tesellation (sp?) of equilateral triangles on a sphere...
Ben Horner
Please direct me to a good one if it is not.
I am trying to create a tesellation of equilateral triangles on a
sphere. My first thought is that it is impossible because six triangles
fit in a circle (on a plane), and then adjusting for the curvature of
the surface of the sphere would "raise" the center of the circle, and
increase the lengths of the radii that form two of the edges of each
triangle... So I may have to use isosceles triangles. I seem to
remember seeing dice having both 20 and 100 sides that were made up of
triangles. What multiples of trianlges can tesellate a sphere? Am I
right in thinking that they must be isosceles? Any help would be
appreciated. It is important for the triangles to have the same area, I
am trying to find densities of points on a sphere, I thought breaking it
up into trianlges and counting the points would be a good way. Other
approaches would be welcome as well.
Thanks in advance!
-Ben
Lutz Tautenhahn
"Ben Horner" <hor...@wernervas.com> schrieb im Newsbeitrag
news:MPG.174ee040c...@news.cis.dfn.de...
> Hello, not sure if this is the correct group for this...
> Please direct me to a good one if it is not.
>
> I am trying to create a tesellation of equilateral triangles on a
equilateral triangle -> 3 angles of 60°
> sphere. My first thought is that it is impossible because six triangles
> fit in a circle (on a plane), and then adjusting for the curvature of
so take 5 (ikosahedron), 4 (octahedron) or 3(tetrahedron) instead
> the surface of the sphere would "raise" the center of the circle, and
> increase the lengths of the radii that form two of the edges of each
> triangle... So I may have to use isosceles triangles. I seem to
a isosceles triangle is in general not a equilateral triangle
> remember seeing dice having both 20 and 100 sides that were made up of
20 sides = ikosahedron, can not imagine that 100 is possible, but you could
get
80 sides, when you devide every triangle of the ikosahedron in 4 triangles
(1 equilateral triangle -> 1 equilateral triangle in the center + 3
isosceles triangles )
> triangles. What multiples of trianlges can tesellate a sphere? Am I
if not required triangles of a special symetry, then any number >4 is
possible
(by iteratively deviding any triangle into two other triangles)
> right in thinking that they must be isosceles? Any help would be
Why should they be isosceles? The only reason is, that it looks better.
> appreciated. It is important for the triangles to have the same area, I
> am trying to find densities of points on a sphere, I thought breaking it
When you want a small number of triangles, then take for instance the
icosahedron.
If you want a large number of triangles, you could use the following method:
Start with any tessellation of N triangles, which have the same area.
while number of triangles is to small loop
devide every triangle (at the longest side), into two triangles of the
same area
end loop
After I loops you have a tessellation of N*2^I triangles, which have all
the same area.
> up into trianlges and counting the points would be a good way. Other
> approaches would be welcome as well.
> Thanks in advance!
> -Ben
If you need the coordinates for a icosahedron (approx.):
0 0 -1
0.894422 0 -0.447212
0.276394 0.85065 -0.447212
-0.723606 0.525732 -0.447212
-0.723606 -0.525732 -0.447212
0.276394 -0.85065 -0.447212
0.723606 0.525732 0.447212
-0.276394 0.85065 0.447212
-0.894432 0 0.447212
-0.276394 -0.85065 0.447212
0.723606 -0.525732 0.447212
0 0 1
Regards,
Lutz Tautenhahn
Jay and Diane Rudin
> I am trying to create a tesellation of equilateral triangles on a
> sphere. My first thought is that it is impossible because six triangles
> fit in a circle (on a plane), and then adjusting for the curvature of
> the surface of the sphere would "raise" the center of the circle, and
> increase the lengths of the radii that form two of the edges of each
> triangle...
Well, yes. In fact, it should be clear that no plane figure can be put on
the surface of a sphere, because a sphere is not a plane.
