**Richard Klitzing's paper**

I am grateful to Dr Richard Klitzing for
allowing me to present his (unpublished) work here (*with some minor amendment*)
(JM):

I have recently looked at the
consequences of the fact that the rhombi forms of the quasitruncated polyhedra,
i.e. those with Dynkin diagram: **(o)- ^{n1}/_{d1}-o-^{n2}/_{d2}-(o) **or vertex
notation [

Only recently Mr. Verheyen informed me about his article "The complete set of Jitterbug transformers and the analysis of their motions". It is concerned about the setups and the motions of pairs of polygons, but any filling of the holes is not explicitly considered therein.

The first thing one would like to do would be to get this topological identity into a continous deformation, i.e. to have a kind of movie which transforms the first state into the second. The easiest way is to discard the rhombisquares for a moment; having in mind only that these holes form in general a nonplanar equilateral tetragon, to which the shorter diagonal will be added as additional edge.

Discarding these rhombisquares of [^{n1}/_{d1},4,^{n2}/_{d2},4], one
would be left with a kind of a net spun around the sphere, consisting of
alternating ^{n1}/_{d1}-gons and ^{n2}/_{d2}-gons, stuck together pairwise at their
vertices. Doing this abstraction for the snub polyhedra as well (discarding the
flag triangles), one would observe that the symmetry axes of the remaining
polygons are identical, the polygons are only twisted around them, say with
angles chi_{1} and chi_{2}.

The condition that the vertices have to be connected would
result in the fact that the sines and the circumradii r_{1} and r_{2}
of the polygons ^{n1}/_{d1} and ^{n2}/_{d2} obey

r

_{1}*sin(chi_{1}) = r_{2}*sin(chi_{2})

or more explicitly

sin(chi

_{1}) / sin(180*^{d1}/_{n1}) = sin(chi_{2}) / sin(180*^{d2}/_{n2})

i.e. these angles chi aren't independent.

Next is, that again by the same condition these polygons
are forced to alter their heights above the center of the polyhedron. (The
height of the ^{n1}/_{d1}-gons would be

h

_{1 }= (r_{2}*cos(chi_{2})-r_{1}*cos(chi_{1})*cos(psi)) / sin(psi)

here psi is the angle between the planes defined by two
connected ^{n1}/_{d1} and ^{n2}/_{d2}-gons.)

Having introduced the angles chi the deformation has
become continous. Now one gets curious to look beyond the starting and the final
points, to see what comes next. Going beyond the starting point (A), that is
having chi_{1}< 0, this would generate nothing new, for chi'_{1}=-chi_{1}
would reproduce the same polyhedra as chi_{1}, only the right chirality
would change into a left chirality.

Going beyond the final point (B) instead would become more
interesting.Here there would
come a point (C) where the deformed squares become degenerate. This would be the
case when the trigons are rotated about 60 degrees or alternatively the
pentagrams are rotated about 72 degrees. Drilling further the net polygons ^{n1}/_{d1}
and ^{n2}/_{d2} do intersect.

If both net polygons have vertex angles which are small
enough, there would be further on another interesting point (D), where the
deformed (and now retrograde) holes become again the form of a pair of
equilateral triangles. But finally you would come to a sudden stop: if the
polygon with the smaller vertex angle, say ^{n1}/_{d1}, has been rotated by chi_{1}
= 90 degrees.

The trial of a rotation still further, that is having chi_{1}
> 90 degrees, could be interpreted in different ways. This is due to the fact
that no inverse function of the sine exists. (One could use the branch of arcsin,
but any other as well.) That is, either (a) you have to make a jump (or
discontinuity) in chi_{2} from chi_{2}=chi_{2}(chi_{1}=90)
to chi_{2}=180-chi_{2}(chi_{1}=90) and rotate then both polygons
in the same direction as before, or (b) you have to rotate chi_{1}
onwards up to 180 degrees and chi_{2} would be reduced backwards down to 0
degrees. (note: in case of ^{n1}/_{d1}=^{n2}/_{d2} the discontinuity in
case (a) degenerates, i.e. you are left with a bifurcating
point.)

