C# 使用NetTopologySuite计算点之间的地理距离

我正在尝试使用NetTopologySuite计算两点之间的距离。由于我参考了Microsoft文档，因此我提出了以下GeometryExtension和GeometryHelper类：

        public static class GeometryExtensions
    {
        private static readonly CoordinateSystemServices _coordinateSystemServices = new CoordinateSystemServices(new Dictionary<int, string>
            {
                // Coordinate systems:

                [4326] = GeographicCoordinateSystem.WGS84.WKT,

                // CRS for Denmark ESPG 25832. Source: https://epsg.io/25832 and https://sdfe.dk/
                [25832] = @"PROJCS[""ETRS89 / UTM zone 32N"",
                                GEOGCS[""ETRS89"",
                                    DATUM[""European_Terrestrial_Reference_System_1989"",
                                        SPHEROID[""GRS 1980"",6378137,298.257222101,
                                            AUTHORITY[""EPSG"",""7019""]],
                                        TOWGS84[0,0,0,0,0,0,0],
                                        AUTHORITY[""EPSG"",""6258""]],
                                    PRIMEM[""Greenwich"",0,
                                        AUTHORITY[""EPSG"",""8901""]],
                                    UNIT[""degree"",0.0174532925199433,
                                        AUTHORITY[""EPSG"",""9122""]],
                                    AUTHORITY[""EPSG"",""4258""]],
                                PROJECTION[""Transverse_Mercator""],
                                PARAMETER[""latitude_of_origin"",0],
                                PARAMETER[""central_meridian"",9],
                                PARAMETER[""scale_factor"",0.9996],
                                PARAMETER[""false_easting"",500000],
                                PARAMETER[""false_northing"",0],
                                UNIT[""metre"",1,
                                    AUTHORITY[""EPSG"",""9001""]],
                                AXIS[""Easting"",EAST],
                                AXIS[""Northing"",NORTH],
                                AUTHORITY[""EPSG"",""25832""]]"
            }
        );
        


        /// <summary>
        /// Projects a geometry to another SRID
        /// </summary>
        /// <param name="geometry"></param>
        /// <param name="targetSrid"></param>
        /// <param name="defaultSourceSrid">If the geometry SRID has not been specified (i.e. equals 0) defaultSourceSrid is used</param>
        /// <returns></returns>
        public static Geometry ProjectTo(this Geometry geometry, int targetSrid, int? defaultSourceSrid = null)
        {
            if (geometry == null)
                throw new Exception("Geometry is null, cannot project");

            var sourceSrid = geometry.SRID == 0 && defaultSourceSrid.HasValue ? defaultSourceSrid.Value : geometry.SRID;
            var transformation = _coordinateSystemServices.CreateTransformation(sourceSrid, targetSrid);

            var result = geometry.Copy();
            result.Apply(new MathTransformFilter(transformation.MathTransform));

            return result;
        }
}




     public  class GeometryHelper
    {
        private static readonly int _longLatSRID = 4326;
        private static readonly int _targetSRID = 25832;

        /// <summary>
        /// In order to get the distance in meters, we need to project to an appropriate
        /// coordinate system. If no SRID is provided 25832, which covers Denmark is used.
        /// If the provided Points have no SRID, 4326 (longitude/latitude) is assumed.
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <param name="targetSrid"></param>
        /// <returns></returns>
        public static double DistanceInMeters(Point a, Point b, int? targetSrid = null)
        {
            targetSrid ??= _targetSRID;

            try
            {
                //If SRID is not set, assume long/lat, ie. 4326
                return a.ProjectTo(targetSrid.Value, _longLatSRID).Distance(b.ProjectTo(targetSrid.Value, _longLatSRID));
            }
            catch (Exception e)
            {
                throw new Exception("Failed to calculate distance", e);
            }
        }
}

---

您需要计算

大圆距离

。 NetTopologySuite

Point.Distance

方法返回

笛卡尔距离

请尝试以下操作：

public static double Radians(double x)
{
    return x * Math.PI / 180;
}

public static double GreatCircleDistance(double lon1, double lat1, double lon2, double lat2)
{
    double R = 6371e3; // m

    double sLat1 = Math.Sin(Radians(lat1));
    double sLat2 = Math.Sin(Radians(lat2));
    double cLat1 = Math.Cos(Radians(lat1));
    double cLat2 = Math.Cos(Radians(lat2));
    double cLon = Math.Cos(Radians(lon1) - Radians(lon2));

    double cosD = sLat1*sLat2 + cLat1*cLat2*cLon;

    double d = Math.Acos(cosD);

    double dist = R * d;

    return dist;
}

您也可以使用内置的，它实现了小距离时更精确的公式。
尽管saeedkazemi的答案是正确的，结果更接近真实值，但我的代码给出的结果也非常接近，但我发现我必须翻转Google maps给出的坐标。所以如果给我55.6765，12.5672，我需要给公式12.5672，55.6765。尽管如此，saeedkazemi的回答提供了更接近实际价值的结果

public static double Radians(double x)
{
    return x * Math.PI / 180;
}

public static double GreatCircleDistance(double lon1, double lat1, double lon2, double lat2)
{
    double R = 6371e3; // m

    double sLat1 = Math.Sin(Radians(lat1));
    double sLat2 = Math.Sin(Radians(lat2));
    double cLat1 = Math.Cos(Radians(lat1));
    double cLat2 = Math.Cos(Radians(lat2));
    double cLon = Math.Cos(Radians(lon1) - Radians(lon2));

    double cosD = sLat1*sLat2 + cLat1*cLat2*cLon;

    double d = Math.Acos(cosD);

    double dist = R * d;

    return dist;
}