quickfur wrote:1) Name a uniform polyhedron with 8 faces, 18 edges, and 12 vertices.
quickfur wrote:2) Can you name a different polyhedron with 8 faces, 18 edges, and 12 vertices? Interesting coincidence, isn't it?
Klitzing wrote:... Just for fun - haven't worked that out so far - we could proceed that idea of yours, and build instead of a ring of 3 trigonal-antiprismatically attached triddips - as a next step - a corresponding ring of 4: xooo3oxoo oxoo3ooxo ooxo3ooox ooox3xooo&#zx.
Hmm, would that figure then be an otherwise already well-known one, too?
Obviously that one ought to be 8D, and as each layer is a triddip, i.e. has 9 vertices each, that figure then has 36 vertices...
... could that one then possibly be rene = o3x3o3o3o3o3o3o ? - At least that one is 8D and it has 36 vertices ...
Klitzing wrote:...
On the other hand, the mere found jak structure looks not being restricted to links being all "3". We also could consider xoo-P-oxo-2-oxo-Q-oox-2-oox-R-xoo&#zx or, more specially, xoo-P-oxo-2-oxo-P-oox-2-oox-P-xoo&#zx for any P>=3. Those (with all "P"s) even ought be at least scaliform then!
Provided that "x" part in the trailing "&#zx" remains applicable - which might impose some additional restriction on P.
...
a x x o 1--b
/ \
b o x x a 2
\ /
c x o x 3--c
1 4 vertex 1ab 2 squares 123a square pyra
2 4 vertice 2bc 2 squares 1abc (nothing)
3 4 vertices 3ca 2 sqiares 1a2c tetrahedron
1c2 4 triangles 1ab2 digon ap.
12 8 edges 1c3 4 triangles
13 8 edges 2a1 4 triangles 123ab tetra prism.
23 8 edges 2a3 4 triangles 12abc tetra || square
1a 4 edges 3b1 4 do
1b 4 edges 3b2 4 do
2a 4 edges 123 8 triangles 123abc itself
2b 4 edges
3a 4 edges
3b 4 edges
Klitzing wrote:For the more general setup we thus just could consider xoo3oxo oxo3oox oox4xoo&#zx at most ...
wendy wrote:For example xxo2xox2oxx&#x would exist in 5D as a separate figure. It would evidently be scaliform, among other things, consisting of loops of tetrahedral prisms.
- Code: Select all
a x x o 1--b
/ \
b o x x a 2
\ /
c x o x 3--c
1 4 vertex 1ab 2 squares 123a square pyra
2 4 vertice 2bc 2 squares 1abc (nothing)
3 4 vertices 3ca 2 sqiares 1a2c tetrahedron
1c2 4 triangles 1ab2 digon ap.
12 8 edges 1c3 4 triangles
13 8 edges 2a1 4 triangles 123ab tetra prism.
23 8 edges 2a3 4 triangles 12abc tetra || square
1a 4 edges 3b1 4 do
1b 4 edges 3b2 4 do
2a 4 edges 123 8 triangles 123abc itself
2b 4 edges
3a 4 edges
3b 4 edges
o3o3o x3o3o
x3o3o o3x3o
T
T=tet
T 4 4=square
wendy wrote:Whence the reference to j5j2j5j. This is a 'gap', in terms of the bipentagonal symmetry.
Evidently j7j2j7j exists, but is not scalaform.
wendy wrote:You would have to consider all six lacing-edges, bot only for one case. In &#x, we arrange the points at a simplex, which is how it comes from the vertex-figure of a wythoff-figure. Of course, this is not &#xt (lace tower), or &#xr (lace ring). For example, the xooo3oxoo oxoo3ooxo ooxo3ooox ooox3xooo&#zr would have 8+2=10 dimensions too, and the same circumdiam as the jak. On the other hand, something like xooo3oxoo oxoo3ooxo ooxo3ooox ooox3xooo&#zx would give 11 dimensions, since there are four points in the altitude, and eight dimensions across the bases, makes 8+4-1=11.
wendy wrote:c for chord, c(x,y) makes the y'th chord of x. default y=2. The function is a real one, so you can evaluate c(2pi, 3).
x(n/d, m) = sin(π md/n)/sin(π d/n)
x(n/d) = x(n/d, 2) = sin(π 2d/n)/sin(π d/n) = 2cos(π d/n)
wendy wrote:I use 'z' in the manner that i tried to figure out what student91 meant. He particularly seemed concerned when the lace tower has a zero height, as some of the lace-compounds do. Of course &# is a branch creating a new axis of symmetry, a new height, and z gets rid of it.
There is room for something that you are meaning by &#.z is something that is best described by using a ÷ sign to 'divide space against this symmetry. It can be used in any place then. So the kiddies making Zome-tool projections of the twelftychoro, are making "x5o3o3o÷" ie, a projection into a lesser space 5,3, by dividing it down that axis. So your &#.z is the same as &x.÷ No formal symbol is created for it. So we can remove a dimension with ÷, it basically allows you to correctly describe a lace city in line by removing axies, but it is not always the case that the lace city has a notation that is a subset of the whole, eg 5,3,3 does not conveniently contain 5,2,5 as a symmetry, but the symmetry can be described as 52÷5÷ or the like.
In 2014 Mrs. Krieger extended her zoo of lace prisms, towers, and simplexes a bit. She introduced a further qualifier "z" (zero). If this qualifier precedes the lacing edge lengths ("...&#zx.."), then the respective segmental heights happen to become zero. (In that this qualifier adds just some optional further information to the reader.) But there is a bit more to that: For lace prisms the bases usually are part of the figure. And for lace towers the extremal layers similarily. But not the intermediate ones. Those happen to be just pseudo facets, sectioning the tower into segments. But for degenerate lace towers there might occur situations, where this splitting of the vertex set into subsets (those co-hyperrealmic "layers") would result in pseudo facets solely. This then is a case, where this preceding "z" qualifier surely would be required! – A simple example here would be the ike, when being represented in briquet symmetry: fxo ofx xof&#zx. (None of those golden rectangles belongs to the elements of ike.)
She then also allowed for a suffixing "z" too. This one would build up the structure as devised without that additional "z" first, and thereafter scales all lacing edges down, such that the segmental heights all become zero. – This probably would be rather seldomly used explicitly. But implicitly it provides a different pictoral representation for the segmentochora, in fact the "telescope view" from infinity. (The pictures provided within this website however usually provide views from nearby, thus relatively scaling the two bases additionally in a perspectivic way.)
A further concept closely related to the former zoo is that of lace cities. Here the name derives from the picturesque description of a city: being a set of towers. So re-consider the just described lace towers. If each of those layers on its own can be described as a tower, if further all those towers of one dimension less are describable within the same symmetry group, then there is an arrangement of the higher tower in the manner of a street, while the lower towers are given as sets of layers at the specific grounds. This yields a 2 dimensional graphical arrangement of dynkin symbols, each of 2 dimensions less. Take for example the octahedron (oct). It can be given as the lace tower (in fact only a lace prism) xo3ox&#x, i.e. x3o || o3x. Now x3o = x o&#x and o3x = o x&#x. Thus we have for lace city
Users browsing this forum: No registered users and 2 guests