In spite of the discontinuity, (a) is in a way the more
natural branch, as both polygons are rotated in the same direction as before.
But, alas, it wouldn't produce anything new: chi'_{1}=180-chi_{1} would
reproduce exactly the same polyhedra as chi_{1}, only the way of looking would be
retrograde: the inside and outside of the 'net polygons' would be interchanged.

Therefore, we are left with the other branch (b). On that
branch other things do occur, but those are mostly less interesting
than the ones of the first quarter rotation of ^{n1}/_{d1}. For the
Platonic
family, for example, only a single highly degenerate intermediate state (B')
with coincident faces would come up. For the Keplerian family there are
more interesting intermediate states (E), producing uniform polyhedra again.
Finally, at rotation chi_{1} = 180 degrees, there would be in any case
the last state (A'), where the holes become squares again.

In the sense of a continuous deformation, keeping the same edges connected while rotating, only cases (A) the exploded form, (B) the outer snub, (C) the closed hole, (D) the inner snub, (A') the imploded form are relevant. Cases (B') and (E) belong to other possible closings of the holes.

Concluding:

In the symmetry group of ** [[2,3,3]]** the
following polyhedra / compounds occur:

(A), chi

_{1}= chi_{2}= 0 deg, [3,4,3,4], cuboctahedron(B), chi

_{1}= chi_{2}= 22.239 deg, [3,3,3,3,3], icosahedron = snub tetrahedron(C), chi

_{1}= chi_{2}= 60 deg, [3,3,3,3], octahedron(D), chi

_{1}= chi_{2}= 82.239 deg, [3,3,3,3,3]/2, great icosahedron(B'), chi

_{1}= 120 deg, chi_{2}= 60 deg, compound of 5 identical tetrahedra [3,3,3](A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, compound of 2 identical tetrahemihexahedra [^{3}/_{2},4,3,4]Animation: (A) (B) (C)

For ** [[2,3,4]]** we have:

(A), chi

_{1}= chi_{2}= 0 deg, [3,4,4,4], rhombicuboctahedron(B), chi

_{1}= 20.315 deg, chi_{2}= 16.468 deg, [3,3,3,3,4], snub cube(C), chi

_{1}= 60 deg, chi_{2}= 45 deg, [3,4,3,4], cuboctahedron(B'), chi

_{1}= 120 deg, chi_{2}= 45 deg, compound of 2 non-identical tetrahemihexahedra [^{3}/_{2},4,3,4] and 3 identical octahedra [3,3,3,3](A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, [^{3}/_{2},4,4,4], great rhombicuboctahedronAnimation: (A) (B) (C)

For ** [[2,3,5]]** we get:

(A), chi

_{1}= chi_{2}= 0 deg, [3,4,5,4], rhombicosidodecahedron(B), chi

_{1}= 19.518 deg, chi_{2}= 13.106 deg, [3,3,3,3,5], snub dodecahedron(C), chi

_{1}= 60 deg, chi_{2}= 36 deg, [3,5,3,5], icosidodecahedron(B'), chi

_{1}= 120 deg, chi_{2}= 36 deg, compound of great dodecahedron [5,5,5,5,5]/2 and 4 identical icosahedra [3,3,3,3,3](A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, semi compound 3[^{3}/_{2},4,5,4] (=[3,5,3,5,3,5]/2 + compound of 5 cubes) - "small complex rhombicosidodecahedron"Animation: (A) (B) (C)

For **[[2, ^{5}/_{2},3]]**:

(A), chi

_{1}= chi_{2}= 0 deg, semi compound 3[^{3}/_{2},4,5,4] (=[3,5,3,5,3,5]/2 + compound of 5 cubes) - "small complex rhombicosidodecahedron"(B), chi

_{1}= 27.108 deg, chi_{2}= 24.515 deg, [3,3,3,3,^{5}/_{2}], great snub icosidodecahedron(C), chi

