计算两个位置坐标之间的轴承

你的数学是正确的,但有以下例外:

1. 一定要转换 , 特隆 在应用任何 罪恶() cos() 对他们来说。除以180.0,再乘以π。

2. 输入之间的增量 和 fLng公司 作为 tLng法兰 ,而不是相反。请注意,此差异在表达式中出现两次。

有了这些变化,我得到了1.18660677830947弧度与双精度数学和问题中的值。

#define degreesToRadians(x) (M_PI * x / 180.0)
#define radiansToDegrees(x) (x * 180.0 / M_PI)

- (float)getHeadingForDirectionFromCoordinate:(CLLocationCoordinate2D)fromLoc toCoordinate:(CLLocationCoordinate2D)toLoc
{
    float fLat = degreesToRadians(fromLoc.latitude);
    float fLng = degreesToRadians(fromLoc.longitude);
    float tLat = degreesToRadians(toLoc.latitude);
    float tLng = degreesToRadians(toLoc.longitude);

    float degree = radiansToDegrees(atan2(sin(tLng-fLng)*cos(tLat), cos(fLat)*sin(tLat)-sin(fLat)*cos(tLat)*cos(tLng-fLng)));

    if (degree >= 0) {
        return degree;
    } else {
        return 360+degree;
    }
}

extension CLLocationCoordinate2D {
    func bearing(to point: CLLocationCoordinate2D) -> Double {
        func degreesToRadians(_ degrees: Double) -> Double { return degrees * Double.pi / 180.0 }
        func radiansToDegrees(_ radians: Double) -> Double { return radians * 180.0 / Double.pi }

        let lat1 = degreesToRadians(latitude)
        let lon1 = degreesToRadians(longitude)

        let lat2 = degreesToRadians(point.latitude);
        let lon2 = degreesToRadians(point.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(radiansBearing)
    }
}

你可以用我的密码。。在我的项目中,微控制器使用GPS获取数据。

#define d2r ((22/7.0)/180.0)
#define r2d (180.0/(22/7.0))

double get_heading1(double lat1, double long1, double lat2, double long2)  
{
    double diff_lat, diff_long;
    double degree;

    diff_long =(double) (((long2*1000000)-(long1*1000000))/1000000) * d2r;
    diff_lat = (double) (((lat2*1000000)-(lat1*1000000))/1000000) * d2r;     

    degree = r2d     (atan2(sin(diff_long)*cos(d2r*lat2),cos(d2r*lat1)*sin(d2r*lat2)-sin(d2r*lat1)*cos(d2r*lat2)    *cos(diff_long)));

    if (degree >= 0) {
        return degree;
    } else {
        return 360+degree;
    }                                                                 
}