LAMINATOPES
Miss Wendy Y Krieger
wykrieger@hotmail.com
2002.02.26
-----------------------------------------------------------------------
A new process for generating frighteningly large masses of
uniform
honeycombs in H^2, and some limited applications in
higher dimensions.
A practical application of polytopes with
infinite sides and ideal
verticies.
Wythoff's
construction has company.......
========================================================================
LAMINATOPES
-----------
A laminatope is a figure
bounded by unbounded sides. The general interest
is in
"closed" ones, which mean, those for which the addition of further
unbounded planes, other than the supplied ones, is not
possible. For
example, the single plane is "open", since
one can place another parallel
plane underneath it. In the
spherical geometry, the same single plane is
"closed".
The best proof of existance of laminatopes in hyperbolic
space, is the
interior planes of regular and iso-face figures
with hyperbolic cells. For
example, the {8,3,4} or {17,4,3} are
both examples of such a figure. Another
class are figues like
bt{p,q,p}, such as bt{19,27,19} which is tiled
completely with
t{27,19}.
One fills a laminatope up with cells, such as one
might a slab with
hexagonal prisms. These cells put an
"etching" on the laminate faces. If
two sets of laminate
etchings match, the two laminatopes can be put together
side to
side, and space filled alternately with these laminatopes.
An example is a strip, filled with triangles or with
squares, both expose an
aperigonal etching. You can stack
strips of triangles and squares in any
order, but in order to
maintain uniformity, the vertecies must be the same.
This gives
either all triangles, all squares, or alternating triangles and
squares.
In higher dimensions, a further element of
sense is added. The one slab
etching that is shared by
different slabs is the triangular lattice {3,6},
which appears
as the sides of an prismatic form or an "oct-tet truss". The
"octtet truss" exists in all dimensions, as a slice of the
lattice made from
the span of the simplex's edges.
There
are five different alternations, because the oct-tet truss is not
symmetric on the etching.
In three or higher dimensions,
the "oct-tet" lattice does not have the
planes of symmetry
running through the face planes. What happens is, is
that
the etchings advance down the long diagonal of the 60 deg rhomb,
through the sequence of cells.
A different lattice is
formed by treating the laminate faces of the octtet
truss as a
mirror. In three dimensions, this yields the verticies at the
"hexagonal close pack" points.
One can for either of
these honeycombs, slip a layer of mirrors in them, and
both of
these will have the same vertex figure (essentially half of a
rr{3..3} and a pyramid on the other side). But only one
will preserve the
mirror plane that runs around the "waist" of
the prisms.
The fifth example is simply a stack of prisms.
In hyperbolic figures, one looks for various forms where one
of the
verticies is in the ideal region. More often than
not, a face forms there,
and this could end up flat.
Let
{p,q,r} be such a group, where {p,q} is spherical and {q,r}
hyperbolic.
now, for the face of {q,r} symmetry to be flattable,
it must not touch
another copy of it, even at the
verticies. Of the 15 possible Wythoff
constructions on
{p,q,r}
{p,q,r} has ideal verticies.
Anything that does not mark the first node will result in the
hyperbolic cells sharing some element, and hence not
flatenable.
This leaves:
etching
etch I've included my "new"
t{p,q,r}
{q,r} xPxQoRo
xQoRo notation so I can follow
c{p,q,r}
r{q,r} xPoQxRo
oQxRo what I'm talking about!
rr{p,q,r}
{r,q} xPoQoRx oQoRx
ct{p,q,r}
t{q,r} xPxQxRo
xQxRo [n] is the sch(pi/n)
rt{p,q,r} rr{q,r}
xPxQoRx xQoRx the base of a
trianlgle
rt{r,q,p}
t{r,q} xPoQxRx
oQxRx bounded by two sides of n.
ot{p,q,r} ot{q,r}
xPxQxRx xQxRx
To complete the picture, you need
to also consider the following four
honeycombs and etchings.
etch
{3,5,3,3}
{3,5,3} x3.3.5.3.
.3.5.3. . is all combos
{5,3,4,3} {5,3,4}
x3.4.3.5.
.4.3.5. of x and o.
{5,3,3,3,3} {5,3,3,3}
x3.3.3.3.5. .3.3.3.5.
