'CLLocation Category for Calculating Bearing w/ Haversine function
I'm trying to write a category for CLLocation to return the bearing to another CLLocation.
I believe I'm doing something wrong with the formula (calculous is not my strong suit). The returned bearing is always off.
I've been looking at this question and tried applying the changes that were accepted as a correct answer and the webpage it references:
Calculating bearing between two CLLocationCoordinate2Ds
http://www.movable-type.co.uk/scripts/latlong.html
Thanks for any pointers. I've tried incorporating the feedback from that other question and I'm still just not getting something.
Thanks
Here's my category -
----- CLLocation+Bearing.h
#import <Foundation/Foundation.h>
#import <CoreLocation/CoreLocation.h>
@interface CLLocation (Bearing)
-(double) bearingToLocation:(CLLocation *) destinationLocation;
-(NSString *) compassOrdinalToLocation:(CLLocation *) nwEndPoint;
@end
---------CLLocation+Bearing.m
#import "CLLocation+Bearing.h"
double DegreesToRadians(double degrees) {return degrees * M_PI / 180;};
double RadiansToDegrees(double radians) {return radians * 180/M_PI;};
@implementation CLLocation (Bearing)
-(double) bearingToLocation:(CLLocation *) destinationLocation {
double lat1 = DegreesToRadians(self.coordinate.latitude);
double lon1 = DegreesToRadians(self.coordinate.longitude);
double lat2 = DegreesToRadians(destinationLocation.coordinate.latitude);
double lon2 = DegreesToRadians(destinationLocation.coordinate.longitude);
double dLon = lon2 - lon1;
double y = sin(dLon) * cos(lat2);
double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
double radiansBearing = atan2(y, x);
return RadiansToDegrees(radiansBearing);
}
Solution 1:[1]
This is a porting in Swift of the Category at the beginning:
import Foundation
import CoreLocation
public extension CLLocation{
func DegreesToRadians(_ degrees: Double ) -> Double {
return degrees * M_PI / 180
}
func RadiansToDegrees(_ radians: Double) -> Double {
return radians * 180 / M_PI
}
func bearingToLocationRadian(_ destinationLocation:CLLocation) -> Double {
let lat1 = DegreesToRadians(self.coordinate.latitude)
let lon1 = DegreesToRadians(self.coordinate.longitude)
let lat2 = DegreesToRadians(destinationLocation.coordinate.latitude);
let lon2 = DegreesToRadians(destinationLocation.coordinate.longitude);
let dLon = lon2 - lon1
let y = sin(dLon) * cos(lat2);
let x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
let radiansBearing = atan2(y, x)
return radiansBearing
}
func bearingToLocationDegrees(destinationLocation:CLLocation) -> Double{
return RadiansToDegrees(bearingToLocationRadian(destinationLocation))
}
}
Solution 2:[2]
Here is another implementation
public func bearingBetweenTwoPoints(#lat1 : Double, #lon1 : Double, #lat2 : Double, #lon2: Double) -> Double {
func DegreesToRadians (value:Double) -> Double {
return value * M_PI / 180.0
}
func RadiansToDegrees (value:Double) -> Double {
return value * 180.0 / M_PI
}
let y = sin(lon2-lon1) * cos(lat2)
let x = (cos(lat1) * sin(lat2)) - (sin(lat1) * cos(lat2) * cos(lat2-lon1))
let degrees = RadiansToDegrees(atan2(y,x))
let ret = (degrees + 360) % 360
return ret;
}
Solution 3:[3]
Working Swift 3 and 4
Tried so many versions and this one finally gives correct values!
extension CLLocation {
func getRadiansFrom(degrees: Double ) -> Double {
return degrees * .pi / 180
}
func getDegreesFrom(radians: Double) -> Double {
return radians * 180 / .pi
}
func bearingRadianTo(location: CLLocation) -> Double {
let lat1 = self.getRadiansFrom(degrees: self.coordinate.latitude)
let lon1 = self.getRadiansFrom(degrees: self.coordinate.longitude)
let lat2 = self.getRadiansFrom(degrees: location.coordinate.latitude)
let lon2 = self.getRadiansFrom(degrees: location.coordinate.longitude)
let dLon = lon2 - lon1
let y = sin(dLon) * cos(lat2)
let x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon)
var radiansBearing = atan2(y, x)
if radiansBearing < 0.0 {
radiansBearing += 2 * .pi
}
return radiansBearing
}
func bearingDegreesTo(location: CLLocation) -> Double {
return self.getDegreesFrom(radians: self.bearingRadianTo(location: location))
}
}
Usage:
let degrees = location1.bearingDegreesTo(location: location2)
Solution 4:[4]
This is an another CLLocation extension can be used in Swift 3 and Swift 4
public extension CLLocation {
func degreesToRadians(degrees: Double) -> Double {
return degrees * .pi / 180.0
}
func radiansToDegrees(radians: Double) -> Double {
return radians * 180.0 / .pi
}
func getBearingBetweenTwoPoints(point1: CLLocation, point2: CLLocation) -> Double {
let lat1 = degreesToRadians(degrees: point1.coordinate.latitude)
let lon1 = degreesToRadians(degrees: point1.coordinate.longitude)
let lat2 = degreesToRadians(degrees: point2.coordinate.latitude)
let lon2 = degreesToRadians(degrees: point2.coordinate.longitude)
let dLon = lon2 - lon1
let y = sin(dLon) * cos(lat2)
let x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon)
let radiansBearing = atan2(y, x)
return radiansToDegrees(radians: radiansBearing)
}
}
Solution 5:[5]
I use the Law of Cosines in Swift. It runs faster than Haversine and its result is extremely similar. Variation of 1 metre on huge distances.
