# NavList:

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

**Re: Azimuth Formula Questions**

**From:**George Huxtable

**Date:**2005 Oct 31, 15:54 -0000

Bill, has just corrected an earlier posting about the final adjustments in the formula for azimuth in HO 229 as follows- > N. Lat > L.H.A. greater than 180.....Zn = Z > L.H.A. less than 180....... Zn = 360-Z > > A. Lat > L.H.A. greater 180..........Zn = 180-Z > L.H.A. less than 180........Zn = 180+Z That correction itself has a typo, in that "A. Lat" has been written, where obviously "S.Lat" was intended. Correcting that, and putting it together with the formula that Bill quoted from HO229, we get- ================ tanZ = (cosd sinLHA)/(cosL sind - sinL cosd cosLHA) L is latitude d is declination > N. Lat > L.H.A. greater than 180.....Zn = Z > L.H.A. less than 180....... Zn = 360-Z > > S. Lat > L.H.A. greater 180..........Zn = 180-Z > L.H.A. less than 180........Zn = 180+Z ============================== But even after those corrections, is that now correct, I wonder? Here, the mystery deepens. I don't think it gives the right answers. Put an observer at, or VERY near the equator, and let him deduce the aximuth of a high star with a declination of 1 degree North, with a LHA of + 1 degree (i.e. to his West). Or, exactly the same problem, but rather easier to visualise: from that same position, deduce the true bearing of another vessel, at 1 degree North, and one degree further West. Let's say the observer was just a few feet North of the Equator, with latitude = 0.001 degrees. Then apply that formula quoted above, with L = +0.001 deg, d = +1 deg, LHA = +1deg That gives me Z = 45.024 degrees. With observer in the North hemisphere (just) and LHA less than 180, applying the relevant quoted rule of Zn = 360 - Z gives Zn = 314.975 degrees,. Just as you would expect, its bearing is nearly 45 nearly Northwest from the observer. Seems entirely reasonable, doesn't it? But now let the observer travel a few feet South, across the equator, to a lat of -.001 degrees South. Plugging in that new value of lat changes Z to 44.967, almost exactly the same as before (jst as we would expect, for such a small movement of the observer). But now, the observer's Lat is in the South hemisphere, so we have to apply a different rule to put azimuth in the right quadrant. With LHA less than 180, we now must apply Zn = 180 + z, , or 224 967 degrees, so its calculated bearing is now SW of the observer. This is obviously quite wrong, if we sketch the picture out. So either I have made a silly error, or HO229 has, or there's a further error in the transcription. Which is it? I don't think that the observer's latitude should be the basis for making that selection of the correct quadrant. Unless we can resolve this matter, I suggest that the HO229 "method" of calculating azimuth should be disregarded, in favour of the formula from Meeus that I quoted earlier. That has the important advantage that it gives the right answers! As a reminder, here it is again- 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 However, I suggest that getting the azimuth using the Atan2 function or POL, if available, is simplest of all. contact George Huxtable at george@huxtable.u-net.com or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.