From: uds@trace.eng.wayne.edu (Seetamraju Udaybhaskar)
Subject: Re: Equispaced points on sphere? ( Repeat?)
Reply-To: uds@trace.eng.wayne.edu (Seetamraju Udaybhaskar)
Organization: Wayne State University, Detroit
Date: Wed, 2 Dec 1992 01:26:01 GMT

In article sshws@convx1.lerc.nasa.gov (Herb Schilling) writes:
>
>am trying to figure out how to determine the locations of n equispaced 
>points on the surface of 3-D sphere. 


I have a algorithmic solution.

The idea is based on this observation.  ANy 3 neighboring points will
form a `triangle' (in quotes, because U dont have a straight line
triangle on a sphere, but I hope U get what I am seeing in my mind.
maybe we shall call it a `bulged triangle' made up of arcs...)

Now since all points are equidistant, clearly this bulged triangle
will be `equilateral' and more importantly, all such triangles are
identical.

Hence if there are n equidistant points on the sphere, there are c(n,3)
such triangles.

Clearly,  if f(l) is the area of each bulged triangle (all are equal in
area), then

                f(l) * c(n,3)  == 4.pi.r^2              ---- (1)

where l is the `equi'-distance between the equidistant points.

Now heres the fuzzy part...  I am giving a not so clear derivation of
f(l)... then since eqn-(1) contains only one variable (i.e., the distance
`l') one can solve for it...!!

After I solve for `l', I use a compass or divider (in software ofcourse)
to `mark-out' all the neighboring points of ANY ARBITRARY INTIAL point
and THEN the NEIGHBORING points of THOSE NEIGHBORING points  and so on...


Well, here's the formula for f(l) ::::

Clearly, each such triangle is HALF the area of a `bulged-square' whose
sides are of length `l'.  The area of the square (I am guessing here!!)

                  (theta)^2
                = _______  * 4 * pi * r^2               ---- (2)
                   (360)^2

where theta is the angle subtended by the arc connecting ANY TWO neighboring
equidistant points.  right ?

Clearly, theta = l / 2pi*r  in radians :: to be converted into degrees...


Hence f(l) =     (l)^2 *  4 * pi * r    * a conversion factor for
radians-to-degrees
                --------------------
                2 * (360)^2 *  2 * pi * r

          =   l^2 / (2 * pi)^2     =   ( l / 2*pi)^2.


This is too simple a formula.   I am not at all confident of it...
But, if U can correctly determine f(l), the problem is solved...


Seetamraju Udaya Bhaskar Sarma
(email : seetam @ ece7 . eng . wayne . edu)




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From: Mark A Stewart <mstewart@spam.maths.adelaide.edu.au>
Message-Id: <9212030138.AA23545@spam.maths.adelaide.edu.au>
To: sshws@convx1.lerc.nasa.gov
Subject: Re: Equispaced points on sphere? ( Repeat?)
Newsgroups: sci.math
Organization: Statistics, Pure & Applied Mathematics, University of Adelaide

Gooday,
        This is just a top of the head response so dont take it as gospel.
But I believe the way to do this is to look at the symetry groups of the
plutonic solids and then look at the finite fixed point sets of these groups.
These points correspond to either vertices, midpoints of the edges, and
midpoints
of the faces all of which may be scaled to fit on the sphere and satisfy your
condition. Of course this somewhat restricts the choices of 'n'.
        If this doesnt help mail me, as I looked at a similar problem for
a geographer some years back now and have the notes stashed away somewhere.

        Mark Stewart

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Date: Thu, 3 Dec 1992 17:19:59 +1100
From: Glen Barnett <barnett@mummy.agsm.unsw.OZ.AU>
Message-Id: <199212030619.AA07927@mummy.agsm.unsw.OZ.AU>
To: sshws@convx1.lerc.nasa.gov
Subject: Re: Equispaced points on sphere? ( Repeat?)
Newsgroups: sci.math
Organization: The Australian Graduate School of Management
Cc: 

In article <1992Dec1.134328.19116@eagle.lerc.nasa.gov> you write:
>
>I apologize if this was just discussed ( I think it was ), but I
>am trying to figure out how to determine the locations of n equispaced 
>points on the surface of 3-D sphere. 

There was a fair discussion of this fairly recently. Strictly, I guess,
only 1,2,3 and 4 points can be placed so all points are equally
distant from all other points (and the placements are obvious).

What you probably refer to would correspond (for >3 points) to 
the vertices of an inscribed Platonic solid - i.e. 4,6,8,12,20
points can be placed 'equidistant' in the same sense that an inscribed
regular hexagon places 6 points 'equidistantly' when inscribed in a circle -
they aren't equidistant, except from their closest neighbours.

The interesting questions come when you try to place, say 5
points "as far apart as possible" on a sphere. In general, I don't
think there is an algorithm for this.

>
> By the way, I searched hi and low for archives of this newsgroup but
>was unable to find one. Is there one ?
>
If its not in the FAQ (posted to sci.math regularly), I doubt it.

Glen


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Date: Thu, 3 Dec 92 14:29:26 -0500
From: Tim Murdoch <tmurdoch@liberty.uc.wlu.edu>
To: sshws@convx1.lerc.nasa.gov

To: sshws@convx1.lerc.nasa.gov (Herb Schilling)
Subject: Re: Equispaced points on sphere? ( Repeat?)
Newsgroups: sci.math
Organization: Washington & Lee University

Hi,

I'm responding to your post about equi-spaced points on a sphere. Do you
mean a set of points such that any pair of points are the same distance
apart? Also, are you talking about a three dimensional sphere or a sphere
in ordinary 3-space? I can answer your question if your answer to my first
question is yes. In fact, I have a paper to appear in the American
Mathematical Monthly that addresses this very question.

I look forward to your reply.

-- <<Tim Murdoch>>

-- 
Herb Schilling , NASA Lewis Research Center , 21000 Brookpark Road, MS 142-5
Cleveland, Ohio 44135 . (216) 433-8955       sshws@convx1.lerc.nasa.gov