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Thread: Question about Archimedean solids

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    Contributor repoman's Avatar
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    Question about Archimedean solids

    The definition I am using from Wikipedia is:

    an Archimedean solid is one of the 13 solids first enumerated by Archimedes. They are the semi-regular convex polyhedra composed of regular polygons meeting in identical vertices, excluding the 5 Platonic solids (which are composed of only one type of polygon) and excluding the prisms and antiprisms. They differ from the Johnson solids, whose regular polygonal faces do not meet in identical vertices


    But looking at the picture of a truncated icosidodecahedron there are light and dark gray vertices that if I use a right hand rule are (4,6,10) for dark and (4,10,6) for light.

    Same for the truncated cuboctahedron, which is listed as have (4,6,8) vertex figure, but seems to have that split between (4,6,8) and (4,8,6).



    Does this silly nitpick I have mean anything?

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    Administrator lpetrich's Avatar
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    Those different orderings are for the different asymmetries of each of those kinds of vertices -- each kind is a mirror image of the other.

    Archimedean solid with duals Catalan solid

    Dual: vertex <-> face and edge <-> edge.
    Dualities of Platonic solids: Tetrahedron: self-dual, cube <-> octahedron, dodecahedron <-> icosahedron.

    Symmetries:
    • Isohedral - face-transitive - faces alike to within reflection
    • Isotoxal - edge-transitive - edges alike
    • Isogonal - vertex-transitive - vertices alike to within reflection

    The Archimedean solids are isogonal, while the Catalan solids are isohedral. The Platonic solids are both.

    Their rotation/reflection symmetry groups are the quasi-spherical ones: tetrahedral, octahedral, icosahedral.

    There are some additional isogonal and isohedral polyhedra, the axially-symmetric ones.

    Prism (barrel with quad sides) - Bipyramid (triangles connected to poles and equator)
    Antiprism (barrel with alternating-direction triangle sides) - Trapezohedron (quads connected to poles and making a zigzag equator)

    I got interested in this issue out of interest in the fair-dice problem: what kinds of dice are there were all the faces behave the same? This includes all the isohedral polyhedra (Platonic, Catalan, bipyramids, trapezohedra), and with ignoring the end-cap faces, the prisms and antiprisms. There are some degenerate cases:
    • Tetrahedron: 2-antiprism
    • Cube: 4-prism, 3-trapezohedron
    • Octahedron: 3-antiprism

    Endcaps are counted with the other faces, except for the 2-antiprism, where they degenerate into lines.

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    Administrator lpetrich's Avatar
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    Next is the question of why there are only 13 Archimedean solids. There are three subsets of them: tetrahedral, octahedral, and icosahedral. The octahedral and icosahedral ones have 6 members, while the tetrahedral one has only 1 member.


    Reference Tetrahedron Octahedron Icosahedron
    Dual (same) Cube Dodecahedron
    Vertex truncated Truncated tetra Truncated octa Truncated icosa
    Vtx trunc of dual (same) Truncated cube Truncated dodeca
    Full vtx trunc Octahedron Cuboctahedron Icosadodecahedron
    Edge truncated Cuboctahedron Rhombicubocta Rhombiicosadodeca
    Vtx edge truncated Truncated octa Truncated cubocta Truncated icisadodeca
    Snub Icosahedron Snub cube Snub dodeca
    Pyramid dual Triakis tetra Tetrakis hexa Pentakis dodeca
    Pyramid (same) Triakis octa Triakis icosa
    Rhombic quad Cube Rhombic dodeca Rhombic triaconta
    Deltoidal quad Rhombic dodeca Deltoidal icosatetra Deltoidal hexeconta
    Split Pyramid Tetrakis hexa Disdyakis dodeca Disdyakis triaconta
    Pentagonal Dodecahedron Pentagonal icosatetra Pentagonal hexeconta
    The colored ones are the Archimedean ones, and then the Catalan ones, in corresponding order.

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    Administrator lpetrich's Avatar
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    List of Euclidean uniform tilings has plane-tiling versions of the Platonic, Archimedean, and Catalan solids, and Uniform tilings in hyperbolic plane has hyperbolic-plane versions of them.

    Which one's which can be distinguished with the help of Euler's theorem. One calculates the Euler characteristic X from number of vertices V, edges E, and faces F:

    X = V - E + F

    The Euler characteristic is a topological invariant, independent of splitting edges with new vertices, splitting faces with edges that connect vertices, and those operations' inverses. For spherical topology, it is 2, for genus g (how many "holes"), it is 2(1-g), for the flat plane, it is 0, and for the hyperbolic plane, it is negative.

