Calculating Distance between two Latitude and Longitude GeoCoordinates

I'm calculating the distance between two GeoCoordinates. I'm testing my app against 3-4 other apps. When I'm calculating distance, I tend to get an average of 3.3 miles for my calculation whereas other apps are getting 3.5 miles. It's a big difference for the calculation I'm trying to perform. Are there any good class libraries out there for calculating distance? I'm calculating it like this in C#:

public static double Calculate(double sLatitude,double sLongitude, double eLatitude, 
                               double eLongitude)
{
    var radiansOverDegrees = (Math.PI / 180.0);

    var sLatitudeRadians = sLatitude * radiansOverDegrees;
    var sLongitudeRadians = sLongitude * radiansOverDegrees;
    var eLatitudeRadians = eLatitude * radiansOverDegrees;
    var eLongitudeRadians = eLongitude * radiansOverDegrees;

    var dLongitude = eLongitudeRadians - sLongitudeRadians;
    var dLatitude = eLatitudeRadians - sLatitudeRadians;

    var result1 = Math.Pow(Math.Sin(dLatitude / 2.0), 2.0) + 
                  Math.Cos(sLatitudeRadians) * Math.Cos(eLatitudeRadians) * 
                  Math.Pow(Math.Sin(dLongitude / 2.0), 2.0);

    // Using 3956 as the number of miles around the earth
    var result2 = 3956.0 * 2.0 * 
                  Math.Atan2(Math.Sqrt(result1), Math.Sqrt(1.0 - result1));

    return result2;
}

What could I be doing wrong? Should I calculate it in km first and then convert to miles?

Nigel Sampson

The GeoCoordinate class (.NET Framework 4 and higher) already has GetDistanceTo method.

var sCoord = new GeoCoordinate(sLatitude, sLongitude);
var eCoord = new GeoCoordinate(eLatitude, eLongitude);

return sCoord.GetDistanceTo(eCoord);

The distance is in meters.

You need to reference System.Device.

GetDistance is the best solution, but in many cases we can't use this Method (e.g. Universal App)

Pseudocode of the Algorithm to calculate the distance between to coorindates:

public static double DistanceTo(double lat1, double lon1, double lat2, double lon2, char unit = 'K')
{
    double rlat1 = Math.PI*lat1/180;
    double rlat2 = Math.PI*lat2/180;
    double theta = lon1 - lon2;
    double rtheta = Math.PI*theta/180;
    double dist =
        Math.Sin(rlat1)*Math.Sin(rlat2) + Math.Cos(rlat1)*
        Math.Cos(rlat2)*Math.Cos(rtheta);
    dist = Math.Acos(dist);
    dist = dist*180/Math.PI;
    dist = dist*60*1.1515;

    switch (unit)
    {
        case 'K': //Kilometers -> default
            return dist*1.609344;
        case 'N': //Nautical Miles 
            return dist*0.8684;
        case 'M': //Miles
            return dist;
    }

    return dist;
}

Real World C# Implementation, which makes use of an Extension Methods

Usage:

var distance = new Coordinates(48.672309, 15.695585)
                .DistanceTo(
                    new Coordinates(48.237867, 16.389477),
                    UnitOfLength.Kilometers
                );

Implementation:

public class Coordinates
{
    public double Latitude { get; private set; }
    public double Longitude { get; private set; }

    public Coordinates(double latitude, double longitude)
    {
        Latitude = latitude;
        Longitude = longitude;
    }
}
public static class CoordinatesDistanceExtensions
{
    public static double DistanceTo(this Coordinates baseCoordinates, Coordinates targetCoordinates)
    {
        return DistanceTo(baseCoordinates, targetCoordinates, UnitOfLength.Kilometers);
    }

    public static double DistanceTo(this Coordinates baseCoordinates, Coordinates targetCoordinates, UnitOfLength unitOfLength)
    {
        var baseRad = Math.PI * baseCoordinates.Latitude / 180;
        var targetRad = Math.PI * targetCoordinates.Latitude/ 180;
        var theta = baseCoordinates.Longitude - targetCoordinates.Longitude;
        var thetaRad = Math.PI * theta / 180;

