IOS根据两个经纬度计算相距距离

时间：2016-01-20 13:01:36 阅读：169 评论：0 收藏：0 [点我收藏+]

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//第一种苹果自带的

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CLLocation *orig=[[[CLLocation alloc] initWithLatitude:[mainDelegate.latitude_self doubleValue] longitude:[mainDelegate.longitude_self doubleValue]] autorelease];
CLLocation* dist=[[[CLLocation alloc] initWithLatitude:[tmpNewsModel.latitude doubleValue] longitude:[tmpNewsModel.longitude doubleValue] ] autorelease];
CLLocationDistance kilometers=[orig distanceFromLocation:dist]/1000;
NSLog(@"距离:",kilometers);

//第二种，手动计算

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#pragma mark - calculate distance 根据2个经纬度计算距离
#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 = 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;
}

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原文地址：http://www.cnblogs.com/lihaibo-Leao/p/5144557.html

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