0 0.... .... ..x.     |  0  0  2   0 |  *  *   *  *   *   * 60   *  *  *  * |  0  0  0  0  0  0  1  0   0  1  0  1  0  0  0 | 0  0  0  1  0  1  1 0..oo3..oo5..oo&#x  |  0  0  1   1 |  *  *   *  *   *   *  * 120  *  *  * |  0  0  0  0  0  0  0  0   1  0  1  1  0  0  0 | 0  1  0  0  0  1  1 0...x .... ....     |  0  0  0   2 |  *  *   *  *   *   *  *   * 60  *  * |  0  0  0  0  0  1  0  0   0  0  0  0  1  1  0 | 0  1  0  0  1  0  0 1.... ...x ....     |  0  0  0   2 |  *  *   *  *   *   *  *   *  * 60  * |  0  0  0  0  0  0  0  0   0  0  1  0  1  0  1 | 0  1  0  0  0  0  1 1.... .... ...x     |  0  0  0   2 |  *  *   *  *   *   *  *   *  *  * 60 |  0  0  0  0  0  0  0  1   0  0  0  1  0  1  1 | 0  0  0  0  1  1  1 1-------------------+--------------+--------------------------------------+-----------------------------------------------+----------------------o...3x... ....     |  3  0  0   0 |  3  0   0  0   0   0  0   0  0  0  0 | 20  *  *  *  *  *  *  *   *  *  *  *  *  *  * | 1  1  0  0  0  0  0 0.... x...5x...     | 10  0  0   0 |  5  5   0  0   0   0  0   0  0  0  0 |  * 12  *  *  *  *  *  *   *  *  *  *  *  *  * | 1  0  0  1  0  0  0 0ox.. .... ....&#x  |  1  2  0   0 |  0  0   2  1   0   0  0   0  0  0  0 |  *  * 60  *  *  *  *  *   *  *  *  *  *  *  * | 0  1  1  0  0  0  0 0.... xfo. ....&#xt |  2  2  1   0 |  1  0   2  0   2   0  0   0  0  0  0 |  *  *  * 60  *  *  *  *   *  *  *  *  *  *  * | 0  1  0  1  0  0  0 0.... .... xo..&#x  |  2  1  0   0 |  0  1   2  0   0   0  0   0  0  0  0 |  *  *  *  * 60  *  *  *   *  *  *  *  *  *  * | 0  0  1  1  0  0  0 0.x.x .... ....&#x  |  0  2  0   2 |  0  0   0  1   0   2  0   0  1  0  0 |  *  *  *  *  * 60  *  *   *  *  *  *  *  *  * | 0  1  0  0  1  0  0 0.... .... .ox.&#x  |  0  1  2   0 |  0  0   0  0   2   0  1   0  0  0  0 |  *  *  *  *  *  * 60  *   *  *  *  *  *  *  * | 0  0  0  1  0  1  0 0.... .... .o.x&#x  |  0  1  0   2 |  0  0   0  0   0   2  0   0  0  0  1 |  *  *  *  *  *  *  * 60   *  *  *  *  *  *  * | 0  0  0  0  1  1  0 0.ooo3.ooo5.ooo&#x  |  0  1  1   1 |  0  0   0  0   1   1  0   1  0  0  0 |  *  *  *  *  *  *  *  * 120  *  *  *  *  *  * | 0  1  0  0  0  1  0 0.... ..o.5..x.     |  0  0  5   0 |  0  0   0  0   0   0  5   0  0  0  0 |  *  *  *  *  *  *  *  *   * 12  *  *  *  *  * | 0  0  0  1  0  0  1 0.... ..ox ....&#x  |  0  0  1   2 |  0  0   0  0   0   0  0   2  0  1  0 |  *  *  *  *  *  *  *  *   *  * 60  *  *  *  * | 0  1  0  0  0  0  1 0.... .... ..xx&#x  |  0  0  2   2 |  0  0   0  0   0   0  1   2  0  0  1 |  *  *  *  *  *  *  *  *   *  *  * 60  *  *  * | 0  0  0  0  0  1  1 0...x3...x ....     |  0  0  0   6 |  0  0   0  0   0   0  0   0  3  3  0 |  *  *  *  *  *  *  *  *   *  *  *  * 20  *  * | 0  1  0  0  0  0  0 1...x .... ...x     |  0  0  0   4 |  0  0   0  0   0   0  0   0  2  0  2 |  *  *  *  *  *  *  *  *   *  *  *  *  * 30  * | 0  0  0  0  1  0  0 1.... ...x5...x     |  0  0  0  10 |  0  0   0  0   0   0  0   0  0  5  5 |  *  *  *  *  *  *  *  *   *  *  *  *  *  * 12 | 0  0  0  0  0  0  1 1-------------------+--------------+--------------------------------------+-----------------------------------------------+----------------------o...3x...5x...     | 60  0  0   0 | 60 30   0  0   0   0  0   0  0  0  0 | 20 12  0  0  0  0  0  0   0  0  0  0  0  0  0 | 1  *  *  *  *  *  * *  tidoxFx3xfox ....&#xt |  3  6  3   6 |  3  0   6  3   6   6  0   6  3  3  0 |  1  0  3  3  0  3  0  0   6  0  3  0  1  0  0 | * 20  *  *  *  *  * *  thawroox.. .... xo..&#x  |  2  2  0   0 |  0  1   4  1   0   0  0   0  0  0  0 |  0  0  2  0  2  0  0  0   0  0  0  0  0  0  0 | *  * 30  *  *  *  * *  tet.... xfo.5xox.&#xt | 10  5  5   0 |  5  5  10  0  10   0  5   0  0  0  0 |  0  1  0  5  5  0  5  0   0  1  0  0  0  0  0 | *  *  * 12  *  *  * *  pero.x.x .... .o.x&#x  |  0  2  0   4 |  0  0   0  1   0   4  0   0  2  0  2 |  0  0  0  0  0  2  0  2   0  0  0  0  0  1  0 | *  *  *  * 30  *  * *  trip.... .... .oxx&#xr |  0  1  2   2 |  0  0   0  0   2   2  1   2  0  0  1 |  0  0  0  0  0  0  1  1   2  0  0  1  0  0  0 | *  *  *  *  * 60  * *  squippy.... ..ox5..xx&#x  |  0  0  5  10 |  0  0   0  0   0   0  5  10  0  5  5 |  0  0  0  0  0  0  0  0   0  1  5  5  0  0  1 | *  *  *  *  *  * 12 *  pecu...x3...x5...x     |  0  0  0 120 |  0  0   0  0   0   0  0   0 60 60 60 |  0  0  0  0  0  0  0  0   0  0  0  0 20 30 12 | *  *  *  *  *  *  * 1  grid`

