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George Huxtable wrote:

> I have been pondering on what would be a meaningful way to define a centre
> of gravity for individuals members scattered over the surface of a sphere.
>
> Somehow, I doubt whether averaging latitudes and then separately averaging
> longitudes provides the best answer. It should be possible to do the job
> properly, for anyone with a bit of time on his hands. Here is a suggested
> mechanical analogy.
>
> Take a globe of the world, that will float. For example, I have a beachball
> printed with the World on its surface, crudely marked with lat and long. If
> you put it into a pool, it will float any way up, like most spheres.
>
> Now glue on to its surface a set of identical coins or weights, one for
> each member, in the spot where they live, and chuck it into the pool again.
> Now, the lowest point of this globe represents in some way a centre of
> concentration of the members.
>
> It should be possible to devise a computer analogy to avoid having to do
> the physical experiment. Would it produce the same answer as Dan Allen's
> method? I wonder...

This problem can be solved mathematically in the following way;

Regard each position on the Earth's surface as a vector and reduce each
longitude-latitude co-ordinate to its Cartesian co-ordinates (x,y,z):

  x = cos(lat) * cos(long)
  y = cos(lat) * sin(long)
  z = sin(lat)

Add all the x's, y's and z's to form the sum vector (X,Y,Z), and perform
the inverse operation:

  mean long = arctan(Y/X)
  mean lat  = arctan(Z/sqrt(X*X+Y*Y))

When the arithmetical signs of X and Y are properly taken into account,
there will be no longitude ambiguity (in FORTRAN one could use the
ATAN2(X,Y) function to circumvent this).

If I have some time later this evening, I will write a quick program and
do some calculations.

My lat-long = +52.086  -5.129

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