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    Contributor repoman's Avatar
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    I just found out through tinkering that the unit cell of FCC (face centered cubic) is a cuboctahedron.

    pretty cool.

    Are any other Archimedean solids real world objects?

    found this:

    https://en.wikipedia.org/wiki/Waterm...imedean_solids

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    Contributor repoman's Avatar
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    Oh and HCP is a triangular orthobicupola.

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    Administrator lpetrich's Avatar
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    I will now find what partially transitive tilings there are.

    Edge-transitive means all the edges look alike. There may be one or two kinds of vertices, and likewise for faces. In general,

    X = V - E + F = E * ( (sum of 1/rk over k) - 1 + (sum of 1/nk over k) )

    where each kind k of face has nk vertices and edges, and each kind k of vertex has r[sub]k[/sup] faces and edges. If there are two kinds of vertices, then every face has even n, while if there are two kinds of faces, then every vertex has even r.

    For all different, X = E * ( 1/r1 + 1/r2 - 1 + 1/n1 + 1/n2 )

    Here, r1, r2, n1, and n2 must be at least 4. For all equal to 4, then X = 0. If any of them are greater than 4, then X < 0 -- hyperbolic.

    For face-transitive (Catalan) or vertex-transitive (Archimedean), one of the pairs is equal, and for convenience, I will consider the Catalan case. The Archimedean case is its dual. r1 and r2 are different, while n1 = n2 = even.

    X = E * ( 1/r1 + 1/r2 - 1 + 2/n )

    Here are all the non-hyperbolic ones:
    r1 = r2 = 3, n = 4 ... o polyhedron
    r1 = r2 = 3, n = 6 ... - planar
    r1 = 3, r2 = 4, n = 4 ... o
    r1 = 3, r2 = 5, n = 4 ... o
    r1 = 3, r2 = 6, n = 4 ... -
    r1 = r2 = 4, n = 4 ... -

    The Platonic solids and their planar and hyperbolic relatives have n1 = n2 and r1 = r2.

    X = E * ( 2/r - 1 + 2/n )

    Non-hyperbolic:
    r = 3, n = 3 ... o ... tetrahedron
    r = 3, n = 4 ... o ... cube
    r = 3, n = 5 ... o ... dodecahedron
    r = 3, n = 6 ... - ... hexagon tiling
    r = 4, n = 3 ... o ... octahedron
    r = 4, n = 4 ... - ... square tiling
    r = 5, n = 3 ... o ... icosahedron
    r = 6, n = 3 ... - ... triangle tiling

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    Administrator lpetrich's Avatar
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    The general face-transitive case (Catalan) satisfies

    X = F * ( (sum of rk over k = 1 to n) - n/2 + 1)

    The vertex-transitive case (Archimedean) is its dual, so solutions automatically carry over.

    I will consider triangles here, n = 3, the r's being r1, r2, r3.

    For all the vertices different, then the r's are even, since for each one vertex, the other two kinds alternate around it. That makes the r values even.

    X = F * ( 1/r1 + 1/r2 + 1/r3 - 1/2 )

    Non-hyperbolic:
    r1 = r2 = 4, r3 = 2m ... o
    r1 = 4, r2 = 6, r3 = 6 ... o
    r1 = 4, r2 = 6, r3 = 8 ... o
    r1 = 4, r2 = 6, r3 = 10 ... o
    r1 = 4, r2 = 6, r3 = 12 ... -
    r1 = 4, r2 = 8, r3 = 8 ... -
    r1 = 6, r2 = 6, r3 = 6 ... -

    Make the second and third vertices alike: r2 = r3. That makes r2 even.

    X = F * ( 1/r1 + 2/r2 - 1/2 )

    Non-hyperbolic:
    r1 = m, r2 = 4 ... o
    r1 = 3, r2 = 6 ... o
    r1 = 3, r2 = 8 ... o
    r1 = 3, r2 = 10 ... o
    r1 = 3, r2 = 12 ... -
    r1 = 4, r2 = 6 ... o
    r1 = 4, r2 = 8 ... -
    r1 = 5, r2 = 6 ... o
    r1 = 6, r2 = 6 ... -

    For all vertices alike, r1 = r2 = r3 = r, and we get the triangle-face Platonic solids and their relatives.

    X = F * ( 3/r - 1/2 )

    Non-hyperbolic:
    r = 3 ... o
    r = 4 ... o
    r = 5 ... o
    r = 6 ... -

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    Administrator lpetrich's Avatar
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    For quadrangles, it's much more difficult. In general, the vertices have r = r1, r2, r3, r4, and

    X = F * ( 1/r1 + 1/r2 + 1/r3 + 1/r4 - 1 )

    If all the vertices are different, then all the r's are even, and the only non-hyperbolic case is for all the r's equaling 4.

