ios – MKMapView MKCircle呈现一个半径太大的圆

private func addCircle() {
    mapView.removeOverlays(mapView.overlays)
    let circle = MKCircle(centerCoordinate: mapCenter,radius: 8500000.0)
    mapView.addOverlay(circle)
}

private func configureGestureRecognizer() {
    doubleTapGestureRecognizer.addTarget(self,action: Selector("handleDoubleTap:"))
    doubleTapGestureRecognizer.numberOfTapsrequired = 2
    if let subview = mapView.subviews.first as? UIView {
        subview.addGestureRecognizer(doubleTapGestureRecognizer)
    }
    else {
        println("Can't add a gesture recognizer")
    }
}

@objc private func handleDoubleTap(sender: UITapGestureRecognizer) {
    let point = sender.locationInView(mapView)
    let location = mapView.convertPoint(point,toCoordinateFromView: mapView)
    mapCenter = location
    addCircles()
}

根据Apple关于MKCircle的文档：“当纬度值远离赤道并向极点移动时,地图点之间的物理距离变小.这意味着需要更多的地图点来表示相同的距离.结果,边界当圆圈的中心点远离赤道并朝向极点移动时,圆形覆盖的矩形变大.“

正如Anna和Warren都提到的那样,这不是一个错误 – 这是预期的行为.但是,在boundingMapRect和radius之间的文档中似乎存在差异.文档表明,半径是距离中心点的度量单位,在您的示例中显然不是这种情况.

我认为这里发生的事情是苹果可能从未打算将MKCircle用于你正在使用它的规模上. MKCircle创建一个2D圆,它既不能是圆形,也不能是投影地图上圆形区域的精确表示.

现在,如果要做的只是创建一个没有扭曲并且相对于赤道长度的半径的均匀圆,可以将赤道上的圆的长度设置为基本半径,然后计算比例当前点的半径如下：

let baseCoord = CLLocationCoordinate2D(latitude: 0,longitude: 0)
let radius: Double = 850000.0

override func viewDidLoad() {
    super.viewDidLoad()
    mapView.region = MKCoordinateRegion(
        center: baseCoord,span: MKCoordinateSpan(
            latitudeDelta: 90,longitudeDelta: 180
        )
    )
    mapCenter = baseCoord
    let circle = MKCircle(centerCoordinate: mapCenter,radius: radius)
    baseRadius = circle.boundingMapRect.size.height / 2

    mapView.delegate = self
    configureGestureRecognizer()
}

private func addCircle() {

    mapView.removeOverlays(mapView.overlays)
    let circle = MKCircle(centerCoordinate: mapCenter,radius: radius)

    var currenTradius = circle.boundingMapRect.size.height / 2
    let factor = baseRadius / currenTradius
    var updatedRadius = factor * radius

    let circletoDraw = MKCircle(centerCoordinate: mapCenter,radius: updatedRadius)
    mapView.addOverlay(circletoDraw)
}

但如果你的计划要准确覆盖点击x米范围内的所有空间,那就有点棘手了.首先,您将获取双击动作中的单击坐标,然后将其用作多边形的中心.

@objc private func handleDoubleTap(sender: UITapGestureRecognizer) {
    let point = sender.locationInView(mapView)
    currentCoord = mapView.convertPoint(point,toCoordinateFromView: mapView)
    mapCenter = currentCoord
    addpolygon()
}

在addpolygon中,获取坐标并设置叠加层：

private func addpolygon() {
    var mapCoords = getCoordinates()
    mapView.removeOverlays(mapView.overlays)

    let polygon = MKpolygon(coordinates: &mapCoords,count: mapCoords.count)
    mapView.addOverlay(polygon)
}

给定一个点,一个方位和一个角距离(坐标之间的距离除以地球的半径),您可以使用以下公式计算另一个坐标的位置.请务必导入Darwin,以便您可以访问三角函数库

let globalRadius: Double = 6371000
let π = M_PI

private func getCoordinates() -> [CLLocationCoordinate2D] {
    var coordinates = [CLLocationCoordinate2D]()

    let lat1: Double = (currentCoord!.latitude)
    let long1: Double = (currentCoord!.longitude) + 180
    let factor = 30

    if let a = annotation {
        mapView.removeAnnotation(annotation)
    }

    annotation = MKPointAnnotation()
    annotation!.setCoordinate(currentCoord!)
    annotation!.title = String(format: "%1.2f°,%1.2f°",lat1,long1)
    mapView.addAnnotation(annotation)

    var φ1: Double = lat1 * (π / 180)
    var λ1: Double = long1 * (π / 180)
    var angulardistance =  radius / globalRadius

    var metersTonorthPole: Double = 0
    var metersToSouthPole: Double = 0

    for i in Int(lat1)..<89 {
        metersTonorthPole = metersTonorthPole + 111132.92 - (559.82 * cos(2 * φ1)) + (1.175 * cos(4 * φ1))
    }

    for var i = lat1; i > -89; --i {
        metersToSouthPole = metersToSouthPole + 111132.92 - (559.82 * cos(2 * φ1)) + (1.175 * cos(4 * φ1))
    }

    var startingbearing = -180
    var endingbearing = 180

    if metersTonorthPole - radius <= 0 {
        endingbearing = 0
        startingbearing = -360
    }

    for var i = startingbearing; i <= endingbearing; i += factor {

        var bearing = Double(i)

        var bearingInradians: Double = bearing * (π / 180)

        var φ2: Double = asin(sin(φ1) * cos(angulardistance)
            + cos(φ1) * sin(angulardistance)
            * cos(bearingInradians)
        )

        var λ2 = atan2(
            sin(bearingInradians) * sin(angulardistance) * cos(φ1),cos(angulardistance) - sin(φ1) * sin(φ2)
        ) + λ1

        var lat2 = φ2 * (180 / π)
        var long2 = ( ((λ2 % (2 * π)) - π)) * (180.0 / π)

        if long2 < -180 {
            long2 = 180 + (long2 % 180)
        }

        if i == startingbearing && metersTonorthPole - radius <= 0 {
            coordinates.append(CLLocationCoordinate2D(latitude: 90,longitude: long2))
        } else if i == startingbearing && metersToSouthPole - radius <= 0 {
            coordinates.append(CLLocationCoordinate2D(latitude: -90,longitude: long2))
        }

        coordinates.append(CLLocationCoordinate2D(latitude: lat2,longitude: long2))
    }

    if metersTonorthPole - radius <= 0 {
        coordinates.append(CLLocationCoordinate2D(latitude: 90,longitude: coordinates[coordinates.count - 1].longitude))
    } else if metersToSouthPole - radius <= 0 {
        coordinates.append(CLLocationCoordinate2D(latitude: -90,longitude: coordinates[coordinates.count - 1].longitude))
    }

    return coordinates
}

在getCoordinates中,我们将度数转换为弧度,然后在我们的半径大于到北极或南极的距离的情况下添加一些锚定坐标.

以下是极点附近曲线的几个例子,半径分别为8500km和850km：

下面是最终输出的示例,其中包含一个额外的MKGeodesicpolyline叠加(Geodesics表示球面上最短的曲线),它显示了曲线的实际构建方式：