计算2个GPS坐标之间的距离

如何计算两个GPS坐标之间的距离(使用纬度和经度)？

按纬度和经度计算两个坐标之间的距离,包括Javascript实现.

西部和南部地区都是负面的.记住分数和秒数都在60以上,所以S31 30'是-31.50度.

不要忘记将度数转换为弧度.许多语言都有这个功能.或者它的简单计算: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;
}

以下是一些用法示例:

distanceInKmBetweenCoordinates(0,0,0,0)//相同点之间的距离应为0 0 distanceInKmBetweenCoordinates(51.5,0,38.8,-77.1)//从伦敦到阿灵顿5918.185064088764

寻找与谷歌的bersine; 这是我的解决方案:

#include 
#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;
}

C#版本的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;
}

这是一个.NET小提琴,所以你可以用你自己的Lat/Longs来测试它.

Java版本的Haversine算法基于Roman Makarov对此主题的回复

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;
    }

}

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

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}

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));
}

$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;

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
}

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))

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