Keiji wrote:Johnson may have excluded prisms and uniforms from his list, but I personally prefer to see them as a series of sets of increasing scope, i.e. Regulars < Uniforms < CRFs. (that's why I talk of CRF polytopes, not Johnsonian polytopes.) That way, the existence of prisms of Johnson solids is not a problem either.
I also had another thought. It looks like all the 3D crown jewels are related to the rhodomorphs, as opposed to the stauromorphs. This makes me wonder whether there will be any 4D "stauromorphic" crown jewels - and if there are not, there may be no crown jewels at all in 5D and up, due to the lack of any rhodomorphs above 4D!
So if we are to find higher dimensional crown jewels, either they would have to be stauromorphic, or somehow based upon the symmetry of rhodomorphic prisms, without being prisms themselves.
Klitzing wrote:Well, Keiji, it highly depends on what you are considering a crown jewel to be.
Look e.g. at my set of convex segmentochora. There are several ones, which have different symmetries for bottom and top layer, even if we neglect all those diminishings etc., i.e. restrict the bases to full higher symmetry groups only. Then you will have to consider the common subgroup of those. That one thus would be the true axial symmetry, Independent on what the base polyhedra would show up themselves. Most often that common subgroup would be itself a known Coxeter group. Then both base polyhedra can be given as Dynkin symbol with respect to that one, and you'll get the lace prism notation for free. Just for the cube||ike that common subsymmetry happens to be the non-Coxeter pyrithohedral group. This is what makes this beasty so special.
But also tet||oct, tet||co, oct||tut, tut||co, and tut||toe all show up different symmetries in their bases (here: tetrahedral vs. cubical). But the common subgroup here is the tetrahedral group itself.
quickfur wrote:Klitzing wrote:Well, Keiji, it highly depends on what you are considering a crown jewel to be.
Look e.g. at my set of convex segmentochora. There are several ones, which have different symmetries for bottom and top layer, even if we neglect all those diminishings etc., i.e. restrict the bases to full higher symmetry groups only. Then you will have to consider the common subgroup of those. That one thus would be the true axial symmetry, Independent on what the base polyhedra would show up themselves. Most often that common subgroup would be itself a known Coxeter group. Then both base polyhedra can be given as Dynkin symbol with respect to that one, and you'll get the lace prism notation for free. Just for the cube||ike that common subsymmetry happens to be the non-Coxeter pyrithohedral group. This is what makes this beasty so special.
But also tet||oct, tet||co, oct||tut, tut||co, and tut||toe all show up different symmetries in their bases (here: tetrahedral vs. cubical). But the common subgroup here is the tetrahedral group itself.
The cuboctahedron, while usually considered to have cubic symmetry, can also be regarded as the tetrahedral equivalent of the rhombi- polyhedra (i.e., x3o3x), and similarly the truncated octahedron can be regarded as the omnitruncated tetrahedron (x3x3x). So in that sense these segmentochora just belong to the tetrahedral family.
quickfur wrote:I think I may have found another possible CRF construction involving J91. Take a look at the following projection of the rectified 120-cell o5x3o3o:
Look at where two o5x3o's touch each other. There are two pentagons straddled by two tetrahedral gaps, which, if you fill in the tetrahedra, results in unit edges from the pentagons to each other. The outline of the two pentagons and two triangles thus introduced corresponds with the outline of a J91. Furthermore, the ends of the J91 outline coincide exactly with where the o5x3o's can be bisected into pentagonal rotunda. So it would appear that we have a CRF fragment here consisting of 2 pentagonal rotundae girded by 5 J91's. I'm not sure yet what would need to be inserted on the far side to close it up, but the construction sure looks rather promising! My initial thought is to put a reflected copy of the same fragment on the other side, in which case we have here another rhombochoron-like construction.
x5x
f5o f5A
o5x o5B
f5o f5A
x5x
x
f f
o o
f f
x
Marek14 wrote:Conversation with Robert Webb (author of Great Stella) told me that there's an easy method to measure dichoral angles in the program. This means that if I get *.off files of the polychora we have so far, I can check them for possible augmentations.
