# Work the Shell - Calculating the Distance between Two Latitude/Longitude Points

Last month, I closed this column with a script that can return latitude/longitude values for two addresses, with the intent ultimately being to have the script calculate the distance between those two points. As an example:

$ farapart.sh "union station, denver co" \ "union station, chicago il" Calculating lat/long for union station, denver co = 39.75288, -105.000473 And calculating lat/long for union station, chicago il = 41.878658, -87.640404

The formula to calculate distance actually is pretty complicated. Here's a JavaScript implementation of the math I showed last month:

var R = 6371; // kilometers var dLat = (lat2-lat1); var dLon = (lon2-lon1); var a = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) * Math.sin(dLon/2) * Math.sin(dLon/2); var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); var d = R * c;

This is going to be a wee bit tricky to convert into a shell script, needless to say, but because we have to use a more sophisticated math tool than the built-in capabilities of Bash anyway, this also means we have a number of options to work with, including Perl, awk and bc. For that matter, we also can just write a quick C program that solves this equation given four variables, but really, why make it easy when I can make it complex? If I wanted easy, I would whip out some Perl, right? Last month, I promised some bc, so let's see if we can make that rusty old app do the heavy lifting.

The first step mathematically is to convert the lat/lon values we get from the mapping system from degrees to radians. This turns out to be straightforward:

Radians = degrees * ( pi / 180 )

Pi, of course, is 3.1415926535897932384.

Given values like:

41.878658, -87.640404

The radians equivalent of those is then:

0.7309204767, -1.529613605

To warm up with bc, here's a simple command-line way to calculate one of these values:

echo "scale=8; -87.640404 * ( 3.14159265 / 180)" | bc

That's all well and good, but it turns out that the different equations I explored for calculating the distance between two points requires the atan2() function, which isn't part of bc.

Rather than beat my head against the old-school wall until the bits are bloodied, I'm going to throw in the towel and admit that this might just be a bit too complex a mathematical problem for a shell script and bc.

Having spent way more hours than I want to admit trying to get this to work properly in bc, I'm going to “cry uncle” and switch temporarily into a different programming language. I'm going to jump into C for a few lines and whip out a simple program that, given two lat/lon pairs in degrees, calculates the distance between them in miles (Listing 1).

**Listing 1. C Distance Program**

#include <stdio.h> #include <math.h> #include <stdlib.h> #define EARTH_RADIUS (6371.0072 * 0.6214) #define TORADS(degrees) (degrees * (M_PI / 180)) main(int argc, char **argv) { double lat1, long1, lat2, long2; double dLat, dLong, a, c, d; lat1 = TORADS(atof(argv[1])); long1 = TORADS(atof(argv[2])); lat2 = TORADS(atof(argv[3])); long2 = TORADS(atof(argv[4])); dLat = lat2 - lat1; dLong = long2 - long1; a = sin(dLat/2) * sin(dLat/2) + cos(lat1) * cos(lat2) * sin(dLong/2) * sin(dLong/2); c = 2 * atan2(sqrt(a), sqrt(1-a)); printf("%g\n", EARTH_RADIUS * c); }

Does it work? Let's find out:

$ distance 39.75288 -105.000473 41.878658 -87.640404 917.984

That seems reasonable. The great circle distance between those two points is 917 miles. Google Maps, of course, shows about 10% greater distance, but perhaps that's because there is no direct-as-the-crow-flies route via roads?

Of course, there also are errors with this formula too, because Earth isn't a perfect sphere but rather an oblate spheroid that has a different diameter depending on where you're measuring. But for our purposes, let's just gloss over that problem. You can Google it to learn about things like the Vincenty formula, but that's beyond the scope of this ridiculous sidetrack.

Now we have all the pieces we need: location to lat/lon and distance between two lat/lon points. Let's put it all together.

Dave Taylor has been hacking shell scripts for over thirty years. Really. He's the author of the popular "Wicked Cool Shell Scripts" and can be found on Twitter as @DaveTaylor and more generally at www.DaveTaylorOnline.com.