{5,3,3,3,4}
{5,3,3,3} x4.3.3.3.5. .3.3.3.5.
Because uniformity implies reduction to one degree of
freedom, and the
flatness uses that degree, all of the like
etchings must be the same size.
So, if {q,r} rises in different
ways, a honeycomb is possible with the two
sides alternating.
For example, {4,5} can arise from t{3,4,5} and
rr{3,5,4}. The vertex
figures of these are a pentagonal
pyramid, and a pentagonal antiprism, where
the top side is
smaller than the equator. Put together, this makes a little
thing not unlike a drop, or an distorted icosahedron, where the
south pole
has been lobbed off, and the tropic of cancer moved
south to the equator.
The array of faces at a vertex is:
equator pointing north = tr {3,5}
equator
pointing south = pentagonal prisms
equator
= {5,4} laminates (a van oss figure)
polar c pointing
north = triangle prisms
polar
plate
= icosahedron.
While all of these yield "possible"
semi-uniform figures, not all are may
become "uniform", since
the uniformity and flatness consume all degrees of
freedom. That is, it is simply a matter of working through
the list of the
figures above to see which are uniform.
It reveals itself as a relatively easy task to see if the
vertex figure is
passable. Firstly, we need to consider
the halves in isolation. That is,
we can leave the {5,4}
in there, and then, if one or more figures use {5,4},
form the
family of laminatopes on {5,4}.
With t{p,q,r}, the cells are
t{p,q}, and {q,r}, the vertex figure being a
{r} prism, the edge
at the base should be [p], and the sloping edge a [2q].
r can be
solved from p,q, since {q,r} and {2p,4} have the same edge. The
trivial case of q=2p, r=4 yields the half-octahedron of
{q,3,4}. The non
trivial cases are {3,6} or {6,3} = {4,4},
which give triangle and hexagonal
prisms, from t{2,4,4},
t{2,3,6} and t{2,6,3}
The other non-trivial case is {3,8} =
{8,4}, ie t{4,3,8}. This produces a
thing with 8 t{4,3}
and a flat {3,8} etching, as uniform.
The other regular
etching is from rr{p,q,r} giving {r,q} as the etching. For
{3,8}
to appear here, we need a rr{p,8,3} with {p,8} as spheric.
This does
not happen, except {2,8,3}. But this has
diverging edges without hope, and
so we abandon that idea
too. It would need faces like {2,8}.
Of other figures,
we note the face is a q-gonal antiprism, with the base as
the
equator, of edge [r], and the top as [p]. The zigzag between
the top
and bottom is sqrt(2), or [4]. You can calculate
the value of r in each of
the cases, since such an antiprism
that has a [p] sized {q} and a zigzag of
sqrt(2) to the equator
defines some definite size.
One can procede through all 35
instances, being the five spherical values of
{p,q} for the
seven examples above, to see if any of these produce an
integeral {r}. This is because the operation of
uniformity, flatness, and
the flatness of form in {q,r} uses one
additional degree of freedom, but the
value of {r} is an
additional degree of freedom. The whole can be solved in
regular Euclidean geometry, since for the figure to exist, the
vertex figure
must be properly formed in Euclidean
geometry. This is not one of my
strengths, though.
Of the cases in the type bt{q,p,q}, none of the hyperbolic
cells may share
elements (since they would not be flattenable,
and so we need something like
rr{P,Q,P} xPoQoPx
Every other form shares more than verticies, but even in
this case, the
sharing of this single vertex means the
hyperbolic faces are not
flattenable.
And this remains
true in higher dimensions, so we can immediately rule out
the
likes of:
rr{5,3,3,3,5} x5o3o3o3o5x
VAN OSS SNUBS
-------------
Another fruitful
source is to look at "Van Oss snubs", and other figures
where
the van Oss polygon appears as a cell. In all of these cases,
the van
Oss figure becomes an etching on a laminatope, and we
are free to tile space
with different laminatopes of the same
polygon.
There is a family of snubs whose verticies fall
into the verticies of a
{3,p}, where the sides are counted thus:
3 snub triangles
0 triangles
for a "2"
1 triangle for a "3"
2
triangles for a "p" (being a 'cap' of triangles removed)
p/2 triangles for a "oo" (in the role of Van Oss polygon)
The individual cases are listed below.