        double dist =
            Math.Sin(baseRad) * Math.Sin(targetRad) + Math.Cos(baseRad) *
            Math.Cos(targetRad) * Math.Cos(thetaRad);
        dist = Math.Acos(dist);

        dist = dist * 180 / Math.PI;
        dist = dist * 60 * 1.1515;

        return unitOfLength.ConvertFromMiles(dist);
    }
}

public class UnitOfLength
{
    public static UnitOfLength Kilometers = new UnitOfLength(1.609344);
    public static UnitOfLength NauticalMiles = new UnitOfLength(0.8684);
    public static UnitOfLength Miles = new UnitOfLength(1);

    private readonly double _fromMilesFactor;

    private UnitOfLength(double fromMilesFactor)
    {
        _fromMilesFactor = fromMilesFactor;
    }

    public double ConvertFromMiles(double input)
    {
        return input*_fromMilesFactor;
    }
}

Here is the JavaScript version guys and gals

function distanceTo(lat1, lon1, lat2, lon2, unit) {
      var rlat1 = Math.PI * lat1/180
      var rlat2 = Math.PI * lat2/180
      var rlon1 = Math.PI * lon1/180
      var rlon2 = Math.PI * lon2/180
      var theta = lon1-lon2
      var rtheta = Math.PI * theta/180
      var dist = Math.sin(rlat1) * Math.sin(rlat2) + Math.cos(rlat1) * Math.cos(rlat2) * Math.cos(rtheta);
      dist = Math.acos(dist)
      dist = dist * 180/Math.PI
      dist = dist * 60 * 1.1515
      if (unit=="K") { dist = dist * 1.609344 }
      if (unit=="N") { dist = dist * 0.8684 }
      return dist
}

For those who are using Xamarin and don't have access to the GeoCoordinate class, you can use the Android Location class instead:

public static double GetDistanceBetweenCoordinates (double lat1, double lng1, double lat2, double lng2) {
            var coords1 = new Location ("");
            coords1.Latitude = lat1;
            coords1.Longitude = lng1;
            var coords2 = new Location ("");
            coords2.Latitude = lat2;
            coords2.Longitude = lng2;
            return coords1.DistanceTo (coords2);
        }

And here, for those still not satisfied, the original code from .NET-Frameworks GeoCoordinate class, refactored into a standalone method:

public double GetDistance(double longitude, double latitude, double otherLongitude, double otherLatitude)
{
    var d1 = latitude * (Math.PI / 180.0);
    var num1 = longitude * (Math.PI / 180.0);
    var d2 = otherLatitude * (Math.PI / 180.0);
    var num2 = otherLongitude * (Math.PI / 180.0) - num1;
    var d3 = Math.Pow(Math.Sin((d2 - d1) / 2.0), 2.0) + Math.Cos(d1) * Math.Cos(d2) * Math.Pow(Math.Sin(num2 / 2.0), 2.0);

    return 6376500.0 * (2.0 * Math.Atan2(Math.Sqrt(d3), Math.Sqrt(1.0 - d3)));
}

Based on Elliot Wood's function, this C function is working...

#define SIM_Degree_to_Radian(x) ((float)x * 0.017453292F)
#define SIM_PI_VALUE                         (3.14159265359)

float GPS_Distance(float lat1, float lon1, float lat2, float lon2)
{
   float theta;
   float dist;

   theta = lon1 - lon2;

   lat1 = SIM_Degree_to_Radian(lat1);
   lat2 = SIM_Degree_to_Radian(lat2);
   theta = SIM_Degree_to_Radian(theta);

   dist = (sin(lat1) * sin(lat2)) + (cos(lat1) * cos(lat2) * cos(theta));
   dist = acos(dist);

//   dist = dist * 180.0 / SIM_PI_VALUE;
//   dist = dist * 60.0 * 1.1515;
//   /* Convert to km */
//   dist = dist * 1.609344;

   dist *= 6370.693486F;

   return (dist);
}

You may change it to double. It returns the value in km.

Earth mean radius = 6,371km = 3958.76 miles

Rather than use var I suggest you use double, just to be explicit.

