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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David Pike
Date: 2023 Nov 21, 03:57 -0800
Roger Errington you wrote: Is the Lat used when calculating the change in longitude (dLon = Dist x sin course / cos Lat) the original latitude (Lat0) or the adjusted latitude after adding the change in latitude (Lat = Lat0 + dLat/60)?
Roger
Your easting is dist gone x sine course made good, and your ch long is your easting/cos lat. Personnally I would use mid-lat i.e. lat0+ (northing/2).
E.g. lat 0 = 53°.10'N, course made good (cmg) = 060°, dist gone (dg) = 60nm.
Northing = dist gone x cos cmg 60= 60nm x cos60 = 30nm.
Therefore L1 = 53°.10' + 30' = 53°.40', and mid-lat = 53°10'+ (30'/2) 53°25'N.
Therefore chlong = easting/cos mid-lat = dist gone x sine cmg/cos mid-lat = 60 x sine 60/cos (53+(25/60))°=60sine60/cos53.417°
= 60 x 0.866/0.596 = 87.2' = 1°27' to the east. I never give final results closer than the thickness of a 2B pencil. I just went out and ran it and it seemed OK. Any mistakes is cos I'm still out of breath. DaveP
