just to throw a cat among the pigeons:

* Has anybody considered the earth's curvature as part of determining the centre of SA? Could well be a few meters underground...

* How much will the "centre" shift by if using different map projections? (http://en.wikipedia.org/wiki/Map_projection for more info)

* Is this centre to be determined by area calculation alone, or should it include mean elevation as a weighting?

Far too many variables for one distinct anwer - Mr Adams might well be technically correct, but for places to seek, visit or explore, this dilemma is a traveller's manna!

Pigeon pie comes to mind

You are spot on in that the projections of maps may be distorted from a representation perspective. That would depend on the projection method used.

If one looks for the shortest distance between two spherical points, then the elevation would become evident (i.e. below the surface).

One should therefore use a spherical surface formulae (ref:

http://en.wikipedia.org/wiki/Geographical_distance}As a result of the Earth not being a sphere and having a bulge around the equator, makes the Earth an ellipsoid. So more accurately one needs to use an Elipsoidal-surface formulae to get the approximated distance over great distances. There are several documented methods - Vincenty's Bowring's and Lambert's formulae.

For interest's sake take a look at Vincenty's formulae -

http://en.wikipedia.org/wiki/Vincenty%27s_formulaeWikipedia then goes further to say ...

The systems are needed because the earth is not a perfect sphere. **Neither is the earth an ellipsoid**. This can be verified by differentiating the equation for an ellipsoid and solving for dy/dx. It is a constant multiplied by x/y. Then derive the force equation from the centrifugal force acting on an object on the earth's surface and the gravitational force. Switch the x and y components and multiply one of them by negative one. This is the differential equation which when solved will yield the equation for the earth's surface. This is not a constant multiplied by x/y. Note that the earth's surface is also not an equal-potential surface, as can be verified by calculating the potential at the equator and the potential at a pole. The earth is an equal force surface. A one kilogram frictionless object on the ideal earth's surface does not have any force acting upon it to cause it to move either north or south. There is no simple analytical solution to this differential equation. A power series solution using three terms when substituted into this differential equation bogs down a TI-89 calculator and yields about three hundred terms after about five minutes.

This is enough to make a monkey fuck a pineapple.

But when one uses Lat and Long coordinates and GPS coordinates, then presentation is irrelevant.

So for the sake of DS motorcycle enjoyment, the simplest method should be applied: (E.point - W.point) /2 + W.point = Mid Horizontal .point

and; (N.point - S.point) /2 + S.point = Mid Vertical .point