C # method for calculating the direction of vehicle driving _c# Tutorial

In this paper, we analyze the method of C # Calculation of vehicle driving direction. Share to everyone for your reference, specific as follows:

1. Scene: A known 2 GPS coordinates point A (N1,E1), B (e) In the course of the vehicle's journey, calculates the direction it travels.

2. Analysis: As shown in the above figure, know two points A, B, you can assume a C point, so that three points to form a right-angled triangle. It is now known that the A,b,c three points of the GPS coordinates can easily find three corners of the a,b,c. According to the cosine set cosb= (A2+C2-B2)/2AC, you can find the COSB value.

3.c# implementation code.

<summary>///Compute Two-point GPS coordinates distance///</summary>///<param name= "N1" > 1th latitude coordinates </param>///< Param name= "E1" > 1th longitude coordinates </param>///<param name= "n2" > 2nd latitude coordinates </param>///<param name= "E2" > 2nd Longitude coordinates </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>///The two GPS points that are known to drive the car, the direction of the car///</summary>///<param name= "N1" > The first GPS point latitude </param >///<param name= "E1" > First GPS point longitude </param>///<param name= "N2" > second GPS point latitude </param>///< param name= "E2" > second GPS point longitude </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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