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

Postby quickfur » Sun Feb 23, 2014 6:36 am

Marek14 wrote:
quickfur wrote:
Keiji wrote:Thanks for the hard work quickfur!

I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.

I didn't build it because basically it's CJ4.5.1 with the top part of the rectified 120-cell glued on both ends. It's just extra vertices (=slower convex hull times, slower rendering, etc.) away from the exciting action (i.e. where the J92's are). You could still create a page for it, and mention the cell counts, etc., which are all already known. It's just that it's not that much more interesting than CJ4.5.1 itself. :)


If the top part of rectified 120-cell is glued on "both ends", does it mean you can get anohter CRF that would count as crown jewel by gluing it on only one end?

Yes, and since the caps glued on are themselves made of two separable layers, you can also glue half a cap on one end or half a cap on the other end, and any combination thereof. Also, any diminishing/augmentation that applies to the upper part of the rectified 120-cell also applies here. I consider these variations as uninteresting, since they don't relate to the region of interest where the J92's are. In that sense, CJ4.5.1 is the most "fundamental" form, and CJ4.5.2 and the other unnumbered possibilities are just cut-n-paste variations on it. I assigned CJ4.5.2 a number because it represents the simplest derivation from the rectified 120-cell that contains the CJ4.5.1 core, but other than that, it's not fundamentally interesting.

quickfur wrote:Hmm. Today I was reading up on the snub disphenoid on Wolfram, and it claims that the analytic solution for its coordinates requires solving a cubic equation?? How did Marek get a solution in terms of square roots??


Hm. Let's have a look.

Snub disphenoid has 8 vertices which can be separated into 4 layers. Let's use edge length 2 here:

1st layer: (1,0,0) and (-1,0,0)
2nd layer: (0,x,y) and (0,-x,y)
3rd layer: (x,0,z) and (-x,0,z)
4th layer: (0,1,w) and (0,-1,w)

Each layer has two vertices and the line of these two is always perpendicular between layers. Further, snub disphenoid has dihedral symmetry, meaning that distance between layers 1 and 2 is the same as distance between 3 and 4, so we can replace w with sum y+z.

Now then, snub disphenoid has four kinds of edges:

1. Edge joining two vertices in layers 1 and 4 -- this is 2 by definition.
2. Edge joining layers 1-2 or 3-4 -- this means distance (1,0,0) and (0,x,y) is 2.
3. Edge joining layers 1-3 or 2-4 -- this means distance (1,0,0) and (x,0,z) is 2.
4. Edge joining layers 2-3 -- this means distance (0,x,y) and (x,0,z) is 2.

So all the conditions mean that it's sufficient to ensure the triangle (1,0,0), (0,x,y), (x,0,z) is equilateral with side 2.

We have equations:

x^2+y^2+1 = 4
(x-1)^2+z^2 = 4
2x^2+(z-y)^2 = 4

By solving these three equations in Mathematica, I now get cubic solutions as well (there are quadratic solutions, but these have z = 0). The positive solution has x = 1.28917, y = 1.15674 and z = 1.97898. Not sure what went wrong before...

So the snub disphenoid has CVP 3 after all?
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Feb 23, 2014 7:06 am

Seems so...
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Re: Johnsonian Polytopes

Postby wendy » Sun Feb 23, 2014 7:12 am

Hi

I found a few more CRF's. Lots more.

If you take a decagon of edge 2, ie, u10o, and rectify it, you get (say) o10G (where G=1.90211303259&c). The vertices of this correspond to the centre of a decagon x10o placed against a central decagon x10o. It also corresponds to the vertices of a pentagon x5o placed against the edge of the inner decagon.

So you have, in theory, xfu P ooo, rectifies to xfoPoox. The only polygon uPo, whose edge midpoints form an oPx, is when P=3. So we have xfo3oox&#x

The pyramid built on top of this would, if it were coplanar with the pentagons, would have a lacing of f, so a pyramid of lacing x would always produce a convex figure too, as long as P had a diameter less than r4.

