Johnsonian Polytopes

[Klitzing]

Alright, I've finally discovered a real, valid, CRF augment of a duoprism. [...]
[...]

I'd say this was analogous to the square orthobicupola, which can be seen as the convex hull of a cuboid and octagon. The fact that the 8-cell is regular in the 4D version would be because of the √4 = 2 coincidence.

Yeah it does resemble the square orthobicupola in some respects. Although in this case, you can't really decompose it into two cupolae. (Or can you??)

Sure you can!

Your finding is exactly what would be got by attaching those 2x K-4.73.1 (i.e. cube || {8}) at their octagons: You then omit all the 4 squacues (and the body of that octagon, which then won't connect to anything any longer). That is, you just keep the top (resp. the mirrored bottom) cube(s), the latterally attached trips, and the tets as partial complexes.

The now open boundaries will then be closed as follows. Note that the heights of K-4.73.1 were 1/2, thus the total height between the opposite cubes here equals 1! That is, you connect the 2 open faces of the original cubes by 2 further cubes, you connect the latterals of these ones to trips (used here as digonal cupolas) the lacings of which connect to the open faces of the trips of the former complex. And, at the lacing edges of the new cubes you attach tets, connecting 2 of their triangles to the new trips, and the other ones to the still open triangles of the tets of the former complex.

Note moreover, that the equatorial (body-less) octagon breaks the symmetry of the cubes in K-4.73.1. They just became 4-prisms. Likewise the 2 added cubes are similar 4-prisms (by their to be attached trips). Thus all 4 cubes (4-prisms) attach into a cycle of 4 (along that axial symmetry). And all fillings, i.e. either former K-4.73.1 and both added ones, are alike. This shows that you just could interchange "old" ones and "added" ones.

Now re-consider what this procedure was - placing it, per analogy, into a better seeable dimension.
Its kind of a figure like J91(= bilbiro = bilunabirotunda) : it re-uses partial shapes of known polyhedra. For J91 those were 2 parts of srid and 2 parts of id. But yours re-uses 4x the same part from just a single segmentochoron!

Nice indeed!

Finally, consider it as a lace city. Then your figure is just:

Code: Select all

x4o     x4o
          
    x4x   
          
x4o     x4o

This now shows, reading it along its diagonal, that it indeed should be describable as an ortho-stack of 2x "{4} || J28" (J28 = squobcu = square orthobicupola = x4o || x4x || x4o)!

You might ask, why then is "{4} || J28" not contained in my paper about segmentochora? Well, it surely is a possible monostratic figure with unit edges only. Just that its base, J28, misses to be orbiform (vertices on a single sphere). And so would your finding thus too miss that attribution - as your provided coordinates did show likewise:

[...]
apacs<1,1,1,1>
<±1, ±(1+sqrt(2)), 0, 0>
<±(1+sqrt(2)), ±1, 0, 0>

--- rk

[Klitzing]

This very lace city moreover shows further-on:

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x4o     x4o
           
    x4x   
           
x4o     x4o

This is quickfurs finding. And, depending on the direction of reading it could be alternatively been called the bistratic stack "cube || pseudo {8} || cube", or the bistratic stack "{4} || pseudo J28 || {4}".

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x4o     x4o
          
    x4x   
          
x4o       

This is that hemi version, which quickfur was asking for. Clearly it can be read as monostratic stack "{4} || J28". In fact it is kind a rotunda of his finding.

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x4o     x4o
           
    x4x   

And this then is again that segmentochoron, I started my description with, i.e. "cube || {8}" (or K-4.73.1).

In the view of our recent discussion, that last segmentochoron after all comes out to be a luna. In fact, without any further calculation of dihedral angles, this 1. city (that of quickfurs finding) shows directly, that the dihedral angle between those J4s (= squacues) of that luna would be 90°. Accordingly we have: K-4.73.1 = 1/4-luna of quickfurs finding!

BTW., "quickfurs finding" is a mere working title. We should find a better name for that CRF!

--- rk

[Marek14]

Hm, from that lace city description I like the name "polytope X", as the diagram is X-shaped. And it's basically (if I understand it correctly) an octagon in xy plane reaching out to four squares in z and w directions (Doctor Octopus?).

