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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: Azimuth Equation
From: Bill B
Date: 2005 Oct 19, 15:04 -0500

```>
> I'm using the azimuth equation out of the Nautical Almanac:
> cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc]
>
> To find Hc: sin-1[(sinDec sinLat)+(cosDec cosLHA cos Lat)]
> In entering the equation, if the Declination or latitude is South, or the
> Longitude is West, the value is negative.

Todd

After a nights sleep, and looking at the almanac, have rethought the
problem.  Your signing of Lat, West Lon, and declination is correct.
Remember LHA must always be a positive number from 0 to 359.xxx.  If GHA+W
Long is negative, add 360. If GHA + E Long is equal to or greater than 360,
subtract 360.  Some will tell you when using a modern calculator, a negative
LHA or LHA greater than 360 will work fine, the calculator knows the
appropriate trig value of those angles.

The formula for Hc looks correct: sinHC=(sinDec sinLat)+(cosDec cosLHA cos
Lat)

cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc] seems strange to me.
Perhaps it folds the formula for Hc in the formula below.  For Z and Zn
(azimuth) I use:

cosZ = (sinDec - (sinLat sinHc))/(cosLat cosHc)

If sinLHA is negative, Z=Zn
If sinLHA is positive, Zn = 360-Z

Purely by inspection if, LHA is >0 to <180, the body will be in the SW or NW
quadrant. If LHA is >180 to <360, the body will be in the NE or SE quadrant,
which tells you whether to use Z as Zn, of subtract it from 360.

Hope that helps,

Bill

```
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