计算两个地理点的距离（以公里为单位）c#

Question

我想计算两个地理点的距离。这些点以经度和纬度给出。

坐标为：

点 1：36.578581，-118.291994

点 2：36.23998，-116.83171

这里有一个比较结果的网站：

http://www.movable-type.co.uk/scripts/latlong.html

这里是我从此链接中使用的代码： 计算 Google 地图 V3 中两点之间的距离

    const double PIx = Math.PI;
    const double RADIO = 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 R = 6371; // km
        double dLat = Radians(lat2 - lat1);
        double dLon = Radians(lon2 - lon1);
        lat1 = Radians(lat1);
        lat2 = Radians(lat2);

        double a = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Sin(dLon / 2) * Math.Sin(dLon / 2) * Math.Cos(lat1) * Math.Cos(lat2);
        double c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
        double d = R * c;

        return d;
    }


Console.WriteLine(DistanceAlgorithm.DistanceBetweenPlaces(36.578581, -118.291994, 36.23998, -116.83171));

问题是我得到两个不同的结果。

我的结果：163,307 公里

网站结果：136 公里

有什么建议吗？？？

托蒂

Answer 1

您的公式几乎是正确的，但您必须交换经度和纬度的参数

Console.WriteLine(DistanceAlgorithm.DistanceBetweenPlaces(-118.291994, 36.578581, -116.83171, 36.23998)); // = 136 km

我使用简化的公式：

// cos(d) = sin(φА)·sin(φB) + cos(φА)·cos(φB)·cos(λА − λB),
//  where φА, φB are latitudes and λА, λB are longitudes
// Distance = d * R
public static double DistanceBetweenPlaces(double lon1, double lat1, double lon2, double lat2)
{
    double R = 6371; // km

    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;
}

测试：（

赤道距离）：经度 0, 100;纬度 = 0,0; DistanceBetweenPlaces(0, 0, 100, 0) = 11119.5 km

（北极距离）：经度 0, 100；纬度 = 90,90；地点之间的距离(0, 90, 100, 90) = 0 公里

经度: -118.291994, -116.83171;纬度：36.578581, 36.23998 = 135.6 公里

经度：36.578581, 36.23998；纬度：-118.291994, -116.83171 =

公里

163.2 在 网站 您用于结果比较，每个点第一个文本框是纬度，第二个是经度

Answer 2

由于您使用的是框架 4.0，我建议使用 GeoCooperative 类。

// using System.Device.Location;

GeoCoordinate c1 = new GeoCoordinate(36.578581, -118.291994);
GeoCoordinate c2 = new GeoCoordinate(36.23998, -116.83171);

double distanceInKm = c1.GetDistanceTo(c2) / 1000;
// Your result is: 136,111419742602

您必须添加对 System.Device.dll 的引用。

Answer 3

在我几年前发表的文章中（链接：http ://www.codeproject.com/Articles/469500/Edumatter-School-Math-Calculators-and-Equation-Sol）我有描述了 3 个有用的函数来计算 2 个地理点之间的距离（换句话说，地球上 2 个地理点之间的大圆（正向）距离），其在精度/性能方面有所不同：

// Haversine formula to calculate great-circle distance between two points on Earth

    private const double _radiusEarthMiles = 3959;
    private const double _radiusEarthKM = 6371;
    private const double _m2km = 1.60934;
    private const double _toRad = Math.PI / 180;

    /// <summary>
    /// Haversine formula to calculate 
    /// great-circle (orthodromic) distance on Earth 
    /// High Accuracy, Medium speed
    /// </summary>
    /// <param name="Lat1">double: 1st point Latitude</param>
    /// <param name="Lon1">double: 1st point Longitude</param>
    /// <param name="Lat2">double: 2nd point Latitude</param>
    /// <param name="Lon2">double: 2nd point Longitude</param>
    /// <returns>double: distance in miles</returns>
    public static double DistanceMilesHaversine(double Lat1,
                                                double Lon1,
                                                double Lat2,
                                                double Lon2)
    {
        try
        {
            double _radLat1 = Lat1 * _toRad;
            double _radLat2 = Lat2 * _toRad;
            double _dLatHalf = (_radLat2 - _radLat1) / 2;
            double _dLonHalf = Math.PI * (Lon2 - Lon1) / 360;

