Frederik Ramm wrote:
Hi,
Sebastian Klein wrote:
Good to know... I think we could get this value down to 0.642cm if use
the full 64 bit and don't skip the range from 180*10^7 to 2^31
(longitude) and from 85.1*10^7 to 2^31 (latitude).
But why stop there and allow ridiculous resolution at the poles? Given
that the earth has a surface of 510,072,000 square kilometres, if you
divide that surface in 2^64 almost-round shapes, then each of those only
has an area of 0.2765 square centimetres, allowing you to address each
place on the surface with 0.52cm precision!
Right, but how is it done in practice? You could map the sphere to a
disk, without area distortion (e.g.[1]). Then cut the disk into pieces
and rearrange them on a square.
Nice puzzle game... What's the minimum size of the square when you limit
the number of straight cuts on the disk? Maybe it gets better if you
allow arc shaped cuts, e.g. slice off crescents and shift them into a
nice pattern... :)
Sebastian
[1] <http://en.wikipedia.org/wiki/Lambert_azimuthal_equal-area_projection>
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