Question

ios知道两点的经纬度怎么算距离

Answer 1

您好，很高兴为你回答。在网上找到一个算法

6378.7 * acos[sin(lat1/57.2958) * sin(lat2/57.2958) + cos(lat1/57.2958) * cos(lat2/57.2958) * cos(lon2/57.2958 -lon1/57.2958)] 具体实现代码如下。

#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 < 0 ) radlat1 = PI/2 + fabs(radlat1);// south

if( radlat1 > 0 ) radlat1 = PI/2 - fabs(radlat1);// north

if( radlong1 < 0 ) radlong1 = PI*2 - fabs(radlong1);//west

if( radlat2 < 0 ) radlat2 = PI/2 + fabs(radlat2);// south

if( radlat2 > 0 ) radlat2 = PI/2 - fabs(radlat2);// north

if( radlong2 < 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;

}

希望对您有所帮助，期待您的采纳。