Snub disphenoid based CRF?

[Klitzing]

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     o4o o4o
  o4o A4o     as A approaches o, this becomes a square :S

        o4o
  o4o B4o
          o4o    as B approaches some value, the distance between the leftmost and rightmost o4o might become one, which may result in something cool?


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      o4o      |     o4o  o4o
  o4o x4o o4o  |  o4o A4o
      o4o      |     o4o  o4o


These then are the 3-4-tegum and the 5-4-tegum. For sure here neither A nor B has unit size. Accordingly both are not CRFs.
--- rk

[student5]

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     o4o o4o
  o4o A4o     as A approaches o, this becomes a square :S

        o4o
  o4o B4o
          o4o    as B approaches some value, the distance between the leftmost and rightmost o4o might become one, which may result in something cool?


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      o4o      |     o4o  o4o
  o4o x4o o4o  |  o4o A4o
      o4o      |     o4o  o4o


These then are the 3-4-tegum and the 5-4-tegum. For sure here neither A nor B has unit size. Accordingly both are not CRFs.
--- rk

I was already wondering if my 3D-to-4D analogy was correct

my mistake was in assuming that you could break the octahedron, just as you can break the pentagon in peppy, but I forgot about edges needing to be unit length
on second thought, the breaking can only be done on none-lacing distances (the q in squippy; the f in peppy), so taking oct as a first experiment was a bit stupid, since it's fully rigid; changing something would directly result in non-unit edges

the cubic pyramid, on the other hand looks rather promising, especially in this orientation:

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x2o
x2q  o2o   the q distance can be changed, without losing any unit edges in the cube :)
x2o

  x2o
  x2A o2o     if A = x, you must get something interesting, with two triangular prisms at an angle, with a point, has this been discovered yet?
    x2o

    x2o
    x2B o2o    if B = sqrt(3), sort of the same situation occurs, does this result in the same polychoron?
  x2o


note: I'm not quite sure about the sqrt(3), I just took the height of two triangles (but they are at dichoral angles, so I don't know if it should be different)
any tower that has more than 3 segments does weird stuff, because changing a (non-symetrical) middle segment, will introduce different changes as to lacing height, resulting in two dichoral angle changes:

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  o3o o3q q3o o3o
        o3o

  o3o       |      o3o
  o3q  o3o  |      o3q
   A3o      |     B3o  o3o
       o3o  | o3o


now, let's take a look at polychora with cubic pyramids and change them!
first: the cubic bipyramid:

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    x2o       |     x2o       |       x2o
o2o x2q o2o   | o2o x2A o2o   |   o2o x2C o2o
    x2o       |   x2o x2o     |       x2x

    x2o       |   x2o
o2o x2q o2o   |   x2B
   (x2o)      | o2o  o2o


the 24-cell might also be a candidate, but it's quite difficult to get a good viewpoint, and I need to go to bed now

[quickfur]

Your lace city with x2C looks interesting. It's basically the bipyramid of a (not necessarily regular) pentagonal prism. Here there are two constraints that we must satisfy: (1) the edges of this non-uniform prism lie on the exterior of the polytope, and so are required to be unit length. However, (2) the vertices of this prism must also be unit distance from the o2o's, otherwise it becomes non-CRF. So the vertices can only lie on a circle, since otherwise the prism is non-corealmar and you wouldn't be able to close it up the way you described it. But these two requirements: vertices on a circle and unit length apart, can only have one solution: a regular pentagon. So C must be equal to f, and we get the (regular) pentagonal prism bipyramid.

Perhaps the middle lace city with x2A might yield something interesting?

[student5]

Your lace city with x2C looks interesting. It's basically the bipyramid of a (not necessarily regular) pentagonal prism. Here there are two constraints that we must satisfy: (1) the edges of this non-uniform prism lie on the exterior of the polytope, and so are required to be unit length. However, (2) the vertices of this prism must also be unit distance from the o2o's, otherwise it becomes non-CRF. So the vertices can only lie on a circle, since otherwise the prism is non-corealmar and you wouldn't be able to close it up the way you described it. But these two requirements: vertices on a circle and unit length apart, can only have one solution: a regular pentagon. So C must be equal to f, and we get the (regular) pentagonal prism bipyramid.

Perhaps the middle lace city with x2A might yield something interesting?

