the center of the world (was Re: Check out O'Reilly's Open Source Convention Highlights)

[Toby Dickenson]

"Alex Martelli" <aleaxit at yahoo.com> wrote:

>"Konrad Hinsen" <hinsen at cnrs-orleans.fr> wrote in message
>news:m33d8j8j2j.fsf at chinon.cnrs-orleans.fr...
>> markus at kepler.ibp.de writes:
>>
>> > Then find first the weighted average as a point in three-dimensional
>> > space, i.e. as (an approximation to) the center of gravity of the
>> > human population. This will be some point deep in the interior of the
>> > Earth, which you could the project back to the surface.
>>
>> But that doesn't minimize the distance on the surface.
>
>Doesn't it (in terms of great-circle distance)?  I can't prove
>it in my head either way, but trying to visualize a few cases
>does suggest the requested surfacepoint IS obtained this way.

Imagine 51 people gathered at the north pole, and 50 people gathered a
very small distance away from the south pole.

Surface-distance-wise the midpoint is roughly on the equator (actually
1% of the half-circumference north of the equator)

However, the center of mass of the people lies only slightly lies off
the polar axis, but northwards of the earths center by a 1% of its
diameter. Projected onto the surface this is a point very close to the
north pole.

Best effort ascii-art: 
* is the earths center
. is center of mass of people
x is that point projected onto the surface

        /                            \
       /                              \
      x                                )
     (            .                     ) 50 
N 51 (                *                 )    S
     (                                  )



Toby Dickenson
tdickenson at geminidataloggers.com