_{1}= 72 deg, chi_{2}= 60 deg, [3,^{5}/_{2},3,^{5}/_{2}], great icosidodecahedron(D), chi

_{1}= 88.953 deg, chi_{2}= 65.566 deg, [3,^{3}/_{2},3,^{5}/_{3},3] (or just easier to understand: [3,3,3,3,^{5}/_{2}]/2), great retrosnub icosidodecahedron)(E), chi

_{1}= 108 deg, chi_{2}= 60 deg, compound of small stellated dodecahedron [^{5}/_{2},^{5}/_{2},^{5}/_{2},^{5}/_{2},^{5}/_{2}] and 4 identical great icosahedra [3,3,3,3,3]/2(B'), chi

_{1}= 133.845 deg, chi_{2}= 41.052 deg, [3,3,3,3,^{5}/_{3}], great inverted snub icosidodecahedron(A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, [^{5}/_{3},4,3,4], great rhombicosidodecahedronAnimation: (A) (B) (C), (C) (D) (A')

For **[[2, ^{5}/_{2},5]]**:

(A), chi

_{1}= chi_{2}= 0 deg, [^{5}/_{2},4,5,4], rhombidodecadodecahedron(B), chi

_{1}= 23.932 deg, chi_{2}= 14.519 deg, [3,3,^{5}/_{2},3,5], snub Dodecadodecahedron(C), chi

_{1}= 72 deg, chi_{2}= 36 deg, [^{5}/_{2},5,^{5}/_{2},5], dodecadodecahedron(D), chi

_{1}= 136.388 deg, chi_{2}= 25.233 deg, [3,3,^{5}/_{3},3,5], vertisnub dodecadodecahedron(A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, semi compound 3[^{5}/_{3},4,5,4] (=[^{5}/_{3},5,^{5}/_{3},5,^{5}/_{3},5] + compound out of 5 cubes) - "complex rhombidodecadodecahedron"Animation: (A) (B) (C), (C) (D) (A')

For the dihedral case, **[[2,2,n]]** (here with n=7):

(A), chi

_{1}= chi_{2}= 0 deg, [4,4,7], heptagonal prism(B) chi

_{1}= 30.854 deg, chi_{2}= 12.857 deg, [3,3,3,7], heptagonal antiprism(C) chi

_{1}= 90 deg, chi_{2}= 25.714 deg, [7,7], double heptagon(B') chi

_{1}= 149.146 deg, chi_{2}= 12,857 deg [3,3,3,7] heptagonal antiprism(A') chi

_{1}= 180 deg, chi_{2}= 0 deg, [4,4,7], heptagonal prism(as psi=90 deg it is completely symmetric)

Note that (A) is always the exploded form, and (A') is the imploded form, and (C) is the quasitruncated form of each dual regular pair.

Finally, applying this idea to plain tesselations, one would get:

For **[[2,3,6]]**:

(A), chi

_{1}= chi_{2}= 0 deg, [3,4,6,4](B), chi

_{1}= 19.107 deg, chi_{2}= 10.893 deg, [3,3,3,3,6](C), chi

_{1}= 60 deg, chi_{2}= 30 deg, [3,6,3,6](B'), chi

_{1}= 120 deg, chi_{2}= 60 deg, [3,3,^{3}/_{2},3,^{6}/_{5}] (degenerate)(A'), chi

_{1}= 180 deg, chi_{2}= 0 deg, [^{3}/_{2},4,^{6}/_{5},4]

For **[[2,4,4]]**:

(A), chi

_{1}= chi_{2}= 0 deg, [4,4,4,4] (black-and-white symmetry)(B), chi

_{1}= chi_{2}= 15 deg, [3,3,4,3,4](C), chi

_{1}= chi_{2}= 45 deg, [4,4,4,4] (total symmetry)(B'), chi

_{1}= chi_{2}= 75 deg, [3,3,^{4}/_{3},3,^{4}/_{3}](A'), chi

_{1}= 90,...,180 deg, chi_{2}= 180-chi_{1}, all squares coincident

(which shows that the branch (a) is the more natural one...)