3,3 0 0+0+0
s{2,2} = tetrahedron as half-cube.
3,4 1
0+0+1 s{2,3} = octahedron
3,5 2
0+0+2 s{2,5} + poles = icosa
0+1+1 s{3,3}
3,6
3 0+0+3 s{2,oo}
triangular strip
0+1+2 s{3,6} s{3,6}
1+1+1 s{3,3,3:} {3,6}
3,7 4 0+2+2
s{7,7} subgroup of {4,7}
1+1+2 s{3,3,7;} subgroup of {3,14}
3,8 5
1+2+2 s{3,8,8;} subgroup of {6,8}
0+1+4 s{3,oo}
3,9
6 2+2+2 s{9,9,9} subgroup
of {3,18}
3,10 7
1+1+5 s{3,3,oo;}
0+2+5 s{10,oo}
3,12
9 1+2+6 s{3,12,oo}
3,14 11 2+2+7 s{14,14,oo}
Another source of laminates is where one can form a laminate
systematically
in a {p,p/2}. One can use any suitable
{p,q}, but this becomes an exercise
in fruitility, as the
etching is almost always unique to that {p,q}, and the
original
figure arises when the laminatopes are tiled. In the case of a
{p,p/2}, the laminatope is the same etching as one derived from
a {3,p}. Of
course, we need not restrict ourselves to this
class, any set that has
matching pairs will do.
3,6 3
0+0+3 s{2,oo} triangular strip
4,4 2
sqra t{2,oo} strip of squares
The only distinct laminate here is alternating squares and
triangles
3,8
5 0+1+4 s{3,oo}
2 octa t{4,oo}
Using either of these alone produces
either the {3,8} or {8,4} alone. But
alternately, these produce
honeycombs with a vertex array 8 8 3 3 3 3. There
are two
different forms, as the s{3,oo} is chiral. One can preserve
the
chirality as one moves across the octagon, or one can
reverse the chirality
(so waist-mirrors are created in the
t{4,oo}'s octagons.
3,10
7 1+1+5 s{3,3,oo;}
0+2+5 s{10,oo}
The first of these is symmetric, and
completely of triangles, so using it
alone will result in the
{3,10} itself. The second is both chiral and
distinct, so
it can be used either in distinct chiral forms, in alternate
forms, or in conjunction with {3,3,oo}
This gives the
admissiable set of
3,10,3,3, + 3,10,3,3
3,10,3,3
+ 3,3,10,3
3,10,3,3, + 3,3,3,3,3
The {5,6} has the same van Oss polygon
as well. So we may be able to put
stable van Oss polygons
that has the vertex figure oo,5,5,5. It does not
happen,
since you can not distribute the five exposed and ten internal sides
around the three pentagons, so that each pentagon has the same
vertex
configuration.
3,12
9 1+2+6 s{3,12,oo}
6-0-6
4-4-4
The 3,12 is of course, chiral, and distinct, since it
has a non-triangular
face. We can then use it either in
the same chiral (with a digonal
rotation at each vertex), or as
a reflected form (where the laminate van-oss
polygon is a
mirror, to give
3,12,3,3,3
+ 3,12,3,3,3 rotated
3,12,3,3,3 +
3,3,3,12,3 reflected
The 12,6 can be
divided into a figure with a vertex figure of oo,12,12,12 in
two
distinct ways. In one way, the dodecagon opposite the aperigon
has an
aperigon opposite every vertex. The other two
dodecagons at a vertex are
from a dodecagon that shares
alternate sides with the middle dodecagon and
an aperigon.
The number of laminagons on each dodecagon around the vertex
is
6,0,6.
The alternate way is that every dodecagon shares four
sides with an
aperigon, and is opposite to four others.
This can only happen if the four
'external' are evenly spaced
(at N, S, E, and W), and the four verticies not
on these sides
(NE, NW, SE SW) are the medial verticies. The number of
laminagons on each dodecahedron is 4,4,4.
Both of these
laminatopes support alternation of the sides (ie they can act
as
waist-mirrors).