So your triangles are not true triangles by definition of the problem. The
modified problem becomes this: how can you divide a sphere's surface into
subsets of three sides, where a side is defined as a segment of a great
circle? (I'm going to call these subsets triangles -- remember that they
are not planar figures.) The essential fact about the sphere-surface
triangles is that their three angles always add up to more than 180 degrees.
How much more is determined by the size of the triangle relative to the
sphere. If it's small, then the angles are very close to 180 degrees. But
consider a triangle with one vertex at the North pole, and the other two on
the equator, at 0 degrees longitude and at 90 degrees of longitude. This
triangle is equilateral (all sides are equal), but each angle is 90
degrees -- an equilateral right triangle.
Let's assume that every side is equivalent to every other side, and every
angle is equivalent to every other angle. Then there are only three ways to
do it.
A. Three triangles come together at every point -- four triangles, six
edges, four vertices.
B. Four triangles come together at every point -- eight triangles, twelve
edges, six vertices. (This uses equilateral right triangles as described
above).
C. Five triangles come together at every point -- twenty triangles, thirty
edges, twelve vertices.
These are the equivalents of a tetrahedron, octahedron, and icosahedron,
respectively. (The cube and dodecahedron are not made of triangles.)
Note that if F is the number of triangles (faces), E is the number of edges,
and V is the number of vertices, then F + V = E + 2. This is always true.
> So I may have to use isosceles triangles. I seem to
> remember seeing dice having both 20 and 100 sides that were made up of
> triangles.
The 100-sided die is not a tesselation. It's built rather like a golf ball,
but with dimples that are bigger and closer together. It has 100 flat
circles that don't quite meet, and always ends up with one of these on top.
It's sold by Gamescience, owned by Lou Zocchi, and a picture of it can be
found at http://membres.lycos.fr/arjan/num100.htm
> What multiples of trianlges can tesellate a sphere?
Any even number, but without distorting it too badly, you're restricted to
4, 6, 8, 12, 20, 24, 48, 60, and 120.
> Am I right in thinking that they must be isosceles?
No. In the 48-sided one, each triangle has:
one vertex in which 4 triangles meet,
one vertex in which 6 triangles meet, and
one vertex in which 8 triangles meet.
This makes it a 90-60-45 triangle. It isn't isosceles, and the angles add
to 195 degrees. Similarly, the 120 has 90-60-36 triangles, and 4, 6, and 10
triangles come together at a point. There's no 4/6/12 equivalent, because
that's a 90-60-30 triangle, which is a plane figure, so the tesselation is a
plane, not a sphere.
Now, picture a long skinny triangle with one vertex at the north pole, and
two more on the equator. The edge on the equator is 1/n of the equator.
It's clear that n of these triangles can cover the northern hemisphere.
Another n of these can cover the southern hemisphere, so the entire sphere
can be covered with 2n triangles. These are 90-90-360/n triangles.
In fact, assuming that the triangles must all be congruent, and that their
neighbors must have the same configuration, you can characterize a
tesselation by the number of faces that come together at each vertex of a
single triangle. Thus, a tetrahedron-equivalent is a 3-3-3 tesselation.
The number of sides must be either be all even, all identical, or an odd
number and two identical even numbers.
Number of sides / tesselation type
4 / 3-3-3
6 / 3-4-4
8 / 4-4-4
10 / 4-4-5
12 / 4-4-6
12 / 3-6-6
20 / 5-5-5
24 / 4-6-6
24 / 3-8-8
48 / 4-6-8
60 / 3-10-10
60 / 5-6-6
120 / 4-6-12
2n / 4-4-n
These do not cover all possibilities -- just the ones in which each face is
identical to every other face in size, shape, and position. You can expand
on these by dividing each triangle into any square number of smaller
triangles. Note, however, that unlike the equivalent operation on a planar
triangle, the subdivided triangle swill not be congruent.
> Any help would be
> appreciated. It is important for the triangles to have the same area, I
> am trying to find densities of points on a sphere, I thought breaking it
> up into trianlges and counting the points would be a good way. Other
> approaches would be welcome as well.
Good luck.
John Rudin