student91 wrote:So far I've got the following tower:.
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x3o5xx3o5fx3f5oF3x5oV3x5oF3f5o(x3F5o+V3o5f)f3x5fo3x5F(???)(V+x)3o5x
I'm not sure what should be inserted into the (???), all other is just the tedrids and the completed dodecahedra.
xxxFVF(xV)f(oB)F(oB)f(xV)FVFxxx3oofxxf(Fo)x(xo)x(xo)x(Fo)fxxfoo5xfoooo(of)f(Fx)x(Fx)f(of)oooofx&#xt
x3o5x  x3o5f  x3f5o  F3x5o  V3x5o  F3f5o  (x3F5o + V3o5f)  f3x5f  (o3x5F + B3o5x)  F3x5x  ...
Klitzing wrote:[...]
Wrangled it out now, that mentioned 42diminishing of sidpith  with 696 cells and 1560 vertices!  is given by
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xxxFVF(xV)f(oB)F(oB)f(xV)FVFxxx3oofxxf(Fo)x(xo)x(xo)x(Fo)fxxfoo5xfoooo(of)f(Fx)x(Fx)f(of)oooofx&#xt
i.e. could be given as the tower stack
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x3o5x  x3o5f  x3f5o  F3x5o  V3x5o  F3f5o  (x3F5o + V3o5f)  f3x5f  (o3x5F + B3o5x)  F3x5x  ...
(where the rightmost given layer is the equatorial one, beyond the former ones reoccur in reversed order). Here B = V+x = F+f = 2f+x.
 rk
quickfur wrote:P.S. but surely you mean 42diminishing of sidpixhi (x5o3o3x), not sidpith (x4o3o3x)?
quickfur wrote:BTW, for the past few weeks I've been working on and off on completing Klitzing's idea of a trigonal J82 polychoron (attach 3 J82's around a triangular prism and fill in the rest). So far I've gotten up to the bottom of the J82's by filling in the gaps between them with mibdies, cuboctahedra, and truncated tetrahedra, but haven't found a way to close up the bottom of the polytope in a CRF way just yet. I'm getting bottom cells with edge length 2, that suggest incomplete hexagons, but there isn't enough space to actually complete the hexagons in a convex way. Any ideas?
# x2o3x
# 
# F2o3f
# / \
# A2x3o \
#  V2o3F
#  / \
# B2x3x F2o3A
#    \
# W2o3x   \
#  D2f3x  \
#  / \  x2o3B
#  /  V2x3F 
# B2o3u  \ 
#   f2x3V
# X2x3x  / 
# A2F3o / 
# \ / 
# F2u3f 
# /\ 
# o2f3F  \ 
# x2F3x \ 
#  \x2o3V
#  \ 
#  u2x3f
#  
#  F2x3o
#  /
# x2u3o
#
# A = f+2x
# B = f^3
# V = 2f
# W = 3f
# u = 2x
1 = x2o3x  r=0,763762616 = sqrt(7/12)
2 = F2o3f  r=1,608167742 = sqrt((33+13*sqrt(5))/24)
3 = A2x3o  r=1,898914379 = sqrt((53+15*sqrt(5))/24)
4 = V2o3F  r=2,214211969 = sqrt((8+3*sqrt(5))/3)
5 = B2x3x  r=2,342235679 = sqrt((39+12*sqrt(5))/12)
6 = F2o3A  r=2,465140051 = sqrt((81+29*sqrt(5))/24)
7 = W2o3x  r=2,494776505 = sqrt((89+27*sqrt(5))/24)
8 = B2f3x  r=2,496282029 = sqrt((39+16*sqrt(5))/12)
9 = x2o3B  r=2,496282029 (same as 8)
10 = V2x3F  r=2,471587743 = sqrt((21+7*sqrt(5))/6)
11 = B2o3u  r=2,412343531 = sqrt((43+12*sqrt(5))/12)
12 = f2x3V  r=2,357380568 = sqrt((73+27*sqrt(5))/24)
13 = A2x3x  r=2,067012938 = sqrt((23+5*sqrt(5))/8)
14 = A2F3o  r=2,357380568 (same as 12)
15 = F2u3f  r=2,235671297 = sqrt((73+21*sqrt(5))/24)
16 = o2f3F  r=2,137615801 = sqrt((7+3*sqrt(5))/3)
17 = x2F3x  r=1,934092031 = sqrt((27+8*sqrt(5))/12)
18 = x2o3V  r=1,934092031 (same as 17)
19 = u2x3f  r=1,656911583 = sqrt((6+sqrt(5))/3)
