# NavList:

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

**Re: Calculate new coordinates given bearing and distance?**

**From:**Frank Reed

**Date:**2015 Nov 25, 08:08 -0800

John Brown, you wrote:

"use the polar to rectangular function on a calculator"

I never recommend this to beginners. First, "polar to rectangular" means little or nothing to many people who have not done this sort of math in a long time (if ever), and they may mistakenly conclude that it's necessary or more accurate or distinct in some way from the simple calculation they've already learned. Second it's an unusual calculator function that one has to "study on" to figure out how to use on a particular device. Third, it saves only a very few keystrokes. Even myself, wise in the ways of the force, I never use the P/R functions on my calculator(s).

You added:

"Using the mean latitude is not strictly accurate"

And you'll note that the instructions that Sean posted, taken from "Sight Reduction Procedures section of the Nautical Almanac," suggested using the latitude of the "last fix". Sounds like different rules! Truth is, at the level of approximation we're dealing with here, any latitude in range will do just fine. Suppose for example my last fix latitude was 41.2334° N and my roughly estimated current latitude is 41.6° N. When I do the detailed calculation, for the longitude-scaling factor (=cos(Lat)), I could use any latitude in the range from 41.2 to 41.6 and the results will be good enough. And if you're working on a calculator, you might as well use a latitude somewhere in the middle rounded to the nearest tenth of a degree. In this example, 41.3 or 41.4 or 41.5 would all be good choices for the latitude entered in the scaling factor. It doesn't really matter much.

Frank Reed