CLLocation distanceFromLocation result is differ when I calculate - ios

Question

I have used CLLocation's distanceFromLocation: method to calculate distance from some location.
But it's result is slightly different when I calculate by using "Haversine formula".
- (CLLocationDistance)distanceFromCoordinate:(CLLocationCoordinate2D)fromCoord {
double earthRadius = 6371.0; // Earth's radius in Kilometers
// Get the difference between our two points then convert the difference into radians
double nDLat = RADIANS((fromCoord.latitude - self.coordinate.latitude));
double nDLon = RADIANS((fromCoord.longitude - self.coordinate.longitude));
double fromLat = RADIANS(self.coordinate.latitude);
double toLat = RADIANS(fromCoord.latitude);
double nA = pow ( sin(nDLat/2.0), 2 ) + cos(fromLat) * cos(toLat) * pow ( sin(nDLon/2.0), 2 );
double nC = 2.0 * atan2( sqrt(nA), sqrt( 1 - nA ));
double nD = earthRadius * nC;
return nD * 1000.0;
}
CLLocation * loc = [[CLLocation alloc] initWithLatitude:[location.latitude doubleValue]
longitude:[location.longitude doubleValue]];
CLLocationDistance dist = [userLocation distanceFromLocation:loc];
CLLocationDistance dist2 = [userLocation distanceFromCoordinate:loc.coordinate];
Why two values are different?
Should I init location object with horizontalAccuracy and verticalAccuracy?

Answer 1

Your results are different because you are using different code.
You don't say how different.
Totally different? There's a bug in your code.
Big differences for places close together? Maybe your formula has problems with rounding errors.
Differences that grow as places are further apart? Maybe your definition of distance is different. Should be the closest distance on a path along earth surface.
In general, Earth is not a sphere but a spheroid. Taking that into account is more difficult but gives more precise results.

Answer 2

It is slightly different because Earth's Radius is not exact 6371 km. Use the correct Earth's radius, may be you can get better results.

Answer 3

Use the official WGS84 earth radius:
6 378 137 meter
I remember that ios delivers exactly the same result.
Should I init location object with horizontalAccuracy and
verticalAccuracy?
No, for sure not. That attributes are hints how acurate the position might be.
Distance is calculated by latitude and longitude only.
There are not much formulas:
- haversine formula (a bit slower than law of cosines, otherwise fine)
- law of cosines (problematic on small distances if not using 64 bit precision)
- vicenties which i smore acurate, it uses an elipsoidal earth model.

Answer 4

I also got the same problem... so i used google webservice to calculate distance. use this method, you will get accurate distance
-(void)calculateDistance()
{
//http://maps.googleapis.com/maps/api/directions/json?origin=41.742964,-87.995971& destination=41.811511,-87.967923&mode=driving&sensor=false
NSString *LocationUrl = [NSString stringWithFormat:#"http://maps.googleapis.com/maps/api/directions/json?origin=%#,%#&destination=%#,%#&mode=driving&sensor=false",origin.latitude,origin.longitude,destination.latitude,destination.latitude];
NSLog(#"Location URL:%#",LocationUrl);
NSURL *finalurl = [NSURL URLWithString: LocationUrl];
NSLog(#"Final URL = %#",finalurl);
NSData *data = [NSData dataWithContentsOfURL:finalurl];
NSLog(#"Data:-%#",data);
NSError *error;
NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data
options:kNilOptions error:&error];
// NSLog(#"josn data of location:%#",[json description]);
NSMutableDictionary *routes = [json objectForKey:#"routes"];
NSMutableArray *legs = [routes valueForKey:#"legs"];
NSMutableDictionary *newDistance =[legs valueForKey:#"distance"];
NSMutableArray distanceList =[[newDistance valueForKey:#"text"]objectAtIndex:0];
distance = [[distanceList objectAtIndex:0]floatValue];
NSLog(#"%.1f",distance);
}
Hope it will help you