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 » Mon Jan 14, 2013 4:57 am

Klitzing wrote:As to your question:
spic = smal prismated icositetrachoron = x3o4o3x.

Hmm. I am having some trouble seeing where there are segments on x3o4o3x that can be gyrated. I see some near candidates, but they produce two square pyramids joined at their bases at 45° -- that is, the result is non-convex. Other potential gyrations would cut some triangular prisms and produce unequal edge lengths. Do you have an example of where gyration is possible? I'd love to see it, 'cos it would give rise to very interesting CRFs!

As to your provided cross-link:
Even so hemidemi makes a nice ringing rhyme it seems inapropriate for those ex-wedges. Hemi is appropriate for the first cut. Hemi has greek root. The same word with latin root is demi. Thus hemidemi would be something which is a quarter. But your angle between the 2 equator planes is meant to vary.

Yeah, basically they are the 600-cell cut by two hyperplanes that pass through the center, with bisected edges deleted and pentagonal pyramids inserted in their place. There are 5 possible cutting angles, from 180° to 36°, in 36° increments, producing a series of 5 CRFs that are slices of the 600-cell. These CRFs can be further diminished in various ways, to produce more CRFs.

Wedges are what those are indeed. But then again wedge alone might be not specific enough (on how those are to become wedges, i.e. where the faceting planes are to be chosen).

So we might try a completely different way. Any such cut results individually in a hemispherical remainder. So you are, more specifically, dealing with differnent possibilities of an intersection of hemispheres. Might this result in a better (more specific) naming advice?

Good idea. I am thinking of how similar slices can be made in an analogous way in 3D. For example, we divide the surface of the Earth with longitudinal lines, so that the area between two longitude lines represents a slice of the Earth in the same sense as the 600-cell slices. So we may, in a crude sense, call the latter "longitudinal slices of the 600-cell".

However, in 4D, we don't have a direct analogue of longitude lines, so this name is somewhat misleading. But maybe we can derive a good name along these lines.

That very building procedure itself is quite intersting! The same could most probably be applied to other figures too. E.g. to ico, which has a co as its equatorial section. Thus you'd get 2 tricues for wedge facets, 2*3 squippies (halved octs) and some remaining octs. (I suppose there should be 3 such wedges, according to neighbouring, orthogonal, and meta positions of normal directions, right?) Or to hex, halving an oct as equatorial section. Thus the single case here would result in the quater hex with 2 squippies for wedge facets and some few remaining tets. (In fact, that latter one is nothing but the pyramid above squippy, i.e. squippypy).

--- rk

The bisected 24-cell is actually just octahedron||cuboctahedron. Can you make a wedge out of it? I'm not sure... since oct||cuboct is a segmentochoron, any wedges would also have to be segmentochora, which should already be in your list of segmentochora. :)

And yeah, the only wedge of the 16-cell would just be squippypy (square pyramid pyramid, or point||4pyr).
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 14, 2013 5:21 am

quickfur wrote:[...]The bisected 24-cell is actually just octahedron||cuboctahedron. Can you make a wedge out of it? I'm not sure... since oct||cuboct is a segmentochoron, any wedges would also have to be segmentochora, which should already be in your list of segmentochora. :) [...]

I built a model of the first wedge and checked the projections. It looks like oct||3cup: It has 2 triangular cupolae, 5 octahedra, and 6 square pyramids.

Based on these projections, it seems like the second (and final) wedge should be triangle||3cup, with 2 triangular cupolae, 1 octahedron, 6 square pyramids. That is, the same as what Keiji calls the triangular gyrobicupolic ring.

This is quite interesting! This means a 24-cell can be cut into 6 triangle||3cup's. Keiji would be elated to hear about that. :)
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Jan 14, 2013 6:47 am

What about a word for "half-moon"?
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 14, 2013 9:59 am

Marek14 wrote:What about a word for "half-moon"?

Good idea! Reminds me to some of the names of johnsonians: ...-luna-...

Thus quickfurs hemidemi wedges then are just lunae. And those I just mentioned too.
So we still need qualifiers for the width, i.e. reminding to te separation of the normals to the respective equatorial sectionhyperplanes. Whether those vertices are one edge length apart or greater.

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Re: Johnsonian Polytopes

Postby wendy » Mon Jan 14, 2013 10:34 am

Crescent for half-moon.
The dream you dream alone is only a dream
the dream we dream together is reality.

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Re: Johnsonian Polytopes

Postby Marek14 » Mon Jan 14, 2013 11:08 am

wendy wrote:Crescent for half-moon.


I think crescent is less than half moon -- it doesn't have the middle line, thus I didn't want to use that.

Klitzing: when thinking about width of these parts, the word "span" comes to mind...monospanning, bispanning, trispanning...
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 14, 2013 11:20 am

quickfur wrote:
Klitzing wrote:As to your question:
spic = smal prismated icositetrachoron = x3o4o3x.

Hmm. I am having some trouble seeing where there are segments on x3o4o3x that can be gyrated. I see some near candidates, but they produce two square pyramids joined at their bases at 45° -- that is, the result is non-convex. Other potential gyrations would cut some triangular prisms and produce unequal edge lengths. Do you have an example of where gyration is possible? I'd love to see it, 'cos it would give rise to very interesting CRFs!
[...]

