ubersketch wrote:I've also been interested in plain old isogonal polytopes (which I just like to call isoforms (semiscaliform also works but the scali part gets cancelled out by semi so it just means "form" and there is an alternative definition where every face just looks like a semiuniform polygon)), especially semiuniforms (uniforms but without the silly restriction of having to have all edges be the same length) which only have documentation for a few examples. I don't really know how to find isogonal polychora nor CRFs however except maybe for faceting vertex figures.
quickfur wrote:If you can compute coordinates for 4D objects you're interested in (convex only, unfortunately), I can make polyview models for them and render them.
quickfur wrote:Hmm. This is a tough one. It looks like it has 14 (irregular) tetrahedral cells, all joining at rather sharp angles, which makes it really hard to find a good 4D viewpoint. I spent about 20 mins trying to find a good viewpoint, but couldn't.
This is the best image I could get after ~20 min's of trying. The yellow cell is the one facing the 4D viewpoint. The faint blue edges lie on the far side of the polytope. There are 4 other cells that lie on this side of the polytope, the rest lie on the far side.
According to the edge length classifier, there are 3 kinds of edges, with lengths approx 2.06, 2.46, and 2.88, respectively. As far as I can tell, all cells are congruent, with all 3 kinds of edges (1 edge of length 2.88, 2 edges of length 2.06, 3 edges of length 2.46). For every cell, there are 2 others that do not share any edges with it (i.e., they only touch at the vertices). Every cell shares at least 1 vertex with every other cell.
If you're interested in step prisms, probably the best bet would be to start with a larger duoprism, so the the result would have blunter angles, as that tends to be easier to visualize via projections.
P.S. It occurred to me that it might be useful to do a vertex-centered projection, so here's one:
Again, the faint blue edges are on the far side of the polytope. This viewpoint isn't that clear either, but at least if you focus on the edges emanating from the central vertex (the vertex the projection is centered on), you can sort of see the tetrahedral cells folded around it. It's not very clear, though. As I said, it has very sharp angles and so it's hard to get a good viewpoint on it.
As for the dual, let me get to that.
Mercurial, the Spectre wrote:[...]
Good, but one thing I noticed is that there are actually two types of tetrahedra here. This can be illustrated by coloring each edge according to the length (red = short, blue = medium, and green = long for example). The tetrahedra are divided into two classes: those with one green edge and those with three green edges.
BTW, it is not identical to the 6-simplex even though they share the same vertex/edge arrangement in 2D, but it appears as a 4D projection of a 6-simplex.
Anyways, I mentioned a class of swirl polychora, also derived from the n-n duoprisms. They are compounds of n-n duoprisms rotated like how three equilateral triangles can be placed in a regular nonagon, so the simplest non-trivial one is the so-called triangular duoprismatic triswirlchoron with 18 triangular antiprism cells (in two perpendicular sets of 9 cells each) and 54 tetrahedra. More about here: viewtopic.php?f=25&t=2231
P.S. the coordinates of the triangular duoprismatic triswirlchoron is the Cartesian product (P1) x (P1) where P1 is a unit equilateral triangle (resulting in a unit 3-3 duoprism), (P2) x (P2) where P2 is a unit equilateral triangle rotated 40 degrees with respect to P1, and (P3) x (P3) where P3 is a unit equilateral triangle rotated 80 degrees with respect to P1.
username5243 wrote:One of these days I might try to come up with a list of types of isogonal polychora. Bowers already has a list of 4D dice (which will be the duals of the isogonals) so I'll start from there.
quickfur wrote:[...]
There's also the pyramid antiprisms (i.e., n-pyramid || dual n-pyramid), which if I didn't make a mistake ought to be vertex-transitive, so their duals also ought to be 4D dice.
quickfur wrote:If you can compute coordinates for 4D objects you're interested in (convex only, unfortunately), I can make polyview models for them and render them.
Mecejide wrote:quickfur wrote:If you can compute coordinates for 4D objects you're interested in (convex only, unfortunately), I can make polyview models for them and render them.
How about the disphenoidal diacosioctacontaoctachoron (288-cell)? Its coordinates are all permutations of sign and location of:
{1, 0, 0, 0}
{√2/2, √2/2, 0, 0}
{1/2, 1/2, 1/2, 1/2}
Mecejide wrote:How about the convex hull of 2 rectified 24-cells?
{0, 1, 1, 1}
{0, √2/2, √2/2, √2}
{1/2, 1/2, 1/2, 3/2}
quickfur wrote:Mecejide wrote:How about the convex hull of 2 rectified 24-cells?
{0, 1, 1, 1}
{0, √2/2, √2/2, √2}
{1/2, 1/2, 1/2, 3/2}
Whoa, this one turned out far more interesting than I expected! It's a pretty little polychoron with 48 cubes and 192 (irregular) square antiprisms:
<rest of post snipped>
quickfur wrote:Mecejide wrote:How about the convex hull of 2 rectified 24-cells?
{0, 1, 1, 1}
{0, √2/2, √2/2, √2}
{1/2, 1/2, 1/2, 3/2}
Whoa, this one turned out far more interesting than I expected! It's a pretty little polychoron with 48 cubes and 192 (irregular) square antiprisms:
Surprisingly, there are no "filler" cells like tetrahedra or triangular prisms; it's just regular cubes and squished square antiprisms. Especially interesting is how at every vertex 2 cubes and 6 square antiprisms meet. A totally unexpected structure (for me!).
Mercurial, the Spectre wrote:[...]
How about the alternated truncated octahedral prism (aka the pyritohedral icosahedral antiprism)? It's not uniform. Coordinates are:
(0, ±1/2, ±0.80901699437494742410, 0.35355339059327376220) (all even permutations for the first three coordinates, for a total of 12 vertices)
(±1/2, 0, ±0.80901699437494742410, -0.35355339059327376220) (same as above for the first three coordinates)
And also its dual.
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