quickfur wrote:Also, the all-important question: what should we call this new CRF?

Would it qualify as a crown jewel too? I'm inclined to say yes, because it resembles J92 itself in some ways: it contains a piece of the rectified x5o...3o, with the bottom shrunk from a phi-scaled truncated icosahedron to something that contains hexagonal faces (i.e. x5x3x), analogous with J92 being a piece of the rectified dodecahedron with its bottom phi-scaled hexagon shrunk to a hexagon. Perhaps it can be considered as a "true" 4D analogue of the J92?

Yes, it is quite this analogue: Just as thawro itself is related to the threefold rotunda of id (oxFf3xfof&#xt), when the bottom base layer f3f is concentrically replaced by x3x, this polychoron too is related to the id-first tristratic sub-cap of rahi (oxFf3xfof5xoxo&#xt), when the bottom base layer f3f5o is concentrically replaced by x3x5x (grid).

I too have no good name for that one so far. Still using the mere descriptive "triaconta-sphenated id-first tristratic sub-cap of rahi" as its working title...

Btw. nice pic - as always!

--- rk
Klitzing
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
quickfur wrote:Also, the all-important question: what should we call this new CRF?

Would it qualify as a crown jewel too? I'm inclined to say yes, because it resembles J92 itself in some ways: it contains a piece of the rectified x5o...3o, with the bottom shrunk from a phi-scaled truncated icosahedron to something that contains hexagonal faces (i.e. x5x3x), analogous with J92 being a piece of the rectified dodecahedron with its bottom phi-scaled hexagon shrunk to a hexagon. Perhaps it can be considered as a "true" 4D analogue of the J92?

Yes, it is quite this analogue: Just as thawro itself is related to the threefold rotunda of id (oxFf3xfof&#xt), when the bottom base layer f3f is concentrically replaced by x3x, this polychoron too is related to the id-first tristratic sub-cap of rahi (oxFf3xfof5xoxo&#xt), when the bottom base layer f3f5o is concentrically replaced by x3x5x (grid).

I too have no good name for that one so far. Still using the mere descriptive "triaconta-sphenated id-first tristratic sub-cap of rahi" as its working title...

Hmm. Unlike J92 itself, where there are three triangle-square-triangle sequences that Johnson calls "lunes" (which leads to the "hebespheno-" name, which he defines as being a series of 3 lunes), here it's not immediately clear what the 4D analogue of Johnson's lunes should be, nor what the 4D equivalent of "hebespheno-" might mean. Arguably, the triangular prism + 2 square pyramids combos can serve as "spheno-" (which in 3D is defined by Johnson as a complex of 2 lunes). This then leads to a possible naming as triacontaspheno-(rectified 120-cell)-rotunda for student91's original, non-diminished version of this CRF (i.e., with the o5x3o cap still attached), or perhaps triacontasphenorotunda for short. Then this diminished form can be named the diminished triacontasphenorotunda. (This naming scheme presumes that this is the only kind of 4D sphenorotunda possible, which unfortunately seems rather shaky in the face of circumstantial evidence that the 600-cell family of uniform polychora may have much more to offer in the way of crown jewel constructions.)

