Perhaps a navigation example would help.
1) Find track and distance from position 30deg 20' S / 20deg 05' W to position 32deg 55' S / 18deg 50' W
d.long = 1deg 15' East = 75'E
d.lat = 2deg 35' South = 155'S
Mean Lat = 32deg 55' + 30deg 20'
___________________ = 31deg 37 1/2'
2
Mid Lat = 31deg 37 1/2 ' + 1 = 31deg 38 1/2'
Dep = 75 Cos 31deg 38 1/2' = 63.85 miles
Log 75 = 1.87506
L Cos 31deg 38 1/2' = 1.93011
L Dep = 1.80517
Log 63.86 = 1.80517
Log 155 = 2.19033
________
1.61484
Tan Tr = 63,85
_____
155
Thus Track = S 22deg 23 1/2' E
= approx 157 1/2 T
Distance:
= 63.85 Cosec 22deg 23 1/2'
=167.65 miles (nautical)
=167 1/2'
Log 63.85 =1.80517
L Cosec 22deg 23 1/2' = 0.41915
_________
2.22432
Last edited by CAPTAIN WOOBLAH; 21st July 2007 at 11:23.