[Klitzing]

Hy Marek, "polytope X" does get the lace city description correctly, but others might have similar forms here (with different shapes in perp space). And it somehow conflicts with being a convex polychoron.

Up to your question: yes, cf. the provided coords (at the end of this post).

Other suggestions?

--- rk

[Marek14]

Hmm, somehow the description evokes in me the image of pinwheel or windmill, with four square "blades" mounted on the central octagon. Latin for "windmill" seems to be... ventimola?

[quickfur]

Visually speaking, it is basically a tesseract with a rim of alternating 3,4-duoprisms and line||squares, sorta like a disc with sharpened edges. Quite similar to the square bicupola in 3D, as Klitzing has noted. Except that due to the extra dimension, this is more of a "tetra-cupola" than a bicupola, as it consists of 4 lunae pasted together.

So maybe "cubical tetracupola"?

[Marek14]

Visually speaking, it is basically a tesseract with a rim of alternating 3,4-duoprisms and line||squares, sorta like a disc with sharpened edges. Quite similar to the square bicupola in 3D, as Klitzing has noted. Except that due to the extra dimension, this is more of a "tetra-cupola" than a bicupola, as it consists of 4 lunae pasted together.

So maybe "cubical tetracupola"?

Disc with sharpened edges? How about "shuriken"?

[quickfur]

Visually speaking, it is basically a tesseract with a rim of alternating 3,4-duoprisms and line||squares, sorta like a disc with sharpened edges. Quite similar to the square bicupola in 3D, as Klitzing has noted. Except that due to the extra dimension, this is more of a "tetra-cupola" than a bicupola, as it consists of 4 lunae pasted together.

So maybe "cubical tetracupola"?

Disc with sharpened edges? How about "shuriken"?

Shuriken are non-convex, though.

[Marek14]

Visually speaking, it is basically a tesseract with a rim of alternating 3,4-duoprisms and line||squares, sorta like a disc with sharpened edges. Quite similar to the square bicupola in 3D, as Klitzing has noted. Except that due to the extra dimension, this is more of a "tetra-cupola" than a bicupola, as it consists of 4 lunae pasted together.

So maybe "cubical tetracupola"?

Disc with sharpened edges? How about "shuriken"?

Shuriken are non-convex, though.

How about "lens"?

[quickfur]

[...]
How about "lens"?

Hmm, I like that! What should the naming convention be, in this case? cubical lens? or tesseractic lens?

[Marek14]

I might actually use the term "octagonal lens"...

[quickfur]

I might actually use the term "octagonal lens"...

The problem with that, is that it doesn't tell you anything about the alternating prism/tetrahedron structure around the rim of the lens.

Anyway, I just thought of another approach in classifying this shape: this particular lens can be thought of as a modified Stott expansion in which the expansion only takes place along some axes but not others. Start with a 16-cell, and expand along 1 axis, and you get the elongated 16-cell (i.e., elongated octahedral bipyramid). Expand along 2 axes, and you get this lens. Expand along 3 axes, and you get a kind of shrunken cantellated tesseract x4o3x3o with its middle section cut out. Expand along 4 axes, and you get the full cantellated tesseract. So you could think of this lens as one of the intermediates between a 16-cell and a cantellated tesseract. A sort of "partially-expanded" 16-cell, if you will.

Maybe the name should be derived along these lines instead?

[Marek14]

I might actually use the term "octagonal lens"...

The problem with that, is that it doesn't tell you anything about the alternating prism/tetrahedron structure around the rim of the lens.

Anyway, I just thought of another approach in classifying this shape: this particular lens can be thought of as a modified Stott expansion in which the expansion only takes place along some axes but not others. Start with a 16-cell, and expand along 1 axis, and you get the elongated 16-cell (i.e., elongated octahedral bipyramid). Expand along 2 axes, and you get this lens. Expand along 3 axes, and you get a kind of shrunken cantellated tesseract x4o3x3o with its middle section cut out. Expand along 4 axes, and you get the full cantellated tesseract. So you could think of this lens as one of the intermediates between a 16-cell and a cantellated tesseract. A sort of "partially-expanded" 16-cell, if you will.

Maybe the name should be derived along these lines instead?

So, if there are two axes of expansion... a dioptric lens?

