最近在做有關(guān)于地圖的APP 需要用到知道兩個(gè)坐標(biāo)如何求兩地的距離
方法有兩種一種是CoreLocation 框架中自帶求兩地距離
-(BOOL)compareTowPlacesDistancewhetherOrNotLessTenMeter:(Annotation *)annotationA another:(Annotation *)anotationB
{
CLLocation *destlocA=[[CLLocation alloc]initWithLatitude:annotationA.coordinate.latitude longitude:annotationA.coordinate.longitude];
CLLocation *destlocB=[[CLLocation alloc]initWithLatitude:anotationB.coordinate.latitude longitude:anotationB.coordinate.longitude];
CLLocationDistance dist=[destlocA distanceFromLocation:destlocB];
NSLog(@"=========多少米%f",dist);
if (dist<=10) {
return YES;
} else{
return NO;
}
}
這種方法我發(fā)現(xiàn)有個(gè)不好的地方會(huì)出現(xiàn)小錯(cuò)，！
如果傳入的coordinate 為NULL值，返回的數(shù)值有可能就不會(huì)正確
so:
有一個(gè)更原始的方法 高中地理學(xué)過(guò)的
直接貼代碼吧
#define PI 3.141592653
-(BOOL)compareTowPlacesDistancewhetherOrNotLessTenMeter:(Annotation )annotationA another:(Annotation )anotationB
{
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 = PIannotationA.coordinate.latitude/180.0f;
double radlat2 = PIanotationB.coordinate.latitude/180.0f;
//now long.
double radlong1 = PIannotationA.coordinate.longitude/180.0f;
double radlong2 = PIanotationB.coordinate.longitude/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;
NSLog(@"=========多少米%f",dist);
if (dist<=10) {
return YES;
} else{
return NO;
} 