CLLocation 类别，用于计算带半正弦函数的轴承

Question

我正在尝试为 CLLocation 编写一个类别以将方位返回到另一个 CLLocation。

我认为我的公式做错了（精打细算不是我的强项）。返回的轴承始终处于关闭状态。

我一直在查看这个问题并尝试应用被接受为正确答案的更改及其引用的网页：

Calculating bearing between two CLLocationCoordinate2Ds

http://www.movable-type.co.uk/scripts/latlong.html

感谢您的任何指点。我已经尝试整合来自其他问题的反馈，但我仍然没有得到任何东西。

谢谢

这是我的类别 -

----- CLLocation+Bearing.h

#import <Foundation/Foundation.h>
#import <CoreLocation/CoreLocation.h>


@interface CLLocation (Bearing)

-(double) bearingToLocation:(CLLocation *) destinationLocation;
-(NSString *) compassOrdinalToLocation:(CLLocation *) nwEndPoint;

@end

---------CLLocation+Bearing.m

#import "CLLocation+Bearing.h"

double DegreesToRadians(double degrees) {return degrees * M_PI / 180;};
double RadiansToDegrees(double radians) {return radians * 180/M_PI;};


@implementation CLLocation (Bearing)

-(double) bearingToLocation:(CLLocation *) destinationLocation {

 double lat1 = DegreesToRadians(self.coordinate.latitude);
 double lon1 = DegreesToRadians(self.coordinate.longitude);

 double lat2 = DegreesToRadians(destinationLocation.coordinate.latitude);
 double lon2 = DegreesToRadians(destinationLocation.coordinate.longitude);

 double dLon = lon2 - lon1;

 double y = sin(dLon) * cos(lat2);
 double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
 double radiansBearing = atan2(y, x);

 return RadiansToDegrees(radiansBearing);
}

Answer 1

在 Swift 5 中实现了这一点。重点是准确性，而不是速度，但它实时运行 np。

let earthRadius: Double = 6372456.7
let degToRad: Double = .pi / 180.0
let radToDeg: Double = 180.0 / .pi

func calcOffset(_ coord0: CLLocationCoordinate2D,
                _ coord1: CLLocationCoordinate2D) -> (Double, Double) {
    let lat0: Double = coord0.latitude * degToRad
    let lat1: Double = coord1.latitude * degToRad
    let lon0: Double = coord0.longitude * degToRad
    let lon1: Double = coord1.longitude * degToRad
    let dLat: Double = lat1 - lat0
    let dLon: Double = lon1 - lon0
    let y: Double = cos(lat1) * sin(dLon)
    let x: Double = cos(lat0) * sin(lat1) - sin(lat0) * cos(lat1) * cos(dLon)
    let t: Double = atan2(y, x)
    let bearing: Double = t * radToDeg

    let a: Double = pow(sin(dLat * 0.5), 2.0) + cos(lat0) * cos(lat1) * pow(sin(dLon * 0.5), 2.0)
    let c: Double = 2.0 * atan2(sqrt(a), sqrt(1.0 - a));
    let distance: Double = c * earthRadius

    return (distance, bearing)
}

func translateCoord(_ coord: CLLocationCoordinate2D,
                    _ distance: Double,
                    _ bearing: Double) -> CLLocationCoordinate2D {
    let d: Double = distance / earthRadius
    let t: Double = bearing * degToRad

    let lat0: Double = coord.latitude * degToRad
    let lon0: Double = coord.longitude * degToRad
    let lat1: Double = asin(sin(lat0) * cos(d) + cos(lat0) * sin(d) * cos(t))
    let lon1: Double = lon0 + atan2(sin(t) * sin(d) * cos(lat0), cos(d) - sin(lat0) * sin(lat1))

    let lat: Double = lat1 * radToDeg
    let lon: Double = lon1 * radToDeg

    let c: CLLocationCoordinate2D = CLLocationCoordinate2D(latitude: lat,
                                                           longitude: lon)
    return c
}

我发现与 CLLocation 的 distance 方法相比，Haversine 确定了距离，但没有提供可与 CL 一起使用的轴承。所以我没有将它用于轴承。这给出了我从我尝试过的所有数学中遇到的最准确的测量结果。 translateCoord 方法还将根据原点、以米为单位的距离和以度为单位的方位精确绘制一个新点。

Answer 2

值得一提的是，如果您使用的是 Google 地图 GMSMapView，则可以使用 GMSGeometryHeading method 的开箱即用解决方案：

GMSGeometryHeading(from: CLLocationCoordinate2D, to: CLLocationCoordinate2D)

