通过两个点的经纬度计算地面距离

通过两个点的经纬度计算地面距离

1:原JAVA

double distanceByLnglat(double _Longitude1,
   double _Latidute1,
   double _Longitude2,
   double _Latidute2)
  {
   
   double radLat1 = _Latidute1 * Math.PI / 180;
   double radLat2 = _Latidute2 * Math.PI / 180;
   double a = radLat1 - radLat2;
   double b = _Longitude1 * Math.PI / 180 - _Longitude2 * Math.PI / 180;
   double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a / 2), 2) + Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Pow(Math.Sin(b / 2), 2)));
   s = s * 6378137.0;// 取WGS84标准参考椭球中的地球长半径(单位:m)
   s = Math.Round(s * 10000) / 10000;

   return s;
  }

 

 

 2:原.Net

public struct EarthPoint
  {
   public const double Ea = 6378137; // 赤道半径 WGS84标准参考椭球中的地球长半径(单位:m) 
   public const double Eb = 6356725; // 极半径  
   public readonly double Longitude,Latidute;
   public readonly double Jd;
   public readonly double Wd;
   public readonly double Ec;
   public readonly double Ed;
   public EarthPoint(double _Longitude,double _Latidute)
   {
    Longitude = _Longitude;
    Latidute = _Latidute;
    Jd = Longitude * Math.PI / 180; //转换成角度
    Wd = Latidute * Math.PI /180; //转换成角度
    Ec = Eb + (Ea - Eb) * (90 - Latidute) / 90;
    Ed = Ec * Math.Cos(Wd);
   }
   public double Distance(EarthPoint _Point)
   {
    double dx = (_Point.Jd - Jd) * Ed;
    double dy = (_Point.Wd - Wd) * Ec;
    return Math.Sqrt(dx * dx + dy *dy);
   }
  }

 

public static Double GetDistance(double _Longitude1,
   double _Latidute1,
   double _Longitude2,
   double _Latidute2)
  {
   EarthPoint p1 = new EarthPoint(_Longitude1,_Latidute1);
   EarthPoint p2 = new EarthPoint(_Longitude2,_Latidute2);
   return p1.Distance(p2);
   
  }

 

测试代码:

DateTime   dt1=DateTime.Now;
   DateTime dt2 = DateTime.Now;
   TimeSpan ts1;
   
   dt1 = DateTime.Now;
   for(int i=0;i<sumCount;i++)
    ss = distanceByLnglat(jd1,wd1,jd2,wd2);
   dt2 = DateTime.Now;
   ts1 = dt2- dt1;
   this.label4.Text =  String.Format("Java:{0:N3}米 计算次数: {1:N:0} 用时: {2:N1} 毫妙",ss,sumCount,ts1.TotalMilliseconds);

   
   dt1=DateTime.Now;
   for(int i=0;i<sumCount;i++)
    ss = GetDistance(jd1,wd1,jd2,wd2);
   dt2 = DateTime.Now;
   ts1 = dt2- dt1;
   this.label3.Text =  String.Format("Net:{0:N3}米 计算次数: {1:N:0} 用时: {2:N1} 毫妙",ss,sumCount,ts1.TotalMilliseconds);

 

 

第一种方法用时基本上是第二种方法的用时的2倍!

此处没有比较2种语言的意思,因为根本没有在2种环境下运行,这里只比较了算法的执行效率而已! 

 

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从google maps的脚本里扒了段代码，没准啥时会用上。大家一块看看是怎么算的。

private const double EARTH_RADIUS = 6378.137;
private static double rad(double d)
{
   return d * Math.PI / 180.0;
}

public static double GetDistance(double lat1, double lng1, double lat2, double lng2)
{
   double radLat1 = rad(lat1);
   double radLat2 = rad(lat2);
   double a = radLat1 - radLat2;
   double b = rad(lng1) - rad(lng2);
   double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a/2),2) +
    Math.Cos(radLat1)*Math.Cos(radLat2)*Math.Pow(Math.Sin(b/2),2)));
   s = s * EARTH_RADIUS;
   s = Math.Round(s * 10000) / 10000;
   return s;
}