如何计算两个GPS坐标之间的距离(使用经纬度)?


当前回答

下面是c#语言(用纬度和弧度表示):

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));
}

如果你的纬度和长度是用角度表示的,那么除以180/PI就可以转换成弧度。

其他回答

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;
    }

}

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;
}

Scala版本

  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)
    )
  }

下面是c#语言(用纬度和弧度表示):

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));
}

如果你的纬度和长度是用角度表示的,那么除以180/PI就可以转换成弧度。

这取决于你需要它有多准确。如果你需要精确到毫米的精度,最好看看使用椭球的算法,而不是球体,比如Vincenty的算法。