Klitzing wrote:Well, in cases where the chopped off caps at most touch, the cap clearly might be re-attached in a gyrated way (if at all possible, cf. below). But surely not if those intersect, what they do here.
Further there then has to be checked additional, whether the dihedral angles at those (re-)connections still are convex, else they disqualify as CRF.
quickfur wrote:Klitzing wrote:Well, in cases where the chopped off caps at most touch, the cap clearly might be re-attached in a gyrated way (if at all possible, cf. below). But surely not if those intersect, what they do here.
Further there then has to be checked additional, whether the dihedral angles at those (re-)connections still are convex, else they disqualify as CRF.
Sorry, I didn't make myself clear again. I didn't mean gyration as cutting off a segmentochoron and reattaching it in rotated orientation; what I meant was, when we pick the two rings of icosahedra to diminish, one of the rings can be rotated relative to the other. That is, since there are 10 icosahedra in each orthogonal ring, and the bistratic diminishing is CRF only when we skip over every other icosahedron, we can only pick 5 icosahedra from each ring. So there are at least two ways to pick them: we can pick all the odd-numbered ones from the first ring and the even-numbered ones from the second ring, or we can pick odd-numbered ones from both rings. The case where both are even-numbered seems to be identical to the case where both are odd-numbered, due to duoprism rotational symmetry; what I haven't verified yet is whether picking odd+even vs. odd+odd is the same or not.
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<(1+sqrt(5))/2, -1, -(3+sqrt(5))/2, -(3+sqrt(5))>
<(1+sqrt(5))/2, -1, -(3+sqrt(5))/2, 3+sqrt(5)>
<(1+sqrt(5))/2, -1, (3+sqrt(5))/2, -(3+sqrt(5))>
<(1+sqrt(5))/2, -1, (3+sqrt(5))/2, 3+sqrt(5)>
<(1+sqrt(5))/2, -1, (5+3*sqrt(5))/2, 0>
<(1+sqrt(5))/2, 0, -1, -(5+3*sqrt(5))/2>
<(1+sqrt(5))/2, 0, -1, (5+3*sqrt(5))/2>
<(1+sqrt(5))/2, 0, 1, -(5+3*sqrt(5))/2>
<(1+sqrt(5))/2, 0, 1, (5+3*sqrt(5))/2>
<(1+sqrt(5))/2, 1, -(5+3*sqrt(5))/2, 0>
<(1+sqrt(5))/2, 1, -(3+sqrt(5))/2, -(3+sqrt(5))>
<(1+sqrt(5))/2, 1, -(3+sqrt(5))/2, 3+sqrt(5)>
<(1+sqrt(5))/2, 1, (3+sqrt(5))/2, -(3+sqrt(5))>
<(1+sqrt(5))/2, 1, (3+sqrt(5))/2, 3+sqrt(5)>
<(1+sqrt(5))/2, 1, (5+3*sqrt(5))/2, 0>
<(1+sqrt(5))/2, (3+sqrt(5))/2, -(3+sqrt(5)), -1>
<(1+sqrt(5))/2, (3+sqrt(5))/2, -(3+sqrt(5)), 1>
<(1+sqrt(5))/2, (3+sqrt(5))/2, -(1+sqrt(5)), -(2+sqrt(5))>
<(1+sqrt(5))/2, (3+sqrt(5))/2, -(1+sqrt(5)), 2+sqrt(5)>
<(1+sqrt(5))/2, (3+sqrt(5))/2, 1+sqrt(5), -(2+sqrt(5))>
<(1+sqrt(5))/2, (3+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5)>
<(1+sqrt(5))/2, (3+sqrt(5))/2, 3+sqrt(5), -1>
<(1+sqrt(5))/2, (3+sqrt(5))/2, 3+sqrt(5), 1>
<(1+sqrt(5))/2, 1+sqrt(5), -(2+sqrt(5)), -(3+sqrt(5))/2>
<(1+sqrt(5))/2, 1+sqrt(5), -(2+sqrt(5)), (3+sqrt(5))/2>
<(1+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5), -(3+sqrt(5))/2>
<(1+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5), (3+sqrt(5))/2>
<(1+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))>
<(1+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, 1+sqrt(5)>
