# NavList:

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

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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

```
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