NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Robin Stuart
Date: 2017 Jul 6, 07:28 -0700
Tony,
This shouldn’t be too much of a surprise if you inspect spherical triangles that these formulas come from and identify the corresponding sides and vertex angles (see Bowditch 2012, p. 348). By analogy it follows from the earlier discussion that you can get both the great circle distance and course using the calculator’s Pol() function.
Pol{ [ cos(Lat1) · tan(Lat2) - sin(Lat1) · cos(λ2 - λ1) ] , [sin(λ2 - λ1) ] }
and then
D = 60 × sin-1( r cos(Lat2) ) and C = θ
where D is the distance in nautical miles and C is measured eastward from north (again you may have to add 360° to bring it into the correct range).
The distance will be correct provided the two places are less than 90°apart on the Earth’s surface. If you know the distance is greater than 90° then
D = 10,800 - 60 × sin-1( r cos(Lat2) )
Regards,
Robin Stuart
