计算两个坐标之间的纬度和经度的距离，包括一个Javascript实现。

西部和南部的位置是负的。 记住，分和秒是60度，所以S31 30'是-31.50度。

别忘了把角度转换成弧度。许多语言都有这个功能。或者它是一个简单的计算:弧度=角度* 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;
}

下面是一些用法的例子:

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

我认为R中的一个算法版本仍然缺失:

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
}

对于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;
}

寻找带谷歌的哈弗辛;以下是我的解决方案:

#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;
}

对于任何寻找Delphi/Pascal版本的人:

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;

我对这个代码没有任何功劳，我最初是在一个公共论坛上发现Gary William发布的。

基于Roman Makarov对这个线程的回复的Java版本的Haversine算法

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;
    }

}