<(1+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, -(1+sqrt(5))>
<(1+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, 1+sqrt(5)>
<(1+sqrt(5))/2, 3+sqrt(5), -1, -(3+sqrt(5))/2>
<(1+sqrt(5))/2, 3+sqrt(5), -1, (3+sqrt(5))/2>
<(1+sqrt(5))/2, 3+sqrt(5), 1, -(3+sqrt(5))/2>
<(1+sqrt(5))/2, 3+sqrt(5), 1, (3+sqrt(5))/2>
<(1+sqrt(5))/2, (5+3*sqrt(5))/2, 0, -1>
<(1+sqrt(5))/2, (5+3*sqrt(5))/2, 0, 1>
Klitzing wrote:Thus the chosen segmentochora seem to be restricted to one of the followings: tet||oct, tet||co, tut||toe, {4}||cube, or line||cube, which then shall occur as caps at the corresponding axes. (Oct||co e.g. disqualifies because the top base likewise has the higher symmetry. And tut||co disqualifies too, as tut has the larger circumradius and thus co is not the bottom base, which has to be incident to the cut.)
--- rk
quickfur wrote:I have discovered another CRF diminishing of the o5oxo: a "meta"-bathodiminishing. This is produced by two deep cuttings of the o5oxo, each of which truncates it up to a truncated dodecahedron cross-section. Unlike the previous "meatball sandwich" cutting I posted earlier, though, in which the two cutting hyperplanes are parallel, in this one the hyperplanes make an angle of 72°. It just so happens that the two resulting truncated dodecahedra meet at a decagonal face, thus producing a peculiar spindle-like wedge structure with 31 icosahedra, 2 truncated dodecahedra, 22 pentagonal cupola, 120 octahedra, 12 pentagonal pyramids, and some other cells (I haven't counted them all yet), for a total of 287 cells. The pentagonal cupola wrap around the decagonal face between the two truncated dodecahedra, and are rimmed by a ring of 10 icosahedra around them.
x5x
x5o o5F
F5o o5V
o5o f5f
o5f x5F
x5f F5x F5x
x5o V5o V5o
x5F x5F
F5o F5o F=ff=x+f=2x+v,
x5x f5f F5F f5f x5x V=F+v=2f=2x+2v
o5F o5F
F5x F5x
o5x o5V o5V
f5x x5F x5F
f5o F5x
o5o f5f
o5F V5o
o5x F5o
x5x
Klitzing wrote:Finally, in order to have at all the possibility of a gyration, the sectioning facet polytope should have a higher symmetry than that of the corresponding axis! Within the Johnsonians this is achieved rather easily: the axis is 3-, 4-, or 5-fold, but the base of the corresponding cupolae (or rotunda) are hexagons, octagons, resp. decagons. For polychora this is much harder. We have axes with tetrahedral, octahedral, or icosahedral symmetry. Thus the only sectioning facet polyhedra with a similar higher symmetry are o3x3o (octahedron within a tetrahedral context), x3o3x (cuboctahedron within a tetrahedral context), and x3x3x (truncated octahedron within a tetrahedral context). That is, we have to look for such polychora where some segmentochoral cap can be chopped off, which uses right these polyhedra for there bottom base and have tetrahedral axial symmetry.
Brique symmetry might look also be possible, but there clearly is just one non-degenerate bottom base which has higher symmetry: x2x2x (cube). But note that such an axis would occure only within (degenerate) 2,P-duoprisms, i.e. either x2x2xPo = x4o2xPo or x2x2xPx = x4o2x2Po.
(Suppose that's all. At least I don't see any further such subsymmetry right now.)