Well, that's easy. You have spic (x3o4o34x) \ 8x oct||sirco = srit (x4o3x3o). And srit \ 2*4 {4}||op = odip (x4x x4x). But you needn't to chop off all 8 oct||sirco caps in order to get a possible {4}||op to be gyrated, you only are asked to chop off 2 at a right angle. Then there are 2 sircoes connecting at a (cubical) square. Thus a gyration of the right that square becomes possible. You then could consider the next right angled pair. That one might coincide with one of the already chosen positions (to form a further ortho or a para positioning) or else those might be independent, say in meta positionings! - Therefore we are not bound to tessic symmetries (or their subsymmetries) any longer (as would be for srit). Any icoic subsymmetry becomes possible here.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 14, 2013 11:58 am

quickfur wrote:[...]
Klitzing wrote:That very building procedure itself is quite intersting! The same could most probably be applied to other figures too. E.g. to ico, which has a co as its equatorial section. Thus you'd get 2 tricues for wedge facets, 2*3 squippies (halved octs) and some remaining octs. (I suppose there should be 3 such wedges, according to neighbouring, orthogonal, and meta positions of normal directions, right?) Or to hex, halving an oct as equatorial section. Thus the single case here would result in the quater hex with 2 squippies for wedge facets and some few remaining tets. (In fact, that latter one is nothing but the pyramid above squippy, i.e. squippypy).

--- rk

The bisected 24-cell is actually just octahedron||cuboctahedron. Can you make a wedge out of it? I'm not sure... since oct||cuboct is a segmentochoron, any wedges would also have to be segmentochora, which should already be in your list of segmentochora. :)

And yeah, the only wedge of the 16-cell would just be squippypy (square pyramid pyramid, or point||4pyr).


Displaying hex within 2D as vertex based tower of sections we have pt||hex||pt. Thus the only possible luna here is using 2 adjacent vertices, which results in squippypy (K-4.4).
Displaying ico as such we'd get pt||cube||co||cube||pt. For 2 adjacent vertices we'd then get as luna the oct||tricu (K-4.30). For to orthogonal vertices we'd get as luna the {3}||gyro tricu (K-4.27 - which, btw., is nothing but {6}||oct !). Wonder whether 2 meta vertices would give something too or whether we here get non-CRF cell intersections.
Displaying ex as such we'd get pt||ike||doe||f-ike||id||f-ike||doe||ike||pt. You already described the first 4 lunae (wedges) up to 2 orthogonal vertices, I suppose. Again wonder whether even farther vertices (0-5, 0-6, 0-7) would derive similar cases or whether those result in non-CRF cell intersections. - Perhaps there is even some general rule, which prohibites farer than orthogonal vertices for that reason?

If there are no further ones here, we could call those the (ortho) hex-lune, the big resp. narrow (= ortho) ico-lune, the huge, big, meta, resp. narrow (= ortho) ex-lune. - What'd you think?

And yes, the big ico-lune also allows for further (non-lunaic) diminishing: squippy||tricu (K-4.32).

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 14, 2013 8:11 pm

Klitzing wrote:[...] Displaying hex within 2D as vertex based tower of sections we have pt||hex||pt. Thus the only possible luna here is using 2 adjacent vertices, which results in squippypy (K-4.4).

Yes, because past adjacent vertices you get antipodes, which results in subdimensional cutting.

Displaying ico as such we'd get pt||cube||co||cube||pt. For 2 adjacent vertices we'd then get as luna the oct||tricu (K-4.30). For to orthogonal vertices we'd get as luna the {3}||gyro tricu (K-4.27 - which, btw., is nothing but {6}||oct !). Wonder whether 2 meta vertices would give something too or whether we here get non-CRF cell
intersections.

Be careful of using the vertex-distance approach, because for ico, bisecting with a hyperplane orthogonal to a vertex produces a non-CRF cutting with a rhombic dodecahedron cell!

To produce CRF luna, you need to cut with a hyperplane parallel to an octahedral cell.

For my derivation, I made a model of bisected 24-cell, cutting with a hyperplane parallel to an oct, which is cuboctahedron||octahedron. Now since cuboctahedron = triangular gyrobicupola, it means it should be possible to bisect cuboct||oct with another hyperplane such that it bisects the cuboctahedron (since any other cutting will cut the cuboctahedron in non-CRF ways). This means the second hyperplane must make a hexagonal cross-section with cuboctahedron, and this hexagon defines a 2D plane. In 4D, fixing a 2D plane constrains the orientation of the hyperplanes that passes through it to a single degree of freedom. So the second hyperplane can be uniquely defined by the angle made with the first hyperplane.

Now, since the second cutting hyperplane, taken alone, must also make a (different) cuboctahedron cross-section with ico, the only way for the result of cutting with both hyperplanes to be CRF is that the result must have two 3-cupolae, joined at their hexagonal face. The angle between these 3-cupolae will be equal to the angle between the cutting hyperplane. So it is sufficient to start from the bisected 24-cell, pick a hexagonal cross-section of the bottom 3-cupola cell, and then inspect all possibilities of inserting a 2nd cupola that deletes some vertices and connects the rest, while keeping CRF cuttings of the lateral cells.

For me, this approach is easy to work with, because I use a convex hull algorithm with an input set of vertices; so starting from the bisected 24-cell, I try to find the next luna by finding the smallest set of vertices to delete to make another 3-cupola cell joined to the bottom cuboctahedron by the hexagonal cross-section. The simplest way is to delete a triangular face (i.e. remove 3 vertices) from the the cuboctahedron. This is the minimum, since otherwise you can't make two 3-cupola cells in the result joined by their hexagonal face. Then I run the convex hull algorithm, and use my program to verify that all edges are unit length, and use projections to determine the overall shape of the result.

It turns out that the result is oct||3cup. So this is the first luna of ico. Now, using the above argument that the second hyperplane can only vary by angle, and the result must have two 3cups joined at the 6gon, it follows that the next luna must be obtained by deleting the 3 vertices of the triangular face of one of the 3cups, opposite the hexagonal face. (Any other cutting in between will produce non-CRFs.) Looking at the projections, I can see that deleting this face will result in trigon||3cup (or, equivalently, 6gon||oct, as you mentioned). So this is the second luna.

Are any others possible? Well, note that in 6gon||oct, the distance between the top triangular faces of both 3cups (i.e. the 3gon opposite the 6gon) is 1 edge length. Again using the observation above that the second hyperplane can only vary in angle and the result must have two 3cups joined at the 6gon, this means that no more CRF lunae are possible, because the next cutting would require deleting the triangular face of one of the 3cups, which results in a subdimensional polytope with two coincident 3cup cells!

