一个经纬度相关计算的C++类

本文介绍了一个用于计算经纬度距离的C++类,并提供了测试程序。通过构造函数和成员函数,可以方便地进行经纬度转换和计算两点之间的距离。

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一个经纬度相关计算的C++类

写了一个经纬度距离计算的类
  --------------CJWD.h-------------- 
  #ifndef __JWD_AND_HELPER_20051005
  #define __JWD_AND_HELPER_20051005 
  #include "stdafx.h"
  #include <math.h>
  #include <iostream>
  using namespace std; 
  #ifndef PI
  #define PI 3.14159265;
  #endif
  static double Rc = 6378137; // 赤道半径
  static double Rj = 6356725; // 极半径 
  namespace CDYW{
  class JWD
  {
  public:
  double m_LoDeg, m_LoMin, m_LoSec; // longtitude 经度
  double m_LaDeg, m_LaMin, m_LaSec;
  double m_Longitude, m_Latitude;
  double m_RadLo, m_RadLa;
  double Ec;
  double Ed;
  public:
  // 构造函数, 经度: loDeg 度, loMin 分, loSec 秒; 纬度: laDeg 度, laMin 分, laSec秒
   JWD(double loDeg, double loMin, double loSec, double laDeg, double laMin, double laSec)
  {
   m_LoDeg=loDeg; m_LoMin=loMin; m_LoSec=loSec; m_LaDeg=laDeg; m_LaMin=laMin; m_LaSec=laSec;
   m_Longitude = m_LoDeg + m_LoMin / 60 + m_LoSec / 3600;
   m_Latitude = m_LaDeg + m_LaMin / 60 + m_LaSec / 3600;
   m_RadLo = m_Longitude * PI / 180.;
   m_RadLa = m_Latitude * PI / 180.;
   Ec = Rj + (Rc - Rj) * (90.- m_Latitude) / 90.;
   Ed = Ec * cos(m_RadLa);
  }
  //!
  JWD(double longitude, double latitude)
  {
   m_LoDeg = int(longitude);
   m_LoMin = int((longitude - m_LoDeg)*60);
   m_LoSec = (longitude - m_LoDeg - m_LoMin/60.)*3600;
   
   m_LaDeg = int(latitude);
   m_LaMin = int((latitude - m_LaDeg)*60);
   m_LaSec = (latitude - m_LaDeg - m_LaMin/60.)*3600;
   
   m_Longitude = longitude;
   m_Latitude = latitude;
   m_RadLo = longitude * PI/180.;
   m_RadLa = latitude * PI/180.;
   Ec = Rj + (Rc - Rj) * (90.-m_Latitude) / 90.;
   Ed = Ec * cos(m_RadLa);
  }
  };
  class CJWDHelper
  {
  public:
  CJWDHelper() {};
  ~CJWDHelper() {};
  //! 计算点A 和 点B的经纬度,求他们的距离和点B相对于点A的方位
  /*! 
   * \param A A点经纬度
   * \param B B点经纬度
   * \param angle B相对于A的方位, 不需要返回该值,则将其设为空
   * \return A点B点的距离
   */
  static double distance(JWD A, JWD B, double *angle)
  {
   double dx = (B.m_RadLo - A.m_RadLo) * A.Ed;
   double dy = (B.m_RadLa - A.m_RadLa) * A.Ec;
   double out = sqrt(dx * dx + dy * dy);
   
   if( angle != NULL)
   {
   *angle = atan(fabs(dx/dy))*180./PI;
   // 判断象限
   double dLo = B.m_Longitude - A.m_Longitude;
   double dLa = B.m_Latitude - A.m_Latitude;
   
   if(dLo > 0 && dLa <= 0) {
   *angle = (90. - *angle) + 90.;
   }
   else if(dLo <= 0 && dLa < 0) {
   *angle = *angle + 180.;
   }
   else if(dLo < 0 && dLa >= 0) {
   *angle = (90. - *angle) + 270;
   }
  }
  return out/1000;
  }
  //! 计算点A 和 点B的经纬度,求他们的距离和点B相对于点A的方位
  /*! 
   * \param longitude1 A点经度
   * \param latitude1 A点纬度
   * \param longitude2 B点经度
   * \param latitude2 B点纬度
   * \param angle B相对于A的方位, 不需要返回该值,则将其设为空
   * \return A点B点的距离
   */
  static double distance(
   double longitude1, double latitude1,
   double longitude2, double latitude2, 
   double *angle)
  {
   JWD A(longitude1,latitude1);
   JWD B(longitude2,latitude2);
   return distance(A, B, angle);
  }
  //! 已知点A经纬度,根据B点据A点的距离,和方位,求B点的经纬度
  /*!
   * \param A 已知点A
   * \param distance B点到A点的距离 
   * \param angle B点相对于A点的方位
   * \return B点的经纬度坐标
   */
  static JWD GetJWDB(JWD A, double distance, double angle)
  { 
   double dx = distance*1000 * sin(angle * PI /180.);
   double dy = distance*1000 * cos(angle * PI /180.);
   
   //double dx = (B.m_RadLo - A.m_RadLo) * A.Ed;
   //double dy = (B.m_RadLa - A.m_RadLa) * A.Ec;
   double BJD = (dx/A.Ed + A.m_RadLo) * 180./PI;
   double BWD = (dy/A.Ec + A.m_RadLa) * 180./PI;
   JWD B(BJD, BWD);
   return B; 
  }
  //! 已知点A经纬度,根据B点据A点的距离,和方位,求B点的经纬度
  /*!
   * \param longitude 已知点A经度
   * \param latitude 已知点A纬度
   * \param distance B点到A点的距离 
   * \param angle B点相对于A点的方位
   * \return B点的经纬度坐标
   */
  static JWD GetJWDB(double longitude, double latitude, double distance, double angle)
  { 
   JWD A(longitude,latitude);
   return GetJWDB(A, distance, angle);
  }
  };
  }
  #endif
  =========== 测试程序==========
  #include "stdafx.h"
  #include <math.h>
  #include <iostream>#include "CJWD.h"
  using namespace std;using namespace CDYW;
  double Rc = 6378137; // 赤道半径
  double Rj = 6356725; // 极半径// 绵阳
  double jd1 = 104.740999999;
  double wd1 = 31.4337;// 成都
  double jd2 = 104.01;
  double wd2 = 30.40; int main(int argc, char* argv[])
  {
  double angle = 0;
  cout << "A(绵阳): JD = " << jd1 << " WD = " << wd1 << endl;
  cout << "B(成都): JD = " << jd2 << " WD = " << wd2 << endl; 
  cout << "--------------------" << endl;
  cout << D_jw(wd1,jd1,wd2,jd2, angle) << endl;
  cout << "angle: " << angle <<endl;
  cout << "==============" <<endl;
  JWD A(jd1,wd1),B(jd2,wd2);
  double distance = CJWDHelper::distance(jd1,wd1,jd2,wd2, &angle);
  //cout << CJWDHelper::distance(A,B, &angle) << endl;
  cout << distance << endl;
  cout << "angle: " << angle <<endl; 
  cout << "==============" <<endl;
  JWD C = CJWDHelper::GetJWDB(A, distance, angle);
  cout << "JD = " << C.m_Longitude << " WD = " << C.m_Latitude << endl;
  cout << "==============" <<endl; 
  cout << A.m_LoDeg << " " << A.m_LoMin << " " << A.m_LoSec << endl; return 0;
  }(王朝网络 wangchao.net.cn)
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