//JAVA
    public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
    final int RADIUS_EARTH = 6371;

    double dLat = getRad(latitude2 - latitude1);
    double dLong = getRad(longitude2 - longitude1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return (RADIUS_EARTH * c) * 1000;
    }

    private Double getRad(Double x) {
    return x * Math.PI / 180;
    }

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;
    }

}

下面是另一个转换为Ruby代码的代码:

include Math
#Note: from/to = [lat, long]

def get_distance_in_km(from, to)
  radians = lambda { |deg| deg * Math.PI / 180 }
  radius = 6371 # Radius of the earth in kilometer
  dLat = radians[to[0]-from[0]]
  dLon = radians[to[1]-from[1]]

  cosines_product = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(radians[from[0]]) * Math.cos(radians[to[1]]) * Math.sin(dLon/2) * Math.sin(dLon/2)

  c = 2 * Math.atan2(Math.sqrt(cosines_product), Math.sqrt(1-cosines_product)) 
  return radius * c # Distance in kilometer
end

要计算球体上两点之间的距离，你需要做大圆计算。

如果你需要将距离重新投影到平面上，MapTools中有许多C/ c++库可以帮助你进行地图投影。要做到这一点，你需要不同坐标系的投影字符串。

你可能还会发现MapWindow是一个可视化点的有用工具。此外，由于它是开源的，它是如何使用project.dll库的有用指南，它似乎是核心的开源投影库。

可能有一个更简单、更正确的解决方案:地球的周长在赤道上是40000公里，在格林威治(或任何经度)周期上约为37000公里。因此:

pythagoras = function (lat1, lon1, lat2, lon2) {
   function sqr(x) {return x * x;}
   function cosDeg(x) {return Math.cos(x * Math.PI / 180.0);}

   var earthCyclePerimeter = 40000000.0 * cosDeg((lat1 + lat2) / 2.0);
   var dx = (lon1 - lon2) * earthCyclePerimeter / 360.0;
   var dy = 37000000.0 * (lat1 - lat2) / 360.0;

   return Math.sqrt(sqr(dx) + sqr(dy));
};

我同意它应该被微调，我自己说过它是一个椭球，所以半径乘以余弦值是不同的。但它更准确一点。与谷歌map相比，误差明显减小。

下面是SQL实现，以km为单位计算距离，

SELECT UserId, ( 3959 * acos( cos( radians( your latitude here ) ) * cos( radians(latitude) ) * 
cos( radians(longitude) - radians( your longitude here ) ) + sin( radians( your latitude here ) ) * 
sin( radians(latitude) ) ) ) AS distance FROM user HAVING
distance < 5  ORDER BY distance LIMIT 0 , 5;

要获得通过编程语言实现的更多细节，您可以浏览这里给出的php脚本