I'm sorry, I'm not that good with the 4D visualizing, with the x2C, I was trying to take a side-view orientation of the x2A-polytope I don't know if side-viewing is possible on 4D with lace cities as it is in 3D, because you already squash two dimensions in a point.

I came around to the math today, and wolframalpha gave A = (something really difficult)

x2o-x2A lacing distance = 1/2*sqrt(4-A^2)
o2o-x2A lacing distance = 1/2*sqrt(2-A^2)
x2o-x2o lacing distance = 1
x2o-o2o lacing distance = 1/2*sqrt(3)

so, we get this construction:

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     d----c    a = x2o   b = x2A   c = x2o   d = o2o
    /\  / |    a/d = d-c = sqrt(3)/2    a-b = b/c = 1/2*sqrt(4-A^2)    d\b = 1/2*sqrt(2-A^2)
  a---b   |    a = (-1/2*sqrt(4-A^2), 0)   b = (0, 0)   c = (sqrt((1/2*sqrt(4-A^2)^2-1/4, 1/2) = (1/2*sqrt(3-A^2), 1/2)


now, we can make circles with the lacing length around the coordinates, to find A (and obtain the coordinates for d for free )
a: (x+ sqrt( 4 -A^2)/ 2)^2+ y^2 = 3/4
b: x^2+ y^2 = 1/2-1/4*A^2
c: (x- sqrt(3- A^2)/2)^2+ (y- 1/2)^2 = 3/4

putting this in wolframalpha gives 2 real and positive, really ugly exact solutions, but hey, they are solutions
exact form:

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A = sqrt(17/11-((1-i sqrt(3)) (-2169+11 i sqrt(8391))^(1/3))/(44 3^(2/3))-(31 (1+i sqrt(3)))/(11 (3 (-2169+11 i sqrt(8391)))^(1/3)))
x = -((-1)^(1/4) sqrt(-1344 i+(46128 i 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(4464 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(1488 i 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+12 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+12 i (3 (-2169+11 i sqrt(8391)))^(1/3)+i (3 (-2169+11 i sqrt(8391)))^(2/3)))/(60 sqrt(22))
y = (264-(46128 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 i 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)+(1116 i 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(372 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)-3 i 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 i 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+3 (3 (-2169+11 i sqrt(8391)))^(1/3)-(3 (-2169+11 i sqrt(8391)))^(2/3))/3960


approximate form:

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A≈0.9834232572780998913369702481822330581806+0.×10^-40 i    the 0.*10^-40 i gets smaller as I ask for more digits, so I guess this becomes 0?
x≈-0.1529927470946949461587582865691637298282+0.×10^-40 i
y≈0.4845749617938918523245446506490979027648+0.×10^-40 i


solution 2, exact:

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A = sqrt(17/11-((1+i sqrt(3)) (-2169+11 i sqrt(8391))^(1/3))/(44 3^(2/3))-(31 (1-i sqrt(3)))/(11 (3 (-2169+11 i sqrt(8391)))^(1/3)))
x = -1/60 (-1)^(3/4) sqrt(1/22 (1344 i-(46128 i 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(4464 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)-(1488 i 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+12 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)-12 i (3 (-2169+11 i sqrt(8391)))^(1/3)-i (3 (-2169+11 i sqrt(8391)))^(2/3)))
y = (264-(46128 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)-(46128 i 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(1116 i 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(372 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+3 i 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)+3 i 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+3 (3 (-2169+11 i sqrt(8391)))^(1/3)-(3 (-2169+11 i sqrt(8391)))^(2/3))/3960


approximate:

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A≈1.3272160632106629924127375166040478892325+0.×10^-40 i
x≈0.08739116013663362659489301265096308769828+0.×10^-41 i
y≈-0.2280069418216704466332648296378194445083+0.×10^-40 i


now is this CRF?
(this has got to be a CRF with one of the ugliest coordinates around if it is )

Edit: on second thought, the second solution can't be correct, since it gives a negative pyramid height

Edit: I just realized, that wolframalpha did give right solutions, it was just that it didn't at first, because I messed up my calculations a bit, the notion that A = -1 is therefore incorrect.