(EDIT: oops, there are 30 such triangular prism + square pyramid combos, not 20, so it should be triaconta-, not icosi-.)

Perhaps it is also possible to call the J92 rhombochoron the sphenorhombochoron, based on the fact that the triangular prism + square pyramids combo also appears there -- and in fact, more extensively than here, because there are 3 square pyramids + 2 tetrahedra there, whereas here it's only 2 square pyramids. Based on this, it may not be remiss to call it the hebesphenorhombochoron, the idea being that just as "hebespheno-" implies a larger triangle+square complex in 3D, in 4D it implies a larger triangular prism + square pyramid complex.

Btw. nice pic - as always!

Thanks!
quickfur
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### Re: Johnsonian Polytopes

Klitzing: the dichoral angles I posted were dichoral angles with cubes, since that was what I was trying to augment; I used Stella to measure them as a first approach.

All right, if octahedron||rhombicuboctahedron exists (I should have looked), the question stands: is it possible to arrange nonorbiform triangular bipyramid/elongated triangular orthobicupola and its pentagonal equivalent into cupolas?
Marek14
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### Re: Johnsonian Polytopes

Note that the [3} + {4} + {3} sequence occurs in thawro where it has the layer radius reduced. The same holds true here for the squippy + trip + squippy sequence in one dimension more. So that is clearly the analogue of the spheno- i.e. lunatic part.

Note moreover that this trip will be used here as a digonal cupola, not as something with full trigonal symmetry! And note too, that the squippies are attached to the triangles of that trip. - In contrast, in thawrorh, that rhombochoron, the squippies are attached with their squares to the trips. The trip's triangles are attached to tets! But still, even there the trips are used as digonal cupolae only (by means of farer surroundings).

--- rk
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### Re: Johnsonian Polytopes

Marek14 wrote:All right, if octahedron||rhombicuboctahedron exists (I should have looked), the question stands: is it possible to arrange nonorbiform triangular bipyramid/elongated triangular orthobicupola and its pentagonal equivalent into cupolas?

Well, I've the fear, that those dipyramids resp. elongated orthobicupolae cannot remain flat or otherwise the lacing cells should get some shear. - But I have not done the calculations here. Just a gut instinct.

So what is going on at oct || sirco?
Here is its lace city:
Code: Select all
`        o4o         o4x         o4o                                                                                                                                         x4o         x4x             x4x         x4o`

That is, in fact it is nothing but a bi-augmented {4} || op. - Your figures then would be similarily bi-augmented {n} || 2n-p. Seen as such, those surely exist. But whether both these sequences pt + {n} + pt resp. {n} + {2n} + {2n} + {n} remain flat, i.e. that the whole thingies for n=3 or 5 remain CRF, at least has to be checked.

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:
Marek14 wrote:All right, if octahedron||rhombicuboctahedron exists (I should have looked), the question stands: is it possible to arrange nonorbiform triangular bipyramid/elongated triangular orthobicupola and its pentagonal equivalent into cupolas?

Well, I've the fear, that those dipyramids resp. elongated orthobicupolae cannot remain flat or otherwise the lacing cells should get some shear. - But I have not done the calculations here. Just a gut instinct.

So what is going on at oct || sirco?
Here is its lace city:
Code: Select all
`        o4o         o4x         o4o                                                                                                                                         x4o         x4x             x4x         x4o`

That is, in fact it is nothing but a bi-augmented {4} || op. - Your figures then would be similarily bi-augmented {n} || 2n-p. Seen as such, those surely exist. But whether both these sequences pt + {n} + pt resp. {n} + {2n} + {2n} + {n} remain flat, i.e. that the whole thingies for n=3 or 5 remain CRF, at least has to be checked.

--- rk

Well, even if those sequences are not flat, the whole thing should still be CRF, shouldn't it?
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### Re: Johnsonian Polytopes

Marek14 wrote:
Klitzing wrote:Here is its lace city:
Code: Select all
`        o4o         o4x         o4o                                                                                                                                         x4o         x4x             x4x         x4o`

That is, in fact it is nothing but a bi-augmented {4} || op. - Your figures then would be similarily bi-augmented {n} || 2n-p. Seen as such, those surely exist. But whether both these sequences pt + {n} + pt resp. {n} + {2n} + {2n} + {n} remain flat, i.e. that the whole thingies for n=3 or 5 remain CRF, at least has to be checked.

--- rk

Well, even if those sequences are not flat, the whole thing should still be CRF, shouldn't it?

No, what I meant is, that it might be either this
Code: Select all
`        oNo                     oNo                            oNx                                                                                                                                                     xNo                                     xNo            xNx             xNx            `

or that
Code: Select all
`                    oNx                            oNo                     oNo                                                                                                                                                     xNx             xNx            xNo                                     xNo`

Both obviously would fail to be convex.

