Calculate distance between 2 GPS coordinates

Asked 13/12, 2008 at 22:12 Answered 23/11, 2023 at 19:38

Solved math geolocation geometry latitude-longitude geography

438

How do I calculate distance between two GPS coordinates (using latitude and longitude)?

Asberry answered 13/12, 2008 at 22:12 Comment(2)

This algorithm is known as the Great Circle distance. – Richter 13/12, 2008 at 22:14

@GregHewgill, the first sentence of that article says "This article is about shortest-distance on a sphere." Ie clearly not applicable to GPS coordinates. – Genesisgenet 11/12, 2021 at 23:56

508

Calculate the distance between two coordinates by latitude and longitude, including a Javascript implementation.

West and South locations are negative. Remember minutes and seconds are out of 60 so S31 30' is -31.50 degrees.

Don't forget to convert degrees to radians. Many languages have this function. Or its a simple calculation: radians = degrees * PI / 180.

function degreesToRadians(degrees) {
  return degrees * Math.PI / 180;
}

function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
  var earthRadiusKm = 6371;

  var dLat = degreesToRadians(lat2-lat1);
  var dLon = degreesToRadians(lon2-lon1);

  lat1 = degreesToRadians(lat1);
  lat2 = degreesToRadians(lat2);

  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
          Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  return earthRadiusKm * c;
}

Here are some examples of usage:

distanceInKmBetweenEarthCoordinates(0,0,0,0)  // Distance between same 
                                              // points should be 0
0

distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
                                                          // to Arlington
5918.185064088764

Gobbledygook answered 13/12, 2008 at 22:22 Comment(10)

In case it's not obvious, the toRad() method is a customization to the Number prototype such as: Number.prototype.toRad = function() { return this * (Math.PI / 180); }; . Or, as indicated below, you can replace (Math.PI/2) with 0.0174532925199433 (...whatever precision you deem necessary) for increased performance. – Coadunate 23/7, 2013 at 6:5

If anyone, specifically those of you who don't look for end of line comments, is staring at this formula and looking for a unit of distance, the unit is km. :) – Trela 27/9, 2013 at 18:1

Why are the cosine of lat1 and lat2 calculated and not over lon1 and lon2? – Bozeman 14/5, 2015 at 11:9

@VinneyKelly Small typo but replace (Math.PI/180) not (Math.PI/2), thanks for everyones help – Cindicindie 21/7, 2015 at 16:44

Does anybody have Delphi or Pascal version of the algorithm? – Linda 17/9, 2015 at 17:24

What units is this displayed in? I printed out d, but I don't know what the number is being printed. I'm trying to figure out how to convert this to miles. – Boffa 11/12, 2015 at 20:10

@ChristianKRider Look at the first line. Think about what R normally means in math, then look up relevant, Earth-related quantities to see if the numbers match. – Petrol 12/2, 2017 at 13:1

@Gobbledygook Is there a faster way to compare which of two points is farther from a third than a formula that requires sqrt and atan2 (even if the former is faster than it used to be) – Petrol 12/2, 2017 at 13:2

For imperial units (miles) you could change earthRadiusKm to be var earthRadiusMiles = 3959;, fyi. – Rubbish 9/8, 2017 at 21:53

Another result in google maps... drive.google.com/… – Dictator 13/2, 2018 at 15:35

Look for haversine with Google; here is my solution:

#include <math.h>
#include "haversine.h"

#define d2r (M_PI / 180.0)

//calculate haversine distance for linear distance
double haversine_km(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 6367 * c;

    return d;
}

double haversine_mi(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 3956 * c; 

    return d;
}

Edveh answered 13/9, 2009 at 5:46 Comment(7)

I used haversine in my projects too. – Footstone 13/9, 2010 at 2:25

You can replace (M_PI / 180.0) with 0.0174532925199433 for better performance. – Anglophile 1/8, 2011 at 9:19

In terms of performance: one could calculate sin(dlat/2.0) only once, store it in variable a1, and instead of pow(,2) it's MUCH better to use a1*a1. The same for the other pow(,2). – Cufic 27/10, 2011 at 23:34

Yeah, or just use a post-’60s compiler. – Cohune 28/1, 2014 at 10:45

There is no need to "optimize" (M_PI / 180.0) to a constant that no one understands without context. The compiler calculates these fixed terms for you! – Hedvah 19/9, 2016 at 5:35

