# NavList:

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

**Re: POL and arctan2**

**From:**Lars Bergman

**Date:**2005 Nov 14, 12:58 +0100

George and Bill have asked whether the resulting R value provide any useful information when doing a rectangular to polar conversion with x = -sin LHA y = tan DEC cos LAT - cos LHA sin LAT Out of the conversion comes two values, one of them being the azimuth AZ. Consider a plane right angled triangle where one of the acute angles is AZ. Let x be the opposite side and y the adjacent side. The hypotenuse R = sqrt(x sq + y sq) and tan AZ = x/y. We find that sin AZ = x/R, thus R = x/sin AZ. As R by definition is positive and x has to be positive (the length of a triangle's side), then sin AZ also has to be positive in this way of looking at things. For simplicity we say that sin LHA is positive and we thus have R = sin LHA/sin AZ The spherical trig sine-formula tells us that we can replace sin LHA/sin AZ with sin ZD/sin PD, where, in the pole-zenith-body nav triangle zenith distance ZD and polar distance PD are the sides (or arcs) opposite to the angles LHA and AZ respectively. Thus R = sin ZD/sin PD = cos ALT/cos DEC and we can solve for altitude ALT = arccos(R cos DEC) The altitude thus found is always positive, an altitude that shall be negative is displayed as positive; this is one drawback of using this method. For "ordinary" sights this is no problem. In a similar way R could be used for calculating great circle distance DGC = arcsin(R cos LAT2) where LAT2 is the latitude of destination. Lars Bergman 59N 18E