Thus the chosen segmentochora seem to be restricted to one of the followings: tet||oct, tet||co, tut||toe, {4}||cube, or line||cube, which then shall occur as caps at the corresponding axes. (Oct||co e.g. disqualifies because the top base likewise has the higher symmetry. And tut||co disqualifies too, as tut has the larger circumradius and thus co is not the bottom base, which has to be incident to the cut.)
Marek14 wrote:I don't think this is entirely true -- in some polychora, you can also cut off segmentochoron with a 2n-gonal prism as a cap. For example, a o3x3o5x has a cap of 5gon||10prism which could be gyrated.
Klitzing wrote:quickfur wrote:I have discovered another CRF diminishing of the o5oxo: a "meta"-bathodiminishing. This is produced by two deep cuttings of the o5oxo, each of which truncates it up to a truncated dodecahedron cross-section. Unlike the previous "meatball sandwich" cutting I posted earlier, though, in which the two cutting hyperplanes are parallel, in this one the hyperplanes make an angle of 72°. It just so happens that the two resulting truncated dodecahedra meet at a decagonal face, thus producing a peculiar spindle-like wedge structure with 31 icosahedra, 2 truncated dodecahedra, 22 pentagonal cupola, 120 octahedra, 12 pentagonal pyramids, and some other cells (I haven't counted them all yet), for a total of 287 cells. The pentagonal cupola wrap around the decagonal face between the two truncated dodecahedra, and are rimmed by a ring of 10 icosahedra around them.
Indeed, here is its lace city display:
[...snipped...]
You can spot at the left (vertical column) the "red" icosahedron and at the right the single decagon. The 2 truncated dodecahedra are the slanting sides of this wedge (ASCII graphic here is not too exact: those should align instead of wobbling around...). Thus it is nothing but a huge decagon-icosahedron-wedge. Its wedge angle clearly would not be 72 but rather 36 degrees.
@quicfur: you should give a try for such a side view display too!
Finally, this lace city also Shows that you even could chopp off that left "red" icosahedron, even in a bistratic manner, i.e. resulting there in a parabidiminished rhombicosidodecahedron!
[...]
Klitzing wrote:[...]@quicfur: you should give a try for such a side view display too![...]
Klitzing wrote:Having now provided (in my recent mail of this thread) that segmentochoral gem, the cube || ike, in a (partially) snubbed lace prism notation, i.e. as os3os4xo&#x, it gets quite immediate where to find higher dimensional CRFs, which would contain right that fellow for subelements.
In fact, today I've evaluated the convex segmentoteron rico || sadi. It turns out, that it not only belongs to the right geometry (all edges can be chosen to have the same unit length, without drifting then into euclidean space (height becoming zero) or even an hyperbolic one (height becoming imaginary)), by construction it additionally solves that mentioned problem as well.
...
How did I get to consider rico || sadi? - Just by extension of the Dynkin symbol of that mentioned segmentochoron: Accordingly its incidence matrix too can be derived from os3os4xo3oo&#x as
...
Its facets thus are: 1x rico as top base, 24x cube || ike, 24x co || tet, 96x trippy (here coming in as {3} || tet, for sure), and 1x sadi as bottom base. - Its polyhedral elements, now given cummulative, are 24x cube, 24x co, 24x ike, 288x squippy, 312x tet, and 240x trip. - It uses 1152 triangles and 432 squares. - It has a total of 1008 edges and 192 vertices.
Fascinating, ain't it?
--- rk
that thing is just awesome. I still don't really understand what polytopes with both an s and an x look like, although I do understand the construction proces.Klitzing wrote:On 18. December I posted my then newest findingKlitzing wrote:[...]
Within the last days I now derived a close relative convex segmentoteron: tico || prissi.
What does this one look like?
[...]
--- rk
thank you!student91 wrote:that thing is just awesome.