Take icosahedral rotunda, for example. Great Stella claims that the dichoral angles are:
pentagonal pyramid/truncated tetrahedron (triangle): 157.761
pentagonal pyramid/truncated icosahedron (pentagon): 36
truncated tetrahedron/icosahedron (triangle): 157.761
truncated tetrahedron/truncated icosahedron (hexagon): 22.2388
Note the very low values for truncated icosahedron, showing that this rotunda is in fact quite shallow (though not as shallow to serve as augmentation of o5x3x3o or o5x3x3x).
Now let's look at something more interesting: the J92-choron.
It has following cells:
12 tetrahedra, each adjoining 1 triangular prisms (172.239), 2 pentagonal pyramids (142.239) and 1 metabidiminished icosahedron (142.239).
12 square pyramids A, each adjoining a triangular prism through its square face (155.905) and 2 pentagonal pyramids (157.761), 1 square pyramid B (167.379) and 1 metabidiminished icosahedron (82.2388) through triangular faces.
12 square pyramids B, each adjoining a J92 through its square face (135) and 1 J92 (82.2388), 2 pentagonal pyramids (157.761) and 1 square pyramid A (167.379) through triangular faces.
6 square pyramids C, each adjoining a triangular prism through its square face (155.905) and 2 metabidiminished icosahedra (157.761) and 2 J92's (142.239) through triangular faces.
6 triangular prisms, each adjoining a two square pyramids A (155.905) and one square pyramid C (155.905) through its square faces and two tetrahedra (172.239) through its triangular faces.
24 pentagonal pyramids, each adjoining another pentagonal pyramid through its pentagonal face (144) and 1 tetrahedron (142.239), 1 square pyramid A (157.761), 1 square pyramid B (157.761), 1 metabidiminished icosahedron (120) and 1 J92 (120) through its triangular faces.
6 metabidiminished icosahedra, each adjoining two J92's through its pentagonal faces (144) and 2 tetrahedra (142.239), 2 square pyramids A (82.2388), 2 square pyramids C (157.761) and 4 pentagonal pyramids (120) through its triangular faces.
4 J92's, each adjoining one J92 through its hexagonal face (60), 3 metabidiminished icosahedra through its pentagonal faces (144), 3 square pyramids B through its square faces (135) and 3 square pyramids C (142.239), 6 pentagonal pyramids (120) and 1 J92 (120) through its triangular faces.
quickfur wrote:Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.
Marek14 wrote:quickfur wrote:Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.
What do you use to look into questions like that?
quickfur wrote:[...]Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.
student91 wrote:quickfur wrote:[...]Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.
I thought of this as well. What happens is that you get an x3o in the middle of the hexagon, surrounded by bigger %3%'s. this means there is some sort of pit in the middle of the polytope (like a red blood cell). you could try to fill this pit up, but that means you have to change the interior of the hexagon, and that probably means there aren't any J92-outlines anymore in the middle of the polytope.
quickfur wrote:[...]
Well, yes, once you join two J92-rhombochora together, two of the J92's would become internal to the polychoron and would no longer appear on the surface. So it's already expected that only six of the J92's would lie on the surface. The concave area around the x3o would need to be filled up by something in order to be CRF, so it's expected that no J92's will be present in the "middle" of the polytope (i.e., on the orthogonal ring to the 6 (J92's). At least, none from the original 3 rhombochora; the thought did occur to me that we might be able to use new J92's to close up the orthogonal ring. If this is possible, we might get a duo-hexagonal-J92-choron.
