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?
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);
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