Why do I use the Law of Cosines:
- Run fast (because there is no sqrt functions)
- Precise enough unless you do some astronomy
- Perfect for a background task
func calculateDistance(from: CLLocationCoordinate2D, to: CLLocationCoordinate2D) -> Double {
let ? = M_PI
let degToRad: Double = ?/180
let earthRadius: Double = 6372797.560856
// Law of Cosines formula
// d = r . arc cos (sin ?A sin ?B + cos ?A cos ?B cos(?B - ?A) )
let ?A = from.latitude * degToRad
let ?B = to.latitude * degToRad
let ?A = from.longitude * degToRad
let ?B = to.longitude * degToRad
let angularDistance = acos(sin(?A) * sin(?B) + cos(?A) * cos(?B) * cos(?B - ?A) )
let distance = earthRadius * angularDistance
return distance
}
Solution 6:[6]
Worth mentioning that if you are using Google map GMSMapView, there's an out-of-the-box solution using the GMSGeometryHeading method:
GMSGeometryHeading(from: CLLocationCoordinate2D, to: CLLocationCoordinate2D)
Returns the initial heading (degrees clockwise of North) at from of the shortest path to to.
Solution 7:[7]
Implemented this in Swift 5. Focus is on accuracy, not speed, but it runs in real time np.
let earthRadius: Double = 6372456.7
let degToRad: Double = .pi / 180.0
let radToDeg: Double = 180.0 / .pi
func calcOffset(_ coord0: CLLocationCoordinate2D,
_ coord1: CLLocationCoordinate2D) -> (Double, Double) {
let lat0: Double = coord0.latitude * degToRad
let lat1: Double = coord1.latitude * degToRad
let lon0: Double = coord0.longitude * degToRad
let lon1: Double = coord1.longitude * degToRad
let dLat: Double = lat1 - lat0
let dLon: Double = lon1 - lon0
let y: Double = cos(lat1) * sin(dLon)
let x: Double = cos(lat0) * sin(lat1) - sin(lat0) * cos(lat1) * cos(dLon)
let t: Double = atan2(y, x)
let bearing: Double = t * radToDeg
let a: Double = pow(sin(dLat * 0.5), 2.0) + cos(lat0) * cos(lat1) * pow(sin(dLon * 0.5), 2.0)
let c: Double = 2.0 * atan2(sqrt(a), sqrt(1.0 - a));
let distance: Double = c * earthRadius
return (distance, bearing)
}
func translateCoord(_ coord: CLLocationCoordinate2D,
_ distance: Double,
_ bearing: Double) -> CLLocationCoordinate2D {
let d: Double = distance / earthRadius
let t: Double = bearing * degToRad
let lat0: Double = coord.latitude * degToRad
let lon0: Double = coord.longitude * degToRad
let lat1: Double = asin(sin(lat0) * cos(d) + cos(lat0) * sin(d) * cos(t))
let lon1: Double = lon0 + atan2(sin(t) * sin(d) * cos(lat0), cos(d) - sin(lat0) * sin(lat1))
let lat: Double = lat1 * radToDeg
let lon: Double = lon1 * radToDeg
let c: CLLocationCoordinate2D = CLLocationCoordinate2D(latitude: lat,
longitude: lon)
return c
}
I found that Haversine nailed the distance versus CLLocation's distance method, but didn't provide a bearing ready-to-use with CL. So I'm not using it for the bearing. This gives the most accurate measurement I've encountered from all the math I've tried. The translateCoord method will also precisely plot a new point given an origin, distance in meters, and a bearing in degrees.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | TofuBeer |
| Solution 2 | Jeef |
| Solution 3 | David Seek |
| Solution 4 | |
| Solution 5 | Romain |
| Solution 6 | |
| Solution 7 | BTR |