Manish sharma

Calculating Distance between Latitude and Longitude points...

        double Lat1 = Convert.ToDouble(latitude);
        double Long1 = Convert.ToDouble(longitude);

        double Lat2 = 30.678;
        double Long2 = 45.786;
        double circumference = 40000.0; // Earth's circumference at the equator in km
        double distance = 0.0;
        double latitude1Rad = DegreesToRadians(Lat1);
        double latititude2Rad = DegreesToRadians(Lat2);
        double longitude1Rad = DegreesToRadians(Long1);
        double longitude2Rad = DegreesToRadians(Long2);
        double logitudeDiff = Math.Abs(longitude1Rad - longitude2Rad);
        if (logitudeDiff > Math.PI)
        {
            logitudeDiff = 2.0 * Math.PI - logitudeDiff;
        }
        double angleCalculation =
            Math.Acos(
              Math.Sin(latititude2Rad) * Math.Sin(latitude1Rad) +
              Math.Cos(latititude2Rad) * Math.Cos(latitude1Rad) * Math.Cos(logitudeDiff));
        distance = circumference * angleCalculation / (2.0 * Math.PI);
        return distance;

You can use this function :

Source : https://www.geodatasource.com/developers/c-sharp

private double distance(double lat1, double lon1, double lat2, double lon2, char unit) {
  if ((lat1 == lat2) && (lon1 == lon2)) {
    return 0;
  }
  else {
    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.Acos(dist);
    dist = rad2deg(dist);
    dist = dist * 60 * 1.1515;
    if (unit == 'K') {
      dist = dist * 1.609344;
    } else if (unit == 'N') {
      dist = dist * 0.8684;
    }
    return (dist);
  }
}

//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//::  This function converts decimal degrees to radians             :::
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
private double deg2rad(double deg) {
  return (deg * Math.PI / 180.0);
}

//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//::  This function converts radians to decimal degrees             :::
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
private double rad2deg(double rad) {
  return (rad / Math.PI * 180.0);
}

Console.WriteLine(distance(32.9697, -96.80322, 29.46786, -98.53506, "M"));
Console.WriteLine(distance(32.9697, -96.80322, 29.46786, -98.53506, "K"));
Console.WriteLine(distance(32.9697, -96.80322, 29.46786, -98.53506, "N"));

Try this:

    public double getDistance(GeoCoordinate p1, GeoCoordinate p2)
    {
        double d = p1.Latitude * 0.017453292519943295;
        double num3 = p1.Longitude * 0.017453292519943295;
        double num4 = p2.Latitude * 0.017453292519943295;
        double num5 = p2.Longitude * 0.017453292519943295;
        double num6 = num5 - num3;
        double num7 = num4 - d;
        double num8 = Math.Pow(Math.Sin(num7 / 2.0), 2.0) + ((Math.Cos(d) * Math.Cos(num4)) * Math.Pow(Math.Sin(num6 / 2.0), 2.0));
        double num9 = 2.0 * Math.Atan2(Math.Sqrt(num8), Math.Sqrt(1.0 - num8));
        return (6376500.0 * num9);
    }

You can use System.device.Location:

System.device.Location.GeoCoordinate gc = new System.device.Location.GeoCoordinate(){
Latitude = yourLatitudePt1,
Longitude = yourLongitudePt1
};

System.device.Location.GeoCoordinate gc2 = new System.device.Location.GeoCoordinate(){
Latitude = yourLatitudePt2,
Longitude = yourLongitudePt2
};

Double distance = gc2.getDistanceTo(gc);

good luck

There's this library GeoCoordinate for these platforms:

Mono
.NET 4.5
.NET Core
Windows Phone 8.x
Universal Windows Platform
Xamarin iOS
Xamarin Android

Installation is done via NuGet:

PM> Install-Package GeoCoordinate

Usage

GeoCoordinate pin1 = new GeoCoordinate(lat, lng);
GeoCoordinate pin2 = new GeoCoordinate(lat, lng);

double distanceBetween = pin1.GetDistanceTo(pin2);

The distance between the two coordinates, in meters.

来源：https://stackoverflow.com/questions/6366408/calculating-distance-between-two-latitude-and-longitude-geocoordinates

标签

windows-phone-7

geocoding

latitude-longitude