            // intermediate result
            double _a = Math.Sin(_dLatHalf);
            _a *= _a;

            // intermediate result
            double _b = Math.Sin(_dLonHalf);
            _b *= _b * Math.Cos(_radLat1) * Math.Cos(_radLat2);

            // central angle, aka arc segment angular distance
            double _centralAngle = 2 * Math.Atan2(Math.Sqrt(_a + _b), Math.Sqrt(1 - _a - _b));

            // great-circle (orthodromic) distance on Earth between 2 points
            return _radiusEarthMiles * _centralAngle;
        }
        catch { throw; }
    }

// Spherical law of cosines formula to calculate great-circle distance between two points on Earth

       /// <summary>
        /// Spherical Law of Cosines formula to calculate 
        /// great-circle (orthodromic) distance on Earth;
        /// High Accuracy, Medium speed
        /// http://en.wikipedia.org/wiki/Spherical_law_of_cosines
        /// </summary>
        /// <param name="Lat1">double: 1st point Latitude</param>
        /// <param name="Lon1">double: 1st point Longitude</param>
        /// <param name="Lat2">double: 2nd point Latitude</param>
        /// <param name="Lon2">double: 2nd point Longitude</param>
        /// <returns>double: distance in miles</returns>
        public static double DistanceMilesSLC(  double Lat1, 
                                                double Lon1, 
                                                double Lat2, 
                                                double Lon2)
        {
            try
            {
                double _radLat1 = Lat1 * _toRad;
                double _radLat2 = Lat2 * _toRad;
                double _radLon1 = Lon1 * _toRad;
                double _radLon2 = Lon2 * _toRad;

                // central angle, aka arc segment angular distance
                double _centralAngle = Math.Acos(Math.Sin(_radLat1) * Math.Sin(_radLat2) +
                        Math.Cos(_radLat1) * Math.Cos(_radLat2) * Math.Cos(_radLon2 - _radLon1));

                // great-circle (orthodromic) distance on Earth between 2 points
                return _radiusEarthMiles * _centralAngle;
            }
            catch { throw; }
        }

// Great-circle distance calculation using Spherical Earth projection formula**

/// <summary>
/// Spherical Earth projection to a plane formula (using Pythagorean Theorem)
/// to calculate great-circle (orthodromic) distance on Earth.
/// http://en.wikipedia.org/wiki/Geographical_distance
/// central angle = 
/// Sqrt((_radLat2 - _radLat1)^2 + (Cos((_radLat1 + _radLat2)/2) * (Lon2 - Lon1))^2)
/// Medium Accuracy, Fast,
/// relative error less than 0.1% in search area smaller than 250 miles
/// </summary>
/// <param name="Lat1">double: 1st point Latitude</param>
/// <param name="Lon1">double: 1st point Longitude</param>
/// <param name="Lat2">double: 2nd point Latitude</param>
/// <param name="Lon2">double: 2nd point Longitude</param>
/// <returns>double: distance in miles</returns>
public static double DistanceMilesSEP(double Lat1,
                                      double Lon1,
                                      double Lat2,
                                      double Lon2)
{
    try
    {
        double _radLat1 = Lat1 * _toRad;
        double _radLat2 = Lat2 * _toRad;
        double _dLat = (_radLat2 - _radLat1);
        double _dLon = (Lon2 - Lon1) * _toRad;

        double _a = (_dLon) * Math.Cos((_radLat1 + _radLat2) / 2);

        // central angle, aka arc segment angular distance
        double _centralAngle = Math.Sqrt(_a * _a + _dLat * _dLat);

        // great-circle (orthodromic) distance on Earth between 2 points
        return _radiusEarthMiles * _centralAngle;
    }
    catch { throw; }
}

函数返回以英里为单位的结果；要查找以公里为单位的距离，请将结果乘以 1.60934（请参阅private const double _m2km = 1.60934）。

