C#计算汽车行驶方向的方法分析

/// <summary>///计算两点GPS坐标的距离/// </summary>/// <param name="n1">第一点的纬度坐标</param>/// <param name="e1">第一点的经度坐标</param>/// <param name="n2">第二点的纬度坐标</param>/// <param name="e2">第二点的经度坐标</param>/// <returns></returns>public static double Distance(double n1, double e1, double n2, double e2){  double jl_jd = 102834.74258026089786013677476285;  double jl_wd = 111712.69150641055729984301412873;  double b = Math.Abs((e1 - e2) * jl_jd);  double a = Math.Abs((n1 - n2) * jl_wd);  return Math.Sqrt((a * a + b * b));}/// <summary>/// 已知汽车行驶的两个GPS点，求汽车行驶的方向/// </summary>/// <param name="n1">第一个GPS点纬度</param>/// <param name="e1">第一个GPS点经度</param>/// <param name="n2">第二个GPS点纬度</param>/// <param name="e2">第二个GPS点经度</param>/// <returns></returns>public static double GetBusDirection( double n1,double e1, double n2, double e2){  double e3 = 0;  double n3 = 0;  e3 = e1 + 0.005;  n3 = n1;  double a = 0;  double b = 0;  double c = 0;  a = Distance(e1, n1, e3, n3);  b = Distance(e3, n3, e2, n2);  c = Distance(e1, n1, e2, n2);  double cosB = 0;  if ((a * c) != 0)  { cosB = (a * a + c * c - b * b) / (2 * a * c);  }  double B = Math.Acos(cosB) * 180 / Math.PI;  if(n2<n1)  { B=180+(180-B);  }  return B;}