oracle中有没有distance,如何使用Oracle的sdo_distance

我试图计算Oracle DB中两点之间的距离,

Point A is 40.716715, -74.033907

Point B is 40.716300, -74.033900

使用这个sql语句：

SELECT sdo_geom.sdo_distance( sdo_geom.sdo_geometry(2001 ,8307 ,sdo_geom.sdo_point_type(40.716715, -74.033907 , NULL) ,NULL ,NULL)

,sdo_geom.sdo_geometry(2001 ,8307 ,sdo_point_type(40.716300,-74.033901, NULL) ,NULL ,NULL) ,0.0001 ,'unit=M') distance_in_m

from DUAL;

结果是12.7646185977151

使用Apple的CoreLocation api进行此操作时：

CLLocation* pa = [[CLLocation alloc] initWithLatitude:40.716715 longitude:-74.033907];

CLLocation* pa2 = [[CLLocation alloc] initWithLatitude:40.716300 longitude:-74.033900];

CLLocationDistance dist = [pa distanceFromLocation:pa2];

结果是46.0888946842423

自己实施

double dinstance_m(double lat1, double long1, double lat2, double long2){

double dlong = (long2 - long1) * d2r;

double dlat = (lat2 - lat1) * d2r;

double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);

double c = 2 * atan2(sqrt(a), sqrt(1-a));

double d = 6367 * c;

return d * 1000.;

}

结果是46.120690774231

Oracle的实现显然是关闭的,但我不明白为什么.任何帮助将不胜感激.