[Klitzing]

my mistake was in assuming that you could break the octahedron, just as you can break the pentagon in peppy, but I forgot about edges needing to be unit length
on second thought, the breaking can only be done on none-lacing distances (the q in squippy; the f in peppy), so taking oct as a first experiment was a bit stupid, since it's fully rigid; changing something would directly result in non-unit edges

the cubic pyramid, on the other hand looks rather promising, especially in this orientation:

Code: Select all

x2o
x2q  o2o   the q distance can be changed, without losing any unit edges in the cube :)
x2o

  x2o
  x2A o2o     if A = x, you must get something interesting, with two triangular prisms at an angle, with a point, has this been discovered yet?
    x2o


...

Well, trying A=x in fact makes this:

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  x2o
x2x o2o
  x2o


Either half (upper or lower) then clearly is a trippy. 2 of these here are to be adjoined at a squippy.

As the squippy-3-squippy dihedral angle = 75.5 degrees, this external blend remains convex at this triangle. But, OTOH, the squippy-3-tet dihedral angle = 104.5 degrees; thus the aimed for external blend will become concave at this (other) triangle!


The other proposed "solution" with A=o would be dimensionally degenerate. As then all right nodes are "o" and the linkage "2" asks for a cross product. So we would do the product with nothing. Thus we simply adjoin 2 squippies at a triangle, such that their squares adjoin a common edge. - Again this would not be convex at the triangle-triangle dihedrals at those lacing adjoining edges.

--- rk

[quickfur]

@student5: you do realize that those coordinates you got from Wolfram Alpha are basically zero, right? Look at the exponent part of the floating point numbers... they are all large negative exponents, meaning the value is very close to zero. It's probably due to roundoff error that they didn't work out to be exactly zero, but I'd imagine the actual solution is zero. :-P

Besides, the "i" in the values is kinda a red flag, because that means it's a value along the imaginary axis. So the real part is probably actually zero, and the non-zero (but practically zero) imaginary part is probably just a side effect of whatever algorithms Wolfram Alpha used to compute the result.

[Klitzing]

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    x2o
    x2B o2o    if B = sqrt(3), sort of the same situation occurs, does this result in the same polychoron?
  x2o


note: I'm not quite sure about the sqrt(3), I just took the height of two triangles (but they are at dichoral angles, so I don't know if it should be different)

That one took me some time to recognize.

First of all I calculated coords for that.
a) The 2 edges "x2o" make up a square. Coords are (1/2, 1/2, 0, 0) & all sign changes.
b) The tip "o2o" then has coords (0, 0, 1/sqrt(2), 0).
c) Finally the rectangle "x2B" has coords (1/2, 0, h, B/2), where the halved elements allow for all sign changes.

Calculating the distances a-c and b-c, which both ought be unit sized, one gets by comparision that h=1/sqrt(8).
Using that and introducing it into one of those distance equations, one gets B=sqrt(5/2).

Still I had to decide: is it CRF or not ... (In contrast to quickfur I've no tool aid for that.)

But after a while of consideration I recognized: Even so these lace cities

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   x2o
x2A   o2o   with A=x
   x2o

x2o
  x2B o2o  with B=sqrt(5/2)
x2o


look different (convex angle at x2A, concave angle at x2B resp., but there then having filled additionally that left undercut), those are actually the same thing, once seen from the side, once from the top! In fact, those 2 "x2o" of the lower pic are the "x2A" of the upper and those 2 "x2o" of the upper are the "x2B" of the lower!

Therefore, the already proven non-CRF property of the former thus holds for the other one too.

--- rk

[student5]

@student5: you do realize that those coordinates you got from Wolfram Alpha are basically zero, right? Look at the exponent part of the floating point numbers... they are all large negative exponents, meaning the value is very close to zero. It's probably due to roundoff error that they didn't work out to be exactly zero, but I'd imagine the actual solution is zero. :-P

Besides, the "i" in the values is kinda a red flag, because that means it's a value along the imaginary axis. So the real part is probably actually zero, and the non-zero (but practically zero) imaginary part is probably just a side effect of whatever algorithms Wolfram Alpha used to compute the result.

yes, that's true, the imaginary part of the solution is basically zero, but due to the roundoff error in wolframalpha, it gives: i*o.*10^-(something getting larger as I ask for more digits)

so the actual approximate part would be:

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A≈0.9834232572780998913369702481822330581806
x≈-0.1529927470946949461587582865691637298282
y≈0.4845749617938918523245446506490979027648


which is something

[quickfur]

Ah, I misread the "+" as "*". Silly me. Nevermind what I said.

