Potential crown jewel verf

[mr_e_man]

I found this 4D vertex configuration, with 2 each of n-gon prisms, trigonal prisms, pentagonal prisms, and pentagonal pyramids. Each edge is surrounded by 4 polyhedra, so we can't easily calculate the dichoral angles.

prismsPolychoron1.png (43.52 KiB) Viewed 18107 times

I have numerically calculated the dichoral angle between the two n-gon prisms:

n=3: 167.093891°
n=4: 166.559686°
n=5: 165.732024°
n=6: arccos((1 - 3φ)/4) = 164.48°
n=7: 162.505191°
n=8: 159.162429°
n=9: no convex solution
n≥10: dihedral angle sum exceeds 360°

From this we can get the other dichoral angles. They are all less than 180°. In the case n=6, I recognized some of the angles, and thereby derived the exact algebraic value given above; the angle between 'trip' and 'peppy' is 150°, and the angle between 'pip' and 'peppy' is 162°. In the other cases, none of the dichoral angles were found in the list of CRF polyhedron dihedral angles.

It remains to be seen whether this can be part of a CRF polychoron.

[quickfur]

Oooh, this is very interesting! Maybe an augmented duoprism candidate? But none of the angles look like they could wrap around into a ring of prisms... so however this closes up, it can't have duoprism symmetry. Would have to be something more irregular...

[mr_e_man]

Potential_crown_jewel_verf_2.png (32.72 KiB) Viewed 18078 times

Here are a few more potential verfs, irregular octahedra. The first two polyhedra can each be split in half by an n.3.4.3 tetragon, giving trivalent vertices, from which the dihedral angles can be easily calculated. In the case n=4, one angle is exactly 90°+90° = 180°, so the two faces combine according to 3.3.4 + 4.3.3 = 3.3.3.3 A, or 4.4.4 + 4.3.3 = 4.4.3.3 B. In the case n=3, that same angle is greater than 180°.

The dihedral angle at the 'n' edge, for the first polyhedron, is 2 arccos( sin(π/n) / √(1 + 2√2 cos(π/n) - 4 cos²(π/n)) ).

n=3: no convex solution
n=4: 90°
n=5: 88.2240°
n=6: 83.5479°
n=7: 75.5561°
n=8: 61.8050°
n=9: 30.6513°
n≥10: angle sum exceeds 360°

For the second polyhedron, the angle at 'n' is 90° + arccos( sin(π/n) / √(1 + 2√2 cos(π/n) - 4 cos²(π/n)) ).

n=3: no convex solution
n=4: 135°
n=5: 134.1120°
n=6: 131.7740°
n=7: 127.7781°
n=8: 120.9025°
n=9: 105.3256°
n≥10: angle sum exceeds 360°

And for the third polyhedron:

n=3: no convex solution
n=4: no convex solution
n=5: 138.480147°
n=6: 137.797469°
n=7: 136.630095°
n=8: arccos(-1/√2) = 135°
n=9: 132.852953°
n=10: 130.086271°
n=11: 126.566789°
n=12: 122.176978°
n=13: no convex solution
n=14: no convex solution
n≥15: angle sum exceeds 360°

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

I tried building the first polychoron you described, and while I didn't complete it, it gave me some ideas.

Pentagonal pyramids in the gap facing the camera are hidden.

Screen Shot 2022-04-06 at 4.57.59 PM.png (24.04 KiB) Viewed 17867 times

It made me think that the strata of pentagonal-rotundaic prisms are in there. Each group of triangular and pentagonal prisms that's connected by square faces should close up into such an arrangement (unless there's actually a non-pentagonal prism at its center). Even though Quickfur said the figure wouldn't close up like a duoprism, I couldn't help but think of an n-10 duoprism with "joints" in between the decagons of the decagonal prism cells.

Here's one of those pieces from a pentagonal-rotundaic prism. The outside ten prisms are the ones seen in the model above, and the inner five triangular prisms fit into the notches seen between pentagonal prisms in the model.

Screen Shot 2022-04-06 at 4.59.26 PM.png (11.5 KiB) Viewed 17867 times

Then I tried connecting two pentagonal rotundae at the base and folding them in towards each other (just like a metabidiminished icosahedron can start with two pentagons folded up towards each other), using the pentagonal pyramids to build around the decagon connection and bridging the gap between the tops of rotundae with a pentagonal antiprism.

