# Calculating lat/long with range bearing

1. May 11, 2012

### michael atlas

All,

I have a latitude/longitude and I want to find another latitude/longitude given a range to the location and the bearing from the latitude and longitude I have.

I have tried using the Haversine formula, which did not work for me because it is for ranges that are in hundreds of kilometers. The ranges that I am looking at are in hundreds or yards or smaller. For example: I have a range and bearing from a given lat long of 300 yards and a bearing of 18.5°

If anyone knows a formula which is more of an approximation (as in it doesnt take into account the Radius of curvature of the earth, etc) it would be greatly appreciated.

R/
Mike

2. May 11, 2012

### Studiot

ΔAt 300 yards the flat earth approximation will suffice.

You need to calculate the latitutes (difference in northings or meridional or N-S distance) and Departures (difference in eastings or parallel or latitudinal distance).

Since your are working in range (R) (metres) and degrees you can calculate this in two stages.

First calculate the latitudes (ΔN) and departures (ΔE) in metres.

Then convert these to the difference in degrees to angular measure to add to your existing lat and long.

Using whole circle bearings (wcb)

ΔN = Rsin(450-wcb)
ΔE = Rcos(450-wcb)

Now all circles of longitude have the same circumference so the change of longitude (θn)) is

$${\theta _n} = \frac{{360\Delta N}}{{2\pi r}}$$

Where r is the radius of the earth (= 6371000 metres)

However the radius of each parallel of latitude varies with latitude , so if your approx latitude is L

$${\theta _E} = \frac{{360\Delta E}}{{2\pi r\cos L}}$$

3. May 13, 2012

### Studiot

It would be nice to know if my last post was of any help.
The equations stated were originally developed for programming a computer/calculator to perform a similar task to yours.

4. May 15, 2012

### michael atlas

The formula seemed to have worked well. I am just making sure though,... Is the answer in lat/long as a decimal? or is it Degrees.minutes.seconds.

5. May 15, 2012

### michael atlas

Also, when you input the latitude L for Theta_E, is that in degrees.minutes.seconds or as a decimal?

6. May 15, 2012

### Studiot

The formulae are for degrees not radians.

It does not matter whether you input DMS or Degrees and decimals of a degree, the output will be in the same format. Obviously in computing when calling your trig functions the parameter you pass ( the angles) have to be suitable for your computer.

This method should be good to 0.1 seconds of arc over about 10 Kilometres range. Beyond this you will need to start adding corrections to the range depending upon how you are measuring it. Distance measuring equipment provides the straight line distance ie the chord, whereas the Lat and Long distances are arcs of a circle.

The 450 degrees is to provide one single formulae for all quadrants - it automatically compensates for the sign of the trig functions.
The sin and cos are switched from convention to compensate for the fact that mathematica (and computers) measure anticlockwise from the horizontal axis, whereas surveyors measure bearings clockwise from the vertical axis.

go well