So the only possible lunae for ico are oct||3cup and oct||6gon (==3gon||3cup).

Displaying ex as such we'd get pt||ike||doe||f-ike||id||f-ike||doe||ike||pt. You already described the first 4 lunae (wedges) up to 2 orthogonal vertices, I suppose.

Actually, the 4 lunae come in increments of 36°, so there are no lunae produced by orthogonal vertices. The 3rd and 4th lunae are actually already past the orthogonal point. The 4th luna I listed is the narrowest one, with dichoral angle between the pentagonal rotundae at 36°, with a single 5-antiprism connecting their top faces.

Again, the situation here is that when you bisect the 600-cell (and replace bisected edges with pentagonal pyramids, since otherwise it's non-CRF), you get a o5x3o, which can be cut into two pentagonal rotundae. When you introduce a second cutting hyperplane, it must bisect the o5x3o with a 10-gon cross section; any other cuttings will cut the o5x3o into non-CRF fragments. So the second hyperplane is constrained by the plane of the 10-gon, which means not all pairs of vertices can be used for determining CRF lunae. It also means that there is only 1 degree of freedom between the two hyperplanes, so they can be defined solely by angle.

Furthermore, the second hyperplane bisecting the 600-cell must make a cross-section that can be CRF-bisected into something with a 10-gon face. But the only 600-cell cross-sections that are both CRF and allows bisection at a 10-gon is o5x3o, which means that the second hyperplane must also bisect the 600-cell at a o5x3o. Therefore, any lunae must consist of two pentagonal rotundae joined at their 10-gon faces.

The 4 lunae that I enumerated were derived in a similar way to the ico lunae; I start with the bisected 600-cell, then incrementally looked for the smallest set of vertices to delete such that it produces a CRF with two pentagonal rotunda cells. It turns out that given a pair of 600-cell vertices that lie on the same 10-gon great circle, you can generate a CRF luna, except for antipodes which produce a subdimensional result with two coincident o5x3o cells, and choosing the same vertex twice, which gives you the bisected 600-cell.

Because of the symmetry of the 10-gon, half of the lunae are identical to the other half, so that leaves only 4 unique possibilities besides the bisected 600-cell itself. Unfortunately, the two cutting hyperplanes are constrained such that the vertices must be selected from a 10-gon great circle (otherwise they produce non-CRFs), and this does not include orthogonal vertices.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 15, 2013 10:08 am

Be careful of using the vertex-distance approach[...]

Hmmm, you seem to be right. Your wedges (of ex) have 210/150/90/30 tets plus 12 peppies each (besides of those 2 rotundae). This makes a net total of 210+12*5/2=240, 150+12*5/2=180, 90+12*5/2=120, 30+12*5/2=60 tets from the 600 tets of ex. Thus your lunae would have dihedral angles of 360° * k * 60 / 600 = k * 36°, k=1,...,4. Mine wedges of ico have 2 tricues, 5 resp. 1 oct, plus 6 squippies. This makes a net total of 5+6/2=8 resp. 1+6/2=4 octs from 24 octs of ico. Thus those tricues would have a dihedral angle of 360° * k * 4 / 24 = k * 60°, k=1 or 2. And that single luna of hex has 2 squippies + 4 tets. As hex itself has 16 tets this results in an dihedral angle of 360° * 4 / 16 = 90°. - Thus that ortho prefix only applies in the latter case, which is singular and therefore does not require any specifying attribute at all. Still we could stay with huge/big/meta/narrow as qualifiers for the ex-lunae, resp. with big/narrow for the ico-lunae.

It just occured to me that you mentioned in that cited post of yours that you even specify how the rotundae are aligned. In W1 and W3 the respective pentagons are aligned at the same decagon edges, while in W2 and W4 they are placed alternatingly. This is, the former lunae do have a mirror symmetry, while the others do have a rotation-reflection. So I looked into mine lunae again with respect to that point of interest. Clearly, the squippy has full rotational symmetry, therefore the mirror symmetry applies. For the big ico-luna one further finds mirror symmetry as well, while for the narrow ico-luna one finds rotation-reflection as well. (The latter becomes obvious btw. in its orientation as oct||6g.)

You could even apply that onto combinations too: your W1 and W4 combine to a half of ex. Same for W2 and W3. Alike for mine big and narrow ico-lunae. And too for 2 copies of the hex luna. Note that the rotundae in both cases would align correctly to rebuild an id, resp. that the tricues would align correctly to rebuild the co. For 2 squippies to rebuild an oct clearly nothing is to be considered.

- Now coming back again to the above intro: None the less, those two used equatorial sectioning hyperplanes surely have normals. And those normals shall mark some specific pair of points on the boundary of the original polychoron. Seems that I was misslead that those are vertices in any case... (Have not looked into that deeper so far. So you might have a run. :) )

[...]I can see that deleting this face will result in trigon||3cup (or, equivalently, 6gon||oct, as you mentioned). So this is the second luna.[...]
[...]So the only possible lunae for ico are oct||3cup and oct||6gon (==3gon||3cup).[...]

You should be careful in that statement likewise, hehe: there are 2 different ways to align a trigon as opposite base to a tricu: para-align or gyro-align. Para-alignment results in 6g||trip, what is not being useful here, while gyro-alignment results in the here required 6g||oct.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 15, 2013 7:00 pm

Klitzing wrote:
Be careful of using the vertex-distance approach[...]