@hlung please test such recommendations before submitting. It does not optimize anything unless -O is disabled. Your recommendation does not work. Even more, it makes previously readable code unreadable. Which is also important. – Parapet 2/1, 2018 at 6:58

@TõnuSamuel Thank you very much for your comment. I really appreciate it. It makes sense that compiler with optimization enabled (-O) can pre-calculate operations of constants, making manual collapsing useless. I will test it out when I have time. – Anglophile 2/1, 2018 at 15:47

C# Version of Haversine

double _eQuatorialEarthRadius = 6378.1370D;
double _d2r = (Math.PI / 180D);

private int HaversineInM(double lat1, double long1, double lat2, double long2)
{
    return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
}

private double HaversineInKM(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * _d2r;
    double dlat = (lat2 - lat1) * _d2r;
    double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
    double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
    double d = _eQuatorialEarthRadius * c;

    return d;
}

Here's a .NET Fiddle of this, so you can test it out with your own Lat/Longs.

Kollwitz answered 29/9, 2011 at 10:27 Comment(2)

I've also added a checky .NET fiddle so people can easily test this out. – Interchange 15/8, 2014 at 5:42

the .Net Framework has a build in method GeoCoordinate.GetDistanceTo. The assembly System.Device has to be referenced. MSDN Article msdn.microsoft.com/en-us/library/… – Endbrain 14/2, 2016 at 18:12

Java Version of Haversine Algorithm based on Roman Makarov`s reply to this thread

public class HaversineAlgorithm {

    static final double _eQuatorialEarthRadius = 6378.1370D;
    static final double _d2r = (Math.PI / 180D);

    public static int HaversineInM(double lat1, double long1, double lat2, double long2) {
        return (int) (1000D * HaversineInKM(lat1, long1, lat2, long2));
    }

    public static double HaversineInKM(double lat1, double long1, double lat2, double long2) {
        double dlong = (long2 - long1) * _d2r;
        double dlat = (lat2 - lat1) * _d2r;
        double a = Math.pow(Math.sin(dlat / 2D), 2D) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r)
                * Math.pow(Math.sin(dlong / 2D), 2D);
        double c = 2D * Math.atan2(Math.sqrt(a), Math.sqrt(1D - a));
        double d = _eQuatorialEarthRadius * c;

        return d;
    }

}

Sunder answered 22/1, 2013 at 13:30 Comment(2)

@Radu make sure you're using it correctly and not exchanging lat/log places when passing them to any method. – Sunder 26/12, 2016 at 17:48

I got a reasonably close answer using this formula. I based the accuracy using this website: movable-type.co.uk/scripts/latlong.html which gave me 0.07149 km whereas your formula gave me 0.07156 which is an accuracy of about 99% – Penetrance 18/9, 2019 at 20:40

This is very easy to do with geography type in SQL Server 2008.

SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm

4326 is SRID for WGS84 elipsoidal Earth model

Laryngeal answered 1/2, 2009 at 17:38 Comment(0)

Here's a Haversine function in Python that I use:

from math import pi,sqrt,sin,cos,atan2

def haversine(pos1, pos2):
    lat1 = float(pos1['lat'])
    long1 = float(pos1['long'])
    lat2 = float(pos2['lat'])
    long2 = float(pos2['long'])

    degree_to_rad = float(pi / 180.0)

    d_lat = (lat2 - lat1) * degree_to_rad
    d_long = (long2 - long1) * degree_to_rad

    a = pow(sin(d_lat / 2), 2) + cos(lat1 * degree_to_rad) * cos(lat2 * degree_to_rad) * pow(sin(d_long / 2), 2)
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    km = 6367 * c
    mi = 3956 * c

    return {"km":km, "miles":mi}

Ettaettari answered 9/8, 2013 at 10:21 Comment(1)

The math module contains a function named radians which converts from degrees to radians. from math import radians – Nicolnicola 7/6, 2022 at 15:57

I needed to calculate a lot of distances between the points for my project, so I went ahead and tried to optimize the code, I have found here. On average in different browsers my new implementation runs 2 times faster than the most upvoted answer.

function distance(lat1, lon1, lat2, lon2) {
  var p = 0.017453292519943295;    // Math.PI / 180
  var c = Math.cos;
  var a = 0.5 - c((lat2 - lat1) * p)/2 + 
          c(lat1 * p) * c(lat2 * p) * 
          (1 - c((lon2 - lon1) * p))/2;

  return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km
}

You can play with my jsPerf and see the results here.

