計算兩個緯度和經度地理座標之間的距離

Question

我正在計算兩個地理座標之間的距離。我正在測試我的應用程序對3-4其他應用程序。當我計算距離時，我的計算結果平均爲3.3英里，而其他應用則爲3.5英里。對於我正在嘗試執行的計算而言，這是一個很大的區別。有沒有好的課程庫用於計算距離？我在C＃中計算它是這樣的：

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; 
} 

我該做什麼錯？我應該先計算公里數，然後轉換爲英里數嗎？

Answer 1

GeoCoordinate類（.NET Framework 4及更高版本）已經有GetDistanceTo方法。

var sCoord = new GeoCoordinate(sLatitude, sLongitude); 
var eCoord = new GeoCoordinate(eLatitude, eLongitude); 

return sCoord.GetDistanceTo(eCoord); 

該距離以米爲單位。

您需要引用System.Device。

Answer 2

地球平均半徑=6371公里=3958.76英里

而不是使用var我建議你使用double，只是要明確。

Answer 3

下面是JavaScript版本球員和加爾斯

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 
} 

Answer 4

試試這個：

public double getDistance(GeoCoordinate p1, GeoCoordinate p2) 
    { 
     double d = p1.Latitude * 0.017453292519943295; 
     double num3 = p1.Longitude * 0.017453292519943295; 
     double num4 = p2.Latitude * 0.017453292519943295; 
     double num5 = p2.Longitude * 0.017453292519943295; 
     double num6 = num5 - num3; 
     double num7 = num4 - d; 
     double num8 = Math.Pow(Math.Sin(num7/2.0), 2.0) + ((Math.Cos(d) * Math.Cos(num4)) * Math.Pow(Math.Sin(num6/2.0), 2.0)); 
     double num9 = 2.0 * Math.Atan2(Math.Sqrt(num8), Math.Sqrt(1.0 - num8)); 
     return (6376500.0 * num9); 
    } 

Answer 5

基於艾略特伍德的功能，這個C函數是工作...

#define SIM_Degree_to_Radian(x) ((float)x * 0.017453292F) 
#define SIM_PI_VALUE       (3.14159265359) 

float GPS_Distance(float lat1, float lon1, float lat2, float lon2) 
{ 
    float theta; 
    float dist; 

    theta = lon1 - lon2; 

    lat1 = SIM_Degree_to_Radian(lat1); 
    lat2 = SIM_Degree_to_Radian(lat2); 
    theta = SIM_Degree_to_Radian(theta); 

    dist = (sin(lat1) * sin(lat2)) + (cos(lat1) * cos(lat2) * cos(theta)); 
    dist = acos(dist); 

// dist = dist * 180.0/SIM_PI_VALUE; 
// dist = dist * 60.0 * 1.1515; 
// /* Convert to km */ 
// dist = dist * 1.609344; 

    dist *= 6370.693486F; 

    return (dist); 
} 

您可以將其更改爲double。它以km爲單位返回值。

Answer 6

getDistance的是最好的解決辦法，但在很多情況下，我們不能使用這個方法（如通用的應用程序）的算法

• 僞到計算之間的距離 coorindates：

  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; 
  } 
  

• 真實世界的C＃實現，這使得使用擴展方法的

  用法：

  var distance = new Coordinates(48.672309, 15.695585) 
         .DistanceTo(
          new Coordinates(48.237867, 16.389477), 
          UnitOfLength.Kilometers 
         ); 
  

  實現：

  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 7

計算緯度和經度點之間的距離...

 double Lat1 = Convert.ToDouble(latitude); 
     double Long1 = Convert.ToDouble(longitude); 

     double Lat2 = 30.678; 
     double Long2 = 45.786; 
     double circumference = 40000.0; // Earth's circumference at the equator in km 
     double distance = 0.0; 
     double latitude1Rad = DegreesToRadians(Lat1); 
     double latititude2Rad = DegreesToRadians(Lat2); 
     double longitude1Rad = DegreesToRadians(Long1); 
     double longitude2Rad = DegreesToRadians(Long2); 
     double logitudeDiff = Math.Abs(longitude1Rad - longitude2Rad); 
     if (logitudeDiff > Math.PI) 
     { 
      logitudeDiff = 2.0 * Math.PI - logitudeDiff; 
     } 
     double angleCalculation = 
      Math.Acos(
       Math.Sin(latititude2Rad) * Math.Sin(latitude1Rad) + 
       Math.Cos(latititude2Rad) * Math.Cos(latitude1Rad) * Math.Cos(logitudeDiff)); 
     distance = circumference * angleCalculation/(2.0 * Math.PI); 
     return distance; 

Answer 8

您可以使用System.device.Location：

System.device.Location.GeoCoordinate gc = new System.device.Location.GeoCoordinate(){ 
Latitude = yourLatitudePt1, 
Longitude = yourLongitudePt1 
}; 

System.device.Location.GeoCoordinate gc2 = new System.device.Location.GeoCoordinate(){ 
Latitude = yourLatitudePt2, 
Longitude = yourLongitudePt2 
}; 

Double distance = gc2.getDistanceTo(gc); 

好運

Answer 9

對於那些誰正在使用Xamarin並沒有進入會有地理座標類，你可以使用Android的位置類，而不是：

public static double GetDistanceBetweenCoordinates (double lat1, double lng1, double lat2, double lng2) { 
      var coords1 = new Location (""); 
      coords1.Latitude = lat1; 
      coords1.Longitude = lng1; 
      var coords2 = new Location (""); 
      coords2.Latitude = lat2; 
      coords2.Longitude = lng2; 
      return coords1.DistanceTo (coords2); 
     }