飞镖版本

Haversine Algorithm。

import 'dart:math';

class GeoUtils {

  static double _degreesToRadians(degrees) {
    return degrees * pi / 180;
  }

  static double distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
    var earthRadiusKm = 6371;

    var dLat = _degreesToRadians(lat2-lat1);
    var dLon = _degreesToRadians(lon2-lon1);

    lat1 = _degreesToRadians(lat1);
    lat2 = _degreesToRadians(lat2);

    var a = sin(dLat/2) * sin(dLat/2) +
        sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
    var c = 2 * atan2(sqrt(a), sqrt(1-a));
    return earthRadiusKm * c;
  }
}

寻找带谷歌的哈弗辛;以下是我的解决方案:

#include <math.h>
#include "haversine.h"

#define d2r (M_PI / 180.0)

//calculate haversine distance for linear distance
double haversine_km(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 6367 * c;

    return d;
}

double haversine_mi(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 3956 * c; 

    return d;
}

在SQL Server 2008中使用地理类型非常容易做到这一点。

SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm

4326是WGS84椭球地球模型的SRID

飞镖版本

Haversine Algorithm。

import 'dart:math';

class GeoUtils {

  static double _degreesToRadians(degrees) {
    return degrees * pi / 180;
  }

  static double distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
    var earthRadiusKm = 6371;

    var dLat = _degreesToRadians(lat2-lat1);
    var dLon = _degreesToRadians(lon2-lon1);

    lat1 = _degreesToRadians(lat1);
    lat2 = _degreesToRadians(lat2);

    var a = sin(dLat/2) * sin(dLat/2) +
        sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
    var c = 2 * atan2(sqrt(a), sqrt(1-a));
    return earthRadiusKm * c;
  }
}

对于任何寻找Delphi/Pascal版本的人:

function GreatCircleDistance(const Lat1, Long1, Lat2, Long2: Double): Double;
var
  Lat1Rad, Long1Rad, Lat2Rad, Long2Rad: Double;
const
  EARTH_RADIUS_KM = 6378;
begin
  Lat1Rad  := DegToRad(Lat1);
  Long1Rad := DegToRad(Long1);
  Lat2Rad  := DegToRad(Lat2);
  Long2Rad := DegToRad(Long2);
  Result   := EARTH_RADIUS_KM * ArcCos(Cos(Lat1Rad) * Cos(Lat2Rad) * Cos(Long1Rad - Long2Rad) + Sin(Lat1Rad) * Sin(Lat2Rad));
end;

我对这个代码没有任何功劳，我最初是在一个公共论坛上发现Gary William发布的。

这取决于你需要它有多准确。如果你需要精确到毫米的精度，最好看看使用椭球的算法，而不是球体，比如Vincenty的算法。