But, you could go checking these dihedral angles - at least numerically - by means of Stella4D, hehe.
(I don't have it.)

--- rk
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### Re: Johnsonian Polytopes

I could check it, but I still don't have the convex hull program and the polytopes are too big to write *.off file by hand. I'd need the x3o o||x3x x, x4o o||x4x x and x5o o||x5x x for basis, and then the augments...
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### Re: Johnsonian Polytopes

Klitzing wrote:
Marek14 wrote:
Klitzing wrote:Here is its lace city:
Code: Select all
`        o4o         o4x         o4o                                                                                                                                         x4o         x4x             x4x         x4o`

That is, in fact it is nothing but a bi-augmented {4} || op. - Your figures then would be similarily bi-augmented {n} || 2n-p. Seen as such, those surely exist. But whether both these sequences pt + {n} + pt resp. {n} + {2n} + {2n} + {n} remain flat, i.e. that the whole thingies for n=3 or 5 remain CRF, at least has to be checked.

--- rk

Well, even if those sequences are not flat, the whole thing should still be CRF, shouldn't it?

No, what I meant is, that it might be either this
Code: Select all
`        oNo                     oNo                            oNx                                                                                                                                                     xNo                                     xNo            xNx             xNx            `

or that
Code: Select all
`                    oNx                            oNo                     oNo                                                                                                                                                     xNx             xNx            xNo                                     xNo`

Both obviously would fail to be convex.

But, you could go checking these dihedral angles - at least numerically - by means of Stella4D, hehe.
(I don't have it.)

--- rk

I wonder if ooxxNoxxo&#xr is possible for N not 4. The thing with N=4 can be seen as two ooxx4oxxo&#xr's placed on the sides of a xxo4xxx&#x. Now ooxx4oxxo&#xr is a true ring (K4.109). For N not 4 such rings don't exist (at least they're not in Klitzings list.) This means that things with N not 4 should have either a connection between xNx and oNo, only possible for N=2, or a connection between oNx and xNo, that would give the first of Klitzings lace-cities.
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### Re: Johnsonian Polytopes

quickfur wrote:Also, the all-important question: what should we call this new CRF?

Would it qualify as a crown jewel too? I'm inclined to say yes, because it resembles J92 itself in some ways: it contains a piece of the rectified x5o...3o, with the bottom shrunk from a phi-scaled truncated icosahedron to something that contains hexagonal faces (i.e. x5x3x), analogous with J92 being a piece of the rectified dodecahedron with its bottom phi-scaled hexagon shrunk to a hexagon. Perhaps it can be considered as a "true" 4D analogue of the J92?

We have found quite some CRF's with a similar buildup: they consist of johnson solids, placed in an symmetrical fashion.
E.g. The ursachora, they can be seen as teddi's placed in a tetrahedral, octahedral resp. icosahedral fashion.
Furthermore you have the castellated prism, that can be seen as bilbro's placed in a icosahedral fashion.
Now we've found another polytope with this buildup, it's J92, oxFx3xfox&#xt, placed in an icosahedral fashion: oxFx3xfox5xoxx&#xt.
I think we will find more polytopes with such a buildup, so maybe we should get a uniform naming convention for those. I would suggest calling the new one thus icosahedral thawrochoron or something like that. I hope we can develop a uniform naming for the group of polytopes I've just described. I think it will be very convienient to describe all those in a similar way .
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### Re: Johnsonian Polytopes

Well, this polytope seems to be a type of rotunda, it's a tristratic polychoron squeezed between two parallel hyperplanes -- calling it a rotunda makes about the same sense as callind the J91 polychoron "prism". A castellated rotunda?

Unfortunately, it can't be made into a birotunda since the dichoral angle between great rhombicosidodecahedron and pentagonal cupola is 108.

By the way, this polychoron is surprisingly thin -- the distance between great rhombicosidodecahedron and truncated dodecahedron is only about 1.309 with unit edge length.
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### Re: Johnsonian Polytopes

I think I agree with Klitzing's reservation of "rotunda" to mean something that tiles half of a 3-sphere, and not just any polystratic CRF, as we've been doing. What we've been calling "rotunda" are really just polystratic caps (or cups, depending on how you look at it) of differing heights and constructions. So maybe it's better to call this new CRF an icosahedral J92 cap? (Or diminished icosahedral J92 cap, for the minimized variant with the top icosidodecahedral cap chopped off).

BTW, I'm wondering if the same cut of o5x3o3o can also produce a "castellated prism"-like CRF, because the cutting hyperplane bisects the 20 icosidodecahedra parallel to their triangular faces, and we have taken the cut-off part and shrunken it into J92's; but I think it should be possible to do this also to the remainder of the o5x3o3o as well. If we do this to both sides of the o5x3o3o, then we should get something with 40 J92's (plus a bunch of pentagonal rotunda or perhaps o5x3o's) sandwiched between two x5x3x's. I have the feeling this should be CRF as well, though I haven't investigated this in detail.

