将CLLocation移动x米

时间:2011-09-02 00:26:22

标签: iphone cocoa-touch core-location latitude-longitude

我定义了一个CLLocation,我想把这个点移到东边x米和南边y米。我怎么能实现这个目标呢?

11 个答案:

答案 0 :(得分:27)

转换为Swift,取自this answer

func locationWithBearing(bearing:Double, distanceMeters:Double, origin:CLLocationCoordinate2D) -> CLLocationCoordinate2D {
    let distRadians = distanceMeters / (6372797.6) // earth radius in meters

    let lat1 = origin.latitude * M_PI / 180
    let lon1 = origin.longitude * M_PI / 180

    let lat2 = asin(sin(lat1) * cos(distRadians) + cos(lat1) * sin(distRadians) * cos(bearing))
    let lon2 = lon1 + atan2(sin(bearing) * sin(distRadians) * cos(lat1), cos(distRadians) - sin(lat1) * sin(lat2))

    return CLLocationCoordinate2D(latitude: lat2 * 180 / M_PI, longitude: lon2 * 180 / M_PI)
}
     

Morgan Chen写道:

     
    

此方法中的所有数学运算均以弧度为单位。在开始时     方法,lon1和lat1为此目的转换为弧度     好。轴承也是弧度。请记住这个方法需要考虑     考虑到地球的曲率,你并不需要这样做     对于小距离。

  

答案 1 :(得分:16)

很棒的帖子,这里是喜欢复制/粘贴的人的Obj-C包装器:

- (CLLocationCoordinate2D) locationWithBearing:(float)bearing distance:(float)distanceMeters fromLocation:(CLLocationCoordinate2D)origin {
    CLLocationCoordinate2D target;
    const double distRadians = distanceMeters / (6372797.6); // earth radius in meters

    float lat1 = origin.latitude * M_PI / 180;
    float lon1 = origin.longitude * M_PI / 180;

    float lat2 = asin( sin(lat1) * cos(distRadians) + cos(lat1) * sin(distRadians) * cos(bearing));
    float lon2 = lon1 + atan2( sin(bearing) * sin(distRadians) * cos(lat1),
                     cos(distRadians) - sin(lat1) * sin(lat2) );

    target.latitude = lat2 * 180 / M_PI;
    target.longitude = lon2 * 180 / M_PI; // no need to normalize a heading in degrees to be within -179.999999° to 180.00000°

    return target;
}

答案 2 :(得分:15)

改进了彼得斯答案的快速解决方案。只有在校正时,轴承应该是弧度。

 func locationWithBearing(bearing:Double, distanceMeters:Double, origin:CLLocationCoordinate2D) -> CLLocationCoordinate2D {
    let distRadians = distanceMeters / (6372797.6)

    var rbearing = bearing * M_PI / 180.0

    let lat1 = origin.latitude * M_PI / 180
    let lon1 = origin.longitude * M_PI / 180

    let lat2 = asin(sin(lat1) * cos(distRadians) + cos(lat1) * sin(distRadians) * cos(rbearing))
    let lon2 = lon1 + atan2(sin(rbearing) * sin(distRadians) * cos(lat1), cos(distRadians) - sin(lat1) * sin(lat2))

    return CLLocationCoordinate2D(latitude: lat2 * 180 / M_PI, longitude: lon2 * 180 / M_PI)
}

答案 3 :(得分:6)

有一个C函数接近你所要求的但它需要一个方位和距离。它位于github上的UtilitiesGeo类中。您可以将CLLocation中的纬度和经度传递给它,然后从返回的结果lat2和lon2创建一个新的CLLocation:

/*-------------------------------------------------------------------------
* Given a starting lat/lon point on earth, distance (in meters)
* and bearing, calculates destination coordinates lat2/lon2.
*
* all params in degrees
*-------------------------------------------------------------------------*/
void destCoordsInDegrees(double lat1, double lon1,
                         double distanceMeters, double bearing,
                         double* lat2, double* lon2);