4OFF
# NumVertices, NumFaces, NumEdges, NumCells
13 43 40 16
# Vertices
1.0 1.0 1.0 0.0
1.0 1.0 -1.0 0.0
1.0 -1.0 1.0 0.0
1.0 -1.0 -1.0 0.0
-1.0 1.0 1.0 0.0
-1.0 1.0 -1.0 0.0
-1.0 -1.0 1.0 0.0
-1.0 -1.0 -1.0 0.0
1.41421356237309504880 0.0 0.0 1.35219344945395667954
-1.41421356237309504880 0.0 0.0 1.35219344945395667954
0.0 1.41421356237309504880 0.0 1.35219344945395667954
0.0 -1.41421356237309504880 0.0 1.35219344945395667954
0.0 0.0 0.0 -1.0
# Faces
4 0 1 3 2
4 0 1 5 4
4 0 2 6 4
4 1 3 7 5
4 2 3 7 6
4 4 5 7 6
4 8 10 9 11
3 0 1 8
3 0 1 10
3 2 3 8
3 2 3 11
3 4 5 9
3 4 5 10
3 6 7 9
3 6 7 11
3 0 2 8
3 1 3 8
3 0 4 10
3 1 5 10
3 2 6 11
3 3 7 11
3 4 6 9
3 5 7 9
3 0 8 10
3 1 8 10
3 2 8 11
3 3 8 11
3 4 9 10
3 5 9 10
3 6 9 11
3 7 9 11
3 0 1 12
3 0 2 12
3 0 4 12
3 1 3 12
3 1 5 12
3 2 3 12
3 2 6 12
3 3 7 12
3 4 5 12
3 4 6 12
3 5 7 12
3 6 7 12
# Cells
10 2 6 15 17 19 21 23 25 27 29 255 0 0
10 3 6 16 18 20 22 24 26 28 30 255 0 0
4 7 8 23 24 0 0 255
4 9 10 25 26 0 0 255
4 11 12 27 28 0 0 255
4 13 14 29 30 0 0 255
5 0 7 9 15 16 127 127 0
5 1 8 12 17 18 127 127 0
5 4 10 14 19 20 127 127 0
5 5 11 13 21 22 127 127 0
5 0 31 32 34 36 0 255 0
5 1 31 33 35 39 0 255 0
5 2 32 33 37 40 0 255 0
5 3 34 35 38 41 0 255 0
5 4 36 37 38 42 0 255 0
5 5 39 40 41 42 0 255 0
# The end.
Marek14 wrote:BTW, I tried to research the augments of square biantiprismatic ring. I'd appreciate if someone checked it, but with dichoral angles from Great Stella, it looks like the cube and both square antiprisms can be independently augmented with pyramids. If both square antiprisms are augmented, two of the pyramids merge into octahedron.
This leads to 2 augmented, 2 biaugmented and 1 triaugmented square biantiprismatic ring.
student91 wrote:quickfur wrote:[...]
Well, yes, once you join two J92-rhombochora together, two of the J92's would become internal to the polychoron and would no longer appear on the surface. So it's already expected that only six of the J92's would lie on the surface. The concave area around the x3o would need to be filled up by something in order to be CRF, so it's expected that no J92's will be present in the "middle" of the polytope (i.e., on the orthogonal ring to the 6 (J92's). At least, none from the original 3 rhombochora; the thought did occur to me that we might be able to use new J92's to close up the orthogonal ring. If this is possible, we might get a duo-hexagonal-J92-choron.
I understand the J92's themselves won't occur on the surface, but that was not what I tried to say. I tried to sa that the x3o itself would become internal, so when you make it convex, the vertices of the x3o can't be used anymore. This isn't very nice, as most of the J92-rhomb's structure will vanish when the x3o is deleted .
Your other idea, the duo-hexagonal J92-choron, does seem interesting, I hope you/we will be able to get a nice CRF out of that.
x3x
F3o F3o
x3f x3f
o3F
o3x o3x
F3x f3x F3x
x3f x3f
x3F x3F
x3o
F3o f3x f3x F3o
o3F o3F
x3F
x3x x3x
F3x
F3o F3o
x3f x3f
o3x
student91 wrote:Marek14 wrote:BTW, I tried to research the augments of square biantiprismatic ring. I'd appreciate if someone checked it, but with dichoral angles from Great Stella, it looks like the cube and both square antiprisms can be independently augmented with pyramids. If both square antiprisms are augmented, two of the pyramids merge into octahedron.
