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计算两个位置坐标之间的轴承

core-location coordinates objective-c ios

38

James J · 技术社区 · 15 年前

非常“简单”的问题:给定两个坐标系,如何得到从第一个坐标系到第二个坐标系的方向角(弧度)?我对此做了很多研究,包括一般问题和Objective-C/Cocoa-Touch/iOS。

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

    return atan2(sin(fLng-tLng)*cos(tLat), cos(fLat)*sin(tLat)-sin(fLat)*cos(tLat)*cos(fLng-tLng));         
}

从50.405018到8.437500

至51.339802、12.403340

我的方法返回:5.918441弧度

应为1.18660576弧度

(见 http://www.movable-type.co.uk/scripts/latlong.html 和 http://www.movable-type.co.uk/scripts/latlong-map.html?lat1=50.405018&long1=8.437500&lat2=51.339802&long2=12.403340 )

有什么想法吗?我的代码有问题吗?也许我误解了数学函数的工作原理?

4 回复 | 直到 8 年前

1

31

Oren Trutner 15 年前

你的数学是正确的,但有以下例外:

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。
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。

2

53

heinst 10 年前

#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;
    }
}

3

10

Valeriy Van 8 年前

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)
    }
}

4

-1

Trianto Kontri 13 年前

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;
    }                                                                 
}