如何计算由经纬度指定的两点之间的距离?

为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。


当前回答

下面是Haversine公式的java实现。

public final static double AVERAGE_RADIUS_OF_EARTH_KM = 6371;
public int calculateDistanceInKilometer(double userLat, double userLng,
  double venueLat, double venueLng) {

    double latDistance = Math.toRadians(userLat - venueLat);
    double lngDistance = Math.toRadians(userLng - venueLng);

    double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
      + Math.cos(Math.toRadians(userLat)) * Math.cos(Math.toRadians(venueLat))
      * Math.sin(lngDistance / 2) * Math.sin(lngDistance / 2);

    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));

    return (int) (Math.round(AVERAGE_RADIUS_OF_EARTH_KM * c));
}

请注意,这里我们将答案四舍五入到最近的km。

其他回答

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。

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;
    }
};

下面是一个c#实现:

static class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIUS = 6378.16;

    /// <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 * RADIUS;
    }

}

要计算球体上两点之间的距离,你需要做大圆计算。

如果你需要将距离重新投影到平面上,MapTools中有许多C/ c++库可以帮助你进行地图投影。要做到这一点,你需要不同坐标系的投影字符串。

你可能还会发现MapWindow是一个可视化点的有用工具。此外,由于它是开源的,它是如何使用project.dll库的有用指南,它似乎是核心的开源投影库。

Java实现在根据哈弗辛公式

double calculateDistance(double latPoint1, double lngPoint1, 
                         double latPoint2, double lngPoint2) {
    if(latPoint1 == latPoint2 && lngPoint1 == lngPoint2) {
        return 0d;
    }

    final double EARTH_RADIUS = 6371.0; //km value;

    //converting to radians
    latPoint1 = Math.toRadians(latPoint1);
    lngPoint1 = Math.toRadians(lngPoint1);
    latPoint2 = Math.toRadians(latPoint2);
    lngPoint2 = Math.toRadians(lngPoint2);

    double distance = Math.pow(Math.sin((latPoint2 - latPoint1) / 2.0), 2) 
            + Math.cos(latPoint1) * Math.cos(latPoint2)
            * Math.pow(Math.sin((lngPoint2 - lngPoint1) / 2.0), 2);
    distance = 2.0 * EARTH_RADIUS * Math.asin(Math.sqrt(distance));

    return distance; //km value
}