## Win an iPhone 6

Enter to Win

## Geek Guides

Practical (and free!) books for the most technical people on the planet. Newly available books include:

**Plus many more.**

## Trending Topics

Microsoft and Linux: True Romance or Toxic Love? | Nov 25, 2015 |

Non-Linux FOSS: Install Windows? Yeah, Open Source Can Do That. | Nov 24, 2015 |

Cipher Security: How to harden TLS and SSH | Nov 23, 2015 |

Web Stores Held Hostage | Nov 19, 2015 |

diff -u: What's New in Kernel Development | Nov 17, 2015 |

Recipy for Science | Nov 16, 2015 |

- Microsoft and Linux: True Romance or Toxic Love?
- Non-Linux FOSS: Install Windows? Yeah, Open Source Can Do That.
- Cipher Security: How to harden TLS and SSH
- Web Stores Held Hostage
- Simple Photo Editing, Linux Edition!
- Firefox's New Feature for Tighter Security
- PuppetLabs Introduces Application Orchestration
- diff -u: What's New in Kernel Development
- It's a Bird. It's Another Bird!
- November 2015 Issue of Linux Journal: System Administration

## Comments

## Problem of distance variation

I have wrote a bash script (using bc + atan2 definition I found online) to calculate both distances between tree points and angle between two distant points using third as a reference point.

Problem is not your C code, problem is number of decimal places used to calculate.

Using 3 point only 0,5-2Km apart and using 4 to 15 decimal places, I got 15-25% difference in distance calculations. So what your code needs are variables with more decimal points.

Hopes this helps.

## in all the code samples

in all the code samples above, "c = 2 * atan2(sqrt(a), sqrt(1-a));" can be written simpler as "c = 2 * asin(sqrt(a));"

## listing 1 link fix

It is actually ftp://ftp.linuxjournal.com/pub/lj/listings/issue188/10606l1.txt and not ftp://ftp.linuxjournal.com/pub/lj/listings/issue188/10606.tgz as mentioned above :)

great article! Thanks! - Israel Torres

## No bugs I believe, just works as it should

Hello,

I was surprised but the answer is:

there is no bug in C program code, the mileage is CORRECT.The shortest distance between 2 points on earth surface of course is not a straight line, that would be impossible unless you are able to dig a tunnel going underneath!

Remember, when talking about earth you have to think 3D!This interesting web page shows how the haversine formula works (http://www.movable-type.co.uk/scripts/latlong.html) and a javascript implements the formula on the web page...

Of course the results are shown in Km and not in Miles because Science is polite with numbers...

Anyway using the coordinates Long Beach and Boston the result is 4180 km (that is ~2597 Mi) that is correct.

The hypothetically shortest way to go from East to West on the Northern Earth surface wouldn't be walking on a Parallel (Meridians and Parallels speaking), but would be walking on an arch pushed towards North (for example it would be a route similar to the following Boston-Michigan-NorthDakota-Utha-California-LongBeach).

Just think about what aiplanes do to save as much fuel as possible. Their routes are very different from what you can imagine...

This is php code I use to calculate (in Km) the shortest distance between 2 points on Earth surface (it is raw but it works):

<?php

function getdistance($lat1,$lon1,$lat2,$lon2)

{

$R = 6371;

$dLat = ($lat2 - $lat1) * 3.14159265359 / 180 ;

$dLon = ($lon2 - $lon1) * 3.14159265359 / 180 ;

$lat1 = $lat1 * 3.14159265359 / 180 ;

$lat2 = $lat2 * 3.14159265359 / 180 ;

$a = sin($dLat/2) * sin($dLat/2) + cos($lat1) * cos($lat2) * sin($dLon/2) * sin($dLon/2);

$c = 2 * atan2(sqrt($a),sqrt(1-$a));

$d = $R * $c;

return $d;

}

$LAT1 = $_REQUEST["lat1"];

$LON1 = $_REQUEST["lon1"];

$LAT2 = $_REQUEST["lat2"];

$LON2 = $_REQUEST["lon2"];

print (getdistance($LAT1,$LON1,$LAT2,$LON2));

?>

You can either call it via POST or via GET.

In fact _REQUEST is magic. :)

take care,

Claudio

## code

hi, I am gayatri i am writing linux c code for distance calculation using haversine formula i am not getting correct if there is any errors plz correct it and mail me plz.