Using the two dodecahedra-only figures alone
yields the {12,6} itself. But
in conjunction with the
snub(3,12,oo:) yields four distinct figures, all
with the same
vertex figure: (3,3,3,12,3,12,12,12)
444 waist, 444 rotate,
606 waist, 606 rotate.
3,14
11 2+2+7 s{14,14,oo}
This reveals
only one new figure, with a vertex figure of
3,14,3,14,3 + 3,14,3,14,3
8,6 4-0-4
4,8
2-0-0-2
There are two van-oss polyhedra. Firstly note
that using two derived from
the {8,6} or from the {4,8} alone
yields the same figure. The two sides of
the van-oss
polygons are different.
The only form that arises from the
{8,6} is where the the octagon opposite
the {oo} is always
opposite {oo}'s, and the octagons that share a side with
the
van-Oss polygon share alternate sides with van-Oss polygons and the
central octagons.
We can make two different figures on
the {4,8} as follows. A strip of {4,8}
squares produces a
cane thingie with right angles along its length. The side
of
this cane thingie is the same as a {4,8}. If we take vast
numbers of
these, we can put another cane so that the
right-angles match, and so on.
When we do the other side of the
cane, we can choose to either match the
same as the first side,
or alternate with it. In the latter case, every
square
shares one side with a van Oss polygon.
Because the {8,6}
and {4,8} have the same edgelength, we can alternate space
with
the laminahedra we have just made. This gives two figures both
with a
vertex figure of (8,8,8,4,4,4,4).
[I have some
reservations about the alternating form, not because it
alternates, but that because it may give a non-through route for
the squares
that arise in it. For example, if we let it
slip along one of the sets of
squares, from the symmetric
pattern, there is no through routes across the
thing.
Either the interconnectors have to be uniformly one (as in brick
work), or some other disruption has to be done to the other
through-routes.]
{5,12} and {12,10}
These
look so tempting. But I think that there are too many internal
edges
for straight lines to meaningfully form here.
All
together, we find 14 uniform honeycombs in hyperbolic space, based
on
laminagons extracted in different regular honeycombs of the
same edge.
Of these, there are four with the vertex
{3,6} and
{4,4} 1 array of (3 3 3 12
3 12 12 12), and two
{3,8} and
{8,4} 2 with the array of
(8 8 8 4 4 4 4)
{3,10}
3
{3,12} and {12,6} 6
{3,14}
1
{4,8} and {8,6}
2* [See note in section]
WAIST
MIRRORS
-------------
Another fruitful source to look
for is in the class of waist-mirrors. Unlike
the laminates,
where the hyperbolic face is used as a boundry, here the
hyperbolic faces are used as a mirror around the waist of the
prisms, the
base of which is a prism.
For example, the
{8,3,4} has cells inscribed in an equidistant surface from
some
interior plane. If we use that plane as a mirror, this
produces two
non-touching {8,3,4}'s. But instead of the
{8,3} faces, we can match the
mirror images as top and bottom of
a prism, with the edges being the ends of
a generally
rectangular margin. [Faces are bounded by margins]
The
octagonal prisms that arise are bisected by a waist-mirror.
This
replaces any hyperbolic face in the vertex by a pyramid of
spheric faces.
For example, the octahedral vertex figure
acquires an additional pyramid on
the faces, where the {8,3}
faces are replaced by a cluster of octagonal
prisms.
For
this to be uniform, the vertex must lie half an edge above the
plane.
It's possible, but how far above the plane the vertex
lies is not an
equation that I have solved as yet.
It's
of cours possible for waist-mirrors to have more than one type of
hyberbolic face. In fact, one can take a tetrahedron, and
label its six
sides with any sorts of integers, and make a
waist-mirror honeycomb from it.
In this case, space is filled
with assorted prism, 12 at any given corner.
But I suspect that
these will never be uniform, because this would make
some of the
sides very crowded.
Waist mirrors function in the same way
as ordinary "mirrors", with the
exception that it's not as easy
to "slide" the faces across it. But if the
thing above the
waist is not symmetrically placed on it, it can be "bumped
around". For example, we were bumping around the different
sides of the
cane thingie under {4,8} above. The sides of
the cane were alternating
canes and laminates.