P.S.: Apparently it is CRF!!!!!!
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### Re: Johnsonian Polytopes

quickfur wrote:I think I agree with Klitzing's reservation of "rotunda" to mean something that tiles half of a 3-sphere, and not just any polystratic CRF, as we've been doing. What we've been calling "rotunda" are really just polystratic caps (or cups, depending on how you look at it) of differing heights and constructions. So maybe it's better to call this new CRF an icosahedral J92 cap? (Or diminished icosahedral J92 cap, for the minimized variant with the top icosidodecahedral cap chopped off).

BTW, I'm wondering if the same cut of o5x3o3o can also produce a "castellated prism"-like CRF, because the cutting hyperplane bisects the 20 icosidodecahedra parallel to their triangular faces, and we have taken the cut-off part and shrunken it into J92's; but I think it should be possible to do this also to the remainder of the o5x3o3o as well. If we do this to both sides of the o5x3o3o, then we should get something with 40 J92's (plus a bunch of pentagonal rotunda or perhaps o5x3o's) sandwiched between two x5x3x's. I have the feeling this should be CRF as well, though I haven't investigated this in detail.

P.S.: Apparently it is CRF!!!!!!

Oh, let me see
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### Re: Johnsonian Polytopes

Alright, here's a render:

I used parallel projection this time, instead of perspective projection, so that you can see the outlines of the equatorial cells clearly -- 30 icosidodecahedra (they straddle each adjacent pair of J92's, which are shown in magenta above). The nearest cell to the 4D viewpoint is the x5x3x in yellow. It's connected to 12 pentagonal rotundae via decagonal prisms, and around the vertices of the J92 where they touch each other is a most fascinating circle of square pyramids. Most interesting!

Note that this is only the near half of the CRF; the other side has exactly the same arrangement of cells. So there are 2 x5x3x's, 40 J92's, 24 pentagonal rotundae, 30 o5x3o's, 24 decagonal prisms, 180(?) square pyramids, and a bunch of tetrahedra (didn't count them).

Anyway, here's the .off file for Marek.

P.S. So basically, here we see that o5x3o3o can be stratified into several layers, the polystratic o5x3o/x5x3o caps on either end, and the middle 3 layers of which can be made CRF via a replacement of o5f3f with x5x3x. This strengthens my suspicion that there are probably a lot of CRFs involving J91 and J92, derived from the 120-cell uniforms. Their having patches of surface compatible with o5x3o makes them easy to connect to 120-cell uniforms with small modifications.

We may have to rethink our naming schemes, as these CRF crown jewels appear to be more numerous than we first thought, and have a stronger connection with the pentagonal polychora than previously suspected.

This particular cutting is now making me wonder if we can have a non-parallel cutting of o5x3o3o, that may produce a CRF mega-wedge with J92 cells, along the same lines as the x5x3o / icosahedron mega-wedge that I discovered recently. The coordinates for that will be much more tricky to compute, though, because I'll have to replace a o5f3f in oblique orientation with x5x3x.
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### Re: Johnsonian Polytopes

Well, Stella built a nice 3D net for this shape which shows its inner structure in an interesting way.

Here's a question: The first layer (made from J92's, decagonal prisms and assorted tets and squippies) has indentations for icosidodecahedra. However, it seems that those indentations are also perfect shape for J91's. Would it be possible to make a CRF polychoron by starting in the same way and then putting J91's in second layer?

Another idea: two pentagonal rotundas joined through their top pentagon, with J91's filling the gaps on the sides.
Marek14
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### Re: Johnsonian Polytopes

Marek14 wrote:Well, Stella built a nice 3D net for this shape which shows its inner structure in an interesting way.

Here's a question: The first layer (made from J92's, decagonal prisms and assorted tets and squippies) has indentations for icosidodecahedra. However, it seems that those indentations are also perfect shape for J91's. Would it be possible to make a CRF polychoron by starting in the same way and then putting J91's in second layer?

So that basically means we cut out the middle layer of icosidodecahedra, and just put the two (x5x3x + J92's) substrata face-to-face with each other? Whoa. I think you might be on to something!!

Lemme see... it should be relatively easy to cut out the middle layer vertices and move the vertices of the outer J92's close together so that they meet. If I'm correct, this should remove the tetrahedra currently sitting on top of the J92's, so that the 20 J92's on this side will touch the J92's on the far side, and their pentagonal faces will be filled by J91's. The tetrahedra currently interfacing the J92's to the pentagonal cupolae should turn into square pyramids, and the pentagonal cupolae should turn into more decagonal prisms perhaps? Not sure...

Another idea: two pentagonal rotundas joined through their top pentagon, with J91's filling the gaps on the sides.