Hmmm, you seem to be right. Your wedges (of ex) have 210/150/90/30 tets plus 12 peppies each (besides of those 2 rotundae). This makes a net total of 210+12*5/2=240, 150+12*5/2=180, 90+12*5/2=120, 30+12*5/2=60 tets from the 600 tets of ex. Thus your lunae would have dihedral angles of 360° * k * 60 / 600 = k * 36°, k=1,...,4. Mine wedges of ico have 2 tricues, 5 resp. 1 oct, plus 6 squippies. This makes a net total of 5+6/2=8 resp. 1+6/2=4 octs from 24 octs of ico. Thus those tricues would have a dihedral angle of 360° * k * 4 / 24 = k * 60°, k=1 or 2. And that single luna of hex has 2 squippies + 4 tets. As hex itself has 16 tets this results in an dihedral angle of 360° * 4 / 16 = 90°. - Thus that ortho prefix only applies in the latter case, which is singular and therefore does not require any specifying attribute at all. Still we could stay with huge/big/meta/narrow as qualifiers for the ex-lunae, resp. with big/narrow for the ico-lunae.

I'm not sure I like using the words "huge" and "big"; maybe we can use lunar phase terminology instead? Gibbous, half, crescent, new.
  • Full luna - bisected 600-cell
  • Gibbous luna - the 144° slice
  • Half luna - the 108° slice (but maybe "half" is not a good word here)
  • Crescent luna - the 72° slice
  • New luna - the 36° slice

It just occured to me that you mentioned in that cited post of yours that you even specify how the rotundae are aligned. In W1 and W3 the respective pentagons are aligned at the same decagon edges, while in W2 and W4 they are placed alternatingly.

You can see this in the ico lunae too. In the bisected ico, the cuboctahedron can be analysed as two 3cups in alternating alignment, in oct||3cup, the 3cups are in ortho alignment, then in 3gon||3cup they are again in alternating alignment.

[...]

The rest I will answer later, I have to go right now.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 15, 2013 8:12 pm

quickfur wrote:[...]
I'm not sure I like using the words "huge" and "big"; maybe we can use lunar phase terminology instead? Gibbous, half, crescent, new.
  • Full luna - bisected 600-cell
  • Gibbous luna - the 144° slice
  • Half luna - the 108° slice (but maybe "half" is not a good word here)
  • Crescent luna - the 72° slice
  • New luna - the 36° slice

Well, being used with small and great qualifiers in polytopal names (small rhombicuboctahedron = x3o4x), great rhombicuboctahedron = x3x4x), english (i.e. native tungue) adjectives are familiar here. But those are somehow pre-used for that purpose, so I don't like to overload them even more. This is why I used big instead of great. And small even could be more precisely given in here by narrow. (The other ones were just fitting in.)

As to your proposed moon-qualifiers marek14 already commented in reply to Wendy.

Sure, we also could well use the very degrees themself for distinction: 144°-luna, 108°-luna, 72°-luna, 36°-luna, 120°-luna, 60°-luna, 90°-luna. (Note that these 7 ones even could survive without the base polychoron name, but I think if context isn't too clear, those should remain in place.)

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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 15, 2013 9:08 pm

Klitzing wrote:
quickfur wrote:[...]
I'm not sure I like using the words "huge" and "big"; maybe we can use lunar phase terminology instead? Gibbous, half, crescent, new.
  • Full luna - bisected 600-cell
  • Gibbous luna - the 144° slice
  • Half luna - the 108° slice (but maybe "half" is not a good word here)
  • Crescent luna - the 72° slice
  • New luna - the 36° slice

Well, being used with small and great qualifiers in polytopal names (small rhombicuboctahedron = x3o4x), great rhombicuboctahedron = x3x4x), english (i.e. native tungue) adjectives are familiar here. But those are somehow pre-used for that purpose, so I don't like to overload them even more. This is why I used big instead of great. And small even could be more precisely given in here by narrow. (The other ones were just fitting in.)

On second thought, I don't really like the lunar phase names anymore, because they are misleading: full luna seems to imply the entire polytope (just as full moon means the entire disk of the moon is visible), but it's actually only half! And gibbous implies the disk is greater than half, but all of these lunae are already less than half the polytope.

OTOH, I think I like "narrow". So maybe a better sequence would be: narrow for the smallest luna of a given polytope (so "narrow 600-cell luna" == 36° luna, "narrow 24-cell luna" == 3gon||3cup, etc.). Wide for the largest luna smaller than the bisected polytope (so wide 600-cell luna == 144° slice, wide 24-cell luna == oct||3cup). Not sure what adjectives to use in-between, though.

Or maybe, to keep names short and to make the exact sequence clear, we can number them in order: 1-luna = smallest luna, up to n-luna for the largest. So for 600-cell:
- 1-luna = 36°
- 2-luna = 72°
- 3-luna = 108°
- 4-luna = 144°
- This also has the nice coincidence of 5-luna = 180° = bisected 600-cell. And the number prefix indicates how many 1-lunae can be glued together to make that luna.

Then 24-cell lunae are:
- 1-luna = 3gon||3cup
- 2-luna = oct||3cup
- 3-luna = oct||cuboct = bisected 24-cell

The full names would be "hex 1-luna", "ico 2-luna", etc..

[...]Sure, we also could well use the very degrees themself for distinction: 144°-luna, 108°-luna, 72°-luna, 36°-luna, 120°-luna, 60°-luna, 90°-luna. (Note that these 7 ones even could survive without the base polychoron name, but I think if context isn't too clear, those should remain in place.)[...]

Yeah I think they should remain in place, otherwise it's unclear what the base polytope is unless you knew beforehand.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 15, 2013 9:17 pm

Klitzing wrote:[...]
- Now coming back again to the above intro: None the less, those two used equatorial sectioning hyperplanes surely have normals. And those normals shall mark some specific pair of points on the boundary of the original polychoron. Seems that I was misslead that those are vertices in any case... (Have not looked into that deeper so far. So you might have a run. :) )

I think there may be other CRFs that can be produced by other cuttings that I haven't considered. So far, the only cuttings covered are those that derive from a bisection that produces a CRF-bisectable cell, like oct (=2*4pyr), cuboct (=2*3cup) and o5x3o (=2*pentagonal rotunda). The idea being that a bisection with a bisectable base cell can be further bisected with another hyperplane, as long as the second hyperplane also produces a bisectable cell, so the resulting luna will have two halves of the two base cells.