Recently I needed to do the same in python, so here is a python implementation:

from math import cos, asin, sqrt
def distance(lat1, lon1, lat2, lon2):
    p = 0.017453292519943295
    a = 0.5 - cos((lat2 - lat1) * p)/2 + cos(lat1 * p) * cos(lat2 * p) * (1 - cos((lon2 - lon1) * p)) / 2
    return 12742 * asin(sqrt(a))

And for the sake of completeness: Haversine on wiki.

Silicate answered 27/12, 2015 at 23:46 Comment(0)

It depends on how accurate you need it to be. If you need pinpoint accuracy, it is best to look at an algorithm which uses an ellipsoid, rather than a sphere, such as Vincenty's algorithm, which is accurate to the mm.

Hollandia answered 13/12, 2008 at 22:28 Comment(2)

Please put all information to your answer instead of linking to external ressources – Heger 15/3, 2021 at 14:12

@NicoHaase Fair call, if perhaps a tad extemporaneous - was over 12 years ago, and this was a slightly different place back then. – Hollandia 16/3, 2021 at 2:10

Here it is in C# (lat and long in radians):

double CalculateGreatCircleDistance(double lat1, double long1, double lat2, double long2, double radius)
{
    return radius * Math.Acos(
        Math.Sin(lat1) * Math.Sin(lat2)
        + Math.Cos(lat1) * Math.Cos(lat2) * Math.Cos(long2 - long1));
}

If your lat and long are in degrees then divide by 180/PI to convert to radians.

Carrasquillo answered 27/5, 2010 at 15:42 Comment(1)

This is the "spherical law of cosines" calculation which is the least accurate and most error-prone method of calculation of a great circle distance. – Benefit 11/1, 2017 at 2:27

PHP version:

(Remove all deg2rad() if your coordinates are already in radians.)

$R = 6371; // km
$dLat = deg2rad($lat2-$lat1);
$dLon = deg2rad($lon2-$lon1);
$lat1 = deg2rad($lat1);
$lat2 = deg2rad($lat2);

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

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

Phase answered 29/1, 2015 at 10:38 Comment(1)

Please change lat1 and lat2 to $lat1 nad $lat2. – Spermophile 31/7, 2017 at 5:56

A T-SQL function, that I use to select records by distance for a center

Create Function  [dbo].[DistanceInMiles] 
 (  @fromLatitude float ,
    @fromLongitude float ,
    @toLatitude float, 
    @toLongitude float
  )
   returns float
AS 
BEGIN
declare @distance float

select @distance = cast((3963 * ACOS(round(COS(RADIANS(90-@fromLatitude))*COS(RADIANS(90-@toLatitude))+ 
SIN(RADIANS(90-@fromLatitude))*SIN(RADIANS(90-@toLatitude))*COS(RADIANS(@fromLongitude-@toLongitude)),15)) 
)as float) 
  return  round(@distance,1)
END

Teteak answered 13/7, 2011 at 12:43 Comment(1)

This is the "spherical law of cosines" calculation which is the least accurate and most error-prone method of calculation of a great circle distance. – Benefit 11/1, 2017 at 2:30

I. Regarding "Breadcrumbs" method

Earth radius is different on different Lat. This must be taken into consideration in Haversine algorithm.
Consider Bearing change, which turns straight lines to arches (which are longer)
Taking Speed change into account will turn arches to spirals (which are longer or shorter than arches)
Altitude change will turn flat spirals to 3D spirals (which are longer again). This is very important for hilly areas.