So you can ursulate any figure made of triangles, up to a diameter of the decagon, ie r10.472, and mount a pyramid if the diameter is less than a hexagon ie r4.00

We then have the following can be ursulated: xfo3oox...

1_k1 as far as 20 dimensions, 2_21, 3_21, 4_21, 2_31, 2_41, and 1_32 (1_42 is too big at r2 = 16).
as far as fourth rectate of the simplex and of the orthotope.
the first rectate of 2_21, id est, 1/3B = o3x3o3o3oBo.

The following can be apiculated as well: ie oxfo3ooox.... and ox3oo...
1_k1 to seven dimensions, 2_21, 3_21, second rectate of the simplex to siven dimensions, the first rectate of the simplex throughout.

The rectate of a n-rectate is a n-cantellate, is also uniform.
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Re: Johnsonian Polytopes

Postby student91 » Sun Feb 23, 2014 2:53 pm

I think I found a new CRF as well. It seems bilbiro || line segment is a possible CRF.

It can be derived from o5x3o||o5o3x, aka oxFf(oX)fFxo2ofxF(Xo)Fxfo&#xt || xofox2oxfxo&#xt, by deleting some layers, so you get oxFxo2ofxfo&#xt || x2o&#xt.
It should have 1 bilbiro, 4 pentagonal pyramids, 4 tetrahedra, 4 square pyramids and 2 triangular prisms. I think it's cool, because it's so simple, and it has every pyramid 4 times :D .
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Re: Johnsonian Polytopes

Postby quickfur » Sun Feb 23, 2014 4:11 pm

student91 wrote:I think I found a new CRF as well. It seems bilbiro || line segment is a possible CRF.

It can be derived from o5x3o||o5o3x, aka oxFf(oX)fFxo2ofxF(Xo)Fxfo&#xt || xofox2oxfxo&#xt, by deleting some layers, so you get oxFxo2ofxfo&#xt || x2o&#xt.
It should have 1 bilbiro, 4 pentagonal pyramids, 4 tetrahedra, 4 square pyramids and 2 triangular prisms. I think it's cool, because it's so simple, and it has every pyramid 4 times :D .

Confirmed! Wonderful!

Image

So this would be the first non-trivial monostratic outside of Klitzing's orbiform segmentochora? :)

Coordinates are dead-easy too:
Code: Select all
# J91
<±1, 0, ±phi^2, 0>
<±phi, ±1, ±1, 0>
<0, ±phi, 0, 0>

# line
<0, 0, ±1, 1/phi>


Edit: here's the .off file.
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Feb 23, 2014 4:23 pm

Very nice! It's a sort of pseudopyramid -- since the base solid isn't orbiform, the "apex" isn't a single point, but a line.

And of course it can exist as bipyramid/elongated bipyramid. Its dichoral angles are very small (largest is 36), but unfortunately it seems that this is still not enough to allow augmenting of J91's in other crown jewels.

So, can a J92 get capped in a similar way? The top and sides of J92 are basically three J91's merged together, so what would happen if the lines from J91 pseudopyramid were added to J92? They would naturally form an equilateral triangle, but what would happen to the vertices in bottom layer?
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Re: Johnsonian Polytopes

Postby Keiji » Sun Feb 23, 2014 6:01 pm

Added to the wiki: bilunabirotunda pseudopyramid (thanks for the unintentional name suggestion, Marek ;) )
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Feb 23, 2014 6:05 pm

Keiji wrote:Added to the wiki: bilunabirotunda pseudopyramid (thanks for the unintentional name suggestion, Marek ;) )


What do you mean, "unintentional"? :)
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Re: Johnsonian Polytopes

Postby student91 » Sun Feb 23, 2014 7:37 pm

quickfur wrote:Confirmed! Wonderful!
[...]
Hurray :D
Marek14 wrote:Very nice! It's a sort of pseudopyramid -- since the base solid isn't orbiform, the "apex" isn't a single point, but a line.

