Regarding D4.8.x

[New Kid on the 4D analog of a Block]

Why can't we join two A2 units together, the same way we join an A2 and a B2 to get a D4.8.4? The edges of the A2 unit that connect with the other unit (the edges on the skew-polyhedron boundary) all have dihedrals of less than 180°, so I think if another A2 was added, those edges would all exhibit positive angle defects (sum of the dihedral angles would be less than 360°).

Also, I think I found a new CRF, but given my track record, it'll turn out to have a flaw.
It has some vague similarities to a D4.8.x, but with a different skew-polyhedron boundary... (Important note - I was only able to find one unit that could fit into this boundary, so I joined two of them together. If the reasons that prevent us from joining two like units of D4.8.x apply here, I guess we can stop this expedition early.) The boundary is best described as a "square rotunda," which I think can be represented by, among others, the lace notations xoM4ofx (M ≈ 0.6702), Nox4ofx (N ≈ 0.7962), or xox4oPx (P ≈ 2.0838). The 4D "top" of the unit is a square, and with two units extending out in 4D from the boundary, the entire polychoron resembles a "dipseudopyramid."
In total, it has 28 cells: 2 cubes and 2 squacus, 16 squippys, and 8 peppys. Of its 24 vertices, 16 are part of the square rotunda, and the other 8 are part of the terminal squares.

The square rotunda viewed from above. Octagonal "base" is not visible.

Screen Shot 2018-09-13 at 6.50.55 PM.png (11.44 KiB) Viewed 17129 times

A projection of one of the units, based in the square rotunda and "rising" towards the square (the dark rhombus at the center). The camera faces a peppy, with 2 squippys on either side. Squacu on the bottom (a triangle faces the camera), cube on the top (vertical edge faces the camera)

Screen Shot 2018-09-13 at 6.52.07 PM.png (32.23 KiB) Viewed 17129 times

Hopefully I learn an important lesson when it turns out not to work, although I wouldn't mind if it actually did work...

[Klitzing]

Well, the connecting surface of A1, A2, or A3 on the one side and B1 or B2 on the other, cf. https://bendwavy.org/klitzing/explain/j ... #tet-axial, is not a flat section. In fact it is described by 3 consecutive vertex layers

Code: Select all

...|Fxo|...3...|oxF|...3...|xfo|...&#xt

with axial non-zero stacking heights.

This surface neither has a prismatic top-down symmetry nor does it show up an antiprismatic rotation symmetry of the Dynkin symbol. Therefore 2 copies of A2 would not match: A mirror symmetrical or prismstic symmetrical arrangement would imply that the deeper vertices would already connect, while the shallower ones are still apart. And for an antiprismatical arrangement the 2 extremal layers of this surface are not vice versas counterparts wrt. tetrahedral inversion nor is the medial one selfsymmetric.

Thus I cannot see how else you would like to match those 2 A2 copies. (Except for pointing into the same direction and thus being completely coincident.)

--- rk

[Klitzing]

Your intended new proposal, New Kid, might be seen this way:
  1) stack a cube onto the top square of a square cupola and then connect either lateral squares by pairs of square pyramids.
(here I neglect your additional pentagonal pyramids with purpose.)

Then you might add a further similar structure, but with gyrated medial square:
  2) stack a square antiprism onto the top of a square cupola and then connect either lateral tip-joining pair of triangles by pairs of edge-joining tetrahedra.

The remaining 4 gaps then would show up 5 triangles on the one side (vertex atop not necessarily flat pentagon) and 5 triangles plus a square (edge atop not necessarily flat pentagon) on the other side.

Code: Select all

   _-|-_         _-+-_   
_-   |   -_   _-  / \  -_
\ --_|_-- /   \ -+---+- /
 \  / \  /     \ |   | /
  \/___\/       \|___|/ 


At least combinatorically such a gap might be filled by 5 tetrahedra plus a square pyramid. - But whether this works out metrically too in a CRF manner, I don't see at the moment.

Might anyone come in here?

--- rk

[New Kid on the 4D analog of a Block]

OK, one last thing. Could you explain how the cells of the A3 unit are arranged? I know it contains 66 vertices and 97 cells, including 2 sets of 4 tricus, a tut at the "center," 12 peppys, and 4 octahedra, but I can't seem to figure the rest out.
After that, I think I'll take a break from D4.8.x. Sorry for bothering you with such trivial matters.

[Klitzing]

OK, one last thing. Could you explain how the cells of the A3 unit are arranged? I know it contains 66 vertices and 97 cells, including 2 sets of 4 tricus, a tut at the "center," 12 peppys, and 4 octahedra, but I can't seem to figure the rest out.
After that, I think I'll take a break from D4.8.x. Sorry for bothering you with such trivial matters.

Code: Select all

of|oxF|..-3-xx|Fxo|..-3-xo|ofx|..-&#xt = A3
======================================
o.|...|..-3-x.|...|..-3-x.|...|..      = 1 tut
..|...|..   xx|...|..-3-xo|...|..-&#x  = 4 tricu
..|...|..   ..|...|..   xo|o..|..-&#x  = 12 tet (a)
..|...|..   xx|.x.|..   ..|...|..-&#x  = 12 trip (b)
o.|ox.|..   ..|...|..   ..|...|..-&#x  = 12 tet (c)
o.|.x.|..-3-x.|.x.|..   ..|...|..-&#x  = 4 tricu
oo|oo.|..-3-oo|oo.|..-3-oo|oo.|..-&#x  = 24 tet (d)
..|...|..   ..|...|..   .o|ofx|..-&#xr = 12 peppy (e)
..|...|..   .x|.xo|..   ..|...|..-&#x  = 12 squippy
..|...|..   .x|..o|..-3-.o|..x|..-&#x  = 4 oct
..|oxF|..-3-..|Fxo|..-3-..|ofx|..-&#xt = (1 wobly connecting surface)


Cf. A3+B1 = https://bendwavy.org/klitzing/incmats/o ... xfo_xt.htm
resp. A3+B2 = https://bendwavy.org/klitzing/incmats/o ... oox_xt.htm

The most difficult cell here probably is the tetrahedron (d), which not is aligned to the overall axially-tetrahedral symmetry. It shows up 4 triangle faces, as obtained by deleting either layer of o's.
If the last o's are deleted, then that triangle connects to one of tetrahedron (a); if the one-but-last o's are deleted, then that triangle connects to one of the triangular prism (b); if the second layer of o's are deleted, then that triangle connects to one of tetrahedra (c); and finally, if the first layer of o's are deleted, then that triangle connects to one of the pentagonal pyramids (e).

--- rk