A different way to expand an ursatope?

[quickfur]

Haha, I made a mistake about the orbiformity of the tetrahedral teddy. My polytope viewer has a command for translating a polytope to be origin-centered, and I usually use that followed by a radius calculation command to see if all vertices have the same distance from the origin. However, the origin-centering calculates the center by averaging vertex coordinates, so it only works correctly if vertices are evenly distributed in the n-sphere! Otherwise it would not completely center the polytope and the radius calculation would wrongly indicate that it is non-orbiform.

However, the fact that the tetrahedral teddy has radius 1 means that the 5D tetrahedral teddy pyramid would be degenerate, since its height would be 0. So we cannot use it for CRF augmentation of the 5D teddies.

[Klitzing]

hmm, still not fully convinced.
Cause if your "x" would be true, then the lace city would look like this

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o o   x x   o o
              
 f o       o f
   o x   x o   
              
   x o   o x   


which shows, that the "o x   x o"-line dimples in wrt. the sloping part.
But I have to admit, that these sloping parts just represent the pentagons ofx&#xt.
So there is a slight - so counterintuitive - chance to be still convex in 4D, just having these vertices projected in an inappropriate orientation...

Provided these vertices still would be true tips (truely bending out), then their coordinates (in the same setup as above) would be:

• (1/sqrt(8), 1/sqrt(8), -1/sqrt(8); -(sqrt(5)-2)/sqrt(8)]) & all even permutations in first 3 coord.s & all changes of sign in first 3 coord.s

(and the lower part of the lace city then indeed describes a hexadecachoron.)

--- rk

[quickfur]

Perhaps it might help to look at it from a different angle:


This is a parallel projection viewed from the tetrahedron at the tip of the teddy (i.e., the x3o3o, shown here in yellow). The f3o3o is indicated by the red vertices. The o3o3x is indicated by the green dots, which in this projection appears at the corners of the cube of edges alternating with the x3o3o's vertices. Of course, in 4D they are actually not corealmar, because the o3o3x is displaced into the 4th direction, but this shows how they protrude outside the tetrahedral shadow of the f3o3o.

The key point here, I suppose, is the fact that the in-radius of the f-tetrahedron is less than the out-radius of the unit dual tetrahedron, which seems to be what you're having difficulty with. So it's entirely possible for o3o3x to fall outside the convex hull of f3o3o, and thereby serve as apices of the teddi pyramids, which are quite shallow.

[quickfur]

To settle this point even more concretely, the inradius of a unit tetrahedron is sqrt(6)/12 = approx 0.2041, and the circumradius is sqrt(6)/4, approx 0.6124. Now, sqrt(6)/12 multiplied by the golden ratio gives the inradius of the f-tetrahedron, which numerically is 0.3303, which is less than the circumradius of the unit tetrahedron. So the vertices of the dual unit tetrahedron would indeed fall outside the convex hull of an f-tetrahedron.

[Klitzing]

you're right, quickfur!
That pic shows how damn flat is this protuding vertex.
But I now managed to find the "other" solution: the cirumcenter itself surely is 1 unit apart from all other vertices. Thus the dimpling in excavations would run here right into the body center!
--- rk

[quickfur]

[...]
Here's the lace tower:

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o4o3x
A4o3o
o4o3f
o4x3o


where A = approx 1.05066753090906. As I said, I don't know the algebraic representation of this number yet. Maybe some other time when I work through the algebra. I got this by manual application of Newton's method on a set of layer height / edge length constraints.

Here's a quick render:

[...]

I have found the algebraic expression for A: 1/√2 + 1/√(2*phi^3), where phi = (1+√5)/2 is the golden ratio.

So here are the full coordinates for the tetraaugmented octahedral teddy:

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apacs<0, 0, √2> ~ <√(2*phi)>
apacs<0, 0, √2*phi> ~ <0>
apacs<0, √2, √2> ~ <-√(2/phi)>
apecs<A, A, A> ~ <C>


where A = 1/√2 + 1/√(2*phi^3), and C = √(2*phi) - √(7 - 3*phi - 1/√phi^3). Numerically, A is approx 1.050667530909060, and C is approx 0.510486017433607.

[Klitzing]

Haha, you well could augment just every alternate teddi, resulting, as you wrote in tips coords

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apecs<A, A, A> ~ <C>

but when augmenting every teddi - as you originally intended when writing A4o3o - then you ought have instead

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apacs<A, A, A> ~ <C>

In fact, the second one is, as you wrote

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x3o4o || o3o4A || f3o4o || o3x4o,

whereas that alternated one would rather be

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o3x3o || (qA)3o3o || o3f3o || x3o3x

---rk

[quickfur]

Except that augmenting every teddi is non-convex. What's the correct symbol for augmenting every other teddi with a node marked A? (I admit my mastery of alternating CD symbols is very weak / non-existent )

[quickfur]

Ah I see it. Haha, just missed your reply when I posted mine. Nevermind what I said.

[quickfur]

Actually, shouldn't the last symbol be o4x3o, not x4o3x?

[Klitzing]

Actually, shouldn't the last symbol be o4x3o, not x4o3x?

No, oct || f-oct || co (the un-augmented one) could be given
in axial [3,4] symmetry as x3o4o || f3o4o || o3x4o, or
in axial [3,3] symmetry as o3x3o || o3f3o || x3o3x.
Augmentation with [3,4] symmetry clearly applies to all teddies, ie. in the former case trivially, in the latter it would apply to both color types.
Alternate augmentation is only possible with [3,3] subsymmetry. It then applies to the black teddies only (and not to the alternate white ones).

You also claim that the [3,4] symmetrical augmentation wouldn't be convex. - Have not checked that. Might be.

Ah sure, the [3,5] case is just related to ex rotunda. Augmenting any teddi is possible there only, when inserting additionally those 5-tet-rosettes in place of the 2-peppy-patches. Without those we could augment here 3 teddies at most. (Furthermore, these rosettes only fix in that special [3,5] symmetry.)

Okay, then the maximal augmented in the octahedral case then is indeed just o3x3o || B3o3o || o3f3o || x3o3x (with B=qA). Esp. not the full augmented one which you originally proclaimed!
--- rk

[quickfur]

Silly me, I did not read your CD symbols carefully. You were using .3.3. symmetry for the alternate augmentation, so what you wrote was actually correct. I misread it as .3.4. symmetry, and I thought you wrote x3o4x rather than x3o3x. Sorry for the mixup.

Also, I'm pretty sure I already said in my original post that full octahedral augmentation is non-convex; only the alternating augmentation is. However, I mixed up my CD symbols and wrote A4o3o, which is wrong as that is clearly non-convex. I really should double-check my CD symbols before posting, but I'm so short of time these days.