Obliczanie odległości między dwoma Geokoordynatami szerokości i długości geograficznej

GetDistance jest najlepszym rozwiązaniem , ale w wielu przypadkach nie możemy użyć tej metody (np.]}

• Pseudokodzie algorytmu Oblicz odległość między koorindatami:

  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;
  }
  

• Implementacja C# w świecie rzeczywistym , która wykorzystuje metody rozszerzenia

  użycie:

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

  realizacja:

  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;
      }
  } 
  

Oto wersja JavaScript guys and gals

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
}

Dla tych, którzy używają Xamarin i nie mają dostępu do klasy GeoCoordinate, możesz użyć klasy lokalizacji Androida:

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);
        }

Średni promień Ziemi = 6,371 km = 3958,76 mil

Zamiast używać var proponuję użyć double, tak dla jasności.

Bazując na funkcji Elliota Wooda, ta funkcja C działa...

#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);
}

Można zmienić na podwójne . Zwraca wartość w km.

Obliczanie odległości między punktami szerokości i długości geograficznej...

        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;

I tutaj, dla tych, którzy nadal nie są zadowoleni, oryginalny kod z klasy. NET-Frameworks GeoCoordinate, przekształcony w samodzielną metodę:

public double GetDistance(double longitude, double latitude, double otherLongitude, double otherLatitude)
{
    var d1 = latitude * (Math.PI / 180.0);
    var num1 = longitude * (Math.PI / 180.0);
    var d2 = otherLatitude * (Math.PI / 180.0);
    var num2 = otherLongitude * (Math.PI / 180.0) - num1;
    var d3 = Math.Pow(Math.Sin((d2 - d1) / 2.0), 2.0) + Math.Cos(d1) * Math.Cos(d2) * Math.Pow(Math.Sin(num2 / 2.0), 2.0);

    return 6376500.0 * (2.0 * Math.Atan2(Math.Sqrt(d3), Math.Sqrt(1.0 - d3)));
}

Spróbuj tego:

    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);
    }

Możesz użyć 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);

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