But maybe there are other possibilities? What about cutting with 3 hyperplanes in a triangular arrangement? Maybe there are no CRF cuttings with 2 hyperplanes in a particular orientation, but it might be possible to make it CRF by introducing a 3rd (or 4th, etc.) hyperplane?

[...]I can see that deleting this face will result in trigon||3cup (or, equivalently, 6gon||oct, as you mentioned). So this is the second luna.[...]
[...]So the only possible lunae for ico are oct||3cup and oct||6gon (==3gon||3cup).[...]

You should be careful in that statement likewise, hehe: there are 2 different ways to align a trigon as opposite base to a tricu: para-align or gyro-align. Para-alignment results in 6g||trip, what is not being useful here, while gyro-alignment results in the here required 6g||oct.[...]

Hehe, you are right. So I should write 6g||oct instead of the ambiguous 3g||3cup. Is there a way to denote gyro alignment of the 3g w.r.t. 3cup?
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 16, 2013 11:39 am

quickfur wrote:[...]
Hehe, you are right. So I should write 6g||oct instead of the ambiguous 3g||3cup. Is there a way to denote gyro alignment of the 3g w.r.t. 3cup?

In my segmentochoron listing I just used 3g||3cup resp. 3g||gyro 3cup.
But you might take refuge to Johnson style naming (cf. J27), i.e. using an ortho prefix in the first case for more specificality.

In Wendy's lace simplex style those are btw. xxx3oxo&#x resp. xxo3oxx&#x.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 16, 2013 11:59 am

quickfur wrote:[...]
Or maybe, to keep names short and to make the exact sequence clear, we can number them in order: 1-luna = smallest luna, up to n-luna for the largest. So for 600-cell:
- 1-luna = 36°
- 2-luna = 72°
- 3-luna = 108°
- 4-luna = 144°
- This also has the nice coincidence of 5-luna = 180° = bisected 600-cell. And the number prefix indicates how many 1-lunae can be glued together to make that luna.

Then 24-cell lunae are:
- 1-luna = 3gon||3cup
- 2-luna = oct||3cup
- 3-luna = oct||cuboct = bisected 24-cell

The full names would be "hex 1-luna", "ico 2-luna", etc..
[...]

This number prefix somehow could be understood to imply a relation towards addition. I.e. a 2-luna might be kind of 2 1-luna combined.
- So I did a quick cross-check whether at least those alignments resp. gyro-alignments of equator-remainders would support this: Odd numbers should relate to gyro, even numbers to ortho (in the reading of J27). And indeed they do! The mere angle sizes do as well. And the total counts of non-equatorial elements (counting as fractional parts if required) likewise.

Thus this seems to work... (Would you mind to re-prove that independantly?)
I.e. the elemental building unit is just the 1-luna each!

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 16, 2013 3:47 pm

Klitzing wrote:
quickfur wrote:[...]
Or maybe, to keep names short and to make the exact sequence clear, we can number them in order: 1-luna = smallest luna, up to n-luna for the largest. So for 600-cell:
- 1-luna = 36°
- 2-luna = 72°
- 3-luna = 108°
- 4-luna = 144°
- This also has the nice coincidence of 5-luna = 180° = bisected 600-cell. And the number prefix indicates how many 1-lunae can be glued together to make that luna.

Then 24-cell lunae are:
- 1-luna = 3gon||3cup
- 2-luna = oct||3cup
- 3-luna = oct||cuboct = bisected 24-cell

The full names would be "hex 1-luna", "ico 2-luna", etc..
[...]

This number prefix somehow could be understood to imply a relation towards addition. I.e. a 2-luna might be kind of 2 1-luna combined.
- So I did a quick cross-check whether at least those alignments resp. gyro-alignments of equator-remainders would support this: Odd numbers should relate to gyro, even numbers to ortho (in the reading of J27). And indeed they do! The mere angle sizes do as well. And the total counts of non-equatorial elements (counting as fractional parts if required) likewise.

Thus this seems to work... (Would you mind to re-prove that independantly?)
I.e. the elemental building unit is just the 1-luna each!
[...]

Yes, I first noticed this when I was studying the 600-cell lunae. If you glue two 600-cell 1-lunae together, you get a non-convex polychoron which can be made CRF by inserting 5-tetrahedron florets (i.e. circles of 5 tetrahedra sharing an edge) where the pentagonal pyramids meet. The result is actually the same as a 600-cell 2-luna. And this construction is general; so 5 1-lunae joined together form the bisected 600-cell, and 10 1-lunae form the entire 600-cell. Conversely, the 600-cell can be cut into 10 1-lunae (after deleting a number of 5-tetrahedron florets).

And from my projections, it seems that this is the case for the 24-cell lunae too. The 2-luna can be cut into two 1-lunae, and in this case no extra insertions are necessary. So the entire 24-cell can be cut exactly into 6 1-lunae, or conversely, 6 copies of oct||6gon can be glued together to form a 24-cell.
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Jan 17, 2013 12:09 pm

quickfur wrote:
Klitzing wrote:
quickfur wrote:[...]
Or maybe, to keep names short and to make the exact sequence clear, we can number them in order: 1-luna = smallest luna, up to n-luna for the largest. So for 600-cell:
- 1-luna = 36°
- 2-luna = 72°
- 3-luna = 108°
- 4-luna = 144°
- This also has the nice coincidence of 5-luna = 180° = bisected 600-cell. And the number prefix indicates how many 1-lunae can be glued together to make that luna.

Then 24-cell lunae are:
- 1-luna = 3gon||3cup
- 2-luna = oct||3cup
- 3-luna = oct||cuboct = bisected 24-cell

The full names would be "hex 1-luna", "ico 2-luna", etc..
[...]