I still don't really understand what polytopes with both an s and an x look like, although I do understand the construction proces.
btw. I just thought of doing this with other snubs as well. Have you thought of demicube-things? e.g. so4ox3ox3oo&#x. This is just a 16-cell atop a o4x3x3o, but the 16-cell is oriented differently. (maybe you can have so4ox3ox3xx&#x etc. as well)
Klitzing wrote:student91 wrote:btw. I just thought of doing this with other snubs as well. Have you thought of demicube-things? e.g. so4ox3ox3oo&#x. This is just a 16-cell atop a o4x3x3o, but the 16-cell is oriented differently. (maybe you can have so4ox3ox3xx&#x etc. as well)
I surely did several such or similar investigations, but for the specific ones I first have to go through my notes first. The main issue here always would be not only that both bases belong to the same symmetry, allow for unit edges only individually, but that a unit edge lacing can be applied in between, which generates a neither imaginary nor degenerate (zero) height.
By view of Stott addition the question about so4ox3ox3xx&#x is not too difficult to answer: it should exist as soon as so4ox3ox3oo&#x does exist. And both then would have the same height.
--- rk
Klitzing wrote:Hmm, sad to disappoint you for those:
R(hex) = 1/sqrt(2) = 0.707, R(tah) = sqrt(9/2) = 2.121.
Thus there cannot be any unit lacing, independent from every orientation!
--- rk
Klitzing wrote:Yes s^{+}s^{-}4oo3oo3oo&#x clearly is possible.
Marek14 wrote:o5oxo
I like that ...wendy wrote:...
But the discussion is interesting, if over my head.
Haha, where elseI've been wrangling the dozenalists at the moment.
Completely correct.Klitzing wrote:Yes s^{+}s^{-}4oo3oo3oo&#x clearly is possible.
Suppose it's s4o3o3o3o.
Me neither, old known stuff, that one.I would not get too excited here.
...
Klitzing wrote:Yes s^{+}s^{-}4oo3oo3oo&#x clearly is possible.
[...]
What becomes different here in these examples to the cases I pointed out, is that you generally would have to apply the snubbing of the bases first, including their relaxations to unit edges, and then would have to apply stacking thereafter. There seems to be no general way for commuting those processes. - But mine examples clearly did allow for a first stacking of the starting figures, and thereafter applying a then D+1-dimensional alternated faceting!
(At least I do not see such a way in general, so far.)
This commutation of operations was what I wanted to underline by calling mine as partial snubs, i.e. choosing to be alternated elements only from one layer, even so the alternated faceting still will be a D+1 dimensional process.
--- rk
Klitzing wrote:Completely correct.wendy wrote:Klitzing wrote:Yes s^{+}s^{-}4oo3oo3oo&#x clearly is possible.
Suppose it's s4o3o3o3o.
Klitzing wrote:Yes s^{+}s^{-}4oo3oo3oo&#x clearly is possible.
Consider the following:
s4o3o...3o is nothing but x3o3o *b3o...3o (using the same number of nodes), i.e. the according demihypercube. (Esp. you'd need at least 3 nodes.) But you always can rewrite any demihypercube (with 4 or more nodes) as a convex segmentotope, in fact as xo3oo3ox *b3oo...3oo&#x (with one node less). The height of that stack then will be dimension independently always 1/sqrt(2).
[...]
--- rk
Klitzing wrote:student91 wrote:Note that there's another (quite obvious) diminishing of o3x3o5o: just cut out a single vertex. The resulting gap will be a pentagonal prism. Maybe we can call this a micro-diminishing?
There might be also some octahedron-centered diminishings. [...]
Is this way of looking at it useful?
student91
Exactly this Investigation has been done already, e.g. cf. http://bendwavy.org/klitzing/incmats/rox.htm, section "Vertex layers": not only the possible uniform polytopes within the section hyperplane are given, in fact all, i.e. using other edge lengths, and for any relevant subsymmetry. Note also the leftmost numbering of layers: within several instances those polytopes cannot given directly, rather they tend to be compounds of several coincident ones.
