If your starting point is (0,0), and your new point is r units away at an angle of θ, you can find the coordinates of that point using the equations x = r cosθ and y = r sinθ.
The task of calculating coordinates of points on offset should start from the coordinates designate the distance between the known points A and B, the following formula: LAB = √________ Δx2 + Δy2, where Δx = XB – XA, Δy = YB – YA.
Unfortunately, there's no such function provided in the API, so you'll have to write your own.
This site gives several calculations involving latitude/longitude and sample JavaScript code. Specifically, the section titled "Destination point given distance and bearing from start point" shows how to calculate what you're asking.
The JavaScript code is at the bottom of that page and here's one possible way to convert it to Objective-C:
- (double)radiansFromDegrees:(double)degrees
{
return degrees * (M_PI/180.0);
}
- (double)degreesFromRadians:(double)radians
{
return radians * (180.0/M_PI);
}
- (CLLocationCoordinate2D)coordinateFromCoord:
(CLLocationCoordinate2D)fromCoord
atDistanceKm:(double)distanceKm
atBearingDegrees:(double)bearingDegrees
{
double distanceRadians = distanceKm / 6371.0;
//6,371 = Earth's radius in km
double bearingRadians = [self radiansFromDegrees:bearingDegrees];
double fromLatRadians = [self radiansFromDegrees:fromCoord.latitude];
double fromLonRadians = [self radiansFromDegrees:fromCoord.longitude];
double toLatRadians = asin( sin(fromLatRadians) * cos(distanceRadians)
+ cos(fromLatRadians) * sin(distanceRadians) * cos(bearingRadians) );
double toLonRadians = fromLonRadians + atan2(sin(bearingRadians)
* sin(distanceRadians) * cos(fromLatRadians), cos(distanceRadians)
- sin(fromLatRadians) * sin(toLatRadians));
// adjust toLonRadians to be in the range -180 to +180...
toLonRadians = fmod((toLonRadians + 3*M_PI), (2*M_PI)) - M_PI;
CLLocationCoordinate2D result;
result.latitude = [self degreesFromRadians:toLatRadians];
result.longitude = [self degreesFromRadians:toLonRadians];
return result;
}
In the JS code, it contains this link which shows a more accurate calculation for distances greater than 1/4 of the Earth's circumference.
Also note the above code accepts distance in km so be sure to divide meters by 1000.0 before passing.
I found one way of doing it, had to dig to find the correct structs and functions. I ended up not using degrees but meters for lat and long instead.
Here's how I did it:
-(CLLocationCoordinate2D)translateCoord:(CLLocationCoordinate2D)coord MetersLat:(double)metersLat MetersLong:(double)metersLong{
CLLocationCoordinate2D tempCoord;
MKCoordinateRegion tempRegion = MKCoordinateRegionMakeWithDistance(coord, metersLat, metersLong);
MKCoordinateSpan tempSpan = tempRegion.span;
tempCoord.latitude = coord.latitude + tempSpan.latitudeDelta;
tempCoord.longitude = coord.longitude + tempSpan.longitudeDelta;
return tempCoord;
}
And of course, if I really need to use degrees in the future, it's pretty easy (I think...) to do some changes to above to get it to work like I actually asked for.
Using an MKCoordinateRegion
has some issues—the returned region can be adjusted to fit since the two deltas may not exactly map to the projection at that latitude, if you want zero delta for one of the axes you are out of luck, etc.
This function uses MKMapPoint
to perform coordinate translations which allows you to move points around in the map projection's coordinate space and then extract a coordinate from that.
CLLocationCoordinate2D MKCoordinateOffsetFromCoordinate(CLLocationCoordinate2D coordinate, CLLocationDistance offsetLatMeters, CLLocationDistance offsetLongMeters) {
MKMapPoint offsetPoint = MKMapPointForCoordinate(coordinate);
CLLocationDistance metersPerPoint = MKMetersPerMapPointAtLatitude(coordinate.latitude);
double latPoints = offsetLatMeters / metersPerPoint;
offsetPoint.y += latPoints;
double longPoints = offsetLongMeters / metersPerPoint;
offsetPoint.x += longPoints;
CLLocationCoordinate2D offsetCoordinate = MKCoordinateForMapPoint(offsetPoint);
return offsetCoordinate;
}
Nicsoft's answer is fantastic and exactly what I needed. I've created a Swift 3-y version which is a little more concise and can be called directly on a CLLocationCoordinate2D
instance:
public extension CLLocationCoordinate2D {
public func transform(using latitudinalMeters: CLLocationDistance, longitudinalMeters: CLLocationDistance) -> CLLocationCoordinate2D {
let region = MKCoordinateRegionMakeWithDistance(self, latitudinalMeters, longitudinalMeters)
return CLLocationCoordinate2D(latitude: latitude + region.span.latitudeDelta, longitude: longitude + region.span.longitudeDelta)
}
}
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