正如指出的那样，精确的计算应该考虑到地球不是一个完美的球体。以下是这里提供的各种算法的一些比较:

geoDistance(50,5,58,3)
Haversine: 899 km
Maymenn: 833 km
Keerthana: 897 km
google.maps.geometry.spherical.computeDistanceBetween(): 900 km

geoDistance(50,5,-58,-3)
Haversine: 12030 km
Maymenn: 11135 km
Keerthana: 10310 km
google.maps.geometry.spherical.computeDistanceBetween(): 12044 km

geoDistance(.05,.005,.058,.003)
Haversine: 0.9169 km
Maymenn: 0.851723 km
Keerthana: 0.917964 km
google.maps.geometry.spherical.computeDistanceBetween(): 0.917964 km

geoDistance(.05,80,.058,80.3)
Haversine: 33.37 km
Maymenn: 33.34 km
Keerthana: 33.40767 km
google.maps.geometry.spherical.computeDistanceBetween(): 33.40770 km

在小范围内，Keerthana的算法似乎与谷歌Maps的算法一致。谷歌Maps似乎没有遵循任何简单的算法，这表明它可能是这里最准确的方法。

不管怎样，这里是Keerthana算法的Javascript实现:

function geoDistance(lat1, lng1, lat2, lng2){
    const a = 6378.137; // equitorial radius in km
    const b = 6356.752; // polar radius in km

    var sq = x => (x*x);
    var sqr = x => Math.sqrt(x);
    var cos = x => Math.cos(x);
    var sin = x => Math.sin(x);
    var radius = lat => sqr((sq(a*a*cos(lat))+sq(b*b*sin(lat)))/(sq(a*cos(lat))+sq(b*sin(lat))));

    lat1 = lat1 * Math.PI / 180;
    lng1 = lng1 * Math.PI / 180;
    lat2 = lat2 * Math.PI / 180;
    lng2 = lng2 * Math.PI / 180;

    var R1 = radius(lat1);
    var x1 = R1*cos(lat1)*cos(lng1);
    var y1 = R1*cos(lat1)*sin(lng1);
    var z1 = R1*sin(lat1);

    var R2 = radius(lat2);
    var x2 = R2*cos(lat2)*cos(lng2);
    var y2 = R2*cos(lat2)*sin(lng2);
    var z2 = R2*sin(lat2);

    return sqr(sq(x1-x2)+sq(y1-y2)+sq(z1-z2));
}

下面是Erlang实现

lat_lng({Lat1, Lon1}=_Point1, {Lat2, Lon2}=_Point2) ->
  P = math:pi() / 180,
  R = 6371, % Radius of Earth in KM
  A = 0.5 - math:cos((Lat2 - Lat1) * P) / 2 +
    math:cos(Lat1 * P) * math:cos(Lat2 * P) * (1 - math:cos((Lon2 - Lon1) * P))/2,
  R * 2 * math:asin(math:sqrt(A)).

非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:

const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km

double convertToRadians(double val) {

   return val * PIx / 180;
}

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

        double dlon = convertToRadians(place2.longitude - place1.longitude);
        double dlat = convertToRadians(place2.latitude - place1.latitude);

        double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
        double angle = 2 * asin(sqrt(a));

        return angle * RADIO;
}

纬度和经度是十进制的。我没有在asin()调用中使用min()，因为我使用的距离非常小，以至于它们不需要min()。

它给出了错误的答案，直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)

额外的更新:

如果你使用的是iOS4或更高版本，那么苹果会提供一些方法来实现相同的功能:

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

    MKMapPoint  start, finish;


    start = MKMapPointForCoordinate(place1);
    finish = MKMapPointForCoordinate(place2);

    return MKMetersBetweenMapPoints(start, finish) / 1000;
}

你可以使用CLLocationDistance中的构建来计算这个:

CLLocation *location1 = [[CLLocation alloc] initWithLatitude:latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:latitude2 longitude:longitude2];
[self distanceInMetersFromLocation:location1 toLocation:location2]

- (int)distanceInMetersFromLocation:(CLLocation*)location1 toLocation:(CLLocation*)location2 {
    CLLocationDistance distanceInMeters = [location1 distanceFromLocation:location2];
    return distanceInMeters;
}

在你的例子中，如果你想要公里，只要除以1000。

FSharp版本，使用里程:

let radialDistanceHaversine location1 location2 : float = 
                let degreeToRadian degrees = degrees * System.Math.PI / 180.0
                let earthRadius = 3959.0
                let deltaLat = location2.Latitude - location1.Latitude |> degreeToRadian
                let deltaLong = location2.Longitude - location1.Longitude |> degreeToRadian
                let a =
                    (deltaLat / 2.0 |> sin) ** 2.0
                    + (location1.Latitude |> degreeToRadian |> cos)
                    * (location2.Latitude |> degreeToRadian |> cos)
                    * (deltaLong / 2.0 |> sin) ** 2.0
                atan2 (a |> sqrt) (1.0 - a |> sqrt)
                * 2.0
                * earthRadius

如果你想要驾驶距离/路线(张贴在这里，因为这是谷歌上两点之间距离的第一个结果，但对大多数人来说，驾驶距离更有用)，你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}