coordinate. The angle from the I<z>-axis is B<phi>, also known as the
I<polar> coordinate. The `North Pole' is therefore I<0, 0, rho>, and
the `Bay of Guinea' (think of the missing big chunk of Africa) I<0,
-pi/2, rho>.
+pi/2, rho>. In geographical terms I<phi> is latitude (northward
+positive, southward negative) and I<theta> is longitude (eastward
+positive, westward negative).
-B<Beware>: some texts define I<theta> and I<phi> the other way round,
+B<BEWARE>: some texts define I<theta> and I<phi> the other way round,
some texts define the I<phi> to start from the horizontal plane, some
texts use I<r> in place of I<rho>.
use Math::Trig 'great_circle_distance'
- $distance = great_circle_distance($theta0, $phi0, $theta1, $phi, [, $rho]);
+ $distance = great_circle_distance($theta0, $phi0, $theta1, $phi1, [, $rho]);
The I<great circle distance> is the shortest distance between two
points on a sphere. The distance is in C<$rho> units. The C<$rho> is
optional, it defaults to 1 (the unit sphere), therefore the distance
defaults to radians.
+If you think geographically the I<theta> are longitudes: zero at the
+Greenwhich meridian, eastward positive, westward negative--and the
+I<phi> are latitudes: zero at North Pole, northward positive,
+southward negative. B<NOTE>: this formula thinks in mathematics, not
+geographically: the I<phi> zero is at the Nort Pole, not on the
+west coast of Africa (Bay of Guinea). You need to subtract your
+geographical coordinates from I<pi/2> (also known as 90 degrees).
+
+ $distance = great_circle_distance($lon0, pi/2 - $lat0,
+ $lon1, pi/2 - $lat1, $rho);
+
=head1 EXAMPLES
To calculate the distance between London (51.3N 0.5W) and Tokyo (35.7N
$km = great_circle_distance(@L, @T, 6378);
-The answer may be off by up to 0.3% because of the irregular (slightly
-aspherical) form of the Earth.
+The answer may be off by few percentages because of the irregular
+(slightly aspherical) form of the Earth.
=head1 BUGS