通过经纬度坐标来计算地表上两点的距离

如果考虑地表的高程变化，使用经纬度结合DEM数据来计算两点之间的地表距离，那是比较复杂的，至少我自己现在不会算。在项目应用中，由于要求不是那么精确，我采用两种思路根据经纬度来计算两点的地表距离。
1.完全把地球假设成一个半径为6378.137公里的正球体，下面给出C#的示例代码。
Class PtDistance
{
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;
}
}

2.如果你想更精确一点，可以按照地球的长短轴把他假设为一个规则的椭球体，下面给出C++（开发环境VC6.0）示例：
————–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;
}