Open GoogleCodeExporter opened 8 years ago
“6371” is the "surface area" radius, based on there being three radii
boundaries: Two equatorial, a_x and a_y, and the polar, b——one for each
axis. This gives an approximation of “(2*a + b)/3”.
However, for a “great circle” circumference approximation, the boundaries
are the meridional circumference and equator.
For the meridional circumference, we will add a third, mean equatorial radius,
a_m, (a_x*a_y)^.5, giving an approximate, mean meridional radius of Mr ≈
.5*(a_m + b).
The equatorial circumference——while considered spherical for Earth and
other spheroids——can be elliptical and therefore requires its
“approximation”: a_e ≈ .5*(a_x + a_y).
Thus the great circle radius approximation is
GCr ≈ .5*(Mr + a_e) = .5*(.5*(a_m + b) + .5*(a_x + a_y)),
= .25*(a_x + a_m + a_y + b),
= .25*(3*a + b);
Given the particular dynamics involved, squaring-rooting gives a closer
approximation of the circumferences, with the “ellipsoidal quadratic mean”
(or just “triaxial quadratic mean”), being the better choice:
GCr ≈ Qr = (.25*(3*a^2 + b^2))^.5;
Both ways give an approximate GCr of 6372.8 km, which is about as precise as
one can get for a universal value, given the variation in the different
ellipsoid models of Earth and the different ways the circumferences can be
differentiated.
There is a wiki article that gives a more in-depth analysis of the concept:
http://math.wikia.com/wiki/Ellipsoidal_quadratic_mean_radius
Therefore, my suggested valuation for “geomath.py” is: RADIUS = 6372800
Original comment by Kaimbri...@gmail.com
on 7 Jun 2013 at 3:12
There's an error in the analysis in
http://math.wikia.com/wiki/Ellipsoidal_quadratic_mean_radius
(the way the vertex latitude was sampled was wrong). When this is corrected
the mean radius becomes R1 = (2*a + b)/3 the IUGG recommended value. So I
propose RADIUS = 6371009.
Original comment by charles....@gmail.com
on 12 Jun 2013 at 9:49
You might also want to consider abandoning the spherical approximation and
solving for the geodesic distance using
http://pypi.python.org/pypi/geographiclib
Original comment by charles....@gmail.com
on 12 Jun 2013 at 10:20
Original issue reported on code.google.com by
david.a....@gmail.com
on 7 Mar 2012 at 5:39