NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/sci/math/11313

Internet Message Format | 1992-09-12 | 3.5 KB

Path: sparky!uunet!usc!zaphod.mps.ohio-state.edu!uwm.edu!rutgers!ub!acsu.buffalo.edu!kriman
From: [e-mail] (Alfred M. Kriman)
Newsgroups: sci.math
Subject: Re: Tiling sphere by triangles (Re: 3 space terahedron-packing)
Message-ID: <[e-mail]>
Date: 12 Sep 92 09:21:04 GMT
References: <f#[e-mail]> <[e-mail]>
Sender: [e-mail]
Organization: UB
Lines: 57
Nntp-Posting-Host: lictor.acsu.buffalo.edu

In article <[e-mail]>
[e-mail] (Albert Goodman) notes the tetrahedron question of:
 [e-mail] (Steven)
 in article <f#[e-mail]>,
and asks:
>
>Instead of 3-space, consider the surface of a sphere (such as [an
>approximation to] the surface of the earth), and suppose we want to
>find a tiling or "grid" of triangles which divide up the surface of
>the sphere into many small pieces all of the same area (and preferably
>all the same shape, i.e. congruent triangles).  (Actually the original
                                   ^
                                   |__ (spherical)
>question didn't have to use triangles, just some constant shape, but
>triangles seems perhaps a nice choice; trying to use squares, as would
>be the obvious solution on a flat surface, doesn't seem to work on the
>sphere to get all of the same area.)

Suggestion:
   Project a (concentric) platonic solid onto the sphere.
That gives you three nice tilings by spherical triangles (from 4-, 8- and
20-gons).  Go with the octohedron; the algorithm for determining which
triangle encloses an arbitrary point is obvious and easy, if you aren't
using a masochistic coordinatization of the sphere.
   The tetrahedron is also fairly easy.  If you prefer the icosahedron,
I suggest using the symmetry group of the icosahedron to simplify the
search.  Rather than storing the descriptions of 30 edges, which lie on
15 (?) great circles, store the rotations that transform those great
circles into meridians (i.e., curves of constant azimuthal angle) or
the equator, so the comparison is easier.

   If all vertices are required to be equivalent, then there are no tilings
of the sphere by congruent regular spherical polygons, other than those
induced by the platonic solids, because such tilings are in one-to-one
correspondence with the regular polyhedra.
   To get the map from the spherical tilings to the platonic solids,
connect vertices on the sphere with chords wherever an edge is present
in the tiling.  The chords corresponding to the edges of a particular
regular spherical polygon are coplanar, and define faces of the regular
polyhedron.

   It may provide insight into the problem to recognize that regular
spherical polygons of different sizes cannot be congruent.  The reason
is that a length scale (the radius of curvature) appears in the metric,
so, roughly speaking, the lengths of a polygon are not homogeneous
functions of the coordinates.  Crudely, if all coordinate values are
scaled by some factor s, not all lengths are scaled exactly by a factor
s.
   More specifically, the sum of the exterior angles of a spherical
triangle (in radians) is pi plus the angular area of the triangle (in
steradians).
   The relevance of these random facts is that once you have, say, a
tiling of the sphere by 8 regular triangles (with _three_ 90 deg.
angles), it is not possible to construct a new tiling by 8 x 4
congruent triangles, as you would in the plane by bisecting each
equilateral triangle's edges to form vertices for four new half-size
triangles.