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I am grateful to Dr Richard Klitzing for allowing me to present his (unpublished) work here (with some minor amendment) (JM):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 [n1/d1,4,n2/d2,4], are topological identical to the snub polyhedra with Dynkin diagram ()-n1/d1-()-n2/d2-() or vertex notation [3,3,n1/d1,3,n2/d2], provided one dissects the squares of the former diagonally following the symmetry group of the second (which is a subgroup of that of the former of index 2).

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 chi1 and chi2.

The condition that the vertices have to be connected would result in the fact that the sines and the circumradii r1 and r2 of the polygons n1/d1 and n2/d2 obey

r1*sin(chi1) = r2*sin(chi2)

or more explicitly

sin(chi1) / sin(180*d1/n1) = sin(chi2) / 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

h1 = (r2*cos(chi2)-r1*cos(chi1)*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 chi1< 0, this would generate nothing new, for chi'1=-chi1 would reproduce the same polyhedra as chi1, 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 chi1 = 90 degrees.

The trial of a rotation still further, that is having chi1 > 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 chi2 from chi2=chi2(chi1=90) to chi2=180-chi2(chi1=90) and rotate then both polygons in the same direction as before, or (b) you have to rotate chi1 onwards up to 180 degrees and chi2 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-chi1 would reproduce exactly the same polyhedra as chi1, 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 chi1 = 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), chi1 = chi2 = 0 deg, [3,4,3,4], cuboctahedron

(B), chi1 = chi2 = 22.239 deg, [3,3,3,3,3], icosahedron = snub tetrahedron

(C), chi1 = chi2 = 60 deg, [3,3,3,3], octahedron

(D), chi1 = chi2 = 82.239 deg, [3,3,3,3,3]/2, great icosahedron

(B'), chi1 = 120 deg, chi2 = 60 deg, compound of 5 identical tetrahedra [3,3,3]

(A'), chi1 = 180 deg, chi2 = 0 deg, compound of 2 identical tetrahemihexahedra [3/2,4,3,4]

Animation:  (A) (B) (C)

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

(A), chi1 = chi2 = 0 deg, [3,4,4,4], rhombicuboctahedron

(B), chi1 = 20.315 deg, chi2 = 16.468 deg, [3,3,3,3,4], snub cube

(C), chi1 = 60 deg, chi2 = 45 deg, [3,4,3,4], cuboctahedron

(B'), chi1 = 120 deg, chi2 = 45 deg, compound of 2 non-identical tetrahemihexahedra [3/2,4,3,4] and 3 identical octahedra [3,3,3,3]

(A'), chi1 = 180 deg, chi2 = 0 deg, [3/2,4,4,4], great rhombicuboctahedron

Animation:  (A) (B) (C)

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

(A), chi1 = chi2 = 0 deg, [3,4,5,4], rhombicosidodecahedron

(B), chi1 = 19.518 deg, chi2 = 13.106 deg, [3,3,3,3,5], snub dodecahedron

(C), chi1 = 60 deg, chi2 = 36 deg, [3,5,3,5], icosidodecahedron

(B'), chi1 = 120 deg, chi2 = 36 deg, compound of great dodecahedron [5,5,5,5,5]/2 and 4 identical icosahedra [3,3,3,3,3]

(A'), chi1 = 180 deg, chi2 = 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), chi1 = chi2 = 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), chi1 = 27.108 deg, chi2 = 24.515 deg, [3,3,3,3,5/2], great snub icosidodecahedron

(C), chi1 = 72 deg, chi2 = 60 deg, [3,5/2,3,5/2], great icosidodecahedron

(D), chi1 = 88.953 deg, chi2 = 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), chi1 = 108 deg, chi2 = 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'), chi1 = 133.845 deg, chi2 = 41.052 deg, [3,3,3,3,5/3], great inverted snub icosidodecahedron

(A'), chi1 = 180 deg, chi2 = 0 deg, [5/3,4,3,4], great rhombicosidodecahedron

Animation:  (A) (B) (C), (C) (D) (A')

For [[2,5/2,5]]:

(A), chi1 = chi2 = 0 deg, [5/2,4,5,4], rhombidodecadodecahedron

(B), chi1 = 23.932 deg, chi2 = 14.519 deg, [3,3,5/2,3,5], snub Dodecadodecahedron

(C), chi1 = 72 deg, chi2 = 36 deg, [5/2,5,5/2,5], dodecadodecahedron

(D), chi1 = 136.388 deg, chi2 = 25.233 deg, [3,3,5/3,3,5], vertisnub dodecadodecahedron

(A'), chi1 = 180 deg, chi2 = 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), chi1 = chi2 = 0 deg, [4,4,7], heptagonal prism

(B) chi1 = 30.854 deg, chi2 = 12.857 deg, [3,3,3,7], heptagonal antiprism

(C) chi1 = 90 deg, chi2 = 25.714 deg, [7,7], double heptagon

(B') chi1 = 149.146 deg, chi2 = 12,857 deg [3,3,3,7] heptagonal antiprism

(A') chi1 = 180 deg, chi2 = 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), chi1 = chi2 = 0 deg, [3,4,6,4]

(B), chi1 = 19.107 deg, chi2 = 10.893 deg, [3,3,3,3,6]

(C), chi1 = 60 deg, chi2 = 30 deg, [3,6,3,6]

(B'), chi1 = 120 deg, chi2 = 60 deg, [3,3,3/2,3,6/5] (degenerate)

(A'), chi1 = 180 deg, chi2 = 0 deg, [3/2,4,6/5,4]

For [[2,4,4]]:

(A), chi1 = chi2 = 0 deg, [4,4,4,4] (black-and-white symmetry)

(B), chi1 = chi2 = 15 deg, [3,3,4,3,4]

(C), chi1 = chi2 = 45 deg, [4,4,4,4] (total symmetry)

(B'), chi1 = chi2 = 75 deg, [3,3,4/3,3,4/3]

(A'), chi1 = 90,...,180 deg, chi2 = 180-chi1, all squares coincident

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