//p1 and p2 are google.maps.LatLng(x,y) objects

function calcDistance(p1, p2) {
          var d = (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
          console.log(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

下面是this公式的c#实现

 public class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIO = 6378.16;

    /// <summary>
    /// This class cannot be instantiated.
    /// </summary>
    private DistanceAlgorithm() { }

    /// <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 * RADIO) * 0.62137;//distance in miles
    }

}    

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

对于谷歌，你可以使用球面api, google.maps.geometry. sphere.computedistancebetween (latLngA, latLngB);

然而，如果球面投影或哈弗辛解的精度对你来说不够精确(例如，如果你靠近极点或计算更长的距离)，你应该使用不同的库。

关于这个主题的大部分信息我都是在维基百科上找到的。

要查看任何给定算法的精度是否足够，一个技巧是填充地球的最大和最小半径，并查看差异是否会给您的用例带来问题。更多细节可以在本文中找到

最后，谷歌api或haversine将毫无问题地满足大多数目的。

首先，你指的是整个路径的长度，还是你只想知道位移(直线距离)?我看没人指出距离和位移的区别。对于距离计算JSON/XML数据给出的每个路由点，至于位移，有一个内置的解决方案使用球面类

//calculates distance between two points in km's
function calcDistance(p1, p2) {
  return (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
}