Below see the function in C which takes #1 and #2 into account:

double   calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
       double rLat2, double rLon2, double rHeading2){
  double rDLatRad = 0.0;
  double rDLonRad = 0.0;
  double rLat1Rad = 0.0;
  double rLat2Rad = 0.0;
  double a = 0.0;
  double c = 0.0;
  double rResult = 0.0;
  double rEarthRadius = 0.0;
  double rDHeading = 0.0;
  double rDHeadingRad = 0.0;

  if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
              || (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
              || (rLon2 > 180.0)) {
        return -1;
  };

  rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
  rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
  rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
  rLat2Rad = rLat2 * DEGREE_TO_RADIANS;

  a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
              rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);

  if (a == 0.0) {
        return 0.0;
  }

  c = 2 * atan2(sqrt(a), sqrt(1 - a));
  rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
              / 2.0));
  rResult = rEarthRadius * c;

  // Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns

  if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
              && (rHeading2 < 360.0)) {
        rDHeading = fabs(rHeading1 - rHeading2);
        if (rDHeading > 180.0) {
              rDHeading -= 180.0;
        }
        rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
        if (rDHeading > 5.0) {
              rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
        } else {
              rResult = rResult / cos(rDHeadingRad);
        }
  }
  return rResult;
}

II. There is an easier way which gives pretty good results.

By Average Speed.

Trip_distance = Trip_average_speed * Trip_time

Since GPS Speed is detected by Doppler effect and is not directly related to [Lon,Lat] it can be at least considered as secondary (backup or correction) if not as main distance calculation method.

Beverly answered 21/1, 2014 at 15:36 Comment(0)

If you need something more accurate then have a look at this.

Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a) They are based on the assumption that the figure of the Earth is an oblate spheroid, and hence are more accurate than methods such as great-circle distance which assume a spherical Earth.

The first (direct) method computes the location of a point which is a given distance and azimuth (direction) from another point. The second (inverse) method computes the geographical distance and azimuth between two given points. They have been widely used in geodesy because they are accurate to within 0.5 mm (0.020″) on the Earth ellipsoid.

Amii answered 22/8, 2011 at 15:35 Comment(0)

If you're using .NET don't reivent the wheel. See System.Device.Location. Credit to fnx in the comments in another answer.

using System.Device.Location;

double lat1 = 45.421527862548828D;
double long1 = -75.697189331054688D;
double lat2 = 53.64135D;
double long2 = -113.59273D;

GeoCoordinate geo1 = new GeoCoordinate(lat1, long1);
GeoCoordinate geo2 = new GeoCoordinate(lat2, long2);

double distance = geo1.GetDistanceTo(geo2);

Melessa answered 19/1, 2017 at 18:11 Comment(1)

This namespace is only available in .NET Framework 4.8.1 and earlier. After that you will have to reinvent the wheel after all :-) – Gurglet 23/11, 2023 at 19:6

here is the Swift implementation from the answer

func degreesToRadians(degrees: Double) -> Double {
    return degrees * Double.pi / 180
}

func distanceInKmBetweenEarthCoordinates(lat1: Double, lon1: Double, lat2: Double, lon2: Double) -> Double {

    let earthRadiusKm: Double = 6371

    let dLat = degreesToRadians(degrees: lat2 - lat1)
    let dLon = degreesToRadians(degrees: lon2 - lon1)

    let lat1 = degreesToRadians(degrees: lat1)
    let lat2 = degreesToRadians(degrees: lat2)

    let a = sin(dLat/2) * sin(dLat/2) +
    sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2)
    let c = 2 * atan2(sqrt(a), sqrt(1 - a))
    return earthRadiusKm * c
}

Jailer answered 14/8, 2017 at 8:26 Comment(0)

This is version from "Henry Vilinskiy" adapted for MySQL and Kilometers:

CREATE FUNCTION `CalculateDistanceInKm`(
  fromLatitude float,
  fromLongitude float,
  toLatitude float, 
  toLongitude float
) RETURNS float
BEGIN
  declare distance float;

  select 
    6367 * ACOS(
            round(
              COS(RADIANS(90-fromLatitude)) *
                COS(RADIANS(90-toLatitude)) +
                SIN(RADIANS(90-fromLatitude)) *
                SIN(RADIANS(90-toLatitude)) *
                COS(RADIANS(fromLongitude-toLongitude))
              ,15)
            )
    into distance;

  return  round(distance,3);
END;

Promisee answered 11/7, 2012 at 9:30 Comment(2)

MySQL said Something is wrong in your syntax near '' on line 8 // declare distance float; – Tiertza 2/12, 2014 at 13:37

This is the "spherical law of cosines" calculation which is the least accurate and most error-prone method of calculation of a great circle distance – Benefit 11/1, 2017 at 2:30

