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}

    private double deg2rad(double deg)
    {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad)
    {
        return (rad / Math.PI * 180.0);
    }

    private double GetDistance(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Distance in Kilo Meter
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
        return (dist);
    }

    private double GetDirection(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Direction in Degrees
        double dlat = deg2rad(lat1) - deg2rad(lat2);
        double dlon = deg2rad(lon1) - deg2rad(lon2);
        double y = Math.Sin(dlon) * Math.Cos(lat2);
        double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
        double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
        if (direct < 0)
            direct = direct + 360;
        return (direct);
    }

    private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
        double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
        return (Speed);
    }

    private double GetDuration(DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
        return (TimeDifferenceInSeconds);
    }

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}

我猜你想让它沿着地球的曲率运动。你的两点和地心在一个平面上。地球的中心是这个平面上的圆心，这两个点(大致)在这个圆的周长上。由此你可以通过求一点到另一点的角度来计算距离。

如果点的高度不一样，或者如果你需要考虑地球不是一个完美的球体，这就有点困难了。

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 }