17

2 つのジオポイントの距離を計算したいと思います。ポイントは経度と緯度で与えられます。

座標は次のとおりです。

ポイント 1: 36.578581、-118.291994

ポイント 2: 36.23998、-116.83171

結果を比較するためのウェブサイトは次のとおりです。

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

ここで、このリンクから使用したコード: Google マップ V3 で 2 点間の距離を計算する

    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));

問題は、2 つの異なる結果が得られることです。

私の結果: 163,307 km

ウェブサイトの結果: 136 km

助言がありますか？？？

トルティ

4

7 に答える 7

29

数式はほぼ正しいですが、経度と緯度のパラメーターを交換する必要があります

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

私は簡略化された式を使用しています：

// 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;
}

テスト:

(赤道での距離): 経度 0、100。緯度 = 0,0; DistanceBetweenPlaces(0, 0, 100, 0) = 11119.5 km

(北極点での距離): 経度 0、100。緯度 = 90,90; DistanceBetweenPlaces(0, 90, 100, 90) = 0 km

経度: -118.291994、-116.83171; 緯度: 36.578581、36.23998 = 135.6 km

経度: 36.578581、36.23998; 緯度: -118.291994、-116.83171 = 163.2 km

よろしくお願いします

PS結果の比較に使用するWebサイトでは、すべてのポイントの最初のテキストボックスは緯度、2番目は経度です

于 2011-07-01T07:42:23.413 に答える
5

数年前に公開された私の記事 (リンク: http://www.codeproject.com/Articles/469500/Edumatter-School-Math-Calculators-and-Equation-SolFunctions ) で、2 つの地理的位置の間の距離を計算するのに役立つ 3 つの方法について説明しました。 -ポイント (つまり、地球上の 2 つの地理的ポイント間の大円 (直交) 距離)。精度/パフォーマンスの点で異なります。

// 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; }
}

関数は結果をマイルで返します。距離を km で求めるには、結果に 1.60934 を掛けます ( を参照private const double _m2km = 1.60934)。

サンプルに関連: 距離 point1 (36.578581, -118.291994) と point2 (36.23998, -116.83171) を見つけます。前述の 3 つの関数は、次の結果 (km) を生成しました。

136.00206654936932
136.00206654937023
136.00374497149613

計算機 (リンク: http://www.movable-type.co.uk/scripts/latlong.html ) の結果: 136.0

これが役立つことを願っています。よろしくお願いします、

于 2015-01-11T04:17:50.093 に答える
1

これを試してみてください...これは以前に使用したアプリです-かなり正確です。これを最初に公開した素晴らしい魂に正当な信用を与えなかったことを許してください。私はそれを Java から C# に置き換えました。

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));
    }
}
}
于 2011-07-01T06:27:45.193 に答える
0

私は GeoDataSource でコーディングしようとしましたが、完全にうまくいきました: http://www.geodatasource.com/developers/c-sharp

于 2014-09-08T09:50:19.047 に答える