计算两个纬度和经度地理坐标之间的距离

Question

我正在计算两个GeoCoordinates之间的距离。我正在针对其他3-4个应用测试我的应用。当我计算距离时，我计算得到的平均值为3.3英里，而其他应用程序的距离为3.5英里。这与我正在尝试执行的计算有很大不同。有没有好的类库来计算距离？我在C＃中计算它是这样的：public static double Calculate(double sLatitude,double sLongitude, double eLatitude,                                double eLongitude){    var radiansOverDegrees = (Math.PI / 180.0);    var sLatitudeRadians = sLatitude * radiansOverDegrees;    var sLongitudeRadians = sLongitude * radiansOverDegrees;    var eLatitudeRadians = eLatitude * radiansOverDegrees;    var eLongitudeRadians = eLongitude * radiansOverDegrees;    var dLongitude = eLongitudeRadians - sLongitudeRadians;    var dLatitude = eLatitudeRadians - sLatitudeRadians;    var result1 = Math.Pow(Math.Sin(dLatitude / 2.0), 2.0) +                   Math.Cos(sLatitudeRadians) * Math.Cos(eLatitudeRadians) *                   Math.Pow(Math.Sin(dLongitude / 2.0), 2.0);    // Using 3956 as the number of miles around the earth    var result2 = 3956.0 * 2.0 *                   Math.Atan2(Math.Sqrt(result1), Math.Sqrt(1.0 - result1));    return result2;}我能做错什么？我应该先以km为单位计算，然后转换为里程？

Answer 1

该会有地理座标类（.NET框架4和更高）已经有GetDistanceTo方法。

var sCoord = new GeoCoordinate(sLatitude, sLongitude);

var eCoord = new GeoCoordinate(eLatitude, eLongitude);

return sCoord.GetDistanceTo(eCoord);

距离以米为单位。

您需要引用System.Device。

Answer 2

GetDistance是最好的解决方案，但在很多情况下我们无法使用此方法（例如通用应用程序）

用于计算 coorindates之间距离的算法的伪代码：

public static double DistanceTo(double lat1, double lon1, double lat2, double lon2, char unit = 'K')

{

    double rlat1 = Math.PI*lat1/180;

    double rlat2 = Math.PI*lat2/180;

    double theta = lon1 - lon2;

    double rtheta = Math.PI*theta/180;

    double dist =

        Math.Sin(rlat1)*Math.Sin(rlat2) + Math.Cos(rlat1)*

        Math.Cos(rlat2)*Math.Cos(rtheta);

    dist = Math.Acos(dist);

    dist = dist*180/Math.PI;

    dist = dist*60*1.1515;

    switch (unit)

    {

        case 'K': //Kilometers -> default

            return dist*1.609344;

        case 'N': //Nautical Miles 

            return dist*0.8684;

        case 'M': //Miles

            return dist;

    }

    return dist;

}

真实世界C＃实现，它使用扩展方法

用法：

var distance = new Coordinates(48.672309, 15.695585)

                .DistanceTo(

                    new Coordinates(48.237867, 16.389477),

                    UnitOfLength.Kilometers

                );

执行：

public class Coordinates

{

    public double Latitude { get; private set; }

    public double Longitude { get; private set; }

    public Coordinates(double latitude, double longitude)

    {

        Latitude = latitude;

        Longitude = longitude;

    }

}

public static class CoordinatesDistanceExtensions

{

    public static double DistanceTo(this Coordinates baseCoordinates, Coordinates targetCoordinates)

    {

        return DistanceTo(baseCoordinates, targetCoordinates, UnitOfLength.Kilometers);

    }

    public static double DistanceTo(this Coordinates baseCoordinates, Coordinates targetCoordinates, UnitOfLength unitOfLength)

    {

        var baseRad = Math.PI * baseCoordinates.Latitude / 180;

        var targetRad = Math.PI * targetCoordinates.Latitude/ 180;

        var theta = baseCoordinates.Longitude - targetCoordinates.Longitude;

        var thetaRad = Math.PI * theta / 180;

        double dist =

            Math.Sin(baseRad) * Math.Sin(targetRad) + Math.Cos(baseRad) *

            Math.Cos(targetRad) * Math.Cos(thetaRad);

        dist = Math.Acos(dist);

        dist = dist * 180 / Math.PI;

        dist = dist * 60 * 1.1515;

        return unitOfLength.ConvertFromMiles(dist);

    }

}

public class UnitOfLength

{

    public static UnitOfLength Kilometers = new UnitOfLength(1.609344);

    public static UnitOfLength NauticalMiles = new UnitOfLength(0.8684);

    public static UnitOfLength Miles = new UnitOfLength(1);

    private readonly double _fromMilesFactor;

    private UnitOfLength(double fromMilesFactor)

    {

        _fromMilesFactor = fromMilesFactor;

    }

    public double ConvertFromMiles(double input)

    {

        return input*_fromMilesFactor;

    }

} 

Answer 3

这是JavaScript版本的男人和女孩

function distanceTo(lat1, lon1, lat2, lon2, unit) {

      var rlat1 = Math.PI * lat1/180

      var rlat2 = Math.PI * lat2/180

      var rlon1 = Math.PI * lon1/180

      var rlon2 = Math.PI * lon2/180

      var theta = lon1-lon2

      var rtheta = Math.PI * theta/180

      var dist = Math.sin(rlat1) * Math.sin(rlat2) + Math.cos(rlat1) * Math.cos(rlat2) * Math.cos(rtheta);

      dist = Math.acos(dist)

      dist = dist * 180/Math.PI

      dist = dist * 60 * 1.1515

      if (unit=="K") { dist = dist * 1.609344 }

      if (unit=="N") { dist = dist * 0.8684 }

      return dist

}