Hervé,

You know, as well as your initial longitude, your initial latitude L , your bearing b, and your run r. Your angular distance from the North Pole is thus (90-L), taking South latitudes as negative; this, r, and b give two sides and the included angle of a spherical triangle with vertices at start, finish, and the North Pole.

This may be solved for the remaining side using the spherical cosine law:

\[\cos x = \cos (90-L) \cos r + \sin (90-L) \sin r \cos b\]

with the new latitude L' equal to $90-x$

or (as $\sin(90-a) = \cos(a)$)

\[\sin(L') = \sin L \cos r + \cos L \sin r \cos b\]

Once you have found L', you need your new longitude, which you find from the spherical sine law. If the angle subtended at the pole by the start and finish points is A, then

\[\frac{\sin(A)}{\sin(r)} = \frac{sin(b)}{sin(x)}.\]

The angle A is also the difference between new and old longitudes.

Good Hunting (fishing?)

RD