    That is also true if two neighbors are the same, and also is the remaining neighbors are the same, making two pairs of neighboring same vertices.

    So I turn to two diagonal vertices being the same: r2 = r4. That makes r2 even. Deltoids or kites.

    X = F * ( 1/r1 + 2/r2 + 1/r3 - 1 )

    Non-hyperbolic:
    r1 = 3, r2 = 4, r3 = 3 ... o
    r1 = 3, r2 = 4, r3 = 4 ... o
    r1 = 3, r2 = 4, r3 = 5 ... o
    r1 = 3, r2 = 4, r3 = 6 ... -
    r1 = 3, r2 = 6, r3 = 3 ... -
    r1 = 4, r2 = 4, r3 = 4 ... -

    Now for each pair of diagonal vertices being the same: r1 = r3, r2 = r4. Rhombi (rhombuses).

    X = F * ( 2/r1 + 2/r2 - 1 )

    Non-hyperbolic:
    r1 = 3, r2 = 3 ... o
    r1 = 3, r2 = 4 ... o
    r1 = 3, r2 = 5 ... o
    r1 = 3, r2 = 6 ... -
    r1 = 4, r2 = 4 ... -

    For three of the four vertices the same: r2 = r3 = r4. Kites again.

    X = F * ( 1/r1 + 3/r2 - 1 )

    Non-hyperbolic:
    r1 = m, r2 = 3 ... o
    r1 = 3, r2 = 4 ... o
    r1 = 4, r2 = 4 ... -

    For all four of the vertices the same:

    X = F * ( 4/r - 1 )

    Non-hyperbolic:
    r = 3 ... o
    r = 4 ... -

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    Pentagons are even more difficult, and the vertices have r = r1, r2, r3, r4, r5 and

    X = F * ( 1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5 - 3/2 )

    In the general case, all the r's are even, making all solutions hyperbolic.

    For two alike (12344), r4 = r5, and at least r1, r2, and r3 are even, making all solutions hyperbolic.

    X = F * ( 1/r1 + 1/r2 + 1/r3 + 2/r4 - 3/2 )

    An alternate for two alike (12343), r3 = r5, and at least r1 and r2 are even.

    X = F * (1/r1 + r/r2 + 2/r3 + 1/r4 - 3/2)

    Non-hyperbolic:
    r1 = r2 = 4, r3 = r4 = 3 ... -

    For three alike (12333), r3 = r4 = r5, and at least r1 and r2 are even.

    X = F * ( 1/r1 + 1/r2 + 3/r3 - 3/2 )

    Non-hyperbolic:
    r1 = r2 = 4, r3 = 3 ... -

    An alternate for three alike (12322), r2 = r4 = 45

    X = F * ( 1/r1 + 3/r2 + 1/r3 - 3/2 )

    Non-hyperbolic:
    r1 = 3, r2 = 3, r3 = 3 ... o
    r1 = 3, r2 = 3, r3 = 4 ... o
    r1 = 3, r2 = 3, r3 = 5 ... o
    r1 = 3, r2 = 3, r3 = 6 ... -
    r4 = 3, r2 = 3, r3 = 4 ... -

    For four alike (12222): r2 = r3 = r4 = r5

    X = F * ( 1/r1 + 4/r2 - 3/2 )

    Non-hyperbolic:
    r1 = 3, r2 = 3 ... o
    r1 = 4, r2 = 3 ... o
    r1 = 5, r2 = 3 ... o
    r1 = 6, r2 = 3 ... -

    For two pairs (12233): r2 = r3, r4 = r5, r1 even

    X = F * ( 1/r1 + 2/r2 + 2/r4 - 3/2 )

    Non-hyperbolic:
    r1 = 4, r2 = r3 = 3 ... o

    For two pairs (12323): r2 = r4, r3 = r5, r1 even

    X = F * ( 1/r1 + 2/r2 + 2/r3 - 3/2 )

    Non-hyperbolic:
    r1 = 4, r2 = r3 = 3 ... o

    For two pairs (12332): r2 = r5, r3 = r4, r2, r3 even

    X = F * ( 1/r1 + 2/r2 + 2/r3 - 3/2 )

    All hyperbolic.

    For three and two (11222): r1 = r2, r3 = r4 = r5, r1 even

    Non-hyperbolic:
    r1 = 4, r3 = 3 ... -

    For three and two (12122): r1 = r3, r2 = r4 = r5

    Non-hyperbolic:
    r1 = 4, r2 = 3 ... -

    For all five equal: r1 = r2 = r3 = r4 = r5

    Non-hyperbolic:
    r = 3 ... -

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