# NavList:

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

**Re: Azimuth Formula Questions**

**From:**Marcel Tschudin

**Date:**2005 Oct 23, 11:58 +0300

George, you suggested this formula to calculate the azimuth > 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 > > If a calculator or computer offers a POL or arctan2 function then you > don't > even need to apply those rules; the azimuth comes out straightaway in its > correct quadrant, from 0 to 360. > For example, the correct angle results from applying- > > POL((tan dec cos lat - cos (hour angle)sin lat, -sin (hour angle)). > Before > applying this formula, however, check whether the term before the comma > and > the term after the comma do not both happen to be zero. If they do, the > angle is indeterminate, and an error may result. > > The formula obtaining Az from its tan has the following advantages- > > It doesn't need the altitude to have been calculated. > > It has no ambiguity, whereas the expression obtaining az from its sin has > an > ambiguity about the East-West line that's very awkward to resolve. > > It is sensitive over the whole angular range, whereas obtaining az from > its > sin is very insensitive and inaccurate near 90 and 270 degrees, and from > its > cos, around 0 and 180 degrees. > > You can find a similar equation for az in Meeus, "Astronomical > Algorithms", > equation 13.5, though there's a slight different in that Meeus, in common > with many astronomers, defines azimuth starting from the South. > > It puzzles me why this arctan formula is so little-known and little-used. Do you know of a formula with the same advantages as the one you mentioned here in order to calculate the azimuth for the direction between two locations on the globe? The spherical triangle formula which I am using at present needs a lot of if-statements for selecting the right quadrant. May be there does also exist a more advantegeous formula to calculate this azimuth. Marcel