Screen Shot 2022-04-06 at 5.05.37 PM.png

10 pentagonal pyramids are yellow, 10+10 tets are blue and purple, and the apex pentagonal antiprism is white. No new vertices are added to the set of those in the pentagonal rotundae. Sum of dihedral angles around an "equatorial" edge (in 4D those edges are either the "bottom" or "top" of the shape) is exactly 180°, but that shouldn't be a deal-breaker.

(31.48 KiB) Not downloaded yet

Of course, the fact that there's no obvious number of n-gonal prisms to use means we don't know what the central prism of the prism arrangement should be (central square prism for 8 n-gonal prisms, central pentagonal prism for 10 n-gonal prisms like I assumed, and so on), so my entire post may be based on unstable ground.

[Klitzing]

Then I tried connecting two pentagonal rotundae at the base and folding them in towards each other (just like a metabidiminished icosahedron can start with two pentagons folded up towards each other), using the pentagonal pyramids to build around the decagon connection and bridging the gap between the tops of rotundae with a pentagonal antiprism.

10 pentagonal pyramids are yellow, 10+10 tets are blue and purple, and the apex pentagonal antiprism is white. No new vertices are added to the set of those in the pentagonal rotundae. Sum of dihedral angles around an "equatorial" edge (in 4D those edges are either the "bottom" or "top" of the shape) is exactly 180°, but that shouldn't be a deal-breaker.

Hi New Kid,

you are indeed on a valid configuration here. It is based on the 1/10-luna of the hexacosachoron, described here: https://bendwavy.org/klitzing/incmats/1 ... -of-ex.htm
Yours is just the diminishing thereof at the singular top vertex, showing up then as the "central" pentagonal antiprism of your description.

--- rk

[mr_e_man]

It made me think that the strata of pentagonal-rotundaic prisms are in there. Each group of triangular and pentagonal prisms that's connected by square faces should close up into such an arrangement (unless there's actually a non-pentagonal prism at its center). Even though Quickfur said the figure wouldn't close up like a duoprism, I couldn't help but think of an n-10 duoprism with "joints" in between the decagons of the decagonal prism cells.

If there were pentagonal rotundaic prisms, the dichoral angle between 'trip' and 'pip' would be 142.6226°.

I calculated those dichoral angles; I didn't bother to write them down, but I did compare them with the list of 3D dihedral angles, and found no matches (except for n=6). The dichoral angle between 'trip' and 'pip' in fact is not 142.6226°.
(For n=6, it is 169.1877°.)

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

Interesting that it's the same angle as is found between triangle and square in elongated pentagonal rotundae.

If there were pentagonal rotundaic prisms, the dichoral angle between 'trip' and 'pip' would be 142.6226°.

But with the pentagonal-rotunda bases removed, could the remaining cells flex to different dichoral angles like the ones you calculated? Like how the triangles and squares in n-gonal (n>3) pyramids, antiprisms, and prisms can be nudged into meeting at different angles (then potentially used in Johnson solids) if the base(s) are removed? Or does this entirely prevent that from happening in 4d?
Wait, no, the triangular-and-pentagonal-prism cell complexes would then be asked to bend in two opposite directions because of the relatively high symmetry of the entire figure. I'd better keep this line of thinking confined to less-symmetric crown-jewel attempts. With no trivalent edges, just to be safe.

[quickfur]

In 4D, edges with degree ≥4 are flexible, the same way vertices of degree ≥3 in 3D are flexible (in both cases assuming rigid facets). However, in 4D, vertices surrounded by rigid facets are not flexible, precisely for the reasons indicated in the link you referenced.

IOW, you could take a configuration of cells in 4D and deform them, but only if there are no vertices that are completely surrounded by cells. Once a vertex has been "closed", i.e., all cells around it are determined, then it cannot be flexible anymore, meaning that all dichoral angles among the cells around that vertex can no longer deform continuously. (There may be more than one set of dichoral angles possible, e.g., alternating between convex/non-convex angles, but continuous deformation will not be possible, unlike the 3D case where the possibility of continuous deformation of e.g., florets of 5 triangles, leads to interesting figures like the Johnson solid crown jewels.)