THE FOUR
DIMENSIONED LAMINATRUNCATES.
-------------------------------------
The pair of
polytopes derived on the {3,4,3,8} were discovered more by
'accident' than anything else. It was a case of sticking
laminatruncate
{4,3,8} on the side of a bt{3,4,3}. With a
little luck, I discovered that
the the figures matched exactly
as follows:
the face of o3m4m3o matches the vertex figure of
o8x2x8o
the face of o8m2m8o matches the vertex figure of o4x3x4o
The figure formed on the tiling of octagonny [o3x4x3o] of
size o3x4x3o
exactly fills space, with a vertex figure being an
octagon-octagon cross, or
o8m8o. The size of the edge
matches an {3,8} or {8,4}, the triangles come
out of the vertex
figures at the octagons, the octagons are the long lines
stretching between the octagons. That this includes two
distinct {8,3,3,4}
is readily seen.
The torus
equidistant from the two octagons cuts the tetrahedral faces of
the vertex figure in squares, this forms an 8x8 torus, such as
the squares
of an {8}{8} prism forms. Any column or row by
itself forms the vertex
figure of a bitruncated {3,4,3,8} =
o3x4x3o8o. The balance of the vertex
figure belongs to two
t{4,3,8} that meet at their common {3,8}. These two
t{4,3,8} show up as the two lines that bound the eight squares
in the vertex
figure.
The dual is a tiling of bioctagon
prisms, o8x2x8o, of edge equal to that of
a {4,8} or
{8,6}. The vertex figure is the o3m4m3o, the convex hull of
two
24-cells in dual positions. The {4,8} appear as the
octagons that girth the
figure, and the {8,6} as the octagons
that pass girth the two separate
24-cells. This figure
contains an two {8,3,4,3}'s.
The faces are squat tetrahedra,
with three of the faces surrounding each
edge of the
{3,4,3}'s. Correspondlingly, there is no simple figure formed
in three dimensions by sections of this figure.
FAT
LAMINAE PIE
---------------
Another frightfully fruitful
source for uniform honeycombs is the {3,p}. You
can remove
selected verticies of this, to leave a tiling consisting entirely
of triangles and p-gons, the extreme case being ONLY p-gons.
This makes a
{p,p/2}.
But the task of selecting
appropiate verticies in something like a {3,127}
may appear
daunting. This is where the fat laminagon comes in.
A
fat laminatope has its verticies equidistant from a laminatope.
Because it
is centred on a laminatope, and not an ideal point,
the thing has multiple
surfaces. (A slab is a fat
laminatope: it has two surfaces.) In spherical
geometry, the
thing has angles greater than 180 deg, the two fat laminatopes
known are those of the prisms and antiprisms.
In hyperbolic geometry, the
angle can be considerably less.
The simplest fat laminatopes are those that lie equidistant
from a plane.
For example, a chain of squares from {4,8} forms a
fat laminatope with
right-angled verticies. Likewise, the
triangles around the petrie polygon
of a {3,p} forms a fat
laminagon. Couppled with two polygons {p}, this
would make
an antiprism. Both the octahedron and pentagonal antiprism can
be made from fat laminahedra in {3,4} and {3,5}.
Ordinary polygons can appear in the sequence as well, giving
the effect of
"traffic islands", similar to the pentagons on the
pentagon-antiprism.
We make the laminagons from equiangled
segments of a {3,p}: the edges are
always equal, and the corners
is something we can control. From a range of
segments (eg
1..7), we can divide the vertex figure into these segments and
construct fat laminahedra in them. Assorted translations
finish off the
task, but we have to watch certian symmetries on
the larger scale.
Here, we see the vertex of a {3,20} divided
5 | 5' .|
5 to segments of 3,5,5 and 7.
All of these are
|
.
| known fat
laminagons. The segments are thus
====+=======+====== populated: 3=(3,3,3),
5=(3,3,3,3,3)
|
| 5'=(3,3,p,3)
and 7=(3,p,3,p,3). Because the
7 | 3 |
7 5' is a chiral snub, the order is effected
a
b by rotation:
ie the {p} falls on the side where
the dot is.
a and b are two adjacent verticies, there are 3 {20}
and 14 triangles around
each vertex, one of which is sandwiched
between the two {20}'s in the "7"
sector. But at vertex a
there are 8 triangles between the two {20} gons via
the [5]
side, and 5 via the [3], whereas in case b there are 7 and 6
respectively. Therefore this one is not uniform.
