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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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Azimith forumula for Great Circle sailings- having problems
From: Samuel L
Date: 2014 Dec 23, 14:41 -0800

I use the following formula to obtain Azimith-

Z = tan-1 (sin (LHA) / (cos (LHA) x sin(AP Latitude) - cos(AP Latitude) x tan(Declination))

If LHA is greater than 180 it's treated as a negative quantity.
If the Azimith angle as calculated is negative, add 180 to it.

In one Great Circle sailing probelm I'm working on to determine Z the answer is negative unless the LHA is entered into the formula as a negative number. The answer should be about 3d 55min. If I add 180 to to the negative answer the wrong Z is obtained.

tan-1((sin(LHA) / (cos(Present Latitude) x tan(Destination Latitude) – (sin(Present Latitude) x cos(LHA))

Here's the figures;

LHA= 0d 9m 40.9sec

Present Latitude- N 39d 58m

Destination Latitude= N 41d 43m 31.8sec

The answer I get is -3d 55min (notice the negative sign.

It would be nice to have an Azimith formula that's independent of Hc and provides the correct figure regardless of LHA greater or less than 180d.

How do I solve this problem?

Thanks,

Sam Lohengrin

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