20 = F2x3o  r=1,430684740 = sqrt((29+9*sqrt(5))/24)
21 = x2u3o  r=1,258305739 = sqrt(19/12)
1+2  h=0,467086179 = sqrt((3+sqrt(5))/24) a = f*b
2+3  h=0,288675135 = sqrt(1/12) b
2+4  h=0,755761314 = sqrt((7+3*sqrt(5))/24) a+b
3+5  h=0,755761314 a+b
4+5  h=0,288675135 b
4+6  h=0,755761314 a+b
5+7  h=0,755761314 a+b
5+8  h=0,934172359 = sqrt((3+sqrt(5))/6) a+a
6+9  h=0,467086179 a
6+10  h=0,755761314 a+b
7+8  h=0,178411045 = sqrt((3sqrt(5))/24) ab
7+11  h=0,755761314 a+b
8+10  h=0,288675135 b
8+11  h=0,577350269 = sqrt(1/3) b+b
8+14  h=0,755761314 a+b
9+12  h=0,755761314 a+b
10+11  h=0,288675135 b
10+12  h=0,467086179 a
10+15  h=0,755761314 a+b
11+13  h=0,755761314 a+b
12+15  h=0,288675135 b
12+16  h=0,467086179 a
12+18  h=0,755761314 a+b
13+14  h=0,577350269 b+b
13+15  h=0,288675135 b
13+19  h=0,467086179 a
14+15  h=0,288675135 b
15+17  h=0,467086179 a
15+19  h=0,755761314 a+b
16+17  h=0,288675135 b
16+18  h=0,288675135 b
17+19  h=0,288675135 b
17+21  h=0,577350269 b+b
18+19  h=0,288675135 b
19+21  h=0,288675135 b
13+20  h=0,645497224 = sqrt(5/12) 2ab = a+ab
20+21  h=0,110264090 = sqrt((73*sqrt(5))/24) 2ba = b+ba
1 = x2o3x  r=0,763762616 

a


2 = F2o3f  r=1,608167742 
b

3 = A2x3o  r=1,898914379 
4 = V2o3F  r=2,214211969
5 = B2x3x  r=2,342235679
6 = F2o3A  r=2,465140051
7 = W2o3x  r=2,494776505
8 = B2f3x  r=2,496282029 + 9 = x2o3B
10 = V2x3F  r=2,471587743
11 = B2o3u  r=2,412343531
14 = A2F3o  r=2,357380568 + 12 = f2x3V
15 = F2u3f  r=2,235671297
16 = o2f3F  r=2,137615801
13 = A2x3x  r=2,067012938
18 = x2o3V  r=1,934092031 + 17 = x2F3x
19 = u2x3f  r=1,656911583
20 = F2x3o  r=1,430684740
21 = x2u3o  r=1,258305739 + 22' = x2(v)3(Vx) : Ansatz 16+17+18+19+21+22'(+23) = co  but see below
22 = o2x3(v)  r=0,504622639 = sqrt((3sqrt(5))/3) : Ansatz 16+17+18+19+21 = tricu and 18+19+21+22 = tut  but see below
quickfur wrote:Klitzing wrote:[...]
Thus, what then would be xfox2oxfoPoooo&#xt (P=3,4,5)?  I will have to take a look into these...
 rk
Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!! The slight adjustment is that the last vertex layer must be x2o3x, not x2o3o, otherwise the mibdies are noncorealmar. So the corrected symbol is xfox2oxfo3ooox&#xt (note the last x in place of o).
This is a truly beautiful little CRF; it has 3 mibdies and 2 tetrahedra on one side, 1 trigonal prism, 2 octahedra(!), 3 square pyramids, and 6 tetrahedra on the other side. Even better yet, it has 17 vertices and 17 cells  the same number of vertices as cells!
Here's a look at the 3 mibdies surrounding an edge:
The 2 tetrahedra that fill in the gaps between the mibdies should be obvious.
Now here's the far side, which has a most interesting configuration of cells:
I highlighted the trigonal prism which lies antipodal to the edge shared by the 3 mibdies. We see a most interesting pattern of tetrahedronsquare pyramidtetrahedron interfacing each mibdi to the trigonal prism.
quickfur wrote:Klitzing wrote:[...]
Got a promising idea! What then about teddi  ...  tedrid?
[...]