This number prefix somehow could be understood to imply a relation towards addition. I.e. a 2-luna might be kind of 2 1-luna combined.
- So I did a quick cross-check whether at least those alignments resp. gyro-alignments of equator-remainders would support this: Odd numbers should relate to gyro, even numbers to ortho (in the reading of J27). And indeed they do! The mere angle sizes do as well. And the total counts of non-equatorial elements (counting as fractional parts if required) likewise.

Thus this seems to work... (Would you mind to re-prove that independantly?)
I.e. the elemental building unit is just the 1-luna each!
[...]

Yes, I first noticed this when I was studying the 600-cell lunae. If you glue two 600-cell 1-lunae together, you get a non-convex polychoron which can be made CRF by inserting 5-tetrahedron florets (i.e. circles of 5 tetrahedra sharing an edge) where the pentagonal pyramids meet. The result is actually the same as a 600-cell 2-luna. And this construction is general; so 5 1-lunae joined together form the bisected 600-cell, and 10 1-lunae form the entire 600-cell. Conversely, the 600-cell can be cut into 10 1-lunae (after deleting a number of 5-tetrahedron florets).

And from my projections, it seems that this is the case for the 24-cell lunae too. The 2-luna can be cut into two 1-lunae, and in this case no extra insertions are necessary. So the entire 24-cell can be cut exactly into 6 1-lunae, or conversely, 6 copies of oct||6gon can be glued together to form a 24-cell.


... the former wedge-facets (peros (pentagonal rotundae, tricues, ...) have to be blended out for sure.
In case of ico-lunae the co-realmic squippies have to be combined to octahedra too (similar to your "florets" re-substitution for ex-lunae).
- But yes, this was what I meant when I stated about the net count on fractional parts of non-equatorial elements.

Good example for a well-chosen nomenclature which even provides additional insight. :)
So we could stay with that for now, I think.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Jan 17, 2013 12:22 pm

Klitzing wrote:So we could stay with that for now, I think.


Minor addendum:
we might even use fractions instead of mere numbers:
(This even codes the wedge face angle additionally and is not too much overdue, I think.)
  • ex 1/10-luna
  • ex 2/10-luna = ex 1/5-luna
  • ex 3/10-luna
  • ex 4/10-luna = ex 2/5-luna
  • ex 5/10-luna = ex 1/2-luna = 1/2 ex
  • ico 1/6-luna
  • ico 2/6-luna = ico 1/3-luna
  • ico 3/6-luna = ico 1/2-luna = 1/2 ico
  • hex 1/4-luna
  • hex 2/4-luna = hex 1/2-luna = 1/2 hex
--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Jan 18, 2013 9:53 am

Just for annotation:
In a different mail list archive (polyhedron list) hedrondude meanwhile accepted the new OBSAs
  • cyted srit (= cyclotetradiminished small rhombitesseract)
  • cyte gysrit (= cyclotetragyrated small rhombitesseract)
  • bicyte gysrit (= bicyclotetragyrated small rhombitesseract)
(cf. this post, where the respective polychora have been described in detail)
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Jan 18, 2013 10:23 am

Klitzing wrote:[...]
This btw. should give rise to a further finding: there should be a compound of 3 odips, which is vertex inscribed into srit.
Further, like there is a compound of 3 tes, vertex inscribed into ico (having 2 such tes incident to any ico vertex), there clearly is also a corresponding compound of 3 srit, vertex inscribed into spic. And by the above there should be then a compound of 3x3=9 odip, vertex inscribable into spic, as well! (Or could you visualize whether some of those odip therein might coincide? - Without going into coordinate calculations: as mentioned above, 2 tes are incident at any ico vertex each. Thus it seems that at most 2 corresponding odip might coincide. But 9 is not dividable by 2. So it looks to me unlikely; i.e. that 9-odip-compound should exist as well.)

--- rk


In a different mail list archive (polyhedron list) I meanwhile discussed with hedrondude about those 2 shapes. Those sadly would come out to be somehow Grünbaumian: i.e. have coincident lower-dimensional elements.

The same can be observed already within 3D: just consider the compound of 3 ops in cubical symmetry. Those can be inscribed into sirco, do not use the triangles thereof, instead add those octagons, but will use the 6 squares in cubical directions twice each. But we could go for a blend instead, i.e. withdrawing the doubled up faces (canceling out), and reconecting the thus open, likewise coincident edges accordingly. This results in 3D in what has as OBSA sroh (small rhombihexahedron).

Hedrondude now annotated that the same rescue could be done in 4D as well:

Thus the 3 odip compound (within the srit regiment) gives rise for a blend which has OBSA garpit (grand rhombic prismatotesseract). That one has 8 srohs and 24 ops for cells. (Cf.: each odip has 16 ops, so the 3 odip compound would have 3*16=48 ops. Each sroh stands for 3 ops. Thus 8*3+24=48 ops again.)

And the 9 odip compound (within spic regiment) gives rise for a blend which has OBSA sirc (small retrotetracontoctachoron). That one has nothing but 48 srohs for cells. (Cf.: the 9 odip compound has 9*16=144 ops. Here we have 3*48=144 ops again.)

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 18, 2013 3:43 pm

Klitzing wrote:
Klitzing wrote:So we could stay with that for now, I think.