And of course it can exist as bipyramid/elongated bipyramid. Its dichoral angles are very small (largest is 36), but unfortunately it seems that this is still not enough to allow augmenting of J91's in other crown jewels.
[...]
Although it can't be used to augment CRF's, it can be cut off of some polytopes. In fact, that's how I found it.
When I was trying to bilbiro o5o3x3o, I concluded the vertices between two o3x5x-cuts should be deleted to make bilbro's. Look at this picture:Imagethe yellow triangles of the ike's form a o3x5x. Now look at the red and orange triangles between the ike's. These triangles, together with pentagonal cuts through the ike's, make a bilbro-shaped depression (it corresponds to an id-cut, centered on an equatorial ike) This means that if we cut the vertices between those away, you get a bilbiro. This was quite interesting, but when I considered the cutout-part, I concluded that was much more interesting. Maybe the o5o3o3x might give a similar thing. Klitzing just pointed out how that should be done. :) :D
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Re: Johnsonian Polytopes

Postby Klitzing » Sun Feb 23, 2014 7:56 pm

Marek14 wrote:Very nice!

Indeed!

I too was quite near to that (just cf. my last post in this thread - I just was having lot more stuff above than a mere line...)

So, can a J92 get capped in a similar way? The top and sides of J92 are basically three J91's merged together, so what would happen if the lines from J91 pseudopyramid were added to J92? They would naturally form an equilateral triangle, but what would happen to the vertices in bottom layer?


I got it!

Student91's line||bilbiro easily provides a very similar figure: {3}||thawro!

Cells then are 1 thawro (for sure) + 1 tricu + 3 trip + 9 tet + 3 squippy + 3 peppy + 1 oct.

--- rk
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Re: Johnsonian Polytopes

Postby student91 » Sun Feb 23, 2014 8:09 pm

Klitzing wrote:[...]
Student91's line||bilbiro easily provides a very similar figure: {3}||thawro!

Cells then are 1 thawro (for sure) + 1 tricu + 3 trip + 9 tet + 3 squippy + 3 peppy + 1 oct.

--- rk

Cool!! :D I'm indeed sometimes feeling like I'm only putting cherries on your cakes, and getting credit for having made the whole cake :\ . Anyway a very beautifull find :)
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Re: Johnsonian Polytopes

Postby Klitzing » Sun Feb 23, 2014 8:40 pm

quickfur wrote:So this would be the first non-trivial monostratic outside of Klitzing's orbiform segmentochora? :)


No!

Remember your own finds:   n-py || inv gyro n-py (2<n<6) and   n-cu || inv gyro n-cu (6/5<n<6) ?
For pyramids only the case n=3 (= hex) and for cupola only the case n=2 (= K-4.13) would be true segmentochora.
(Others need for shifts of non-degenerate bases.)

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Re: Johnsonian Polytopes

Postby Keiji » Sun Feb 23, 2014 9:16 pm

Just worked out the incidence matrix for the bilunabirotunda pseudopyramid and added it to the wiki.

In doing so, I found another pair of neat non-CRF polyhedra:
Image

The hexagonal cupolawedge (as I've named it) appears as two of the cells in the dual to the bilbiro pseudopyramid - it appears in the positions corresponding to the apices! As the dual isn't yet on the wiki, you can see its appearances on its polytope explorer page instead. (The hexagonal pyrawedge, if you haven't guessed, is the dual of the hexagonal cupolawedge.)

I kinda like these "non-CRF polyhedra that appear in duals of CRF polychora" - they remind me of the Catalans in a way. This is the second such pair I've discovered - the first being the BXD's dual's cells (which are self-dual!)
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 5:30 am

Klitzing wrote:
Marek14 wrote:Very nice!

Indeed!

I too was quite near to that (just cf. my last post in this thread - I just was having lot more stuff above than a mere line...)

So, can a J92 get capped in a similar way? The top and sides of J92 are basically three J91's merged together, so what would happen if the lines from J91 pseudopyramid were added to J92? They would naturally form an equilateral triangle, but what would happen to the vertices in bottom layer?


I got it!

Student91's line||bilbiro easily provides a very similar figure: {3}||thawro!

Cells then are 1 thawro (for sure) + 1 tricu + 3 trip + 9 tet + 3 squippy + 3 peppy + 1 oct.

