Geographical Center Problem

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...


Just a thought,


George Huxtable.
george@huxtable.u-net.com
George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
Tel. 01865 820222 or (int.) +44 1865 820222.

The floating world-globe with glued coins idea is clever.  The lat-lon of
the lowest point when the globe is thrown into the pool can be computed as
follows:


For each coin, compute the cartesian position x,y,z from:


         x = cos(lat)cos(lon)
         y = cos(lat)sin(lon)
         z = sin(lat)


Next, separately average the x, y, and z values to get xAv, yAv, zAv.  This
is the position of the center of mass of the coins.  The (lat, lon) of the
center of mass is computed from:


         lat = asin(zAv/r)  where r = sqrt(xAv^2 + yAv^2 + zAv^2)
         lon = atan2(yAv, xAv)


So, assuming my math and logic are correct, this is the lat and lon of the
lowest point of the globe when thrown back into the pool.

Solution

Here is the current dataset and the results:

The center is in Canada, in northern Quebec, south of Hudson Bay:


  206.3 nmi  43° NE of Pickle Lake, Canada. (nearest airport)
  258.5 nmi 310° NW of Moosonee, Canada. (nearest seaport)
  263.0 nmi  16°  N of Katatota Island, Canada.


The weighted center moves southeast 109 nmi from the unweighted center:


  159.5 nmi 297° NW of Moosonee, Canada. (nearest seaport)
  156.9 nmi  31° NE of NAKINA, CA. (nearest airport)
  216.6 nmi  40° NE of Katatota Island, Canada.