# NavList:

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

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Re: Curve fitting for noon Sun
From: Frank Reed
Date: 2020 Jan 14, 15:42 -0800

Rafael, you wrote:
"Ahhh. Thank you very much.  This is precisely the answer to my question, which I am not sure I stated very clearly.  I had an inkling that the formula for altitude vs. time might involve a solution for a spherical triangle..."

Oh great. Glad it's what you were looking for.

It's not difficult to relate that equation to something more familiar. No doubt you have seen the spherical trig equation for great circle distance on the (assumed spherical) globe. It is:
cos(D) = sin(Lat1) · sin(Lat2) + cos(Lat1) · cos(Lat2) · cos(Lon2 - Lon1)
where D is the distance in degrees between two places on the globe with coordinates Lat1, Lon1 and Lat2, Lon2. The key principle of celestial navigation is that the measured zenith distance (ZD = 90° - corrected altitude) of a star is equal to the distance from the observer to the subStar point. The subStar point is the single location on the Earth. at any instant of time where the star is at the zenith. Further the latitude and longitude of the subStar point are exactly the Dec and GHA from almanac data. Dec is latitude. GHA is longitude. So we have a situation with a known location and an observed distance off, but otherwise it is just a great circle distance problem. In the standard great circle distance relationship, we replace D by ZD, Lat1, Lon1 can be just Lat, Lon (of the observer), and Lat2, Lon2 become the coordinates of the subStar point, Dec, GHA. Then we have
cos(ZD) = sin(Lat) · sin(Dec) + cos(Lat) · cos(Dec) · cos(GHA - Lon).
One last step: cos(90°-x) is identical to sin(x) for any x, so by the definition of ZD, we get:
sin(Alt) = sin(Lat) · sin(Dec) + cos(Lat) · cos(Dec) · cos(GHA - Lon).
And that's how celestial altitudes are calculated. It really is just the great circle distance between two points.

Frank Reed

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