Wait, this polychoron cannot possibly be CRF, at least not as described. Proof: the height of the triangular antiprism (octahedron) of edge length 2 is 2/√3, and the height of the triangular cupola of edge length 2 = height of tetrahedron of edge length 2 = 4/√6, whereas the height of J92 of edge length 2 = 2φ2/√3 = approx 3.02305. But 2/√3 + 4/√6 = approx 2.78769 < 3.02305, so it is not possible for the octahedron and the triangular cupola to connect the top triangle of J92 to its bottom hexagon! :(

(I spent all afternoon trying to compute the coordinates for it and keep getting inconsistent answers, until I checked the existence of the J92-oct-tricu triangle, and realized that it was impossible!)


EDIT: OK, I just totally made a fool of myself... a bug in my program computed the wrong height for the triangular antiprism (octahedron). :oops: :oops: :oops: Actually, the polytope does exist, and is definitely CRF. Here's a render:

Image

Here are the coordinates:
Code: Select all
#
# J92
#

# x3o:
<0, 2/√3, 2*phi^2/√3, 0>
<±1, -1/√3, 2*phi^2/√3, 0>

# f3x:
<±1, phi^3/√3, 2*phi/√3, 0>
<±phi^2, -1/(phi*√3), 2*phi/√3, 0>
<±phi, -(phi+2)/√3, 2*phi/√3, 0>

# o3F:
<±phi^2, phi^2/√3, 2/√3, 0>
<0, -2*phi^2/√3, 2/√3, 0>

# Hexagonal base:
<±1, ±√3, 0, 0>
<±2, 0, 0, 0>

#
# Triangle
#

# o3x:
<0, -2/√3, phi^2/√3, 1/phi>
<±1, 1/√3, phi^2/√3, 1/phi>


Here's the .off file.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 6:26 am

Here's another render, from a side-view:

Image

The green part shows the J92, and the yellow triangle is the one above it. As you can see from here, and also from the coordinates, this is a very shallow pseudo-pyramid (1/(2*phi), approx 0.30902 for edge length 1). This means that probably many CRFs containing J92 cells can be augmented with this pseudo-pyramid to make CRFs with J92 cross-sections. Same thing goes for line||J91 (bilunabirotunda pseudo-pyramid).
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Re: Johnsonian Polytopes

Postby Keiji » Mon Feb 24, 2014 6:41 am

quickfur wrote:Image


Don't suppose you can render it from a slightly different angle (3D projection rotated in vertical axis a bit)?

As it is in that image, two vertices of the triangle (and two lacing edges) are almost right on top of another, and it's impossible (for me at least) to see the triangle and triangular cupola without stereo viewing.

(Even with stereo viewing I only got it when I knew what I was looking for, by looking at your side-view projection first!)
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Feb 24, 2014 7:10 am

It's great that it exists -- I'm still not able to find and confirm CRFs properly on my own, but at least I can contribute ideas :)

The dichoral angles for this polychoron (thawro wedge?) are:

Upper tetrahedron (6 in total):
* one triangle (J92) - 22.2388
* second triangle (lower tetrahedron) - 164.478
* third triangle (triangular prism) - 172.239
* fourth triangle (pentagonal pyramid) - 172.239

Lower tetrahedron (3 in total):
* one triangle (J92) - 22.2388
* second triangle (upper tetrahedron) - 164.478
* third triangle (triangular cupola) - 164.478

Square pyramid (3 in total):
* one triangle (J92) - 22.2388
* opposite triangle (octahedron) - 164.478
* 2 side triangles (pentagonal pyramid) - 157.761
* square (triangular prism) - 155.905

Triangular prism (3 in total):
* 2 triangles (upper tetrahedron) - 172.239
* one square (J92) - 20.9052
* second square (square pyramid) - 155.905
* third square (triangular cupola) - 155.905

Pentagonal pyramid (3 in total):
* one triangle (octahedron) - 157.761
* 2 triangles adjacent to it (square pyramid) - 157.761
* remaining 2 triangles (upper tetrahedron) - 172.239
* pentagon (J92) - 36

