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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 Formula Questions
From: Bill B
Date: 2005 Oct 28, 17:37 -0500

```George wrote:
> Here's the expression to use to obtain azimuth from its Tan. It presumes
> that southerly latitudes and declinations are negative, that local hour
> angle is measured positively Westwards 0 to 360 degrees, and that azimuths
> are measured clockwise from North, 0 to 360 degrees.
>
> az = arctan (sin (hour angle) / (cos (hour angle) sin lat - cos lat tan
> dec))
>
> Then to put az into the right quadrant, apply the following rules-
>
> if tan az was negative, add 180 degrees to az.
> if hour angle was less than 180 degrees (i.e., to the West of the observer),
> add another 180 degrees to az

I notice another (longer) formula in the HO229 background section that also
does not require Hc:

tanZ = (cosd sinLHA)/(cosL sind - sinL cosd cosLHA)

L is latitude
d is declination

Zn is determined as per 229 usual rules:
North lat
Z > 180, Zn = Z
Z < 180, Zn = 360-Z
South lat
Z > 180, Zn = 180-Z
Z < 180, Zn = 180-Z

For the mathematically inclined, are there pros and cons for either formula?

Bill

```
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