根据2个经纬度计算距离

两种方法:
(1)
CLLocation *location1 = [[CLLocation alloc] initWithLatitude:_userLocation.location.coordinate.latitude longitude:_userLocation.location.coordinate.longitude];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:mapView.centerCoordinate.latitude longitude:mapView.centerCoordinate.longitude];
CLLocationDistance distance = [location1 getDistanceFrom:location2]/1000;//距离
if (distance>5) {
[_mapView setCenterCoordinate:_userLocation.location.coordinate animated:YES];
}
(2)

define PI 3.1415926

+(double) LantitudeLongitudeDist:(double)lon1 other_Lat:(double)lat1 self_Lon:(double)lon2 self_Lat:(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 = PIlat1/180.0f;
double radlat2 = PIlat2/180.0f;
//now long.
double radlong1 = PIlon1/180.0f;
double radlong2 = PIlon2/180.0f;
if( radlat1 < 0 ) radlat1 = PI/2 + fabs(radlat1);// south
if( radlat1 > 0 ) radlat1 = PI/2 - fabs(radlat1);// north
if( radlong1 < 0 ) radlong1 = PI2 - 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 = PI2 - fabs(radlong2);// west
//spherical coordinates x=rcos(ag)sin(at), y=rsin(ag)sin(at), z=rcos(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((erer+erer-dd)/(2erer));
double dist = theta*er;
return dist;
}