与示例相关：找到距离点1（36.578581，-118.291994）和点2（36.23998，-116.83171）上述三个函数产生以下结果（公里）：

136.00206654936932
136.00206654937023
136.00374497149613

和计算器（链接：http://www.movable-type.co.uk/scripts/latlong.html）给出的结果：136.0

希望这可能有所帮助。此致，

Answer 4

我使用了 Wikipedia 中的公式并将其放入 lambda 函数中：

Func<double, double, double, double, double> CalcDistance = (lat1, lon1, lat2, lon2) =>
        {
            Func<double, double> Radians = (angle) =>
            {
                return angle * (180.0 / Math.PI);
            };

            const double radius = 6371;

            double delataSigma = Math.Acos(Math.Sin(Radians(lat1)) * Math.Sin(Radians(lat2)) +
                Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * Math.Cos(Math.Abs(Radians(lon2) - Radians(lon1))));

            double distance = radius * delataSigma;

            return distance;
        };

Answer 5

试试这个...我以前用过这个应用程序——它非常准确。请原谅我没有给予最初发表此文章的杰出灵魂应有的荣誉，我将其从 java 转换为 C#：

namespace Sample.Geography
{
    using System;

public class GeodesicDistance
{
    private static double DegsToRadians(double degrees)
    {
        return (0.017453292519943295 * degrees);
    }

    public static double? GetDistance(double lat1, double lon1, double lat2, double lon2)
    {
        long num = 0x615299L;
        double num2 = 6356752.3142;
        double num3 = 0.0033528106647474805;
        double num4 = DegsToRadians(lon2 - lon1);
        double a = Math.Atan((1 - num3) * Math.Tan(DegsToRadians(lat1)));
        double num6 = Math.Atan((1 - num3) * Math.Tan(DegsToRadians(lat2)));
        double num7 = Math.Sin(a);
        double num8 = Math.Sin(num6);
        double num9 = Math.Cos(a);
        double num10 = Math.Cos(num6);
        double num11 = num4;
        double num12 = 6.2831853071795862;
        int num13 = 20;
        double y = 0;
        double x = 0;
        double num18 = 0;
        double num20 = 0;
        double num22 = 0;
        while ((Math.Abs((double) (num11 - num12)) > 1E-12) && (--num13 > 0))
        {
            double num14 = Math.Sin(num11);
            double num15 = Math.Cos(num11);
            y = Math.Sqrt(((num10 * num14) * (num10 * num14)) + (((num9 * num8) - ((num7 * num10) * num15)) * ((num9 * num8) - ((num7 * num10) * num15))));
            if (y == 0)
            {
                return 0;
            }
            x = (num7 * num8) + ((num9 * num10) * num15);
            num18 = Math.Atan2(y, x);
            double num19 = ((num9 * num10) * num14) / y;
            num20 = 1 - (num19 * num19);
            if (num20 == 0)
            {
                num22 = 0;
            }
            else
            {
                num22 = x - (((2 * num7) * num8) / num20);
            }
            double num21 = ((num3 / 16) * num20) * (4 + (num3 * (4 - (3 * num20))));
            num12 = num11;
            num11 = num4 + ((((1 - num21) * num3) * num19) * (num18 + ((num21 * y) * (num22 + ((num21 * x) * (-1 + ((2 * num22) * num22)))))));
        }
        if (num13 == 0)
        {
            return null;
        }
        double num23 = (num20 * ((num * num) - (num2 * num2))) / (num2 * num2);
        double num24 = 1 + ((num23 / 16384) * (4096 + (num23 * (-768 + (num23 * (320 - (175 * num23)))))));
        double num25 = (num23 / 1024) * (256 + (num23 * (-128 + (num23 * (74 - (47 * num23))))));
        double num26 = (num25 * y) * (num22 + ((num25 / 4) * ((x * (-1 + ((2 * num22) * num22))) - ((((num25 / 6) * num22) * (-3 + ((4 * y) * y))) * (-3 + ((4 * num22) * num22))))));
        return new double?((num2 * num24) * (num18 - num26));
    }
}
}

Answer 6

我刚刚尝试在 GeoDataSource 上编码，它运行得很好：
http://www.geodatasource.com/developers/c-sharp

Answer 7

我认为您正在交换纬度和经度值。尝试更正这些或更改参数顺序。