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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: A triangle on the equator
From: Lars Bergman
Date: 2025 Oct 9, 07:17 -0700
From: Lars Bergman
Date: 2025 Oct 9, 07:17 -0700
DaveP,
A few days ago you referred to Bodger's Method, of which I have never heard, and cannot find any information of. Anyway, your value of the earth's circumference through the poles seems to be some 62 km too small. My calculation gives
Cp = 4·a·(1 - e2/4 - 3/64·e4 - 5/256·e6 - ...)·π/2 = 40007.86 km for WGS84, with equatorial radius a and eccentricity e. With this value I get the angle at A to be 53°10.6'.
Furthemore, I wouldn't rank this method too high, as it does not take into account the different arc lengths for some unit of arc along a meridian which depends on the latitude. On the upside is its simplicity compared with e.g. Vincenty's.
Lars
