经纬度之间的距离和角度算法

原创
2011/09/01 09:49
阅读数 9K
经纬度之间的距离和角度算法。对于做3S的人来说应该是会碰到的问题,所以暂且记下,以备后用
来源: http://blog.csdn.net/fdnike/archive/2007/07/18/1696603.aspx  (由张hx提供给我,谢谢其帮助)

根据两站点的经纬度求两站点间的距离
/**** 根据两站点的经纬度求两站点间的距离 ****/
double D_jw(double wd1,double jd1,double wd2,double jd2)
{
    double x,y,out;
    double PI=3.14159265;
    double R=6.371229*1e6;

    x=(jd2-jd1)*PI*R*cos( ((wd1 wd2)/2) *PI/180)/180;
    y=(wd2-wd1)*PI*R/180;
    out=hypot(x,y);
    return out/1000;
}

==


 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;
}

=====
通过两个点的经纬度计算距离
关键词: gis                                          

从google maps的脚本里扒了段代码,没准啥时会用上。大家一块看看是怎么算的。

private const double EARTH_RADIUS = 6378.137;
private static double rad(double d)
{
   return d * Math.PI / 180.0;
}

public static double GetDistance(double lat1, double lng1, double lat2, double lng2)
{
   double radLat1 = rad(lat1);
   double radLat2 = rad(lat2);
   double a = radLat1 - radLat2;
   double b = rad(lng1) - rad(lng2);
   double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a/2),2)
    Math.Cos(radLat1)*Math.Cos(radLat2)*Math.Pow(Math.Sin(b/2),2)));
   s = s * EARTH_RADIUS;
   s = Math.Round(s * 10000) / 10000;
   return s;
}


非常感谢,帮了我大忙了:)
虽然我也没看明白到底原理是什么,但验算了A(60,30),B(60,90)两点之间,此段代码和我用余弦定理算出来的结果很一致。
余弦定理的步骤是:1、算A、B弦长:地球半径R*cos(经度差60)=R/2;
2、算角AOB,O为地球圆心,利用余弦定理,
cosAOB=(2R*R-(R/2)^2) /2*R*R=7/8;
3、弧AB的长为:R*arc cos(7/8);求毕 。

http://www.01buxi.com

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