Because of this, and the certian possibility that the same
shape may be a
laminatope in several different ways, the list
should be regarded as an
outer limit, not an inner limit.
That is, it provides a list that contains
the possible
laminatopes.
It should be remembered that there may well be
other constructions, such as
the yickle, that will produce other
classes of uniform figures. That is,
this list does not
bring closure, but a new opening.
To form a meaningfully
different uniform figure, it must have at least one
{p} and one
{3} at each vertex. Otherwise, it degenerates into a
{p,p/2}
or {3,p}.
KNOWN FAT LAMINAGONS in {3,p}.
------------------------------
1: Triangles can
occur in the figure, either loosely, or as a 'snub'
feature.
When they occur loosely, the
figures on either side must be the
same, since the triangle does not support alternation.
In the 'snub' mode, the vertex is
regarded as a snub of three
segments,
the consist being 1+a+1+b+1+c = p.
A 'snub' can be turned on itself to give a retrosnub. That is,
instead of dividing p into 1+a+1+b+1+c,
it divides into r+c, and
a new
r-triangled laminagon is born.
2: Simple polygons can
occur three different ways.
þ
They can occur loose, with all faces symmetric. The same shape
and context must
occur on each side.
þ They can
occur with alternating faces, this happens when the
figure is even.
Different things occur on different sides.
þ They can occur where snub
triangles occur. The sides are numbered
from 1 to 3 over and
again, and these function as they would
for a snub triangle.
3: The sole example here is the s{2,oo}, a strip of
triangles
surrounding a petrie polygon.
| 2 2 oo oo
3 3 3
4: Four can be made from either triangles:
s{3,oo} or p-gons t{p/2,oo}.
| 2 3
oo oo 3 3 3 3
2 oo p/2 oo p
p 2 divides p
5: There are several fat lamina made of five
triangles:
| 3 3
oo oo 3 3 3 3 3
| 2 p oo oo 3 3
p 3 2 divides p
3 oo | p/2 oo p 3 p
6: There are these:
|
3 p oo oo 3 3 3 p 3
p oo | p/2 oo p p
p 2 divides p
|| 2 2 oo oo p p
p 3 divides p
Only oo p | p/2 has a {p}
surrounded entirely by p-gons.
7: These cases
exist for 7
| p p
oo oo 3 p 3 p 3 3
divides p
s| 2 3 oo
oo p 3 p p 3 divides p
8: These have 8:
s| 2 p oo oo p p
p p 3 divides p
s| 3 3 oo oo p 3 p 3
p 3 divides p
9: This
has 9:
s| 3 p oo oo p
p p 3 p 3 divides p
10: This
one has 10:
s| p p
oo oo p p p p p 3 divides
p
SPECIFIC EXAMPLES of {3,p}
--------------------------
Let's look at some specific cases.
5=2+3. Here '2' is the pentagonal face, and
'3' is the strip around
the equator.
6=2+4 Here, 2 is a hexagon, and 4 is the
triangles. You can see
both that the
array of4 triangles is assymeteric, and that it
encompasses a lot of hexagons.
7=2+5. This produces the regular snub, | 3 3
7. Nothing new here.
7=3+4 Both 3
and 4 are triangles-only and this has no heptagon face
it degenerates to a {3,7}. The case
where 4 has p requires p even.
8=4+4 This
is dealt with under the van Oss versions. Essentially,
one of the 4's is a snub {3,oo} and the other a
t{p/2,oo}. As we
have already noted, the
snubs can be preserved or reflected around
the
t{p/2, oo} The vertex array is 4 4 + 3 3 3 3
8=3+5 This produces two possible forms,
both of which feature petrie
corridors. The
first has the same vertex figure as | 8 8 3, but the
| 8 8 3 groups its triangles into clusters of four,
not long corridors
as found in 3 3 3 + 8 3 8.
3,3,3 + 8,3,8
3,3,3 + 3,3,8,3
Of this last, there may be two distinct forms,
depending on whether
the 3 3 8 3 reverses or
rotates across the 3 3 3.
