下面是Haversine公式的typescript实现

static getDistanceFromLatLonInKm(lat1: number, lon1: number, lat2: number, lon2: number): number {
    var deg2Rad = deg => {
        return deg * Math.PI / 180;
    }

    var r = 6371; // Radius of the earth in km
    var dLat = deg2Rad(lat2 - lat1);   
    var dLon = deg2Rad(lon2 - lon1);
    var a =
        Math.sin(dLat / 2) * Math.sin(dLat / 2) +
        Math.cos(deg2Rad(lat1)) * Math.cos(deg2Rad(lat2)) *
        Math.sin(dLon / 2) * Math.sin(dLon / 2);
    var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    var d = r * c; // Distance in km
    return d;
}

这个脚本[在PHP中]计算两点之间的距离。

public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
        $lat1 = $source[0];
        $lon1 = $source[1];
        $lat2 = $dest[0];
        $lon2 = $dest[1];

        $theta = $lon1 - $lon2;
        $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
        $dist = acos($dist);
        $dist = rad2deg($dist);
        $miles = $dist * 60 * 1.1515;
        $unit = strtoupper($unit);

        if ($unit == "K") {
            return ($miles * 1.609344);
        }
        else if ($unit == "M")
        {
            return ($miles * 1.609344 * 1000);
        }
        else if ($unit == "N") {
            return ($miles * 0.8684);
        } 
        else {
            return $miles;
        }
    }

PIP安装haversine

Python实现

原产地是美国毗连的中心。

from haversine import haversine, Unit
origin = (39.50, 98.35)
paris = (48.8567, 2.3508)
haversine(origin, paris, unit=Unit.MILES)

要得到以千米为单位的答案，只需设置unit= unit。千米(这是默认值)。

在提供的代码中有一些错误，我在下面修复了它。

以上所有答案都假定地球是一个球体。然而，更精确的近似是扁球体。

a= 6378.137#equitorial radius in km
b= 6356.752#polar radius in km

def Distance(lat1, lons1, lat2, lons2):
    lat1=math.radians(lat1)
    lons1=math.radians(lons1)
    R1=(((((a**2)*math.cos(lat1))**2)+(((b**2)*math.sin(lat1))**2))/((a*math.cos(lat1))**2+(b*math.sin(lat1))**2))**0.5 #radius of earth at lat1
    x1=R1*math.cos(lat1)*math.cos(lons1)
    y1=R1*math.cos(lat1)*math.sin(lons1)
    z1=R1*math.sin(lat1)

    lat2=math.radians(lat2)
    lons2=math.radians(lons2)
    R2=(((((a**2)*math.cos(lat2))**2)+(((b**2)*math.sin(lat2))**2))/((a*math.cos(lat2))**2+(b*math.sin(lat2))**2))**0.5 #radius of earth at lat2
    x2=R2*math.cos(lat2)*math.cos(lons2)
    y2=R2*math.cos(lat2)*math.sin(lons2)
    z2=R2*math.sin(lat2)
    
    return ((x1-x2)**2+(y1-y2)**2+(z1-z2)**2)**0.5

FSharp版本，使用里程:

let radialDistanceHaversine location1 location2 : float = 
                let degreeToRadian degrees = degrees * System.Math.PI / 180.0
                let earthRadius = 3959.0
                let deltaLat = location2.Latitude - location1.Latitude |> degreeToRadian
                let deltaLong = location2.Longitude - location1.Longitude |> degreeToRadian
                let a =
                    (deltaLat / 2.0 |> sin) ** 2.0
                    + (location1.Latitude |> degreeToRadian |> cos)
                    * (location2.Latitude |> degreeToRadian |> cos)
                    * (deltaLong / 2.0 |> sin) ** 2.0
                atan2 (a |> sqrt) (1.0 - a |> sqrt)
                * 2.0
                * earthRadius

非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:

const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km

double convertToRadians(double val) {

   return val * PIx / 180;
}

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

        double dlon = convertToRadians(place2.longitude - place1.longitude);
        double dlat = convertToRadians(place2.latitude - place1.latitude);

        double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
        double angle = 2 * asin(sqrt(a));

        return angle * RADIO;
}

纬度和经度是十进制的。我没有在asin()调用中使用min()，因为我使用的距离非常小，以至于它们不需要min()。

它给出了错误的答案，直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)

额外的更新:

如果你使用的是iOS4或更高版本，那么苹果会提供一些方法来实现相同的功能:

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

    MKMapPoint  start, finish;


    start = MKMapPointForCoordinate(place1);
    finish = MKMapPointForCoordinate(place2);

    return MKMetersBetweenMapPoints(start, finish) / 1000;
}