//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)).

我在这里发布了我的工作示例。

在MySQL中列出表中指定点(我们使用一个随机点- lat:45.20327, long:23.7806)之间距离小于50 KM的所有点(表中字段为coord_lat和coord_long):

列出所有距离<50，单位:公里(地球半径6371公里):

SELECT denumire, (6371 * acos( cos( radians(45.20327) ) * cos( radians( coord_lat ) ) * cos( radians( 23.7806 ) - radians(coord_long) ) + sin( radians(45.20327) ) * sin( radians(coord_lat) ) )) AS distanta 
FROM obiective 
WHERE coord_lat<>'' 
    AND coord_long<>'' 
HAVING distanta<50 
ORDER BY distanta desc

上面的例子是在MySQL 5.0.95和5.5.16 (Linux)中测试的。

//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;
    }

对于那些寻找基于WGS-84和GRS-80标准的Excel公式的人:

=ACOS(COS(RADIANS(90-Lat1))*COS(RADIANS(90-Lat2))+SIN(RADIANS(90-Lat1))*SIN(RADIANS(90-Lat2))*COS(RADIANS(Long1-Long2)))*6371

源

我已经创建了这个小Javascript LatLng对象，可能对某人有用。

var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2); 

代码:

/**
 * latLng point
 * @param {Number} lat
 * @param {Number} lng
 * @returns {LatLng}
 * @constructor
 */
function LatLng(lat,lng) {
    this.lat = parseFloat(lat);
    this.lng = parseFloat(lng);

    this.__cache = {};
}

LatLng.prototype = {
    toString: function() {
        return [this.lat, this.lng].join(",");
    },

    /**
     * calculate distance in km to another latLng, with caching
     * @param {LatLng} latLng
     * @returns {Number} distance in km
     */
    distanceTo: function(latLng) {
        var cacheKey = latLng.toString();
        if(cacheKey in this.__cache) {
            return this.__cache[cacheKey];
        }

        // the fastest way to calculate the distance, according to this jsperf test;
        // http://jsperf.com/haversine-salvador/8
        // http://stackoverflow.com/questions/27928
        var deg2rad = 0.017453292519943295; // === Math.PI / 180
        var lat1 = this.lat * deg2rad;
        var lng1 = this.lng * deg2rad;
        var lat2 = latLng.lat * deg2rad;
        var lng2 = latLng.lng * deg2rad;
        var a = (
            (1 - Math.cos(lat2 - lat1)) +
            (1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
            ) / 2;
        var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)

        // cache the distance
        this.__cache[cacheKey] = distance;

        return distance;
    }
};