# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Take two: Latitude and longitude around noon**

**From:**Robin Stuart

**Date:**2017 Apr 6, 12:48 -0700

I improved on my rough and ready quadratic fitting calculation with a more rigorous approach. First each observation is corrected by its own corresponding refraction and parallax correction plus semi-diameter (dip=0). Then using the standard altitude with linear time-dependence in GHA and declination

sin *H*_{0} = sin *L* sin(*δ*_{0} + d/60 ΔT) + cos *L* cos(δ_{0} + d/60 ΔT) cos( GHA_{0} + *λ* + 15 ΔT )

where GHA_{0}, *δ*_{0} are the GHA and declination at some convenient time, T_{0}, and ΔT = UT - T_{0}.

After an initial estimate of the latitude, *L*, and longitude, *λ*, you can calculate the sum of squares of differences between the formula and observations and then obtain the least squares fit to *L* and *λ*but numerical methods. This approach has a number of advantages

1) The fit is logically correct and doesn’t rely on the assumption that the altitude as a function of time approximates a parabola. It should therefore work even when the Sun transits at a high altitude or the observations are taken over quite a long time interval. In principle the observations don’t even need to bracket noon.

2) It’s a 2 parameter fit rather than 3 in the case of the quadratic

3) No need for the equation of time or a correction for the difference between LAN and time of maximum altitude. It’s all automatically incorporated in the fit

This method is equivalent to the multi-LoP method given in the Nautical Almanac but without the locally flat Earth approximation. You immediately get don’t get the time of maximum but you don’t need it.

The answer I get is 66°25.4′S 81°37.8ʹE which is on the coast Mikhaylov Island.

Just for fun I applied this approach to the observations at 6:30:30 UT and earlier only. I get 66°25.6′S 81°44.5ʹE

Robin Stuart