matlab根据经纬度测角度,经纬度之间的距离和角度算法
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
#include
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
#include #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 <
cout << "==============" <
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 <
cout << "==============" <
JWD C = CJWDHelper::GetJWDB(A, distance, angle);
cout << "JD = " << C.m_Longitude << " WD = " << C.m_Latitude << endl;
cout << "==============" <
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 。