[student5]

...
Well, trying A=x in fact makes this:

Code: Select all

  x2o
x2x o2o
  x2o


Either half (upper or lower) then clearly is a trippy. 2 of these here are to be adjoined at a squippy.

As the squippy-3-squippy dihedral angle = 75.5 degrees, this external blend remains convex at this triangle. But, OTOH, the squippy-3-tet dihedral angle = 104.5 degrees; thus the aimed for external blend will become concave at this (other) triangle!
...
--- rk

I'm sorry, but I don't really understand why the 104.5 degree angle makes this polytope concave, is it because it is bigger than the 75.5 degree angle? between which lines could you draw a line that goes outside the polytope?

I had the same lace city in my notebook, but I couldn't figure out if it was CRF or not, I also took a look at the coordinates in your other post, but I still don't fully understand where it is concave, I mean, the circumference of the middle B2x is bigger than the square and point, just like the x2x in the different orientation has a smaller circumfere than the x2o's?

I do think the fact that the two orientations are actually the same polytope is really cool thanks for finding out does that also mean you could build the following lace tower?

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x2x2o
x2o2B B = sqrt(5/2)
o2o2o


now, if I understand correctly, this is concave, because the top x2x2o laces both to the middle and bottom layer, while the middle layer is bigger, and you'll get a dent in that line e.g. the line between the (non-lacing) points of the x2o2B goes outside the polytope?

(excuse me for my slow understanding, I'm sort of having difficulty with the 4D-visualization thing )

[Klitzing]

...
Well, trying A=x in fact makes this:

Code: Select all

  x2o
x2x o2o
  x2o


Either half (upper or lower) then clearly is a trippy. 2 of these here are to be adjoined at a squippy.

As the squippy-3-squippy dihedral angle = 75.5 degrees, this external blend remains convex at this triangle. But, OTOH, the squippy-3-tet dihedral angle = 104.5 degrees; thus the aimed for external blend will become concave at this (other) triangle!
...
--- rk

I'm sorry, but I don't really understand why the 104.5 degree angle makes this polytope concave, is it because it is bigger than the 75.5 degree angle? between which lines could you draw a line that goes outside the polytope?
...

Not that 104.5 itself is already larger than 180, but that is just the angle towards the blended out equatorial squippy. Thus in your external blend (both halves, the upper and the lower) you'll get twice that angle. And 2 x 104.5 = 209 obviously is larger than 180, right?

... but I still don't fully understand where it is concave, ...

As I said, the angle of 104.5 degrees occurs in the upper half (say) at the triangle

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  .2.
.2x .2o
  .2.


that is between the upper tetrahedron

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 .2o
.2x .2o
 .2.


and the medial (to be blended out) squippy

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  .2.
x2x o2o
  .2.


Thus putting both halves together, at that triangle you'll get the concave angle of 209 degrees.

Similarily btw. the angle of 75.5 degrees occurs at the following triangle

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  .2.
x2. o2.
  .2.


between the upper squippy

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 x2.
x2. o2.
 .2.


and that to be blended out medial one (same as above). Accordingly, the external blend of both halves (upper and lower) here would use a dihedral angle of 2 x 75.5 = 151 degrees. As 151 < 180, the angle at this type of triangle still is convex. (But those at the other type not...)

I do think the fact that the two orientations are actually the same polytope is really cool thanks for finding out does that also mean you could build the following lace tower?

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x2x2o
x2o2B B = sqrt(5/2)
o2o2o


For sure, that is still the same ting. But still, the dihedral at that specific triangle (between those 2 tetrahedra, one on either halve) will be concave...

now, if I understand correctly, this is concave, because the top x2x2o laces both to the middle and bottom layer, while the middle layer is bigger, and you'll get a dent in that line e.g. the line between the (non-lacing) points of the x2o2B goes outside the polytope?

Yes, you could it phrase that way too.