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x3o f3o o3x teddi
o3x F3o f3x x3o
F=ff=x+f=2x+v,
V=F+v=2f=2x+2v
x3o x3f F3x V3x F3f Vx3oF f3x o3x tedrid
[...]
 rk
Finally got around to this today. Very nice!!!! It has a beautiful arrangement of pentagonal rotundae connecting the teddi to the tetrid, and a fascinating cluster of octahedra around the bottom.
Here's a projection centered on the top teddi, with other teddies highlighted in yellow and midbies in magenta:
Klitzing wrote:And here finally is its full Dynkin symbol:
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xxFVF(Vx)fox3ofxxf(oF)xxx3xoooo(xo)xfo&#xt
(Cf. esp. the layers which contained those unknown quantities B,C,D.)
 rk
quickfur wrote:Klitzing wrote:[...]
Got a promising idea! What then about teddi  ...  tedrid?
[...]
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x3o f3o o3x teddi
o3x F3o f3x x3o
F=ff=x+f=2x+v,
V=F+v=2f=2x+2v
x3o x3f F3x V3x F3f Vx3oF f3x o3x tedrid
[...]
 rk
Finally got around to this today. Very nice!!!! It has a beautiful arrangement of pentagonal rotundae connecting the teddi to the tetrid, and a fascinating cluster of octahedra around the bottom.
[...]
Here are the pentagonal rotundae wrapping around the teddi:
Each pair of rotundae shares a triangular face.
o5o
o5o x5o o5x o5o
diminish:
o5o
o5o x5o o5x
o5o
o5o (x)5f o5x
partial stott:
x5o
x5o o5f x5x
x3o o3f f3o o3x ike
o3x x3f F3o f3f o3F f3x x3o id
F=ff=x+f=2x+v,
V=F+v=2f=2x+2v
x3o x3f oF3Vx f3F V3x x3V F3f Vx3oF f3x o3x srid
diminish:
x3o o3f f3o o3x ike
o3x x3f F3o f3f o3F f3x x3o id
F=ff=x+f=2x+v,
V=F+v=2f=2x+2v
x3o x3f F3x V3x F3f Vx3oF f3x o3x srid
Klitzing wrote:quickfur wrote:[...]
Here's a look at the 3 mibdies surrounding an edge:
I further considered the position of these 3 additional vertices. The former polychoron has been the tower x o3o  f x3o  o f3o  x o3x. Obviously the additional layer then has to be something of the form o o3#. The lacing conditions then brought up that # has to be x. Thus we'd get x o3o  o o3x  f x3o  o f3o  x o3x here.
But then it might be possible too that we don't have pentagonwise connected pairs of peppies, but rather 5fold rosettes of tets. Accordingly the total count well could be instead 15+30+6+2 tets + 2 octs + 3 squippies + 1 trip. (Depends on what would be the convex case.)
 rk
Klitzing wrote:quickfur wrote:Klitzing wrote:[...]
Thus, what then would be xfox2oxfoPoooo&#xt (P=3,4,5)?  I will have to take a look into these...
 rk
[...]
Here's a look at the 3 mibdies surrounding an edge:
The 2 tetrahedra that fill in the gaps between the mibdies should be obvious.
Now here's the far side, which has a most interesting configuration of cells:
I highlighted the trigonal prism which lies antipodal to the edge shared by the 3 mibdies. We see a most interesting pattern of tetrahedronsquare pyramidtetrahedron interfacing each mibdi to the trigonal prism.
Isn't it that the angles between the mibdi and any of the other cells are quite small here? Then one might consider to augment that fellow e.g. by 3 mibdipyramids.
[...]
(Suppose quickfur could manage the needed convexity tests quite easy  and potentially even provide renderings of these up to 3 augmentation cases...)
 rk
# x2o3o
<±1, 0, 0, phi^2/√3>
# f2x3o
<±phi, ±1, 1/√3, 0>
<±phi, 0, 2/√3, 0>
# o2f3o
<0, ±phi, phi/√3, phi/√3>
<0, 0, 2*phi/√3, phi/√3>
# x2o3x
<±1, ±1, 1/√3, phi^2/√3>
<±1, 0, 2/√3, phi^2/√3>
# o2o3x (the 3 augments)
<0, ±1, 1/√3, (phi2)/√3>
<0, 0, 2/√3, (phi2)/√3>
x2o3o
o2o3x
f2x3o
o2f3o
x2o3x
Klitzing wrote:Splendid pics, as usual!
That "B" shouldn't have been too hard  as I already provided to you its solution.
