C#编程练习（02）：大地坐标系（LBH）向空间直角坐标系（XYZ）的转换及其逆转换

需求说明：以WGS-84软件为例，实现大地坐标系（LBH）向空间直角坐标系（XYZ）的转换及其逆转换

原理说明：

程序源码：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;namespace XYZ2BLH
{class Program{//输出为度格式的字符串 （度分秒.秒的小数部分）//例如：35度23分32.9秒    表述为352332.9static string[] XYZ2LBH(double x,double y,double z)   {double[] blh = {0.0, 0.0, 0.0}; double a = 6378137;double b = 6356752.3142;double e2 = 0.0066943799013;double ePie2 = 0.00673949674227;double xita = Math.Atan(z * a / (Math.Sqrt(x * x + y * y) * b));double xitaSin3 = Math.Sin(xita) * Math.Sin(xita) * Math.Sin(xita);double xitaCos3 = Math.Cos(xita) * Math.Cos(xita) * Math.Cos(xita);double B = Math.Atan((z + ePie2 * b * xitaSin3) / ((Math.Sqrt(x * x + y * y) - e2 * a *  xitaCos3)));double L = Math.Atan(y / x);double N = a / Math.Sqrt(1 - e2 * Math.Sin(B) * Math.Sin(B)); ;double H = Math.Sqrt(x * x + y * y) / Math.Cos(B) - N;//转化为角度B = 180 * B / Math.PI;if (B < 0.0) { B = B + 90; }                                 L = 180 * L / Math.PI;if (L < 0.0) { L = L + 180; }return new string[] {rad2dms(B), rad2dms(L), H.ToString()};  }//获取一个小数的整数部分和小数部分static double[] modf(double t){t += 1.0e-14;//避免角度转换过程产生无效近似double t_integer = Math.Floor(t);return new double[] { t_integer, t - t_integer };      }//角度转弧度static double turn1(double t){double[] t1 = modf(t);double[] t2 = modf(t1[1] * 100);double x1 = t1[0];double x2 = t2[0];double x3 = t2[1] * 100;        //Console.Write("{0:N} \t {1:N}\t {2:N}\r\n", x1, x2, x3);return (x1 + x2 / 60 + x3 / 3600) / 180 * Math.PI;}//弧度转角度static string rad2dms(double rad){int du, fen;double miao;du = (int)rad;fen = (int)((rad - (double)du) * 60);miao = ((rad - (double)du) * 60 - (double)fen) * 60;return du.ToString() + fen.ToString().PadLeft(2,'0') + miao.ToString();        }//经度、纬度和大地高转向空间直角坐标static double[] LBH2XYZ(double[] lbh){double B = turn1(lbh[1]);double L = turn1(lbh[0]);double H = lbh[2];double A = 1 / 298.257223563;double a = 6378137.0000000000;double E = 2 * A - A * A;double W = Math.Sqrt(1 - E * Math.Sin(B) * Math.Sin(B));double N = a / W;double[] xyz = { 0.0, 0.0, 0.0 };xyz[0] = (N + H) * Math.Cos(B) * Math.Cos(L);xyz[1] = (N + H) * Math.Cos(B) * Math.Sin(L);xyz[2] = (N * (1 - E) + H) * Math.Sin(B);return xyz;}static void Main(string[] args){double[] lbh = { 108.5743, 32.2352, 100.59 };double[] xyz = LBH2XYZ(lbh);Console.WriteLine("{0:N6} {1:N6} {2:N6}", xyz[0], xyz[1], xyz[2]);string[] lbhTmp = XYZ2LBH(xyz[0], xyz[1], xyz[2]);Console.WriteLine("{0:N6} {1:N6} {2:N6}", double.Parse(lbhTmp[0]), double.Parse(lbhTmp[1]), double.Parse(lbhTmp[2]));         }}
}

测试结果：

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