That's the idea I had before, but I got stuck trying to compute the coordinates for it, so I don't know if it's possible or not. Maybe you can have a go at it?
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### Re: Johnsonian Polytopes

quickfur wrote:That's the idea I had before, but I got stuck trying to compute the coordinates for it, so I don't know if it's possible or not. Maybe you can have a go at it?

Still no convex hull algorithm...
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### Re: Johnsonian Polytopes

Marek14 wrote:
quickfur wrote:That's the idea I had before, but I got stuck trying to compute the coordinates for it, so I don't know if it's possible or not. Maybe you can have a go at it?

Still no convex hull algorithm...

Send me the vertices.
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### Re: Johnsonian Polytopes

quickfur wrote:
Marek14 wrote:[...] Here's a question: The first layer (made from J92's, decagonal prisms and assorted tets and squippies) has indentations for icosidodecahedra. However, it seems that those indentations are also perfect shape for J91's. Would it be possible to make a CRF polychoron by starting in the same way and then putting J91's in second layer?

So that basically means we cut out the middle layer of icosidodecahedra, and just put the two (x5x3x + J92's) substrata face-to-face with each other? Whoa. I think you might be on to something!!

Lemme see... it should be relatively easy to cut out the middle layer vertices and move the vertices of the outer J92's close together so that they meet. If I'm correct, this should remove the tetrahedra currently sitting on top of the J92's, so that the 20 J92's on this side will touch the J92's on the far side, and their pentagonal faces will be filled by J91's. The tetrahedra currently interfacing the J92's to the pentagonal cupolae should turn into square pyramids, and the pentagonal cupolae should turn into more decagonal prisms perhaps? Not sure...
[...]

Now I'm pretty certain that this is possible. Check out this side-view of the current CRF:

Sorry, this looks kinda messy because there are too many edges... but if you look at the nearest part to the 3D (not 4D) viewpoint, you'll see a flattened projection image of an icosidodecahedron that looks like two rhombuses sharing a vertex, with kite-like shapes around it. Now imagine if we "squish" this polychoron together, such that the two layers of purple J92 cells touch each other, this will cause the two rhombuses to merge into one, that is, they become the projection of a J91. If you look on the right side of the image, you can sorta see that some of the current lacing cells will disappear, and the gaps in the squished polychoron should be fillable by square pyramids and more decagonal prisms.

Going to compute the coordinates now...
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### Re: Johnsonian Polytopes

quickfur wrote:
quickfur wrote:
Marek14 wrote:[...] Here's a question: The first layer (made from J92's, decagonal prisms and assorted tets and squippies) has indentations for icosidodecahedra. However, it seems that those indentations are also perfect shape for J91's. Would it be possible to make a CRF polychoron by starting in the same way and then putting J91's in second layer?

So that basically means we cut out the middle layer of icosidodecahedra, and just put the two (x5x3x + J92's) substrata face-to-face with each other? Whoa. I think you might be on to something!!

Lemme see... it should be relatively easy to cut out the middle layer vertices and move the vertices of the outer J92's close together so that they meet. If I'm correct, this should remove the tetrahedra currently sitting on top of the J92's, so that the 20 J92's on this side will touch the J92's on the far side, and their pentagonal faces will be filled by J91's. The tetrahedra currently interfacing the J92's to the pentagonal cupolae should turn into square pyramids, and the pentagonal cupolae should turn into more decagonal prisms perhaps? Not sure...
[...]

Now I'm pretty certain that this is possible. Check out this side-view of the current CRF:

Sorry, this looks kinda messy because there are too many edges... but if you look at the nearest part to the 3D (not 4D) viewpoint, you'll see a flattened projection image of an icosidodecahedron that looks like two rhombuses sharing a vertex, with kite-like shapes around it. Now imagine if we "squish" this polychoron together, such that the two layers of purple J92 cells touch each other, this will cause the two rhombuses to merge into one, that is, they become the projection of a J91. If you look on the right side of the image, you can sorta see that some of the current lacing cells will disappear, and the gaps in the squished polychoron should be fillable by square pyramids and more decagonal prisms.

Going to compute the coordinates now...

I played with it in Stella and found I can mold the projection into outer shape of truncated cube. The idea is to remove its middle layer

Can't come up with good coordinates for that pentagonal rotunda shape. I haven't tried coordinate computing for icosahedral things yet
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### Re: Johnsonian Polytopes

Aaaand here it is, it is CRF!!

As I expected, the lacing cells became decagonal prisms and square pyramids. The orange cells above are J91's, and there are 30 of them. Now let's take a look at the J92's:

This image is kinda cluttered because the J92's are too close to each other, but I wanted to show how they fit in with the J91's in the previous image. The remaining gaps are filled by square pyramids and decagonal prisms.