Minor addendum:
we might even use fractions instead of mere numbers:
(This even codes the wedge face angle additionally and is not too much overdue, I think.)
  • ex 1/10-luna
  • ex 2/10-luna = ex 1/5-luna
  • ex 3/10-luna
  • ex 4/10-luna = ex 2/5-luna
  • ex 5/10-luna = ex 1/2-luna = 1/2 ex
  • ico 1/6-luna
  • ico 2/6-luna = ico 1/3-luna
  • ico 3/6-luna = ico 1/2-luna = 1/2 ico
  • hex 1/4-luna
  • hex 2/4-luna = hex 1/2-luna = 1/2 hex
--- rk

I like the idea of using fractional notation, but I'm not sure about using n/10 for hex lunae and n/6 for ico lunae, because the notation suggests that perhaps 9/10 lunae are possible, whereas n only spans up to 10/2=5 for hex and 6/2=3 for ico. Would it be too misleading to write them as 1/5-luna, 2/5-luna, etc., instead? I.e., reduce the "denominator" by half.
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Jan 18, 2013 5:05 pm

Not at all missleading, quickfur!
A 9/10-luna of ex surely exists. Just that it no longer is a convex one. (It is just the ex minus one 1/10-luna. I.e. its dihedral angle thus would be 360°-36° = 324° = 9/10 x 360°.) :P

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 18, 2013 5:28 pm

Klitzing wrote:Not at all missleading, quickfur!
A 9/10-luna of ex surely exists. Just that it no longer is a convex one. (It is just the ex minus one 1/10-luna. I.e. its dihedral angle thus would be 360°-36° = 324° = 9/10 x 360°.) :P

--- rk

You're right, I was thinking of CRFs, but non-convex lunae are also valid shapes, indeed. So n/m lunae are CRF only when n ≤ m/2.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 22, 2013 4:54 pm

Klitzing wrote:[...]
Hedrondude now annotated that the same rescue could be done in 4D as well:

Thus the 3 odip compound (within the srit regiment) gives rise for a blend which has OBSA garpit (grand rhombic prismatotesseract). That one has 8 srohs and 24 ops for cells. (Cf.: each odip has 16 ops, so the 3 odip compound would have 3*16=48 ops. Each sroh stands for 3 ops. Thus 8*3+24=48 ops again.)

And the 9 odip compound (within spic regiment) gives rise for a blend which has OBSA sirc (small retrotetracontoctachoron). That one has nothing but 48 srohs for cells. (Cf.: the 9 odip compound has 9*16=144 ops. Here we have 3*48=144 ops again.)

--- rk


Got just aware of the related non-convex conjugates as well - just to be mentioned here for record (though not really being concerned with that thread topic):

There is a Grünbaumian compound of 3 stodips, where again some squares would be coincident. That one would be in the wavitoth regiment. Blending out those doubled up squares would result in what is known to be the gaqript (cf. the same link to hedrondudes website, as given for the convex one). That blend would consist of 24 stop and 8 groh.

And there is a similar Grünbaumian compound of 9 stodips, that one in giddic regiment. Blending out those doubled up squares would result in what is known to be the girc (cf. the same link to hedrondudes website, as given for the convex one). That blend would consist of 48 groh only. (I.e. like its conjugate, sirc, a noble polychoron.)

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Sat Feb 09, 2013 6:58 pm

One more thought on lunae.

(Cf. that post and that post for introduction and pictures of the 4 lunae of the hexacosachoron. Cf. additionally that post for their current naming.)

Consider the lace city display of the hexacosachoron (ex).
Code: Select all
                 o5o           o5o                 
                        o5x                       
                                                   
            x5o                     x5o           
                                                   
     o5o                f5o                o5o     
                 o5f           o5f                 
                                                   
     o5x                                   o5x     
            f5o                     f5o           
                                                   
o5o                     x5x                     o5o
                                                   
            o5f                     o5f           
     x5o                                   x5o     
                                                   
                 f5o           f5o                 
     o5o                o5f                o5o     
                                                   
            o5x                     o5x           
                                                   
                        x5o                       
                 o5o           o5o                 


Thus it would be obvious how the several lunae would be displayed as lace cities:
1/10-luna of ex
Code: Select all
    o5o       
           o5x
              
x5o           
              
           f5o
    o5f       
              
              
              
              
           x5x
Here both, the bistratic pentagonal rotundae (pero = x5o || o5f || x5x = xox5ofx&#xt), as well as the dihedral angle between those (here 36°) can be seen very well.

2/10-luna of ex
Code: Select all
            o5o       
                   o5x
                      
       x5o           
o5o                   
                   f5o
            o5f       
                      
o5x                   
       f5o           
                      
                   x5x
(Dihedral angle between those peros obviously is 72°.)

3/10-luna of ex
Code: Select all
                 o5o       
                        o5x
                          
            x5o           
     o5o                   
                        f5o
                 o5f       
                          
     o5x                   
            f5o           
                          
o5o                     x5x
                          
            o5f           
     x5o                   
(Dihedral angle between those peros obviously is 108°.)

4/10-luna of ex
Code: Select all
                 o5o       
                        o5x
                          
            x5o           
     o5o                   
                        f5o
                 o5f       
                          
     o5x                   
            f5o           
                          
o5o                     x5x
                          
            o5f           
     x5o                   
                          
                 f5o       
                          
     o5o                   
            o5x           
(Dihedral angle between those peros obviously is 144°.)

Similarily the vertex-first rotunda of ex, i.e. the 5/10-luna of ex, could be displayed:
Code: Select all
                 o5o       
                        o5x
                          
            x5o           
     o5o                   
                        f5o
                 o5f       
                          
     o5x                   
            f5o           
                          
o5o                     x5x
                          
            o5f           
     x5o                   
                          
                 f5o       
                        o5f
     o5o                   
            o5x           
                          
                        x5o
                 o5o       
Here the dihedral angle between the bounding peros would amount to 180°. This displays moreover how these 2 peros would become co-realmic and will be unified to an icosidodecahedron (id), the equatorial section of ex. In the display this id is shown as multistratic stack o5x || f5o || x5x || o5f || x5o = ofxox5xoxfo&#xt.