This leads to 2 augmented, 2 biaugmented and 1 triaugmented square biantiprismatic ring.
According to my calculations, the 4-biantipr. ring has dichoral angle 53.5156 between cube and antipr., and 72.96875 between two antipr. Dichoral angle of cube-pyr at square is 53.5156 too, and for antipr.-pyr it is 69.295. This means the cube and the antipr. can be independently augmented. The two antiprisms however, can't be similtaiously augmented while retaining convexity.
I based my conclusions on my calculated list of dichoral angles of the segmentochora, here.
I would really like to know whether you have the same angles, I don't want my list to have more wrong numbers
Marek14 wrote:[..]
Great Stella shows smaller angle for antiprism/pyramid, 53.5156 which makes the dichoral angle when both antiprisms are augmented exactly 180 degrees, thus allowing the augmentation while fusing two square pyramids into an octahedron. My results also have dichoral angle of cube-pyr at square 45, not 53.5156.
Basically, if I get the polychora in appropriate form, the software can measure the dichoral angles directly.
Marek14 wrote:quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.
Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?
Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.
In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.
EDIT: I basically say the same thing as Keiji, just in a bit different words.
quickfur wrote:Marek14 wrote:quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.
Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?
Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.
In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.
EDIT: I basically say the same thing as Keiji, just in a bit different words.
After some further thoughts about this, I have come to agree with Marek. Consider, for example, the pentagon. Its coordinates when laid on the 2D plane involves nested square roots; yet in 3D, when it occurs as a cross-section of the icosahedron, its coordinates involve only the Golden Ratio, which has no nested square roots. Obviously, it makes no sense to say that the pentagon has CVP >2 in 2D, and CVP=2 in 3D. So this is strong proof that it's not about how many square roots you take, but what degree of polynomial you need to solve in order to find the values.
Marek14 wrote:quickfur wrote:Marek14 wrote:quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.
Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?
Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.
In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.
EDIT: I basically say the same thing as Keiji, just in a bit different words.
After some further thoughts about this, I have come to agree with Marek. Consider, for example, the pentagon. Its coordinates when laid on the 2D plane involves nested square roots; yet in 3D, when it occurs as a cross-section of the icosahedron, its coordinates involve only the Golden Ratio, which has no nested square roots. Obviously, it makes no sense to say that the pentagon has CVP >2 in 2D, and CVP=2 in 3D. So this is strong proof that it's not about how many square roots you take, but what degree of polynomial you need to solve in order to find the values.
And consider equilateral triangle: in 2D it will always involve square roots, while in 3D it can be constructed between points (1,0,0),(0,1,0) and (0,0,1).
quickfur wrote:Or, for that matter, the n-simplex, which has integer coordinates in (n+1)-dimensions, but in n-dimensions involves square roots of triangular numbers (so not just roots of the same number, like √5 in things involving the Golden Ratio like the pentagonal polytopes).
Marek14 wrote:OK, I explicitly constructed all augments of square biantiprismatic ring. And guess what? If you augment both antiprismatic cell, the result will be just a cube||octahedron antiprism Didn't expect that, though Klitzing probably knows that
So we have 6 polychora in this group:
Square biantiprismatic ring (also bidiminished cube||octahedron)
Cube-augmented square biantiprismatic ring (also augmented bidiminished cube||octahedron)
Antiprism-augmented square biantiprismatic ring (also diminished cube||octahedron)
Cube-antiprism-biaugmented square biantiprismatic ring (also augmented diminished cube||octahedron)
Antiprism-biaugmented square biantiprismatic ring (also cube || octahedron)
Triaugmented square biantiprismatic ring (also augmented cube||octahedron)
EDIT: Of course, in hindsight it's quite obvious that cube||gyrated square would have connection to cube||octahedron...
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