[quickfur]

[...]
So, if there are two axes of expansion... a dioptric lens?

Haha, nice pun! Might sound too confusing for someone who doesn't know what it is, though!

[Klitzing]

Just re-read your post from over a year back, where you described it first. There it is introduced as a partial augmentation of the tesseract. And as a tesseract is nothing but a 4,4-duoprism, this observation is more than correct here: it is that one ring of cubes gets augmentet (and thus is blended out) while the other ring of 4 cubes remains intact.

So we should have a fair try for cyclo-augmented square-duoprism. - What's your opinion?

--- rk

[quickfur]

Just re-read your post from over a year back, where you described it first. There it is introduced as a partial augmentation of the tesseract. And as a tesseract is nothing but a 4,4-duoprism, this observation is more than correct here: it is that one ring of cubes gets augmentet (and thus is blended out) while the other ring of 4 cubes remains intact.

So we should have a fair try for cyclo-augmented square-duoprism. - What's your opinion?

--- rk

That's what I was thinking as well, except that we have an ambiguity with the 24-cell diminishing created by augmenting a tesseract with 4 cube-pyramids. The problem is that this particular augmentation is not the same as your traditional augmentation, because adding the augments by themselves produces a non-convex shape, so you have to insert the tetrahedral pieces between the square-pyramid-prism augments to make the result convex again.

But since a ring is added, maybe we can call it a ring-augmented tesseract?

Interestingly enough, it's actually possible to add a second ring of augments to the 4 unaugmented cubes; I'm pretty sure the result is non-convex, but it could qualify as a bi-ring-augmented tesseract.

Also, IIRC, I did do some research on whether other duoprisms can be augmented in this way, but I think my conclusion was that they are either non-CRF or coincide with existing uniforms. But I only tried it with a few augment shapes; maybe there's a way to make an interesting CRF using an unusual augment shape.

[quickfur]

[...]
But since a ring is added, maybe we can call it a ring-augmented tesseract?
[...]

Or maybe a rimmed tesseract, since we're adding a rim around it?

[Klitzing]

Though I understand, what you are after by "rim" here, in the context of the tesseract this is not the obvious understanding of that word.
The "ring" is clearly understood as being one of its orthogonal cycles, I suppose. - Thus I'd vote for "ring-augmented tesseract".

In fact, "ring" here is not too different from my "cyclo". And I used "4-duoprism" rather than "tesseract" intentionally, to push that idea of 2 orthogonal rings, hehe. - But I understand your intend to distinguish a mere augmentation of any individual facet of a cycle from such an aumentation, which produces a complete ring.

--- rk

[Marek14]

Can other duoprisms be augmented in the same way?

[quickfur]

Can other duoprisms be augmented in the same way?

I've tried to look for them, but I was unsuccessful. But I'm not 100% sure that they don't exist, just that I was unable to find them.

[Klitzing]

The answer here should be quite easy.

Consider the heights H_n of these 2n-g || n-p wedges on the one hand, and consider the inradii rho_m of the regular m-gons on the other hand. According to that lace city display these should match, if an n,m-duoprism would be augmentable in that very sense of consideration.

We know already that H_4 = 1/2 = rho_4, and so were talking about that ring-augmented 4,4-duoprism.

A quick search braught up the following:
H_5 = sqrt[(5-2*sqrt(5))/20] = rho_5/2; therefore there should be a similar ring-augmented 5,5/2-duoprism:

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       x5o       
                
x5o           x5o
       x5x       
                
                
   x5o     x5o   

That one should work quite similarily as the 4,4-duoprism ring-augmentation.
Just that it most probably, right because of that {5/2} would not be convex any more.

Note that in the former case the x4o || x4o (cubes) were arranged at the hull of that square display. In this lace city the x5o || x5o (pips) are arranged along the edges of an inscribed pentagram.

You could even spot the trips as

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x_.           x_.
       x_.       

respectively the tets as

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._o           ._o
       ._x       

Returning to the lune concept in that case, we would spot the segmentochoron K-4.154.1 (i.e. {10} || pip) in here at

Code: Select all

       ._.       
                 
x5o           x5o
       x5x       
                 
                 
   ._.     ._.   