But there is a different lace city display of ex possible, derived by projection not along the orthogonal 5-fold symmetry, but along an orthogonal 3-fold one:
Code: Select all
                o3o               
                                  
        x3o   o3f f3o   o3x       
                                  
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
   f3o       o3F   F3o       o3f   
                                  
                                  
o3x   x3f F3o   f3f   o3F f3x   x3o
                                  
                                  
   f3o       o3F   F3o       o3f   
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
                                  
        x3o   o3f f3o   o3x       
                                  
                o3o               
(Here I used F = ff.) Again the equatorial id can be seen very well: o3x || x3f || F3o || f3f || o3F || f3x || x3o = oxFfofx3xfofFxo&#xt. Likewise it is visible that there would be similar 1/6- and 2/6-lunae of ex too. - But a second thought reveals that those won't be CRF (as the formers were): The intersecting equatorial hyperplanes would cut those ids into o3x || x3f || F3o || f3f = oxFf3xfof&#xt. That is, the bottom face would be a tau-scaled hexagon.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 12, 2013 10:10 am

Mrrl wrote:Looks like there is complete family of skew 3,n-duoprisms: you make combine two n-antiprisms with 1 n-prism, n tetrahedra and n square pyramids and get CRF body. [...]
Yep, those figures described in Mrrl's post from Tue Nov 22, 2011 2:26 am just are K.4.6 (n=3), K.4.14 (n=4), K.4.22 (n=5), K.4.46 (n=6), K.4.58 (n=8), K.4.93 (n=10), and K.4.174 (other n) in my paper on convex segmentochora from 2000.

[...] Contructions based on 4,n rings give nothing new: if we take 2 prisms and 2 antiprsms, we get antiprismatic prism that is uniform, [...]
Those are K.4.9 (n=2), K.4.11 (n=3), K.4.19 (n=4), K.4.39 (n=5), K.4.53 (n=6), K.4.65 (n=8), K.4.96 (n=10), and K.4.176 (other n).

[...] and for 4 antiprisms I guess that resulting body should be either uniform or non-CRF. But we don't have such uniform polychora in the list, so there is no new CRF in that side. [...]
This is true. Only a single value of n results in an uniform figure: n=2, which is nothing but the well-known hexadecachoron.

--- rk
Last edited by Klitzing on Thu Feb 14, 2013 9:08 am, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 12, 2013 10:17 am

quickfur wrote:[...]
Mrrl wrote:But what is skew 3,3? It has one triangular prism, 3 square pyramids attached to its sides, 3 tetrahedra between them and the rest is filled by two octahedra. 6 cells, 9 vertices. Does it really exist?

I think so! Your cell list seems correct, and from what I can tell, it should be possible to make all edge lengths equal. But there's a danger that the square pyramids may be trapezoidal pyramids though. So this is not 100% confirmed yet.

And this post of Tue Nov 22, 2011 3:02 am just deals with K.4.6. It is CRF.

--- rk
Last edited by Klitzing on Thu Feb 14, 2013 9:09 am, edited 2 times in total.
Klitzing
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Feb 14, 2013 8:55 am

quickfur wrote:
quickfur wrote:I'm thinking about a possible CRF, but not sure if it's valid: start with a tetrahedron, then attach 4 elongated triangular cupola, and insert triangular prisms at the elongated part of the cupolae to close up the shape, then the bottom cell is a truncated octahedron. [...]

Actually, I just realized that a truncated octahedron is tiled by a tetrahedron and 4 triangular cupola (plus a few other pieces), so the cupola (elongated or otherwise) lie on a flat hyperplane. :(


Hy Quickfur, that post of yours from Fri Nov 25, 2011 7:21 am looks wrong.

In fact, it is oct || toe, which is flat: i.e. it has a squared height which calculates to be 0. Then the other cells will be 8 tricues and 6 squippies. - In fact, this is just the monostratic oct-first cap of the euclidean tiling of octs and coes, i.e. of o4x3o4o.

If you would stack tet || toe you would get a squared height, which calculates to be negative. - Nothing wrong with that. It just is a thingy of hyperbolic space. In fact, it looks to be the monostratic tet-cap of a non-compact hyperbolic tiling: It uses tets, completely surrounded by coes. The triangles of those coes are alternately adjoint to further tets of the same symmetry equivalence class (one tetrahedral class of triangles), respectively by infinite, euclidean triangle tilings (at the remaining 4 triangles). The squares of those coes adjoin to trips. The triangles of these trips again adjoin to those tilings. (If I'm not too wrong, this should be the hyperbolic tesselation x3o3x3o3*b.) - Accordingly that t be considered cap would deal with 1 tet, 4 tricues, 6 trips, and 4 flat hexagonal pyramids.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 14, 2013 5:17 pm

Klitzing wrote:
quickfur wrote:
quickfur wrote:I'm thinking about a possible CRF, but not sure if it's valid: start with a tetrahedron, then attach 4 elongated triangular cupola, and insert triangular prisms at the elongated part of the cupolae to close up the shape, then the bottom cell is a truncated octahedron. [...]

Actually, I just realized that a truncated octahedron is tiled by a tetrahedron and 4 triangular cupola (plus a few other pieces), so the cupola (elongated or otherwise) lie on a flat hyperplane. :(


Hy Quickfur, that post of yours from Fri Nov 25, 2011 7:21 am looks wrong.

In fact, it is oct || toe, which is flat: i.e. it has a squared height which calculates to be 0. Then the other cells will be 8 tricues and 6 squippies. - In fact, this is just the monostratic oct-first cap of the euclidean tiling of octs and coes, i.e. of o4x3o4o.

If you would stack tet || toe you would get a squared height, which calculates to be negative. - Nothing wrong with that. It just is a thingy of hyperbolic space. In fact, it looks to be the monostratic tet-cap of a non-compact hyperbolic tiling: It uses tets, completely surrounded by coes. The triangles of those coes are alternately adjoint to further tets of the same symmetry equivalence class (one tetrahedral class of triangles), respectively by infinite, euclidean triangle tilings (at the remaining 4 triangles). The squares of those coes adjoin to trips. The triangles of these trips again adjoin to those tilings. (If I'm not too wrong, this should be the hyperbolic tesselation x3o3x3o3*b.) - Accordingly that t be considered cap would deal with 1 tet, 4 tricues, 6 trips, and 4 flat hexagonal pyramids.

--- rk

I see. Well, I guess I didn't check whether the height was exactly zero, or negative. It was, in any case, not CRF in Euclidean space. :)
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