Now let's go back to the usual viewpoint, centered on a x5x3x:

So here the 20 J92's are clearly seen. The purple triangles are where they touch the other 20 J92's. Now let's see the J91's:

Exactly where we expect them.

Here's the .off file for Marek to play with.

Here are the coordinates:
Code: Select all
`# Two sandwiching x5x3x's:<±(1+phi)> ~ apacs<1, 1, 4*phi+1><±(1+phi)> ~ epacs<1, phi^3, 3+2*phi><±(1+phi)> ~ epacs<2, phi^2, phi^4><±(1+phi)> ~ epacs<phi^2, 3*phi, 2*phi^2><±(1+phi)> ~ epacs<2*phi, 1+3*phi, 2+phi># o3F vertices of J92's<±phi, 0, 0, ±(4+2*sqrt(5))><±phi, 0, ±(4+2*sqrt(5)), 0><±phi, ±(4+2*sqrt(5)), 0, 0><±phi, ±(3+sqrt(5))/2, ±(2+sqrt(5)), ±(7+3*sqrt(5))/2><±phi, ±(2+sqrt(5)), ±(7+3*sqrt(5))/2, ±(3+sqrt(5))/2><±phi, ±(7+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(2+sqrt(5))># f3x vertices of J92's<±1, ±1, ±(4+2*sqrt(5)), ±(1+sqrt(5))/2><±1, ±(1+sqrt(5))/2, ±1, ±(4+2*sqrt(5))><±1, ±(4+2*sqrt(5)), ±(1+sqrt(5))/2, ±1><±1, ±1, ±(3+sqrt(5)), ±(7+3*sqrt(5))/2><±1, ±(3+sqrt(5)), ±(7+3*sqrt(5))/2, ±1><±1, ±(7+3*sqrt(5))/2, ±1, ±(3+sqrt(5))><±1, ±(3+sqrt(5))/2, ±(3+sqrt(5))/2, ±(9+3*sqrt(5))/2><±1, ±(3+sqrt(5))/2, ±(9+3*sqrt(5))/2, ±(3+sqrt(5))/2><±1, ±(9+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(3+sqrt(5))/2><±1, ±(3+sqrt(5))/2, ±(5+3*sqrt(5))/2, ±(5+3*sqrt(5))/2><±1, ±(5+3*sqrt(5))/2, ±(3+sqrt(5))/2, ±(5+3*sqrt(5))/2><±1, ±(5+3*sqrt(5))/2, ±(5+3*sqrt(5))/2, ±(3+sqrt(5))/2><±1, ±(1+sqrt(5)), ±(7+3*sqrt(5))/2, ±(2+sqrt(5))><±1, ±(2+sqrt(5)), ±(1+sqrt(5)), ±(7+3*sqrt(5))/2><±1, ±(7+3*sqrt(5))/2, ±(2+sqrt(5)), ±(1+sqrt(5))># x3o vertices of J92's<0, 0, ±(2+sqrt(5)), ±(9+3*sqrt(5))/2><0, ±(2+sqrt(5)), ±(9+3*sqrt(5))/2, 0><0, ±(9+3*sqrt(5))/2, 0, ±(2+sqrt(5))><0, ±1, ±(3+sqrt(5))/2, ±(4+2*sqrt(5))><0, ±(3+sqrt(5))/2, ±(4+2*sqrt(5)), ±1><0, ±(4+2*sqrt(5)), ±1, ±(3+sqrt(5))/2><0, ±(2+sqrt(5)), ±(5+3*sqrt(5))/2, ±(3+sqrt(5))><0, ±(3+sqrt(5)), ±(2+sqrt(5)), ±(5+3*sqrt(5))/2><0, ±(5+3*sqrt(5))/2, ±(3+sqrt(5)), ±(2+sqrt(5))>`

The cell counts are: 2 x5x3x's, 40 J92's (two layers of 20 each), 30 J91's, 36 decagonal prisms, 240 square pyramids (2*3*30 around the J92's o3F vertices, 60 around decagonal prisms), total 348 cells. There are exactly 600 vertices (nice number again!) and 1740 edges. (Too lazy to count the polygons...)

This has got to be the best crown jewel we've found yet; it has both J91 and J92 cells!!
quickfur
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### Re: Johnsonian Polytopes

Well, do you know what I consider most interesting part of this polychoron?

It contains no tetrahedra. Tetrahedra are the most basic struts of polychora, like triangles in 3D -- how many Johnson solids lack triangles? None!

Instead, the smallest struts here are square pyramids.

Second interesting feature are those decagonal prisms which appear in towers of 3. They look like elevator shafts.

And then there's the combination of J91 and J92 in a single shape. In short, this is really a great shape!

So, how to call it? Maybe we could call it a CRUSHED or COMPACTED castellated prism. It looks like the middle layer of previous shape was crushed into thinner layer, but the rest of polychoron stayed unharmed.