--- rk

[student5]

thank you very much for explaining Klitzing, it has helped me a lot with 4D visualization

[Klitzing]

You are welcome!
OTOH I'd encourage you to search on within this current dynamical deformation process, initiated by Quickfur.
Not that I'm believing there much to be found, because the additional dimension does introduce additional restrictions too (by means of lacing edge size, planarity of faces, johnsonianity of cells - at least when considering CRFs). It's more that this track so far hasn't been looked at systematically.
--- rk

[quickfur]

I'm actually increasingly skeptical anything will be found in this direction, as you say, Klitzing, in 4D there are simply too many constraints from lacing edges, planarity of cells, etc., that any 4D polytope will need quite a number of things coinciding in order to close up in a CRF way.

However, I still haven't convinced myself that there isn't some hitherto unthought-of construction that might still work, and perhaps can only be found by exploring this "dynamical deformation process". Perhaps some construction that has an unusual way of meeting the CRF requirement.

One thought the comes to mind is "bad" lacing edges (i.e., with non-unit length) may be "covered over" by some kind of augment with a CRF surface, so that the "bad edge" is internal to the polytope. Meaning that some cell clusters that would otherwise be non-CRF, may be possible to made part of a larger CRF polytope where the troublesome parts are hidden inside the polytope.

One interesting construction that seems to hint at some such construction is the 16-cell augmented with a 5-cell. This is non-convex, because the dichoral angle between the tetrahedra in the 5-cell and the adjacent tetrahedra in the 16-cell is less than 180°. However, the fascinating thing is, the distance between the apex of the 5-cell and the far-side vertices of the 16-cell (i.e., not the vertices shared between them) is exactly the golden ratio phi, which suggests that this cell complex could potentially be closed up in a CRF way by adding some pentagonal pieces of some kind. Thus far, however, I haven't found a way to do this yet.

[student5]

today I investigated the cuboctahedral bipyramid, (which is coplanar without modification, bu I thought it might give something when modified) and it gave negative edge length :'(

another thought also occured to me, which I'll have to investigate:
how bent is the hebesphenomegacorona actually if you were to make a bipyramid from it, and would (which'd be really lucky) it make some nice angle, so it laces to itself, when bent the other way?
I'll keep you guys updated

[Klitzing]

One interesting construction that seems to hint at some such construction is the 16-cell augmented with a 5-cell. This is non-convex, because the dichoral angle between the tetrahedra in the 5-cell and the adjacent tetrahedra in the 16-cell is less than 180°. However, the fascinating thing is, the distance between the apex of the 5-cell and the far-side vertices of the 16-cell (i.e., not the vertices shared between them) is exactly the golden ratio phi, which suggests that this cell complex could potentially be closed up in a CRF way by adding some pentagonal pieces of some kind. Thus far, however, I haven't found a way to do this yet.

Thus you consider oxo3ooo3oox&#xt. The resulting equatorial concavity then provides the same concave dihedral angle as the convex dihedral angle of ex would be.

Therefore you could introduce (as smallest parts) ikepies (pt||ike) into any of those 4 cavets (situated at the triangles of the equatorial tet).
Having a tet = 2-ap, these 4 then could be aligned such that their ikes come in pairs atop those 2 line-bases.
So we thus would have to see, what gap then occurs between these ike pairs - and how that could be closed, when it would turn out to be concave in turn.

And further we would have to consider, how these added pyramidal parts mutually interrelate in turn already...
(In fact, I already have some problems in imaginating that configuration so far obtained ... )

--- rk

[quickfur]

[...] The resulting equatorial concavity then provides the same concave dihedral angle as the convex dihedral angle of ex would be.
[...]

Now, that's interesting. Does this mean that there's a tessellation of 4-space that consists of 600-cells, 16-cells, and 5-cells (perhaps with a few other pieces)?

[wendy]

I evaluated the vertex-angle of the {3,3,5} from dissecting s3s4o3o3o, you get one {3,3,4}, one {3,5,5/2}, one {3,3,5} and five {3,3,3} That makes 5 + 76 + i + 5t. This gives with t = margin angle of simplex - 1/5, you get t = 1:20.V8, and i = 39-5t = 33.15:60.

You would need five times as many pentachora as {3,3,5}, and this makes 0:39 or 13/40 of the circle. But because you can't get 1/5 out of 1/8, 1/24 and 1/16, you would have to multiply the lot by 5, giving 5 {3,3,5} + 25 {3,3,3} = 1;75. You could add, 9 {3,3,4} to get double-density cover.