oA
Bo oE
Co oD
AB BC CD DE EA AC CE EB BD DA
ooooo*a*c*e*b*d*a
o : (real) node
 : link
*a : virtual node, representing the leftmost real node
*b : virtual node, representing the second real node from the left
etc.
x3o5x = rhombicosidodecahedron
. o5x = pentagons
x . x = squares
x3o . = triangles
x x : this is the alignment of the 4
lateral edges of a cube; in fact
x x it is the stack of 2 towers, a square
atop a square; the square in turn is
a stack of 2 edges.
x x : this is the alignment 8 edges of size x=1
x w w x and 4 edges of size w=1+sqrt(2), or, as
towers, a square (x  x), a regular octagon
x w w x (x  w  w  x), an other octagon, and an
x x other square, i.e. the description of sirco.
o5o o5o
o5x
x5o x5o
o5o f5o o5o
o5f o5f
o5x o5x
f5o f5o
o5o x5x o5o
o5f o5f
x5o x5o
f5o f5o
o5o o5f o5o
o5x o5x
x5o
o5o o5o
naq = 3_1,2:
h h* h =x3o3o3o3o (hix)
h* d h h*=o3o3o3o3x
h h* d =o3o3x3o3o (dot)
laq = 2_1,3:
o
o +  o + = o3x3o3o3o (rix)
 = o3o3o3x3o
 # + # = x3o3o3o3x (scad)
o +  o
o
lin = 1_2,3:
= = = x3o3o3o3o (hix)
1 1 # = o3o3o3o3x (hix)
# + # + = x3o3o3x3o (spix)
   = o3x3o3o3x (spix)
1 0 1 1 = o3o3x3o3o (dot)
+ + 0 = xo3xo3oo3ox3ox ((tix, inv tix)compound)
=  =
1 1
#
wendy wrote:Stott Expansion
One of the earliest discoveries of the modern era is due to Mrs Alice Boole Stott. This is that one can convert various polytopes into each other by expanding things.
[...]
One can build these polytopes as vectors, if one has a suitable coordinate system. The t_0,1 becomes not just two operations, but the coordinate (1,1,0), whose mirroredge reflection in the walls of an oblique coordinate system gives the truncated cube.
The vectors are for the cross polytope these. q = sqrt(2), The resulting edge is 2. The primary cell is one where all values are sorted from greatest to least, and made positive. This means, eg for (1, 5,3), the primary coordinate is (5,3,1).
0 (q,0,0,...) = cross
1 (q,q,0,...)
2 (q,q,q,...0)
n1 (1,1,1,1,1,1) = cube
The generalised cubesymmetry uniforms of edge 2 has the coordinates A.q + B, where all values of 1 to A occur, and B is 0 always or 1 always. So, we find, a figure 2q1, 1q1, 0q1 corresponds to q,0,0 + q,q,0 + 1,1,1 = x3x4x.
[...]
Klitzing wrote:Hy Wendy,
do I get your lace cone concept correctly, if I would translate it into what elsewhere is called wedges?
I.e. the top base (say) has to be degenerate?
So, the tetrahedron, positioned as point  triangle, is a lace cone with respect to its top base (vertex only), thus it is a pyramid in fact. But it is not a lace cone with respect to its bottom base, as the triangle already is full dimensional.
The tetrahedron, positioned as line  gyro line, is a lace cone with respect to either base (both are not full dimensional).
Correct?
 rk
wendy wrote:[...]
lace tegums
The dual of a lace prism is a lace tegum. Just as a lace prism is a progression from base to base, the lace tegum is the intersection of lace cones.
[...]
antitegum
The dual of an antitegum is an antiprism. It's the lacetegum of a figure and its dual. All surtopes of the antitegum are antitegums. The prismatotope is a simplexantitegum, for example. The Hass antitegum is the diagram of connections of the surtopes of a polytope. It is in essence, the antitegum of the polytope itself. The "antitegmal sequence" is the sequence of 'decent of the faces of the dual', where one considers at various points the intersection of a cube of size x, and a octahedron of size ax. This corresponds to parallel sections of the corresponding antitegum. In the case of the regulars, the centres of surtopes of every order occur at the same point, so the antitegmal sequence passes through the truncates and rectates in order.
So, the tetrahedron, positioned as point  triangle, is a lace cone with respect to its top base (vertex only), thus it is a pyramid in fact. But it is not a lace cone with respect to its bottom base, as the triangle already is full dimensional.
The tetrahedron, positioned as line  gyro line, is a lace cone with respect to either base (both are not full dimensional)
x = edge of size 1 (corresponding to the verf. of a regular triangle)
q = edge of size sqrt(2) (corresponding to the verf. of a reg. square)
f = edge of size tau (corresp. to the verf. of a reg. pentagon)
v = edge of size 1/tau (corresp. to the verf. of a reg. pentagram)
h = edge of size sqrt(3) (corresp. to the verf. of the reg. hexagon)
fx v x q f h u x r
  V  Q F H U S R
5  52  4 5 6 oo 3 2
         
oo o o o o o o o o
$A vert $A vert
4/ A 1 4 / A
/ B 1 / B
B$$C C 1 1o$B
edges edges
x4x3x AB x4x = r(3.414) x4o3x AB = r(2,000)
AC x2x = r(2.000) A1 = 1.41421
BC x3x = r(3.000) B1 = 1.00000
student91 wrote:One thing I'd like to be highlighted are the incidence matrices. As far as I foud out from klitzing's site, these list the incidences of parts of polytopes. But still there are two things unclear to me:
1 what is exactly ment by incidense, is there a clear definiton for this, and how is it quantified in the matrix?
2 what do the coloums stand for? Are these the same as the rows, but then why is a_ij not equal to a_ji?
student91
v e1 e2 h1 h2
v e h c
v 24 2 1 1 2 v 600 4 6 4
e 2 1200 3 3
e1 38 2 24  1 1 h 5 5 720 2
e2 88 2  12 0 2 c 20 30 12 120
h1 3 3 3 0 8 
h2 8 8 4 4  6
4 8 6 6
tetra (/3) rCO (/4) cube
 12
(/2/) rectangle

=
4B1/ = (72)/4B =====(27)===== 4/B 72
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