下面是一个c#实现:

static class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIUS = 6378.16;

    /// <summary>
    /// Convert degrees to Radians
    /// </summary>
    /// <param name="x">Degrees</param>
    /// <returns>The equivalent in radians</returns>
    public static double Radians(double x)
    {
        return x * PIx / 180;
    }

    /// <summary>
    /// Calculate the distance between two places.
    /// </summary>
    /// <param name="lon1"></param>
    /// <param name="lat1"></param>
    /// <param name="lon2"></param>
    /// <param name="lat2"></param>
    /// <returns></returns>
    public static double DistanceBetweenPlaces(
        double lon1,
        double lat1,
        double lon2,
        double lat2)
    {
        double dlon = Radians(lon2 - lon1);
        double dlat = Radians(lat2 - lat1);

        double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
        double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
        return angle * RADIUS;
    }

}

下面是一个c#实现:

static class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIUS = 6378.16;

    /// <summary>
    /// Convert degrees to Radians
    /// </summary>
    /// <param name="x">Degrees</param>
    /// <returns>The equivalent in radians</returns>
    public static double Radians(double x)
    {
        return x * PIx / 180;
    }

    /// <summary>
    /// Calculate the distance between two places.
    /// </summary>
    /// <param name="lon1"></param>
    /// <param name="lat1"></param>
    /// <param name="lon2"></param>
    /// <param name="lat2"></param>
    /// <returns></returns>
    public static double DistanceBetweenPlaces(
        double lon1,
        double lat1,
        double lon2,
        double lat2)
    {
        double dlon = Radians(lon2 - lon1);
        double dlat = Radians(lat2 - lat1);

        double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
        double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
        return angle * RADIUS;
    }

}

这是我的java实现计算距离经过一些搜索。我用的是世界平均半径(来自维基百科)，单位是千米。İf你想要的结果英里，然后使用世界半径英里。

public static double distanceLatLong2(double lat1, double lng1, double lat2, double lng2) 
{
  double earthRadius = 6371.0d; // KM: use mile here if you want mile result

  double dLat = toRadian(lat2 - lat1);
  double dLng = toRadian(lng2 - lng1);

  double a = Math.pow(Math.sin(dLat/2), 2)  + 
          Math.cos(toRadian(lat1)) * Math.cos(toRadian(lat2)) * 
          Math.pow(Math.sin(dLng/2), 2);

  double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

  return earthRadius * c; // returns result kilometers
}

public static double toRadian(double degrees) 
{
  return (degrees * Math.PI) / 180.0d;
}

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)。

哈弗辛公式在大多数情况下都是很好的公式，其他答案已经包含了它所以我就不占用空间了。但重要的是要注意，无论使用什么公式(是的，不仅仅是一个)。因为可能的精度范围很大，以及所需的计算时间。公式的选择需要更多的思考，而不是简单的无脑答案。

这个帖子来自nasa的一个人，是我在讨论这些选项时发现的最好的一个

http://www.cs.nyu.edu/visual/home/proj/tiger/gisfaq.html

例如，如果您只是在100英里半径内按距离对行进行排序。地平公式比哈弗辛公式快得多。

HalfPi = 1.5707963;
R = 3956; /* the radius gives you the measurement unit*/

a = HalfPi - latoriginrad;
b = HalfPi - latdestrad;
u = a * a + b * b;
v = - 2 * a * b * cos(longdestrad - longoriginrad);
c = sqrt(abs(u + v));
return R * c;

注意这里只有一个余弦和一个平方根。在哈弗辛公式中有9个。

这是一个简单的javascript函数，从这个链接可能是有用的。不知何故相关，但我们使用谷歌地球javascript插件而不是地图

function getApproximateDistanceUnits(point1, point2) {

    var xs = 0;
    var ys = 0;

    xs = point2.getX() - point1.getX();
    xs = xs * xs;

    ys = point2.getY() - point1.getY();
    ys = ys * ys;

    return Math.sqrt(xs + ys);
}

单位不是距离，而是相对于坐标的比率。还有其他相关的计算，你可以在这里代替getApproximateDistanceUnits函数链接

然后我使用这个函数来查看经纬度是否在半径内

function isMapPlacemarkInRadius(point1, point2, radi) {
    if (point1 && point2) {
        return getApproximateDistanceUnits(point1, point2) <= radi;
    } else {
        return 0;
    }
}

点可以定义为

 $$.getPoint = function(lati, longi) {
        var location = {
            x: 0,
            y: 0,
            getX: function() { return location.x; },
            getY: function() { return location.y; }
        };
        location.x = lati;
        location.y = longi;

        return location;
    };

然后你可以做你的事情，看看一个点是否在一个半径范围内，比如:

 //put it on the map if within the range of a specified radi assuming 100,000,000 units
        var iconpoint = Map.getPoint(pp.latitude, pp.longitude);
        var centerpoint = Map.getPoint(Settings.CenterLatitude, Settings.CenterLongitude);

        //approx ~200 units to show only half of the globe from the default center radius
        if (isMapPlacemarkInRadius(centerpoint, iconpoint, 120)) {
            addPlacemark(pp.latitude, pp.longitude, pp.name);
        }
        else {
            otherSidePlacemarks.push({
                latitude: pp.latitude,
                longitude: pp.longitude,
                name: pp.name
            });

        }