我的一个 iPhone 应用程序的一部分需要显示一个指向某个位置的指南针。它不是北方 - 它是一个位置（使用纬度和经度）。
我想出了下面的解决方案，但想问你这是否是制作指南针应用程序的最佳方式，是否可以更精确，是否有什么我没有考虑到的？
我制作了这个示例应用程序来向您展示我的代码以及它是如何工作的。样本上传到http://codeinacup.com/pub/POICompass.tgz
这是源代码的中心：
- (void)viewDidLoad
{
[super viewDidLoad];
locationManager = [[CLLocationManager alloc] init];
[locationManager setDelegate:self];
[locationManager setDesiredAccuracy:kCLLocationAccuracyBest];
[locationManager setDistanceFilter:kCLDistanceFilterNone];
[locationManager startUpdatingLocation];
[locationManager startUpdatingHeading];
// Getting the user's location
CLLocation *location = [locationManager location];
CLLocationCoordinate2D user = [location coordinate];
[self calculateAngle:user];
}
- (void)calculateAngle:(CLLocationCoordinate2D)user {
/*
We imagine an x,y-coordinate system with the user as center (Q)
At this point we do not know in which direction the user faces - we assume he faces north.
The decided location (the Eiffel Tower in Paris for this example) is plotted into the coordinate system as L.
A third point is made. This is called P. This must be on one of the axis and be perpendicular with the location (L)
Now we decide if L is in the first (NE), second (SE), third (SW) or fourth (NW) quadrant.
This is done by comparing the user's latitude (x) and longitude (y) with the latitude (x) and longitude (y) of the location.
If P is in the first quadrant, we assume that the user is facing north.
If P is in the second quadrant, we assume that the user is facing east.
If P is in the third quadrant, we assume that the user is facing south.
If P is in the fourth quadrant, we assume that the user is facing west.
Now we can make two vectors: QP (from the user's location (Q) to the point (P)) and QL (from the user's location (Q) to the decided location (L))
When having two vectors we can calculate the angle (V)
Now we know the angle between one of the axis to the decided location (L).
Whenever the iPhone rotates an amount of degrees, we rotate the needle the same amount of degrees in the other direction.
Check ILLUSTRATION.PNG to see the above illustrated.
*/
/*
Q is the location of the user.
L is the decided location (in this example, the Eiffel Tower)
P is a point on the axis which is perpendicular with L.
*/
/*
(x, y) Coordinates
x = latitude
y = longitude
*/
float locLat = 48.512972;
float locLon = 2.174017;
float pLat;
float pLon;
if(locLat > user.latitude && locLon > user.longitude) {
/*
L is in the first quadrant
NE
*/
pLat = user.latitude;
pLon = locLon;
degrees = 0;
}
else if(locLat > user.latitude && locLon < user.longitude) {
/*
L is in the second quadrant
SE
*/
pLat = locLat;
pLon = user.longitude;
degrees = 45;
}
else if(locLat < user.latitude && locLon < user.longitude) {
/*
L is in the third quadrant
SW
*/
pLat = locLat;
pLon = user.latitude;
degrees = 180;
}
else if(locLat < user.latitude && locLon > user.longitude) {
/*
L is in the fourth quadrant
NW
*/
pLat = locLat;
pLon = user.longitude;
degrees = 225;
}
else {
NSLog(@"Failed locating.");
}
// Vector QP (from user's location to point)
float vQPlat = pLat - user.latitude;
float vQPlon = pLon - user.longitude;
// Vector QL (from user's location to decided location)
float vQLlat = locLat - user.latitude;
float vQLlon = locLon - user.longitude;
// Angle (V) between QP and QL
float cosDegrees = (vQPlat * vQLlat + vQPlon * vQLlon) / sqrt((vQPlat * vQPlat + vQPlon*vQPlon) * (vQLlat * vQLlat + vQLlon * vQLlon));
degrees = degrees + acos(cosDegrees);
}
- (void)locationManager:(CLLocationManager *)manager didUpdateToLocation:(CLLocation *)newLocation fromLocation:(CLLocation *)oldLocation {
/*
Everytime the user's location updates we need to call - (void)calculateAngle:user to calculate the degrees
between the user (Q) and the decided location (L)
*/
[self calculateAngle:newLocation.coordinate];
}
- (void)locationManager:(CLLocationManager *)manager didUpdateHeading:(CLHeading *)newHeading {
/*
When the phone is rotated, the needle is rotated the same amount of degrees in the other direction
*/
[needle setTransform:CGAffineTransformMakeRotation((degrees - newHeading.magneticHeading) * M_PI / 180)];
}