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

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Re: Which calculator to use for "arctan(tan(269))"?
From: Robin Stuart
Date: 2017 Feb 1, 08:40 -0800

Tony,

I went back and rederived the altitude and azimuth formulas in using 3D methods (see the summary at the end of this post). The results are necessarily the same as what Lars gave but inn case it’s not clear I want to emphasize how very useful the interpretation to r is in LoP sight reductions.

The procedure becomes:

Calculate

Pol{ [ cos(Lat) · sin(Dec) - sin(Lat) · cos(Dec) · cos(LHA) ] , [-cos(Dec) · sin(LHA) ] }

and then

Hc = cos-1 r and θ = Zn

or possibly more efficiently in terms of keystrokes

Pol{ [ cos(Lat) · tan(Dec) - sin(Lat) · cos(LHA) ] , [-sin(LHA) ] }

and then

Hc = cos-1( r cos(Dec) ) and θ = Zn

There is no need for a separate calculation of the altitude!!! One proviso, as Lars indicated, is that it only gives the correct answer for objects above the horizon.

Regards,

Robin Stuart

Summary of key results

Define topopcentric coordinates x,y,z in which the x-axis is horizontal pointing East, the y-axis is is horizontal pointing North and the z-axis is along the vertical. Relating the global Greenwich hour and declination GHA, δ coordinates to local azimuth and altitude Zn, Hc (details on request) gives

x = sin Zn cos Hc = -cos δ sin LHA

y = cos Zn cos Hc= cos δ cos L - cos δ sin L cos LHA

z = sin Hc= sin δ sin L + cos δ cos L cos LHA

which satisfy x2+y2+z2 = 1 and hence x2+y2 = 1- z2 = cos2 Hc Browse Files

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