# NavList:

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

**Re: Sight Reduction Formula Question**

**From:**George Huxtable

**Date:**2005 Jan 9, 16:45 +0000

Paul Hirose pointed out that it's rather easy to plot out the azimuth onto a chart, without really having to think too hard about putting it in the right sector, and I agree. But it's if you want to feed the information into a computer, to make (for example) a least-squares analysis of a set of observations, that the azimuths need to be defined in a formal way that the computer will understand. And that means 0 deg to 360 deg, increasing clockwise from North. Note that astronomers, such as Meeus, like to define their azimuths as clockwise from due South, and beware. ===================== Trevor Kenchington asked- >Could someone either explain how the expressions for the sine, cosine >and tangent of Z (as discussed on this thread) are derived or else point >me to a published explanation? I have been trying to equate the three >expressions but the trig I was taught as a 16 year-old is far too rusty >for me to make sense of them. Nor can I find an explanation in any of >the texts I have to hand. Yes. My 2-volume edition of Bowditch (vol.2 is 1981) states and derives all three expressions. In vol 2 it's in "part 3, Navigational calculations", in chapter 7, "Calculations of Celestial Navigation", paras 707-710. Let's see if I can pass on the gist of it. Start by drawing a spherical triangle with vertices A,B,and C in clockwise order, for which the sides opposite those vertices are a, b, c. Bowditch then quotes one case of the "five parts formula" as sin a cos B = cos b sin c- sin b cos c cos A and from the law of sines- sin a / sin A = sin b / sin B from which he gets sin a = sin A sin b / sin B and substitutes that in the first equation to get sin A cot B = sin c cot b - cos c cos A. Now he relates that triangle to the PZX triangle in which A corresponds to the pole P, B corresponds to the observer's zenith Z, and X is the direction of the body. In that case LHA is the angle at P, Z is the azimuth angle at the observer Z, the side opposite Z is (90 - dec), and the side opposite X is (90 - lat). So we can write- sin LHA cot Z = cos lat tan dec - sin lat cos LHA from which tan Z = sin LHA / (cos lat tan dec - sin lat cos LHA) ================ Now for getting the azimuth from its cosine. Bowditch, using the same ABC triangle, writes "the law of cosines for sides" as- cos b = cos c cos a + sin c sin a cos B. This is equivalent to the same PZX triangle as before, but this time we will also need the side opposite to the pole P, which is (90 - alt). substituting these quantities for a, b, c, and B, we get- cos (90 - dec) = cos (90 - lat) cos (90 - alt) + sin (90 - lat) sin (90 - alt) cos Z which gives cos Z = (sin dec - sin lat sin alt) / cos lat cos alt =============== What seems at first sight to be the simplest method, getting Z from its sine, is- take the sine formula sin b / sin B = sin a / sin A Make the same substitutions as before to give= sin (90-d) / sin Z = sin (90 - alt) / sin LHA or sin Z = sin LHA cos dec / cos alt. but this is a lot more complex than it seems, because of the difficulty of deciding, just knowing the sine of Z (when it's near due East or West) whether it's a bit North, or a bit South, of due East or West. Bowditch provides another trig expression to test which, but by the time you have added that test, it's no longer the simplest method. =============== Note that Bowditch refers to his azimuths as being so many degrees East or West of North or South, in the quadrantal notation you used to find on old compasses. To correspond with a modern compass, or to pass data into a computer program, a conversion is required to 0deg to 360 deg, clockwise from North. Trevor refers to "the complications of "ATAN2" calculator buttons or other ways of applying George's tangent approach.". However, there's no need to have an "ATAN2" or "POL" routine available, though it does simplify matters; the ordinary inverse tan will do the job perfectly well. And the arc tan approach applies in every case, with just one formula to learn, and no ambiguities. It's the one formula that you will find for azimuth in Meeus, though as mentioned earlier, Meeus' formula differs slightly from what we navigators need because he defines azimuths from the South. George. ================================================================ contact George Huxtable by email at george---.u-net.com, by phone at 01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. ================================================================