8=2+6 This
yields | 3 8 8, since the 'oo' is '8'.
8=2+3+3 This
yields an 8+3,3,3+3,3,3 Every octagon here is surrounded
by 8 petrie-polygons, rushing in, along one face, and
back out again.
(8,1)
8 3 3 3 3 3 3
(8,2) 8 8 3 3 3 3
8 3 8 3 3 3
(8,3) -
9=8+1 This is p 3 p 3 p + 3, which
is 9 3 | 3
9=7+2 The 3 p 3 p 3 + p is just 9 3 | 3
again.
The p p
3 p + p is a different creature all together. And yes,
it does
exist. In essence, its vertex figure is read as a
s 2 s 3 s 9,
where s is 9 rather than '3'.
The
vertex figure can be seen as a girl with her arms out, and
the triangle as
her dress. The head of one girl represents the
left arm of the
next, and the right arm of one girl runs to a
right arm of
the next. This places the triangles.
9=6+3
This produces two:
9 9 9 + 3 3
3
3 3 3 9 3 + 3
3 3
9=5+4 3 3 9 3 + 3 3 3 3 We have
no indication on whether this is the
same as the 3 3
3 9 3 + 3 3 3.
9=3+3+3 This has no {p} and degenerates to
{3,p}.
(9,1) 9 3 3 3 3 3 3
(9,2)
(9,3) 9 9
9 3 3 3
9 3 9 3 9 3
(9,4) 9 9 9 9 3
10=5+5 3 3 10 3 + 3 3 10 3
3 3 10 3 + 3 10
3 3
3 3 3 3 3 +
3 10 3 3 3
3 3
3 3 3 + 10 3 10
3 3 10 3 + 10 3 10
10 3 10 + 10 3
10
The
first three were discussed under the van-Oss section.
But I had not
explored 10 3 10 as a possibility, and so it
makes three
more. We can not reverse the 3 3 10 3 over the
3 10 3 3 as the
latter divides into an odd number.
10=6+4 10 10
10 + 3 3 3 3
10
10 + 3 3 3 10 3
10=7+3 3 10 3 10 3 + 3 3 3
10=3+3+4 no figures as no {p}
(10,1) 10 3
3 3 3 3 3 3 3
(10,2) 10 3 10 3 3 3 3 3
10 3 3 3 10 3 3 3
(10,3) 10 10 10
3 3 3 3
10 10 3 10 3 3 3
(10,4) 10 10 3
10 10 3
12 Egads,
there ARE a lot here!!!!!!!
= s4s4s1
Putting '4' as
the 'p p' we have a laminagon that passes lots
of little
clusters of triangles
s=3: s1s3s5 s1s4s4 s2s3s4 s3s3s3
s=p: s1s1s4 s1s2s3
Vf 66 75 84
93 10.2 11.1 444 336 345 255
s1s3s5 s 3 s 3 3 3 s 3 3 12 3
s1s4s4 s 3 s 12 12 s 12 12
s 3 s 3 3 3 3 s 12 12
s2s3s4 s 12 s 3 3 3 s 3 3 3 3
s3s3s3 = {3,12}
s1s1s4 12 3 12 3 12 3 3 3 3
12 3 12 3 12 12 12
s1s2s3 12 3 12 12 12 3 3 3
66 see the van oss versions
75 3 12 3 12 3 . 12 3
12 = 4 | 3 12
3 12 3 12 3 . 3 3 12 3
84 12 3 12 3 12 . 3 3 3 3
12 12 12 12 . 3 3 3 3
12 3 12 3 12 . 12 12
93 12 12 12 3 12 . 3 3 3
6.3.3 van oss polygon.