如果您无法使用它,请查看从herehere派生的算法,也许您可​​以对其进行修改,或者这些网站可能更接近您的需求。

答案 4 :(得分:4)

对@CocoaChris的轻微调整回答:现在是CLLocation上的一个类别,并使用内置单元。

#import <CoreLocation/CoreLocation.h>


@interface CLLocation (Movement)

- (CLLocation *)locationByMovingDistance:(double)distanceMeters withBearing:(CLLocationDirection)bearingDegrees;

@end


@implementation CLLocation (Movement)

- (CLLocation *)locationByMovingDistance:(double)distanceMeters withBearing:(CLLocationDirection)bearingDegrees
{
    const double distanceRadians = distanceMeters / (6372797.6); // earth radius in meters
    const double bearingRadians = bearingDegrees * M_PI / 180;

    float lat1 = self.coordinate.latitude * M_PI / 180;
    float lon1 = self.coordinate.longitude * M_PI / 180;

    float lat2 = asin(sin(lat1) * cos(distanceRadians) + cos(lat1) * sin(distanceRadians) * cos(bearingRadians));
    float lon2 = lon1 + atan2(sin(bearingRadians) * sin(distanceRadians) * cos(lat1),
                              cos(distanceRadians) - sin(lat1) * sin(lat2) );

    return [[CLLocation alloc] initWithLatitude:lat2 * 180 / M_PI
                                      longitude:lon2 * 180 / M_PI];
}

@end

答案 5 :(得分:3)

使用Measurement结构进行Swift实现来进行度和弧度之间的转换。

class GPSLocation {

public class func degreesToRadians(degrees: Double) -> Double {
        return Measurement(value: degrees, unit: UnitAngle.degrees).converted(to: .radians).value
    }

    public class func radiansToDegrees(radians: Double) -> Double {
        return Measurement(value: radians, unit: UnitAngle.radians).converted(to: .degrees).value
    }

    public class func location(location: CLLocation, byMovingDistance distance: Double, withBearing bearingDegrees:CLLocationDirection) -> CLLocation {
        let distanceRadians: Double = distance / 6372797.6
        let bearingRadians: Double = GPSLocation.degreesToRadians(degrees: bearingDegrees)

        let lat1 = GPSLocation.degreesToRadians(degrees: location.coordinate.latitude)
        let lon1 = GPSLocation.degreesToRadians(degrees: location.coordinate.longitude)

        let lat2 = GPSLocation.radiansToDegrees(radians:asin(sin(lat1) * cos(distanceRadians) + cos(lat1) * sin(distanceRadians) * cos(bearingRadians)))
        let lon2 = GPSLocation.radiansToDegrees(radians:lon1 + atan2(sin(bearingRadians) * sin(distanceRadians * cos(lat1)), cos(distanceRadians) - sin(lat1) * sin(lat2)))

        return CLLocation(latitude: lat2, longitude: lon2)
    }

}

答案 6 :(得分:3)

更简单的解决方案是使用MKMapPoints。

使用以下方法转换原始坐标以及MKMapPoints所需的任何偏移距离:

let coordinatesInMapPoints = MKMapPointForCoordinate(CLLocationCoordinate2D)
let distancesInMapPoints = yourDistanceInMeters * MKMapPointsPerMeterAtLatitude(CLLocationDegrees) // Do this for both x and y directions if needed.

然后通过简单地将偏移距离添加到原始坐标来创建一个新的MKMapPoint:

let newCoordinatesInMapPoints = MKMapPointMake(coordinatesInMapPoints.x + distancesInMapPoints, coordinatesInMapPoints.y)

最后,将新坐标从MKMapPoint转换回CLLocationCoordinate2D:

let newCoordinate = MKCoordinateForMapPoint(newCoordinatesInMapPoints)

无需复杂的转换计算。

答案 7 :(得分:1)