This Lua code is adapted from stuff found on Wikipedia and in Robert Lipe's GPSbabel tool:

local EARTH_RAD = 6378137.0 
  -- earth's radius in meters (official geoid datum, not 20,000km / pi)

local radmiles = EARTH_RAD*100.0/2.54/12.0/5280.0;
  -- earth's radius in miles

local multipliers = {
  radians = 1, miles = radmiles, mi = radmiles, feet = radmiles * 5280,
  meters = EARTH_RAD, m = EARTH_RAD, km = EARTH_RAD / 1000, 
  degrees = 360 / (2 * math.pi), min = 60 * 360 / (2 * math.pi)
}

function gcdist(pt1, pt2, units) -- return distance in radians or given units
  --- this formula works best for points close together or antipodal
  --- rounding error strikes when distance is one-quarter Earth's circumference
  --- (ref: wikipedia Great-circle distance)
  if not pt1.radians then pt1 = rad(pt1) end
  if not pt2.radians then pt2 = rad(pt2) end
  local sdlat = sin((pt1.lat - pt2.lat) / 2.0);
  local sdlon = sin((pt1.lon - pt2.lon) / 2.0);
  local res = sqrt(sdlat * sdlat + cos(pt1.lat) * cos(pt2.lat) * sdlon * sdlon);
  res = res > 1 and 1 or res < -1 and -1 or res
  res = 2 * asin(res);
  if units then return res * assert(multipliers[units])
  else return res
  end
end

Trimer answered 13/12, 2008 at 22:26 Comment(0)

    private double deg2rad(double deg)
    {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad)
    {
        return (rad / Math.PI * 180.0);
    }

    private double GetDistance(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Distance in Kilo Meter
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
        return (dist);
    }

    private double GetDirection(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Direction in Degrees
        double dlat = deg2rad(lat1) - deg2rad(lat2);
        double dlon = deg2rad(lon1) - deg2rad(lon2);
        double y = Math.Sin(dlon) * Math.Cos(lat2);
        double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
        double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
        if (direct < 0)
            direct = direct + 360;
        return (direct);
    }

    private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
        double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
        return (Speed);
    }

    private double GetDuration(DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
        return (TimeDifferenceInSeconds);
    }

Bauer answered 22/7, 2011 at 11:53 Comment(2)

I think your function GetDistance returns value in meters – Flavory 21/12, 2014 at 14:11

Is this correct? GetDirection() doesn't make use of 'dlat'. – Hargeisa 28/1, 2017 at 12:52

i took the top answer and used it in a Scala program

import java.lang.Math.{atan2, cos, sin, sqrt}

def latLonDistance(lat1: Double, lon1: Double)(lat2: Double, lon2: Double): Double = {
    val earthRadiusKm = 6371
    val dLat = (lat2 - lat1).toRadians
    val dLon = (lon2 - lon1).toRadians
    val latRad1 = lat1.toRadians
    val latRad2 = lat2.toRadians

    val a = sin(dLat / 2) * sin(dLat / 2) + sin(dLon / 2) * sin(dLon / 2) * cos(latRad1) * cos(latRad2)
    val c = 2 * atan2(sqrt(a), sqrt(1 - a))
    earthRadiusKm * c
}

i curried the function in order to be able to easily produce functions that have one of the two locations fixed and require only a pair of lat/lon to produce distance.

Jubilee answered 13/1, 2018 at 21:25 Comment(0)

Here's a Kotlin variation:

import kotlin.math.*

class HaversineAlgorithm {

    companion object {
        private const val MEAN_EARTH_RADIUS = 6371.008
        private const val D2R = Math.PI / 180.0
    }

    private fun haversineInKm(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double {
        val lonDiff = (lon2 - lon1) * D2R
        val latDiff = (lat2 - lat1) * D2R
        val latSin = sin(latDiff / 2.0)
        val lonSin = sin(lonDiff / 2.0)
        val a = latSin * latSin + (cos(lat1 * D2R) * cos(lat2 * D2R) * lonSin * lonSin)
        val c = 2.0 * atan2(sqrt(a), sqrt(1.0 - a))
        return MEAN_EARTH_RADIUS * c
    }
}

Emerson answered 24/5, 2020 at 18:54 Comment(4)

Why did you use equatorial radius instead of mean Earth radius? – Dag 25/5, 2020 at 0:47