Thus we could read off that in this segmentochoron the dihedral angle between the 5-cupolae (pecues) is 360° * 2/5 = 144°. (For sure, the pecues again will be blended out in this ring-augmented duoprism, just as the squacues were in that 4,4-duoprism one.)

--- rk

[quickfur]

There is also the possibility of more complex ring structures, i.e., not necessarily just using n-prism||line augments, but some other kind of augments, with other shapes (possibly more than one) inserted in between to make the result convex.

Marek & myself have independently enumerated all CRF duoprism augmentations with n-prism pyramids, but we didn't consider the possibility that some of the non-convex augmentations might possibly be made CRF by adding additional pieces. There might be some hitherto unsuspected CRFs out there in this direction.

I've also realized that every (n-prism pyramid)-augmented m,n-duoprism, there exists a corresponding augment of the m,2n-duoprism with 2n-prism||n-prism augments, and many more besides (because of the new possibility of augments separated by odd numbers of duoprism cells, which were not possible in the m,n-duoprism).

But still, there are plenty of other augmentation possibilities, esp. if you consider that some of them might be bridgeable by other pieces to make them CRF. Which, btw, leads to another possible name for the ring-augmented tesseract: we could call it the bridged cyclo-tetraaugmented tesseract, since extra pieces are inserted to bridge the augments into a ring.

[Klitzing]

Your bridged augmentation idea is cool. But the name then becomes again longuish. Further it would not be clear how to get that into a Bowers style acronym - perhaps "bri-cy-t-au tes"? Esp. that "bri" part is not much of an acronym any more.

But just came up with a quite different description. In fact one which is starting from the lunes thereof:

Consider the complex of 5 tets around an edge (having the same hyperspherical curvature than ex), being then coverable by 2 peppies. (As segmentochoron nothing but: line||ortho {5}.) That several times was informaly just been called a 5 tet rosette. Even so we here do not need to go up one dimension to embed that spherical structure into an euclidean space, as lunes already do connect in the same euclidean spaces, we could lend that term: rosettes.

Thus your "cube||{8}||cube" would be nothing but a "4 {8}||cube rosette" or just a "quadratic octagon-cube-wedge rosette". Esp. this description lends itself to my just found bridged augmentation of the starpedip (= 5,5/2-duoprism): "pentagrammic decagon-pip-wedge rosette" (pip = 5-prism).

Assuming that octagon-cube part resp. that decagon-pip part being implicite, we could even reduce that to "quadratic wedge rosette" resp. "pentagrammic wedge rosette". Thus being straight forward for lending an OBSA: "(s)quawros" and "stawros", 2 nice new (I suppose) scaliforms for Jonathans listing.

Edit: clearly not scaliforms.

--- rk

[Marek14]

I thought of this possible CRF -- not sure if anyone mentioned this before, maybe *I* did, heck if I know:

Take prip (prismatotruncated pentachoron). It consists of 5 truncated tetrahedra, 10 triangular prisms, 10 hexagonal prisms, and 5 cuboctahedra. A truncated tetrahedron cap (truncated tetrahedron || truncated octahedron) can be cut off. This cap has, as lateral sides, four hexagonal prisms, four triangular prisms and four triangular cupolas.

Now, it should be possible to gyrate this cap and glue it back wrong, switching the two types of hexagons on the truncated octahedral cut. This should make four cuboctahedra and eight hexagonal prisms into eight elongated triangular cupolas, right?

[Klitzing]

I thought of this possible CRF -- not sure if anyone mentioned this before, maybe *I* did, heck if I know:

Take prip (prismatotruncated pentachoron). It consists of 5 truncated tetrahedra, 10 triangular prisms, 10 hexagonal prisms, and 5 cuboctahedra. A truncated tetrahedron cap (truncated tetrahedron || truncated octahedron) can be cut off. This cap has, as lateral sides, four hexagonal prisms, four triangular prisms and four triangular cupolas.

Now, it should be possible to gyrate this cap and glue it back wrong, switching the two types of hexagons on the truncated octahedral cut. This should make four cuboctahedra and eight hexagonal prisms into eight elongated triangular cupolas, right?

Haha, yes you did it yourself way back on Sat Nov 26, 2011 9:58 pm!