将Swift 4.2作为CGPoint扩展

源自Peter O.的解决方案

FloatingPoint扩展名:感谢https://stackoverflow.com/a/29179878/2500428

extension FloatingPoint
{
    var degreesToRadians: Self { return self * .pi / 180 }
    var radiansToDegrees: Self { return self * 180 / .pi }
}

extension CGPoint
{
    // NOTE: bearing is in radians
    func locationWithBearing(bearing: Double, distanceMeters: Double) -> CGPoint
    {
        let distRadians = distanceMeters / (6372797.6) // earth radius in meters

        let origLat = Double(self.y.degreesToRadians)
        let origLon = Double(self.x.degreesToRadians)

        let newLat = asin(sin(origLat) * cos(distRadians) + cos(origLat) * sin(distRadians) * cos(bearing))
        let newLon = origLon + atan2(sin(bearing) * sin(distRadians) * cos(origLat), cos(distRadians) - sin(origLat) * sin(newLat))

        return CGPoint(x: newLon.radiansToDegrees, y: newLat.radiansToDegrees)
    }
}

用法:

let loc = CGPoint(x: lon, y: lat)
let newLoc = loc.locationWithBearing(bearing: 90.degreesToRadians, distanceMeters: 500.0)

答案 8 :(得分:0)

快捷键4

extension CLLocationCoordinate2D {

    /// Get coordinate moved from current to `distanceMeters` meters with azimuth `azimuth` [0, Double.pi)
    ///
    /// - Parameters:
    ///   - distanceMeters: the distance in meters
    ///   - azimuth: the azimuth (bearing)
    /// - Returns: new coordinate
    func shift(byDistance distanceMeters: Double, azimuth: Double) -> CLLocationCoordinate2D {
        let bearing = azimuth
        let origin = self
        let distRadians = distanceMeters / (6372797.6) // earth radius in meters

        let lat1 = origin.latitude * Double.pi / 180
        let lon1 = origin.longitude * Double.pi / 180

        let lat2 = asin(sin(lat1) * cos(distRadians) + cos(lat1) * sin(distRadians) * cos(bearing))
        let lon2 = lon1 + atan2(sin(bearing) * sin(distRadians) * cos(lat1), cos(distRadians) - sin(lat1) * sin(lat2))
        return CLLocationCoordinate2D(latitude: lat2 * 180 / Double.pi, longitude: lon2 * 180 / Double.pi)
    }
}

用法

    let point: CLLocationCoordinate2D!
    let north100 = point.shift(byDistance: 100, azimuth: 0) // 100m to North
    let south100 = point.shift(byDistance: 100, azimuth: Double.pi) // 100m to South

答案 9 :(得分:0)

奇怪的是,没有人想到使用MapKit中的MKCoordinateRegion自动计算出来。

import MapKit

extension CLLocation {
    func movedBy(latitudinalMeters: CLLocationDistance, longitudinalMeters: CLLocationDistance) -> CLLocation {
        let region = MKCoordinateRegion(center: coordinate, latitudinalMeters: abs(latitudinalMeters), longitudinalMeters: abs(longitudinalMeters))

        let latitudeDelta = region.span.latitudeDelta
        let longitudeDelta = region.span.longitudeDelta

        let latitudialSign = CLLocationDistance(latitudinalMeters.sign == .minus ? -1 : 1)
        let longitudialSign = CLLocationDistance(longitudinalMeters.sign == .minus ? -1 : 1)

        let newLatitude = coordinate.latitude + latitudialSign * latitudeDelta
        let newLongitude = coordinate.longitude + longitudialSign * longitudeDelta

        let newCoordinate = CLLocationCoordinate2D(latitude: newLatitude, longitude: newLongitude)

        let newLocation = CLLocation(coordinate: newCoordinate, altitude: altitude, horizontalAccuracy: horizontalAccuracy, verticalAccuracy: verticalAccuracy, course: course, speed: speed, timestamp: Date())

        return newLocation
    }
}

答案 10 :(得分:0)

我发布了有关测量问题的更新答案,其中包括对此绘图问题的答案。此处:CLLocation Category for Calculating Bearing w/ Haversine function