你可以在f#的fssnip中找到这个实现(有一些很好的解释)

以下是重要的部分:

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

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

    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

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;
    }

}

如果你使用的是。net，不要重新启动轮子。看到System.Device.Location。在另一个答案的评论中赞扬fnx。

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);

我需要在PowerShell中实现这个，希望它可以帮助其他人。 关于这种方法的一些注意事项

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 }

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。

下面是我在Python中使用的Haversine函数:

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}