計算兩個經緯度之前的距離
阿新 • • 發佈:2019-02-16
之前有用到,沒有記錄 後來有問到 就記下 以後方便
方法一
方法二- (double)caluteDistance:(double)lat1 :(double)lat2 :(double)lon1 :(double)lon2 { CLLocation *location1 = [[CLLocation alloc] initWithLatitude:lat1 longitude:lon1]; CLLocation *location2 = [[CLLocation alloc] initWithLatitude:lat2 longitude:lon2]; CLLocationDistance distance = [location1 distanceFromLocation:location2]; return distance; }
經緯度計算距離彙總: http://www.cnblogs.com/pengyingh/articles/2373246.html#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; }