需求
设计
因地图暴露的方法中只有设置地图相对于正北的方向角的方法。因此，需要实现“根据两点经纬度坐标计算指南针方位角”的算法，这样在每次切换路段时，调用算法计算新路段指南针方位角，然后设置地图相对于正北的方向角即可实现需求。
示意图如下:
代码实现
新建CLLocation 分类方法
#import <CoreLocation/CoreLocation.h>
+ (doubleca_getCompassAngleFromCoor1:(CLLocationCoordinate2Dcoor1 coor2:(CLLocationCoordinate2Dcoor2 {
double long1 = coor1.longitude;
double lat1 = coor1.latitude;
double long2 = coor2.longitude;
double lat2 = coor2.latitude;
double φ1 = [CLLocation toRadius:lat1];
double φ2 = [CLLocation toRadius:lat2];
double Δλ = [CLLocation toRadius:(long2 - long1];
double x = cos(φ1 * sin(φ2 - sin(φ1 * cos(φ2 * cos(Δλ;
double y = sin(Δλ * cos(φ2;
double θ = atan2(y, x;
double bearing = [CLLocation toDegrees:θ];
return bearing;
}
+ (doubletoDegrees:(doubleradius {
return radius * 180.0 / M_PI;
}
+ (doubletoRadius:(doubledegree {
return degree * M_PI / 180.0;
}
调用示例
double bearing = [CLLocation ca_getCompassAngleFromCoor1:(CLLocationCoordinate2DMake(20, 20 coor2:(CLLocationCoordinate2DMake(20, 140];
NSLog(@"bearing:%.2f", bearing;
//设置地图方位角...
结论
经测试，上面算法可以满足需求，且效果正确！
https://www.movable-type.co.uk/scripts/latlong.html