[...] truncated tetrahedron||truncated octahedron (truncated tetrahedron cut off prismatorhombated pentachoron)
the rest of the body gives a shape made of 1 cuboctahedron, 4 truncated tetrahedra, 6 hexagonal prisms, 4 triangular prisms, 4 triangular cupolas and 1 truncated cuboctahedron. I suspect it might be possible to glue these two parts of prismatotruncated pentachoron together the "wrong way", so that the the triangular cupolas will be joined to hexagonal prisms and truncated tetrahedra to triangular cupolas instead of combining the cupolas to form cuboctahedra. Not sure if the joins won't blend (into augmented truncated tetrahedra and/or elongated triangular cupolas), but either way, it should be a possible shape! [...]

Did you meanwhile check, whether this gyration is truely convex? And that the hips and tricues moreover become co-realmic, so to combine into etcues?

--- rk

[Marek14]

Didn't check it, unfortunately... Not sure how -- I'm more of a combinatory kind of person than calculatory one

[Klitzing]

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

Comes a bit late, I think, my answer to your question of Tue Mar 27, 2012 11:10 pm

Never mind: sure it is!

--- rk

[Klitzing]

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

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

As for "prissi" (prismatorhombated snub icositetrachorn), this was one of mine findings too
It was in September 2005 that I applied my extended snubbing (aka alternated faceting) onto "prico" (prismatorhombated icositetrachoron), alternatingly maintaining resp. rejecting (i.e. chopping off) its triangles. Therefore prissi has the Dynkin diagram s3s4o3x.

Doing so directly, using the coordinates of prico, you would get prissi in a topological variant with edges of sizees 1 : sqrt(2) : sqrt(3). Even so it lends to resizement, and then becomes uniform again.

Wendy later came up with the Stott addition. Then it could be understood as s3s4o3o + o3o4o3x, that is, the icosahedra of sadi will be shifted 1 edge length apart.

Both advises give clues for the coordinates. The first one uses those of prico, applies that alternated faceting, and then has to feed the coords into some heating/cooling procedure like Jim McNeills Hedron software (sadly only for 3D). - The other one uses the exact coords of sadi and ico, and then has to combine (add) those appropriately. (So this is theoretically evident, as these symbols provide a vector basis, the actual application, i.e. which vector has to be added to which other one, can be rather teddious.)

Jonathan much later mentions (at his webpage) that prissi should be derivable from a non-convex uniform polychoron named padohi (x3o5o5/2x) by chopping off all (1440) but 288 vertices, if that helps...

I can provide you the link to my incmats page. But coords I've not calculated so far. At least, its radius is given.

--- rk

[quickfur]

Your bridged augmentation idea is cool. But the name then becomes again longuish. Further it would not be clear how to get that into a Bowers style acronym - perhaps "bri-cy-t-au tes"? Esp. that "bri" part is not much of an acronym any more.

I've just thought of another CRF that is trivially derived from my cube||octagon||cube: instead of starting from the tesseract as a 4,4-duoprism, start with an 8,4-duoprism. Then augment the 4 octagonal prisms with 8prism||cube, and bridge them with 4cup||square pieces. This is just the (modified) Stott expansion of cube||octagon||cube around the axis of one of the duoprism rings, so it retains all previous dichoral angles, and is therefore CRF (I think?).

If I'm not wrong, I believe the cells should be:
- 8 cubes in one ring
- 4*4 triangular prisms + 4*(4+1) cubes for each octagonal prism augment = 16 triangular prisms + another 20 cubes
- 4*1 cubes + 4*4 tetrahedra + 4*4 triangular prisms for each bridging piece = 4 more cubes + 16 tetrahedra + 16 more triangular prisms
So the total should be 32 cubes, 32 triangular prisms, and 16 tetrahedra.

How should we name this shape?

[Klitzing]

[...]
How should we name this shape?

Extended it is, kind of:

Code: Select all

     x4o    x4o     
                   
x4o  x4x    x4x  x4o
                   
                   
                   
x4o  x4x    x4x  x4o
                   
     x4o    x4o     

That one reminded me of having already seen ...
I searched a bit, having the hint of 16 tets, and found where it was:

We should name that new finding of yours simply

• "sidpith" or
• "small disprismatotesseractihexadecachoron" or
• "runcinated tesseract" or
• "x3o3o4x",
• ...

--- rk