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From: Keith Thompson <kst-u@mib.org>
Newsgroups: comp.lang.c
Subject: Re: Choosing a random point on a sphere surface
Date: Tue, 11 Jun 2019 18:33:45 -0700
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Clifford Heath <no.spam@please.net> writes:
> On 11/6/19 1:54 am, Stefan Ram wrote:
>>    The following picture shows a unit ball (sphere) with a width
>>    and a height of «1».
>
> I just came up with a very nice alternate way to uniformly choose a 
> point on a unit sphere, with arbitrary granularity. Use a hierarchical 
> triangular mesh with a random binary bit-string of 3+2n bits. Convert 
> the HTM value into your chosen coordinate system, job done.
>
> HTM is a generalised coordinate system for points on the globe, based on 
> an iterated triangular subdivision of the sides of an octahedron.
>
> <http://www.skyserver.org/htm/>
> <https://www.microsoft.com/en-us/research/publication/indexing-the-sphere-with-the-hierarchical-triangular-mesh/>

I don't believe the selection is uniform.

Quoting from your second link (emphasis added):

    We describe a method to subdivide the surface of a sphere into
    spherical triangles of similar, *but not identical*, shapes
    and sizes.

We start with the surface of the sphere partitioned into 8 identical
spherical triangles, with 90-degree vertices.  Considering one
of the 8 triangles, we draw great circle segments connecting the
midpoints of the edges, partitioning it into 4 smaller triangles.
I believe the outer 3 triangles are congruent to each other, but
the central one is not.

It looks like a useful way to define a coordinate system on the
surface of a sphere, but it doesn't produce equal-area triangles
(and it's not intended to).

I don't quite have the math to confirm or refute this.  Anyone else?

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
Will write code for food.
void Void(void) { Void(); } /* The recursive call of the void */