在 from of 处返回初始航向（北顺时针度数） 到的最短路径。

Answer 3

这是另一个 CLLocation 扩展，可用于 Swift 3 和 Swift 4

public extension CLLocation {

    func degreesToRadians(degrees: Double) -> Double {
        return degrees * .pi / 180.0
    }

    func radiansToDegrees(radians: Double) -> Double {
        return radians * 180.0 / .pi
    }

    func getBearingBetweenTwoPoints(point1: CLLocation, point2: CLLocation) -> Double {
        let lat1 = degreesToRadians(degrees: point1.coordinate.latitude)
        let lon1 = degreesToRadians(degrees: point1.coordinate.longitude)

        let lat2 = degreesToRadians(degrees: point2.coordinate.latitude)
        let lon2 = degreesToRadians(degrees: point2.coordinate.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(radians: radiansBearing)
    }

}

Answer 4

使用 Swift 3 和 4

尝试了这么多版本，这个终于给出了正确的值！

extension CLLocation {


    func getRadiansFrom(degrees: Double ) -> Double {

        return degrees * .pi / 180

    }

    func getDegreesFrom(radians: Double) -> Double {

        return radians * 180 / .pi

    }


    func bearingRadianTo(location: CLLocation) -> Double {

        let lat1 = self.getRadiansFrom(degrees: self.coordinate.latitude)
        let lon1 = self.getRadiansFrom(degrees: self.coordinate.longitude)

        let lat2 = self.getRadiansFrom(degrees: location.coordinate.latitude)
        let lon2 = self.getRadiansFrom(degrees: location.coordinate.longitude)

        let dLon = lon2 - lon1

        let y = sin(dLon) * cos(lat2)
        let x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon)

        var radiansBearing = atan2(y, x)

        if radiansBearing < 0.0 {

            radiansBearing += 2 * .pi

        }


        return radiansBearing
    }

    func bearingDegreesTo(location: CLLocation) -> Double {

        return self.getDegreesFrom(radians: self.bearingRadianTo(location: location))

    }


}

用法：

let degrees = location1.bearingDegreesTo(location: location2)

Answer 5

这是对开头的 Category 在 Swift 中的一个移植：

import Foundation
import CoreLocation
public extension CLLocation{

    func DegreesToRadians(_ degrees: Double ) -> Double {
        return degrees * M_PI / 180
    }

    func RadiansToDegrees(_ radians: Double) -> Double {
        return radians * 180 / M_PI
    }


    func bearingToLocationRadian(_ destinationLocation:CLLocation) -> Double {

        let lat1 = DegreesToRadians(self.coordinate.latitude)
        let lon1 = DegreesToRadians(self.coordinate.longitude)

        let lat2 = DegreesToRadians(destinationLocation.coordinate.latitude);
        let lon2 = DegreesToRadians(destinationLocation.coordinate.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 radiansBearing
    }

    func bearingToLocationDegrees(destinationLocation:CLLocation) -> Double{
        return   RadiansToDegrees(bearingToLocationRadian(destinationLocation))
    }
}

Answer 6

我在 Swift 中使用余弦定律。它比 Haversine 运行得更快，其结果非常相似。远距离变化 1 米。

我为什么使用余弦定理：

• 跑得快（因为没有sqrt函数）
• 足够精确，除非你会做一些天文学
• 非常适合后台任务

func calculateDistance(from: CLLocationCoordinate2D, to: CLLocationCoordinate2D) -> Double {

    let π = M_PI
    let degToRad: Double = π/180
    let earthRadius: Double = 6372797.560856

    // Law of Cosines formula
    // d = r . arc cos (sin ?A sin ?B + cos ?A cos ?B cos(?B - ?A) )

    let ?A = from.latitude * degToRad
    let ?B = to.latitude * degToRad
    let ?A = from.longitude * degToRad
    let ?B = to.longitude * degToRad

    let angularDistance = acos(sin(?A) * sin(?B) + cos(?A) * cos(?B) * cos(?B - ?A) )
    let distance = earthRadius * angularDistance

    return distance

}

Answer 7

这是另一个实现

public func bearingBetweenTwoPoints(#lat1 : Double, #lon1 : Double, #lat2 : Double, #lon2: Double) -> Double {

func DegreesToRadians (value:Double) -> Double {
    return value * M_PI / 180.0
}

func RadiansToDegrees (value:Double) -> Double {
    return value * 180.0 / M_PI
}

let y = sin(lon2-lon1) * cos(lat2)
let x = (cos(lat1) * sin(lat2)) - (sin(lat1) * cos(lat2) * cos(lat2-lon1))

let degrees = RadiansToDegrees(atan2(y,x))

let ret = (degrees + 360) % 360

return ret;

}

Answer 8

您的代码对我来说似乎很好。算计没有错。你没有指定你的结果有多远，但你可以尝试调整你的弧度/度数转换器：

double DegreesToRadians(double degrees) {return degrees * M_PI / 180.0;};
double RadiansToDegrees(double radians) {return radians * 180.0/M_PI;};

如果您得到负方位角，请将2*M_PI 添加到弧度轴承的最终结果中（如果在转换为度数后添加，则添加 360）。 atan2 返回-M_PI 到M_PI（-180 到 180 度）范围内的结果，因此您可能希望使用类似于以下代码的内容将其转换为罗盘方位

if(radiansBearing < 0.0)
    radiansBearing += 2*M_PI;