Analysis of the method to calculate the driving direction of a car in C#

Content

1This article analyzes the method of calculating the car's travel direction in C#. Share it with everyone for reference, as follows:2．　 Scenario: Given the car's travel process,1a GPS coordinate point A (n1,e

2），B(e)，calculate the direction it travels.2+c2-b2)/2. The analysis: As shown in the figure above, knowing two points A, B, we can assume a point C, so that the three points form a right triangle. Now it is easy to calculate the GPS coordinates of points A, B, and C, and the lengths of the opposite sides a, b, and c of the three angles. According to the cosine rule, CosB = (a

3.C# implementation code.

/// <summary>
///Calculate the distance between two GPS coordinates.
/// </summary>
/// <param name="n1">the latitude coordinate of the first point</param>
/// <param name="e1">the longitude coordinate of the first point</param>
/// <param name="n2">the latitude coordinate of the second point</param>
/// <param name="e2">the longitude coordinate of the second point</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>
/// Given two GPS points of a car's travel, calculate the direction of the car's travel.
/// </summary>
/// <param name="n1">the latitude of the first GPS point</param>
/// <param name="e1">the longitude of the first GPS point</param>
/// <param name="n2">the latitude of the second GPS point</param>
/// <param name="e2">the longitude of the second GPS point</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;
}

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I hope this article will be helpful to everyone's C# program design.

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