Octahedron:
* 3 upper side triangles (pentagonal pyramid) - 157.761
* top triangle (J92) - 22.2388
* 3 bottom side triangles (square pyramid) - 164.478
* bottom triangle (triangular cupola) - 172.239

Triangular cupola:
* 3 side triangles (lower tetrahedron) - 164.478
* top triangle (octahedron) - 172.239
* 3 squares (triangular prism) - 155.905
* hexagon (J92) - 22.2388

J92:
* 3 side triangles on top (square pyramid) - 22.2388
* 3 triangles adjacent to the hexagon (lower tetrahedron) - 22.2388
* 6 luna triangles (upper tetrahedron) - 22.2388
* top triangle (octahedron) - 22.2388
* 3 squares (triangular prism) - 20.9052
* 3 pentagons (pentagonal pyramid) - 36
* hexagon (triangular cupola) - 22.2388

Once again, two copies can be glued together and/or elongated by thawro prism.
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Feb 24, 2014 7:20 am

By the way, about ursachora.

I have mentioned before a question about ursachoral expansion of things like triangular bipyramid.

The way I see it, ursachora are a step in sequence:

icosahedral pyramid is an icosahedron with a TRIANGLE built over every edge.
icosahedral prism is an icosahedron with a SQUARE built over every edge.
icosahedral ursachoron is an icosahedron with a PENTAGON built over every edge.
icosahedral rotunda is an icosahedron with a HEXAGON built over every edge.

Now, every polyhedron has a prism, second member in this sequence, and we have completely catalogized the pyramids, but which polyhedra have the third and fourth members?

EDIT: Do you have off files for the pyramid antiprisms?

EDIT2: I built square pyramid antiprism by myself -- now I see it's just square antiprism bipyramid which explains why it's not orbiform.

But, it can be also thought of as a ring formed by sequence x4o-o4o-o4x-o4o. I tried what would happen if I replaced the points by octagons. Unfortunately, then the distance between octagons is not the same length and it creates new edges which prevent it from being CRF. Maybe it would work with another cupola?

EDIT3: Though that seems a bit strange -- shouldn't it just create a square||octagonal prism gyrobicupola? Maybe I just chose wrong coordinates...
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Re: Johnsonian Polytopes

Postby wendy » Mon Feb 24, 2014 8:55 am

More on ursulae.

Some of the ursulate figures might have a base so large that all of the top half forms a convex shell. This would allow back-to-back ursulates (ie xfofx3ooxoo...), and prismated ones (ie xfoofx3ooxxoo...). It seems that the 2_41 and 1_32 may well be big enough to let this happen.
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Re: Johnsonian Polytopes

Postby student91 » Mon Feb 24, 2014 12:16 pm

quickfur wrote:
Klitzing wrote:[...]
[...]
Actually, the polytope does exist, and is definitely CRF. Here's a render:

Image
[...]

And this one can be found by thawro-ing the o5o3x3o too. (the bilbiro-pseudopyramid can be found by bilbiro-ing the o5o3x3o).
a part of the o5o3x3o looks like xoxFof.....3oxoofo.....5ooxoox.....&#xt. Now a "thawro-ing" is done by deleting the o3f5o-layer of vertices, and scaling down the f-hexagons that can be found in the f3o5x-layer (three f's, together with three diagonals of o5x make an f-hexagon). This is done by changing f3o5x to x3x3o. This gives us oxFx3xoox5oxoo&#xt. Special about this polystratic stack is that if you delete a triangle of the x3o5x-layer, a thawro occurs. The maximal way this can be done is with 8 thawro's (deleting the triangles corresponding to the vertices of a cube inscribed in a dodecahedron). This gives a polytope with 8 thawro's, 16 metabidiminished ikes, and some other cells. It should look really cool.
Last edited by student91 on Mon Feb 24, 2014 9:45 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 4:10 pm

Keiji wrote:[...]
Don't suppose you can render it from a slightly different angle (3D projection rotated in vertical axis a bit)?

As it is in that image, two vertices of the triangle (and two lacing edges) are almost right on top of another, and it's impossible (for me at least) to see the triangle and triangular cupola without stereo viewing.