The same analysis is provided for lots of other polytopes as well, so not all
(Feel free to mail me further additions.)
--- rk
student91 wrote:Besides, could you tell me why snubs aren't wythhoffian, while they do have a wythoff-symbol. it seems counterntuitive to me. was the |p q r introduced by someone else or something?)
student91 wrote:besides, you once said this:Klitzing wrote:student91 wrote:Note that there's another (quite obvious) diminishing of o3x3o5o: just cut out a single vertex. The resulting gap will be a pentagonal prism. Maybe we can call this a micro-diminishing?
There might be also some octahedron-centered diminishings. [...]
Is this way of looking at it useful?
student91
Exactly this Investigation has been done already, e.g. cf. http://bendwavy.org/klitzing/incmats/rox.htm, section "Vertex layers": not only the possible uniform polytopes within the section hyperplane are given, in fact all, i.e. using other edge lengths, and for any relevant subsymmetry. Note also the leftmost numbering of layers: within several instances those polytopes cannot given directly, rather they tend to be compounds of several coincident ones.
The same analysis is provided for lots of other polytopes as well, so not all
(Feel free to mail me further additions.)
--- rk
could you tell me what polytopes haven't got such an ionvestigation?
Klitzing wrote:student91 wrote:[...]
That term does not refer to the W. symbol, it rather refers to the kaleidoscopical construction device of W. As you can see from your pic, the snubs do not allow for such a construction, they do need for 2 elementary cells. This is why.
Btw. I simply use that term ("Wythoffian" polytopes) as a synonyme to the much longer explanation that I'd consider all those polytopes, which allow for Dynkin symbols which use only x and o for node symbols. As those such restricted Dynkin symbols readily describe nothing but the application of that kaleidoscopical construction...
--- rk
Klitzing wrote:[...]
But one should stay fair and try to not disappoint newcommers overly much when they rediscover things on their own ...
[...]
student91 wrote:wythoff-symbol. it seems counterntuitive to me. was the |p q r introduced by someone else or something?)Image
wendy wrote:Wythoff is not guilty of the notation this. The notation is actually a form of 'decorated schwarz triangle'. [...] since Coxeter was not really interested in the non-uniform ones, things like s3s3s3s went begging.
s3s3s3s is actually quite interesting. It's got 60 vertices, 60 tetrahedra, 20 octahedra, and 10 icosahedra for its faces. But unlike 3d, where every snub must exist with uniform edges, in four dimensions, it's more luck than good management.
See, if you take something like xPxQx, and turn it into sPsQs, you get three different edge lengths in xPxQx, and you have a triangle with three edges, so it's down to solving (in general) a cubic, and pulling a cubic-like solution. s3s5s solves equations of the sixth degree (ie x^6+...), and the s3s4s solves a cubic (ie x^3+...)
In four dimensions, you get xPxQxRx, which means you can set four different edges freely. But the tetrahedron has six edges, and you generally can't solve six equations in four variables. When you do an amount of symmetrising, you reduce things somewhat.
xPyQyPx has two degrees of freedom (x, y), but the tetrahedron has three kinds of edges. No go.
xPxPyAPx or xPyPx4o or yPx4o3o. Here you have two degrees of freedom, but the vertex figure is a triangle pyramid, the triangle is equalateral. There are two variables, and two kinds of edge, so a solution always arise (ie P=3, x=1, y=1.618033 or something). This has a solution always, for yPx4o3o.....
x2Po3o3o3... Here the vertex figure is a simplex, there is one edge, and one degree of freedom. It is solved always.
s3s3s3s4o no doubt exists. It would be a quite interesting figure made of the s3s3s4o = s3s4o3o 'snub 24choron', and the s3s3s3s. But like s3s3s3s (which is one of its faces), it's not uniform
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