In fact I even provided your last stacking, as my subject line of that PM read:
x o3o  o o3x  f x3o  o f3o  x o3x.
 rk
I just calculated the dichoral angles of the bases of the ursachora, and it turns out all of them can be augmented: if my calculations are correct(EDIT:They're not, look at the bottom of this post), the tetrahedral one has a dichoral angle of exactly 72 degrees, thus adding a pen (dichoral angle of 75.5224) does not make the total exeed 180.quickfur wrote:This makes me wonder, do augmented ursatopes exist too? Ostensibly, the tetrahedral teddy should be augmentable by a 5cell on the top tetrahedron (analogous to the 3D augmented teddi, J64), or up to 3 teddi pyramids on one of the side cells. If these augmentations exist, perhaps augmentations of the expanded ursachora might exist too?
x3o3o
o3o3x
f3o3o
o3x3o
# x3o3o
apecs<1/√2, 1/√2, 1/√2> ~ <phi^2/√2>
# f3o3o
apecs<phi/√2, phi/√2, phi/√2> ~ <0>
# o3x3o
apacs<√2, 0, 0> ~ <phi/√2>
# o3o3x: teddi pyramid apices
apecs<1/√2, 1/√2, 1/√2> ~ <1/(phi*√2)>
# x3o3o
apecs<1/√2, 1/√2, 1/√2> ~ <phi^2/√2>
# f3o3o
apecs<phi/√2, phi/√2, phi/√2> ~ <0>
# o3x3o
apacs<√2, 0, 0> ~ <phi/√2>
# o3o3o: 5cell on top of the top cell
<0, 0, 0, phi^2/√2 + (√10)/2>
o4o3x
A4o3o
o4o3f
o4x3o
# o4o3o: top augment with octahedral pyramid
<0, 0, 0> ~ <√(2*phi) + √2>
# o4o3x
apacs<0, 0, √2> ~ <√(2*phi)>
# o4o3f
apacs<0, 0, √2*phi> ~ <0>
# o4x3o
apacs<0, √2, √2> ~ <√(2/phi)>
quickfur wrote:@student91: are you sure your calculations are correct? 'cos I just built a model of the tetraaugmented tetrahedral teddy, and AFAICT, it's CRF!!!
The lace tower is:
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x3o3o
o3o3x
f3o3o
o3x3o
The apices of the augments are, obviously, o3o3x. And obviously, there are lacing edges from x3o3o to f3o3o. Each augment introduces 3 pentagonal pyramid cells, so there are 12 pentagonal pyramids in 6 pairs. The octahedron and 4 tetrahedra at the bottom are unchanged.
o o x x o o
f o o f
x o o x
quickfur wrote:... Anyway, when I get some free time again, I'll see if I can construct the augmentations of the octahedral/icosahedral teddies. Well, the icosahedral teddy is for sure augmentable, because it's just diminishing of the 600cell. That seems to suggest that the octahedral teddy ought to be augmentable too. We'll see!
quickfur wrote:Oh, and the octahedral teddy can be augmented with an octahedral pyramid ...
Klitzing wrote:quickfur wrote:@student91: are you sure your calculations are correct? 'cos I just built a model of the tetraaugmented tetrahedral teddy, and AFAICT, it's CRF!!!
The lace tower is:
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x3o3o
o3o3x
f3o3o
o3x3o
The apices of the augments are, obviously, o3o3x. And obviously, there are lacing edges from x3o3o to f3o3o. Each augment introduces 3 pentagonal pyramid cells, so there are 12 pentagonal pyramids in 6 pairs. The octahedron and 4 tetrahedra at the bottom are unchanged.
Sorry quickfur, but it seems that you considered the tetraexcavated tetrahedral teddi!
Cause the "mere" (i.e. unaugmented / unexcavated) tetrahedral teddi looks in appropriate lace city like this:
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o o x x o o
f o o f
x o o x
Accordingly the augmentation ought be some #3o3o (or rather: o # # o), with "#" most probably even larger than f!
On the other hand, your proclaimed "tips" provide a dual tet of the same size and situated parallel to the bottom tet. Therefore the convex hull of just these 2 is nothing but the hex. So this, after all, provides a further interesting interrelation to the "mere" tetrahedral teddi!
 rk
In fact I found o3o3x by solving for apex coordinates that are equidistant from each of the 3 cell layers, and yield unit edge lengths. Wouldn't this be the same thing as erecting CRF pyramids on the teddies?
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