(Even with stereo viewing I only got it when I knew what I was looking for, by looking at your side-view projection first!)

Sorry, that was a quick and rough render just to get it out there, because it was bedtime for me and I didn't have time to tweak things. :P

Here's a much better render:

Image

The green triangle shows the "apex" of this pseudo-pyramid.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 4:13 pm

wendy wrote:More on ursulae.

Some of the ursulate figures might have a base so large that all of the top half forms a convex shell. This would allow back-to-back ursulates (ie xfofx3ooxoo...), and prismated ones (ie xfoofx3ooxxoo...). It seems that the 2_41 and 1_32 may well be big enough to let this happen.

Also, don't forget non-convex back-to-back ursulates that can be made convex by unit lacing edges over the non-convex parts, like in the icosahedral ursachoron (which is basically a diminished pseudo-bisected 600-cell). Are there any other ursulates that have this phenomenon?
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Re: Johnsonian Polytopes

Postby Keiji » Mon Feb 24, 2014 7:26 pm

I suppose in addition to the bilbiro and thawro pseudopyramids, there are also pseudobipyramids of them both which would also be CRF?

The bilbiro/thawro base would then become internal, but presumably all of their faces would be kept and would lie on the equator?
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 7:34 pm

Keiji wrote:I suppose in addition to the bilbiro and thawro pseudopyramids, there are also pseudobipyramids of them both which would also be CRF?

The bilbiro/thawro base would then become internal, but presumably all of their faces would be kept and would lie on the equator?

Of course. That's what Marek was trying to say. :) And you can elongate them with J91/J92 prisms too. So basically you have the CRF combinations {J91,J92} x {pseudopyramid, psuedobipyramid, elongated pseudopyramid, elongated pseudobipyramid}. What's interesting, is that the J91's and J92's disappear in the bipyramids, so you're left with quite unique combinations of prisms, pyramids, and antiprisms cells. :)
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Feb 24, 2014 7:41 pm

quickfur wrote:
Keiji wrote:I suppose in addition to the bilbiro and thawro pseudopyramids, there are also pseudobipyramids of them both which would also be CRF?

The bilbiro/thawro base would then become internal, but presumably all of their faces would be kept and would lie on the equator?

Of course. That's what Marek was trying to say. :) And you can elongate them with J91/J92 prisms too. So basically you have the CRF combinations {J91,J92} x {pseudopyramid, psuedobipyramid, elongated pseudopyramid, elongated pseudobipyramid}. What's interesting, is that the J91's and J92's disappear in the bipyramids, so you're left with quite unique combinations of prisms, pyramids, and antiprisms cells. :)


Yes, though I wonder if the J92 thing shouldn't be called a wedge instead.

I guess their wiki pages should eventually include data for all four figures, as they should be easy to compile and they are not overly numerous.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 7:42 pm

student91 wrote:
quickfur wrote:
Klitzing wrote:[...]
[...]
Actually, the polytope does exist, and is definitely CRF. Here's a render:

Image
[...]

And this one can be found by thawro-ing the o5o3x3o too. (the bilbiro-pseudopyramid can be found by bilbiro-ing the o5o3x3o).
a part of the o5o3x3o looks like xoxFof.....3oxoofo.....5ooxoox.....&#xt. Now a "thawro-ing" is done by deleting the o3f5o-layer of vertices, and scaling down the f-hexagons that can be found in the f3o5x-layer (three f's, together with three diagonals of o5x make an f-hexagon). This is done by changing f3o5x to x3x3o. This gives us oxFx3xoox5oxox&#xt.

I think you have a typo there, it should be oxFx3xoox5oxoo&#xt. ;) What you wrote is non-CRF because the bottom x3x5x is too big, and requires either non-CRF lacing edges or self-intersection with the other layers.

Special about this polystratic stack is that if you delete a triangle of the x3o5x-layer, a thawro occurs. The maximal way this can be done is with 8 thawro's (deleting the triangles corresponding to the vertices of a cube inscribed in a dodecahedron). This gives a polytope with 8 thawro's, 16 metabidiminished ikes, and some other cells. It should look really cool.