@user13044086 Good question. It's because I derived this from Paulo Miguel Almeida's Java version. Looks like the C# version is also using that distance. Other versions here have 6371, but then you have to realize that all these algorithms may not perfectly handle the Earth's geoid shape. Feel free to modify this and use 6371. If you tell me that leads to more precise values I'll change my answer. – Emerson 25/5, 2020 at 6:34

6371.008 is commonly used because it minimizes relative error of the formula as explained in notes on page movable-type.co.uk/scripts/latlong.html#ellipsoid – Dag 25/5, 2020 at 7:20

@user13044086 Thanks for the link, I edited my answer a while ago based on that – Emerson 8/6, 2020 at 0:45

you can find a implementation of this (with some good explanation) in F# on fssnip

here are the important parts:


let GreatCircleDistance<[<Measure>] 'u> (R : float<'u>) (p1 : Location) (p2 : Location) =
    let degToRad (x : float<deg>) = System.Math.PI * x / 180.0<deg/rad>

    let sq x = x * x
    // take the sin of the half and square the result
    let sinSqHf (a : float<rad>) = (System.Math.Sin >> sq) (a / 2.0<rad>)
    let cos (a : float<deg>) = System.Math.Cos (degToRad a / 1.0<rad>)

    let dLat = (p2.Latitude - p1.Latitude) |> degToRad
    let dLon = (p2.Longitude - p1.Longitude) |> degToRad

    let a = sinSqHf dLat + cos p1.Latitude * cos p2.Latitude * sinSqHf dLon
    let c = 2.0 * System.Math.Atan2(System.Math.Sqrt(a), System.Math.Sqrt(1.0-a))

    R * c

Tap answered 22/8, 2011 at 15:40 Comment(0)

I needed to implement this in PowerShell, hope it can help someone else. Some notes about this method

Don't split any of the lines or the calculation will be wrong
To calculate in KM remove the * 1000 in the calculation of $distance
Change $earthsRadius = 3963.19059 and remove * 1000 in the calculation of $distance the to calulate the distance in miles

I'm using Haversine, as other posts have pointed out Vincenty's formulae is much more accurate

Function MetresDistanceBetweenTwoGPSCoordinates($latitude1, $longitude1, $latitude2, $longitude2)  
{  
  $Rad = ([math]::PI / 180);  

  $earthsRadius = 6378.1370 # Earth's Radius in KM  
  $dLat = ($latitude2 - $latitude1) * $Rad  
  $dLon = ($longitude2 - $longitude1) * $Rad  
  $latitude1 = $latitude1 * $Rad  
  $latitude2 = $latitude2 * $Rad  

  $a = [math]::Sin($dLat / 2) * [math]::Sin($dLat / 2) + [math]::Sin($dLon / 2) * [math]::Sin($dLon / 2) * [math]::Cos($latitude1) * [math]::Cos($latitude2)  
  $c = 2 * [math]::ATan2([math]::Sqrt($a), [math]::Sqrt(1-$a))  

  $distance = [math]::Round($earthsRadius * $c * 1000, 0) #Multiple by 1000 to get metres  

  Return $distance  
}

Erikerika answered 25/10, 2012 at 13:5 Comment(0)

Scala version

  def deg2rad(deg: Double) = deg * Math.PI / 180.0

  def rad2deg(rad: Double) = rad / Math.PI * 180.0

  def getDistanceMeters(lat1: Double, lon1: Double, lat2: Double, lon2: Double) = {
    val theta = lon1 - lon2
    val dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) *
      Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta))
    Math.abs(
      Math.round(
        rad2deg(Math.acos(dist)) * 60 * 1.1515 * 1.609344 * 1000)
    )
  }

Flavory answered 21/12, 2014 at 14:18 Comment(0)

Here's my implementation in Elixir

defmodule Geo do
  @earth_radius_km 6371
  @earth_radius_sm 3958.748
  @earth_radius_nm 3440.065
  @feet_per_sm 5280

  @d2r :math.pi / 180

  def deg_to_rad(deg), do: deg * @d2r

  def great_circle_distance(p1, p2, :km), do: haversine(p1, p2) * @earth_radius_km
  def great_circle_distance(p1, p2, :sm), do: haversine(p1, p2) * @earth_radius_sm
  def great_circle_distance(p1, p2, :nm), do: haversine(p1, p2) * @earth_radius_nm
  def great_circle_distance(p1, p2, :m), do: great_circle_distance(p1, p2, :km) * 1000
  def great_circle_distance(p1, p2, :ft), do: great_circle_distance(p1, p2, :sm) * @feet_per_sm