As for the counts, Stella does that automagically So I can mention that the square pyramids are divided in groups of 120 + 60 + 60 (Stella divides them like this because the second group of 120 is chiral) and the decagonal prism are in two groups as well, 24+12 -- 24 directly adjacent to the x5x3x's and 12 in the middle layer.

As for polygons, there are 1488 in total: 860 triangles, 420 squares, 120 pentagons, 40 hexagons and 48 decagons.
Marek14
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### Re: Johnsonian Polytopes

Marek14 wrote:Well, do you know what I consider most interesting part of this polychoron?

It contains no tetrahedra. Tetrahedra are the most basic struts of polychora, like triangles in 3D -- how many Johnson solids lack triangles? None!

Instead, the smallest struts here are square pyramids.

Yeah, this part is quite amazing. Also surprising is the appearance of square pyramids in icosahedral symmetry, many of them not aligned with the square faces of a great rhombicosidodecahedron! This structure is indeed marvelous.

Second interesting feature are those decagonal prisms which appear in towers of 3. They look like elevator shafts.

You know what my thought is about them? I think they are segments of rings of decagonal prisms with swirlprism symmetry! There may exist a modified diminishing of o5x3o3o that contains rings of these decagonal prisms. But currently I don't think I can imagine how that would look like yet, nor whether it will be CRF.

And then there's the combination of J91 and J92 in a single shape. In short, this is really a great shape!

So, how to call it? Maybe we could call it a CRUSHED or COMPACTED castellated prism. It looks like the middle layer of previous shape was crushed into thinner layer, but the rest of polychoron stayed unharmed.

I was thinking that maybe this CRF should be the castellated prism, and the previous structure the inflated (or augmented) castellated prism. Edit: Other possible adjectives: susconated (from Gk. souskōnō: to inflate).

As for the counts, Stella does that automagically So I can mention that the square pyramids are divided in groups of 120 + 60 + 60 (Stella divides them like this because the second group of 120 is chiral) and the decagonal prism are in two groups as well, 24+12 -- 24 directly adjacent to the x5x3x's and 12 in the middle layer.

As for polygons, there are 1488 in total: 860 triangles, 420 squares, 120 pentagons, 40 hexagons and 48 decagons.

Yeah, of the 3 square pyramids around the J92's o3F vertices, one lines up with the square face of the x5x3x, but the other two don't.

And yes, my polygon counts match yours, now that you prodded me to go count them.
quickfur
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### Re: Johnsonian Polytopes

quickfur wrote:I was thinking that maybe this CRF should be the castellated prism, and the previous structure the inflated (or augmented) castellated prism. Edit: Other possible adjectives: susconated (from Gk. souskōnō: to inflate).

Well, probably not "augmented" given that this word has a history of meaning specifically augmentation by gluing things together. The process here is more complex.
Marek14
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### Re: Johnsonian Polytopes

Marek14 wrote:
quickfur wrote:I was thinking that maybe this CRF should be the castellated prism, and the previous structure the inflated (or augmented) castellated prism. Edit: Other possible adjectives: susconated (from Gk. souskōnō: to inflate).

Well, probably not "augmented" given that this word has a history of meaning specifically augmentation by gluing things together. The process here is more complex.

True.

I'm also thinking we need to revisit the "castellated prism" name as well. It was appropriate when there was only a single castellated prism, but now that these things are turning out to be rather common, especially here where they are based of various strata of the o5x3o3o, I think we need a more generally-applicable naming scheme, as I'm expecting more CRFs to turn up with similar constructions.
quickfur
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### Re: Johnsonian Polytopes

Marek14 wrote:Well, this polytope seems to be a type of rotunda, it's a tristratic polychoron squeezed between two parallel hyperplanes -- calling it a rotunda makes about the same sense as callind the J91 polychoron "prism". A castellated rotunda?

Once the term cupola in this forum was used quite often rather freely to nearly all neither pyramidal nor prismatic lace prisms. I then argued that cupola is a term being used for 3D in a very specific manner. That one allows for 2 different extensions to 4D. One being the monostratic caps of small prismated polychora. The other one has been prefered in my introductury article on convex segmentochora. There I had discussed that decisission of mine of full. One of the arguments were that the former caps are described just by that property already and thus do not need for that additional naming. Further not any application is possible for those (becoming hyperbolic else). - Thus I finally managed to get These choice of mine being at least respected in this forum.

In this run, so on a different issue, I also pointed out that rotunda so far occurs in 3D only for the half of the icosidodecahedron. Sure, that can be seen as just one instance of multistratic caps. And I observe that you use that term in that way. - I for one so did reserve that term solely for half-polychora. Any other multistratic section is being called either a cap or a diminishing thereof.

Cf. also
http://bendwavy.org/klitzing/explain/johnson.htm#cap
http://bendwavy.org/klitzing/explain/johnson.htm#rotunda
http://bendwavy.org/klitzing/explain/axials.htm#cupola

--- rk
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