Originally Posted by gAMbl3
1) r/l track btw a(45n 10w) and b(48'30n 15w) is
RL track from A to B is 270 + x deg
In triangle AOB
departure dist AO = d long in min x cos 45 = 5 x 60 x cos 45 = 212 nm.
great circle dist BO = (48.5 - 45) x 60 = 210 nm
tan x = BO/AO = 210/212
x = tan inv 210/212 = 44.7 deg
RL track = 270 + 44.7 = 314.7 deg
The reason AB is not a Great Circle is that in a Mercator graticule such as this, a great circle is more often a curve projected concave to the equator.
To find GC track between these two:
Conversion Angle= dLong x Sin (Mean Lat)/2 = 5 x 60 x Sin(46°45')/2 = 109°15'
In the N.H., westerly track, GC>RL. So, GC track at A= 314.7 + 109.25= 64°.
Please correct if I am wrong.