  @doc """
  Calculate the [Haversine](https://en.wikipedia.org/wiki/Haversine_formula)
  distance between two coordinates. Result is in radians. This result can be
  multiplied by the sphere's radius in any unit to get the distance in that unit.
  For example, multiple the result of this function by the Earth's radius in
  kilometres and you get the distance between the two given points in kilometres.
  """
  def haversine({lat1, lon1}, {lat2, lon2}) do
    dlat = deg_to_rad(lat2 - lat1)
    dlon = deg_to_rad(lon2 - lon1)

    radlat1 = deg_to_rad(lat1)
    radlat2 = deg_to_rad(lat2)

    a = :math.pow(:math.sin(dlat / 2), 2) +
        :math.pow(:math.sin(dlon / 2), 2) *
        :math.cos(radlat1) * :math.cos(radlat2)

    2 * :math.atan2(:math.sqrt(a), :math.sqrt(1 - a))
  end
end

Wieland answered 2/11, 2018 at 8:20 Comment(0)

Dart Version

Haversine Algorithm.

import 'dart:math';

class GeoUtils {

  static double _degreesToRadians(degrees) {
    return degrees * pi / 180;
  }

  static double distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
    var earthRadiusKm = 6371;

    var dLat = _degreesToRadians(lat2-lat1);
    var dLon = _degreesToRadians(lon2-lon1);

    lat1 = _degreesToRadians(lat1);
    lat2 = _degreesToRadians(lat2);

    var a = sin(dLat/2) * sin(dLat/2) +
        sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
    var c = 2 * atan2(sqrt(a), sqrt(1-a));
    return earthRadiusKm * c;
  }
}

Apparitor answered 7/6, 2019 at 19:2 Comment(0)

In Python, you can use the geopy library to compute the geodesic distance using the WGS84 ellipsoid:

from geopy.distance import geodesic
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(geodesic(newport_ri, cleveland_oh).km)

Medullated answered 30/6, 2021 at 13:33 Comment(0)

TypeScript Version

export const degreeToRadian = (degree: number) => {
  return degree * Math.PI / 180;
}

export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => {
    const earthRadiusInKm = 6371;

    const dLat = degreeToRadian(lat2 - lat1);
    const dLon = degreeToRadian(lon2 - lon1);

    lat1 = degreeToRadian(lat1);
    lat2 = degreeToRadian(lat2);

    const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
    const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));

    return earthRadiusInKm * c;
}

Sarrusophone answered 12/5, 2022 at 3:33 Comment(0)

I think a version of the algorithm in R is still missing:

gpsdistance<-function(lat1,lon1,lat2,lon2){

# internal function to change deg to rad

degreesToRadians<- function (degrees) {
return (degrees * pi / 180)
}

R<-6371e3  #radius of Earth in meters

phi1<-degreesToRadians(lat1) # latitude 1
phi2<-degreesToRadians(lat2) # latitude 2
lambda1<-degreesToRadians(lon1) # longitude 1
lambda2<-degreesToRadians(lon2) # longitude 2

delta_phi<-phi1-phi2 # latitude-distance
delta_lambda<-lambda1-lambda2 # longitude-distance

a<-sin(delta_phi/2)*sin(delta_phi/2)+
cos(phi1)*cos(phi2)*sin(delta_lambda/2)*
sin(delta_lambda/2)

cc<-2*atan2(sqrt(a),sqrt(1-a))

distance<- R * cc

return(distance)  # in meters
}

Malediction answered 6/4, 2020 at 11:32 Comment(0)

For java

public static double degreesToRadians(double degrees) {
    return degrees * Math.PI / 180;
}

public static double distanceInKmBetweenEarthCoordinates(Location location1, Location location2) {
    double earthRadiusKm = 6371;

    double dLat = degreesToRadians(location2.getLatitude()-location1.getLatitude());
    double dLon = degreesToRadians(location2.getLongitude()-location1.getLongitude());

    double lat1 = degreesToRadians(location1.getLatitude());
    double lat2 = degreesToRadians(location2.getLatitude());

    double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
            Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    return earthRadiusKm * c;
}

Alleenallegation answered 3/3, 2021 at 19:16 Comment(0)