I've successfully constructed your (corrected) lace tower, and verified that the edge lengths are all equal. Now I'm going to graph the vertex connectivity (wink at Keiji) of the said vertex layer, to pick out the 8 triangles to delete...
Last edited by quickfur on Mon Feb 24, 2014 7:45 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 7:44 pm

Marek14 wrote:
quickfur wrote:
Keiji wrote:I suppose in addition to the bilbiro and thawro pseudopyramids, there are also pseudobipyramids of them both which would also be CRF?

The bilbiro/thawro base would then become internal, but presumably all of their faces would be kept and would lie on the equator?

Of course. That's what Marek was trying to say. :) And you can elongate them with J91/J92 prisms too. So basically you have the CRF combinations {J91,J92} x {pseudopyramid, psuedobipyramid, elongated pseudopyramid, elongated pseudobipyramid}. What's interesting, is that the J91's and J92's disappear in the bipyramids, so you're left with quite unique combinations of prisms, pyramids, and antiprisms cells. :)


Yes, though I wonder if the J92 thing shouldn't be called a wedge instead.

I thought about that too. But I decided that a wedge should have two identical cells that meet at a ridge, rather than just any triangular configuration of cells, 'cos otherwise a lot of things would be called wedges.

I guess their wiki pages should eventually include data for all four figures, as they should be easy to compile and they are not overly numerous.

Feel free to contribute. ;)
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Re: Johnsonian Polytopes

Postby Keiji » Mon Feb 24, 2014 7:52 pm

quickfur wrote:
I guess their wiki pages should eventually include data for all four figures, as they should be easy to compile and they are not overly numerous.

Feel free to contribute. ;)


Just a note: We should not put multiple explicit polytopes on one page. It's fine to have many small pages - one for each member of a family - then a page for the family as a whole, as is done with the bicupolic rings. However, you cannot display more than one incidence matrix on the same page - well you could, but doing so would mess up the links to that polytope from other polytopes' incidence matrices.
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Feb 24, 2014 8:30 pm

Well, for now here are the off files:

bilbro||line

bi: http://hddb.teamikaria.com/dl/KD3FPBBJG ... M5ED9K.off

elongated: http://hddb.teamikaria.com/dl/MPR62E6B1 ... PCHJQW.off

bielongated: http://hddb.teamikaria.com/dl/5D76CD6RG ... ASNBZQ.off

J92||triangle:

bi: http://hddb.teamikaria.com/dl/NHEKTMYV1 ... R052XG.off

elongated: http://hddb.teamikaria.com/dl/K4GMCTGVD ... AETRAD.off

bielongated: http://hddb.teamikaria.com/dl/2C58W6YTR ... J1G61Z.off

Um, why do I have this strange notion that it should be possible to have J91||(J91 in perpendicular orientation) so gyroelongation would be possible...?

Replaced your code sections with links, there's really no need to copypaste when you can just link the file! :) ~Keiji
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 24, 2014 8:37 pm

quickfur wrote:
student91 wrote:[...] This gives us oxFx3xoox5oxox&#xt.

I think you have a typo there, it should be oxFx3xoox5oxoo&#xt. ;) What you wrote is non-CRF because the bottom x3x5x is too big, and requires either non-CRF lacing edges or self-intersection with the other layers.

Special about this polystratic stack is that if you delete a triangle of the x3o5x-layer, a thawro occurs. The maximal way this can be done is with 8 thawro's (deleting the triangles corresponding to the vertices of a cube inscribed in a dodecahedron). This gives a polytope with 8 thawro's, 16 metabidiminished ikes, and some other cells. It should look really cool.

[...]

:o_o: This has got to be one of the most beautiful crown jewels we've found yet!! Check it out:

Image

The J92's (shown in burgundy and blue) are actually touching each other at an edge, with triangular prisms (green) straddling the larger gaps between them, with the most fascinating combinations of cells around the triangular prisms.

I'm in awe at this amazing shape. Kudos to student91 for finding it!

.off file
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