实际上GMap3中似乎有一个方法。它是google。maps。geometry。spherical命名空间的静态方法。

它以两个LatLng对象作为参数，并将使用默认的地球半径6378137米，尽管在必要时可以使用自定义值覆盖默认半径。

确保你包括:

<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>

在你的头部部分。

该呼吁将是:

google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);

实际上GMap3中似乎有一个方法。它是google。maps。geometry。spherical命名空间的静态方法。

它以两个LatLng对象作为参数，并将使用默认的地球半径6378137米，尽管在必要时可以使用自定义值覆盖默认半径。

确保你包括:

<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>

在你的头部部分。

该呼吁将是:

google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);

使用PHP，你可以使用这个简单的函数来计算距离:

// to calculate distance between two lat & lon

function calculate_distance($lat1, $lon1, $lat2, $lon2, $unit='N') 
{ 
  $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 ($unit == "N") {
      return ($miles * 0.8684);
    } else {
        return $miles;
      }
}

// function ends here

示例使用GPS的纬度/经度2个点。

var latitude1 = 39.46;
var longitude1 = -0.36;
var latitude2 = 40.40;
var longitude2 = -3.68;

var distance = google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(latitude1, longitude1), new google.maps.LatLng(latitude2, longitude2));       

不得不这么做……动作脚本方式

//just make sure you pass a number to the function because it would accept you mother in law...
public var rad = function(x:*) {return x*Math.PI/180;}

protected  function distHaversine(p1:Object, p2:Object):Number {
    var R:int = 6371; // earth's mean radius in km
    var dLat:Number = rad(p2.lat() - p1.lat());
    var dLong:Number = rad(p2.lng() - p1.lng());

    var a:Number = Math.sin(dLat/2) * Math.sin(dLat/2) +
                Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) * Math.sin(dLong/2) * Math.sin(dLong/2);
    var c:Number = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    var d:Number = R * c;

    return d;
}

离线解-哈弗辛算法

在Javascript中

var _eQuatorialEarthRadius = 6378.1370;
var _d2r = (Math.PI / 180.0);

function HaversineInM(lat1, long1, lat2, long2)
{
    return (1000.0 * HaversineInKM(lat1, long1, lat2, long2));
}

function HaversineInKM(lat1, long1, lat2, long2)
{
    var dlong = (long2 - long1) * _d2r;
    var dlat = (lat2 - lat1) * _d2r;
    var a = Math.pow(Math.sin(dlat / 2.0), 2.0) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r) * Math.pow(Math.sin(dlong / 2.0), 2.0);
    var c = 2.0 * Math.atan2(Math.sqrt(a), Math.sqrt(1.0 - a));
    var d = _eQuatorialEarthRadius * c;

    return d;
}

var meLat = -33.922982;
var meLong = 151.083853;


var result1 = HaversineInKM(meLat, meLong, -32.236457779983745, 148.69094705162837);
var result2 = HaversineInKM(meLat, meLong, -33.609020205923713, 150.77061469270831);

C#

using System;

public class Program
{
    public static void Main()
    {
        Console.WriteLine("Hello World");

        var meLat = -33.922982;
        double meLong = 151.083853;


        var result1 = HaversineInM(meLat, meLong, -32.236457779983745, 148.69094705162837);
        var result2 = HaversineInM(meLat, meLong, -33.609020205923713, 150.77061469270831);

        Console.WriteLine(result1);
        Console.WriteLine(result2);
    }

    static double _eQuatorialEarthRadius = 6378.1370D;
    static double _d2r = (Math.PI / 180D);

    private static int HaversineInM(double lat1, double long1, double lat2, double long2)
    {
        return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
    }

    private 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;
    }
}

参考: https://en.wikipedia.org/wiki/Great-circle_distance