For anyone searching for a Delphi/Pascal version:

function GreatCircleDistance(const Lat1, Long1, Lat2, Long2: Double): Double;
var
  Lat1Rad, Long1Rad, Lat2Rad, Long2Rad: Double;
const
  EARTH_RADIUS_KM = 6378;
begin
  Lat1Rad  := DegToRad(Lat1);
  Long1Rad := DegToRad(Long1);
  Lat2Rad  := DegToRad(Lat2);
  Long2Rad := DegToRad(Long2);
  Result   := EARTH_RADIUS_KM * ArcCos(Cos(Lat1Rad) * Cos(Lat2Rad) * Cos(Long1Rad - Long2Rad) + Sin(Lat1Rad) * Sin(Lat2Rad));
end;

I take no credit for this code, I originally found it posted by Gary William on a public forum.

Dialecticism answered 8/4, 2021 at 13:1 Comment(0)

Unity Version C#

Haversine Algorithm.

public float Distance(float lat1, float lon1, float lat2, float lon2)
{
    var earthRadiusKm = 6371;

    var dLat = (lat2 - lat1) * Mathf.Rad2Deg;
    var dLon = (lon2 - lon1) * Mathf.Rad2Deg;

    var a = Mathf.Sin(dLat / 2) * Mathf.Sin(dLat / 2) +
            Mathf.Sin(dLon / 2) * Mathf.Sin(dLon / 2) * 
            Mathf.Cos(lat1 * Mathf.Rad2Deg) * Mathf.Cos(lat2 * Mathf.Rad2Deg);

    var c = 2 * Mathf.Atan2(Mathf.Sqrt(a), Mathf.Sqrt(1 - a));
    return earthRadiusKm * c;
}

Feldman answered 12/11, 2021 at 15:16 Comment(0)

All these solutions work OK for very long distances, but won't work when you need accurate distances of points of only a few meters apart. Te root of the problem there is that for short distances (lat2 - lat1) and (lon2 - lon1) both resolve to 0 even with long-double floating point values.

In this case the best solution is to convert the points to UTM and use just 2D trigonometry to calculate flat distances, earth curvature is not an issue here. with this method you get sub-millimeter accuracy, provided that you use GeographicLib or your own accurate UTM conversion formulas.

There is no free lunch though, this won't work at boundary UTM zones where each point lays on different UTM zone.

Here are some test results, GcDistance uses the Haverstine formula, the test distance where generated with GeographicLib at 1,2,3,5,10,20,and 40 meters from point 100:

*********** Lat/Lon to UTM test  *********
Sample id 200 - X: 640559.4334   y: 3941554.2882 | Hemisphere: NORTH - Zone: 7
Test   id 200 - X: 640559.4334   y: 3941554.2882 | Hemisphere: NORTH - Zone: 7
Errors:         X:       0.0000  y:       0.0000

*********** Inverse test UTM back to Lat/Lon 
Sample id 200 - Lat:  35.60777125   Lon: -139.44814996
Test   id 200 - Lat:  35.60777125   Lon: -139.44814996
Errors:         Lat:   0.0000       Lon:    0.0000

*********** Short Distance Test *******************
p id 100 - X: 677561.844   y: 7177360.087 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 100 -    0.000
GcDistance 100  - 100 -    0.095

p id 101 - X: 677561.437   y: 7177361.000 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 101 -    1.000
GcDistance 100  - 101 -    0.095

p id 102 - X: 677561.030   y: 7177361.913 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 102 -    2.000
GcDistance 100  - 102 -    0.095

p id 103 - X: 677560.622   y: 7177362.826 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 103 -    3.000
GcDistance 100  - 103 -    0.095

p id 105 - X: 677559.808   y: 7177364.653 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 105 -    5.000
GcDistance 100  - 105 -    0.095

p id 110 - X: 677557.771   y: 7177369.219 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 110 -   10.000
GcDistance 100  - 110 -    0.095

p id 120 - X: 677553.698   y: 7177378.352 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 120 -   20.000
GcDistance 100  - 120 -    0.095

p id 140 - X: 677545.553   y: 7177396.618 | Hemisphere: SOUTH - Zone: 22
Distance   100  - 140 -   40.000
GcDistance 100  - 140 -    0.095

Gurglet answered 23/11, 2023 at 19:38 Comment(0)

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