根据经纬度计算两点之间的距离代码

September 9th, 2009 |

今天看到cocoachina有人问起这个问题，正好把好久以前的一段代码发了上来，这段代码已经经过验证，精度还是可以保障的，对于不同的应用需要更换调整地球半径值，先看代码

#define PI 3.1415926
 
double LantitudeLongitudeDist(double lon1,double lat1,
							  double lon2,double lat2)
{
	double er = 6378137; // 6378700.0f;
	//ave. radius = 6371.315 (someone said more accurate is 6366.707)
	//equatorial radius = 6378.388
	//nautical mile = 1.15078
	double radlat1 = PI*lat1/180.0f;
	double radlat2 = PI*lat2/180.0f;
 
	//now long.
	double radlong1 = PI*lon1/180.0f;
	double radlong2 = PI*lon2/180.0f;
 
	if( radlat1 &lt; 0 ) radlat1 = PI/2 + fabs(radlat1);// south
	if( radlat1 &gt; 0 ) radlat1 = PI/2 - fabs(radlat1);// north
	if( radlong1 &lt; 0 ) radlong1 = PI*2 - fabs(radlong1);//west
 
	if( radlat2 &lt; 0 ) radlat2 = PI/2 + fabs(radlat2);// south
	if( radlat2 &gt; 0 ) radlat2 = PI/2 - fabs(radlat2);// north
	if( radlong2 &lt; 0 ) radlong2 = PI*2 - fabs(radlong2);// west
 
	//spherical coordinates x=r*cos(ag)sin(at), y=r*sin(ag)*sin(at), z=r*cos(at)
	//zero ag is up so reverse lat
	double x1 = er * cos(radlong1) * sin(radlat1);
	double y1 = er * sin(radlong1) * sin(radlat1);
	double z1 = er * cos(radlat1);
 
	double x2 = er * cos(radlong2) * sin(radlat2);
	double y2 = er * sin(radlong2) * sin(radlat2);
	double z2 = er * cos(radlat2);
 
	double d = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2));
 
	//side, side, side, law of cosines and arccos
	double theta = acos((er*er+er*er-d*d)/(2*er*er));
	double dist  = theta*er;
 
	return dist;
}