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

下面是移植到Java的已接受的答案实现，以备任何人需要。

package com.project529.garage.util;


/**
 * Mean radius.
 */
private static double EARTH_RADIUS = 6371;

/**
 * Returns the distance between two sets of latitudes and longitudes in meters.
 * <p/>
 * Based from the following JavaScript SO answer:
 * http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
 * which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
 */
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
    double dLat = toRadians(lat2 - lat1);
    double dLon = toRadians(lon2 - lon1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
            Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
                    Math.sin(dLon / 2) * Math.sin(dLon / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    double d = EARTH_RADIUS * c;

    return d;
}

public double toRadians(double degrees) {
    return degrees * (Math.PI / 180);
}

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。

function distance($lat1, $lon1, $lat2, $lon2) { 
    $pi80 = M_PI / 180; 
    $lat1 *= $pi80; $lon1 *= $pi80; $lat2 *= $pi80; $lon2 *= $pi80; 
    $dlat = $lat2 - $lat1; 
    $dlon = $lon2 - $lon1; 
    $a = sin($dlat / 2) * sin($dlat / 2) + cos($lat1) * cos($lat2) * sin($dlon / 2) * sin($dlon / 2);  
    $km = 6372.797 * 2 * atan2(sqrt($a), sqrt(1 - $a)); 
    return $km; 
}

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;
}

function getDistanceFromLatLonInKm(position1, position2) {
    "use strict";
    var deg2rad = function (deg) { return deg * (Math.PI / 180); },
        R = 6371,
        dLat = deg2rad(position2.lat - position1.lat),
        dLng = deg2rad(position2.lng - position1.lng),
        a = Math.sin(dLat / 2) * Math.sin(dLat / 2)
            + Math.cos(deg2rad(position1.lat))
            * Math.cos(deg2rad(position2.lat))
            * Math.sin(dLng / 2) * Math.sin(dLng / 2),
        c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return R * c;
}

console.log(getDistanceFromLatLonInKm(
    {lat: 48.7931459, lng: 1.9483572},
    {lat: 48.827167, lng: 2.2459745}
));