5.5.2 12 3 12 . 12 3 12 . 12
12 3 12 . 3 3 12 3 . 12
3 3 12 3 . 3 3 12 3 . 12
3 3 12 3 . 3 12 3 3 . 12
5.4.3 12 3 12 . 3 3 3 3 . 3 3 3
4 4
4 12 12 . 3 3 3 3 . 3 3 3 3
Giving,
one or more forms of
1{12} 12 3 3 3 3 3 3
3 3 3 3
2{12} 12 12 3 3 3 3 3 3 3 3
12 3 12 3
3 3 3 3 3 3
12 3 3 12
3 3 3 3 3 3
12 3 3 3
12 3 3 3 3 3
12 3 3 3
3 12 3 3 3 3
3{12} 12 12 12 3 3 3 3 3 3
12 3 12 3
12 3 3 3 3
12 3 12 3
3 3 12 3 3
12 3 3 12
3 3 12 3 3
4{12} 12 12 12 3 12 3 3
3 4
distinct forms known
12 12 3
12 12 3 3
3 2
distinct forms known
12 12 3
12 3 3 12 3
12 3 12 3
12 3 12
3 = 4 |
3 12
5{12} 12 12 12 12 3 12 3
12 12 12
3 12 12 3
ON THE {5,12} and {12,10}
-------------------------
If a uniform figure is to be
made from these two equal-edged honeycombs,
it will no doubt
take this form.
1. It has 6 pentagons and 5
dodecagons at each corner.
2. Five of the pentagons
are probably a snub polygon:
ie
we're seeing five different corners of a pentagon that differs
from corner to corner.
3. The remaining pentagon has pentagonal symmetry around its
environ.
4. Some
or all of the dodecahedra are alternating, and three could
even be themselves snub.
Because of what we have just said, we should note, that
except the 1 penta
there are 5 pentagons and 5 dodecagons, this
does not have a symmetry in
an arc of a circle.
Let F
and D be the standard polygons, and f and d the snub faces.
then: DFD fffff ddd could well
be a possibility.
fffff exists. This can be made from
a pentagon, by rotation around
four of its five sides. The
fifth side forms the side of the fat
laminagon. In the
present case, the 5 pentagons make up 50/120 of
the vertex
angle.
In the process of rotation, we note that an edge AB
becomes BA. Then
if all five corners arise at a point,
then all five edge types are
there too. But we do not
allow rotation past one particular edge, so
this forming of a
face by rotation of its neighbour turns the marked
edge into a
fat laminagon.
We know ddd exists: we have met it in both
the 444 form and the 606
form in the van-Oss {3,12}. The
total angle here is 36/120.
The DfD makes up 12+10+12=34,
and 34+50+36=120, the whole circle.
Since there is no means to
simultaneously force the order of two
verticies of the central
5, we conclude that the the pentagonal
symmetry of the central
pentagon would need to retain at least the
rotational
aspect. Of the dodecagons, these alternate to the central
pentagon, or the external features.
Therefore we
conclude that 12 5 12 5 5 5 5 5 12 12 12 exists in two forms.
the two forms being the two implementations of 12 12 12.
Of course, once we realise that the five of the pentagons
are 'snub'
faces, we can easily populate any of the spaces of
.5.5.5.5.5 with 5, 12,
or any combination of 12's or 12-5-12 or
12-12-5-12-12, and produce a
valid snub, by:
1:
rotation of the non-snub faces
2: preserving the order
around the snub pentagons.
That means: lots of snubs from
the shared {5,12} and {12,10}
EXTENDED SNUBS.
---------------
The notion of using odd numbers to form
large extended snubs should not
be passed up on. For
example, take the {3,15}. It is possible that the
array 15
15 15 15 15 3 3 3 3 3
15 is a multiple of 3 and 5.
Lets number the sides from 1 to 15 in order.
One kind of snub is
formed by forming a laminagon on the sides that are
multiples of
5. Each cell in the laminatope is exposed at three different
points.
The other is to preserve the multiple of three,
so each cell has five
different exposures. We have met
this snub earlier. The difference here
is that instead of
rotating around an edge, we use another {15}, which we
give a
different colour to.
And because outside the this laminagon
packed full of {15}'s is clearly
defined by the triangles
outside it, the number of triangles that the
five {15} around
each vertex goes either 3.3.3.3.3. or 5.0.5.0.5. That
is,
there is a version of 15 15 15 15 15 3 3 3 3 3 that has {15} not
adjacent to triangles, and one that doesn't.
CONCLUSION
----------
We found lots of different uniform
honeycombs, and there are lots more
out there.
The mind
boggles.
Wendy Krieger
_________________________________________________________________
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