以km c为单位计算两个地理点的距离＃
我想计算两个地理点的距离。 这些点以经度和纬度给出。
坐标是：
第1点：36.578581，-118.291994
第2点：36.23998，-116.83171
这里有一个网站来比较结果：
http://www.movable-type.co.uk/scripts/latlong.html
这里我从这个链接使用的代码： 计算谷歌地图V3中两点之间的距离
const double PIx = Math.PI; const double RADIO = 6378.16; /// /// Convert degrees to Radians /// /// Degrees /// The equivalent in radians public static double Radians(double x) { return x * PIx / 180; } /// /// Calculate the distance between two places. /// /// /// /// /// /// 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)); 问题是我得到了两个不同的结果。
我的结果：163,307公里
网站结果：136公里
有什么建议？？？
TORTI
你的公式几乎是正确的，但你必须交换纬度经度的参数
Console.WriteLine(DistanceAlgorithm.DistanceBetweenPlaces(-118.291994, 36.578581, -116.83171, 36.23998)); // = 136 km
我正在使用简化的公式：
// 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; DistanceBetweenPlaces（0,90,100,90）= 0 km
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 = 163.2公里
最好的祝福
PS在您用于结果比较的网站上 ，对于每个点，第一个文本框是纬度，第二个经度
当您使用框架4.0时，我建议使用GeoCoordinate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。
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 ）我已经描述了3个有用的Functions来计算2之间的距离地理点（换句话说，地球上2个地理点之间的大圆（顺时针）距离），在准确性/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; /// /// Haversine formula to calculate /// great-circle (orthodromic) distance on Earth /// High Accuracy, Medium speed /// /// double: 1st point Latitude /// double: 1st point Longitude /// double: 2nd point Latitude /// double: 2nd point Longitude /// double: distance in miles 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 /// /// 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 /// /// double: 1st point Latitude /// double: 1st point Longitude /// double: 2nd point Latitude /// double: 2nd point Longitude /// double: distance in miles 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** /// /// 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 /// /// double: 1st point Latitude /// double: 1st point Longitude /// double: 2nd point Latitude /// double: 2nd point Longitude /// double: distance in miles 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; } }
函数返回里程数; 找到以km为单位的距离乘以1.60934的结果（参见private const double _m2km = 1.60934 ）。
相关样本：找到距离point1（36.578581，-118.291994）和point2（36.23998，-116.83171）三个上述函数产生以下结果（km）：
136.00206654936932 136.00206654937023 136.00374497149613
和计算器（链接： http ： //www.movable-type.co.uk/scripts/latlong.html ）给出了结果：136.0
希望这可能有所帮助。 最好的祝福，
试试这个…我之前使用过这个应用程序 – 它非常准确。 原谅我没有给予最初发表这篇文章的聪明人的应有的信任，我将它从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)); } } }
我只是尝试在GeoDataSource编写代码，它运行得非常好： http ： //www.geodatasource.com/developers/c-sharp
我使用维基百科中的公式并将其放在lambda函数中：
Func CalcDistance = (lat1, lon1, lat2, lon2) => { Func 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; };