ios - 获取iOS中两个坐标之间距离的垂直和水平分量

Question

我正在尝试在 swift 中创建一个函数，准确计算任何位置距地球原点的垂直和水平距离(纬度:0，经度:0)。我知道 iOS 有 distanceFromLocation 函数，但这给了我直接的位置。我正在寻找的是该方向的水平和垂直分量。我尝试提出自己的解决方案，但是当我根据得到的水平和垂直分量测试直接距离时，它与实际距离不匹配。这是我的功能:

func distanceFromOrigin(location:CLLocation) {

    let lat = location.coordinate.latitude
    let lon = location.coordinate.longitude

    let earthOriginLocation = CLLocation(coordinate: CLLocationCoordinate2DMake(0.0, 0.0), altitude: CLLocationDistance(0.0), horizontalAccuracy: kCLLocationAccuracyBestForNavigation, verticalAccuracy: kCLLocationAccuracyBestForNavigation, timestamp: NSDate())

    var horDistance = earthOriginLocation.distanceFromLocation(CLLocation(latitude: 0.0, longitude: location.coordinate.longitude))
    var verDistance = earthOriginLocation.distanceFromLocation(CLLocation(latitude: location.coordinate.latitude, longitude: 0.0))

    let overallDistance = earthOriginLocation.distanceFromLocation(location)

    if lat < 0 {
        print("Object is South of Equator")
        verDistance *= -1
    } else if lat > 0 {
        print("Object is North of Equator")
    } else {
        print("Object is at the Equator")
    }
    if lon < 0 {
        print("Object is West of Prime Meridian")
        horDistance *= -1
    } else if lon > 0 {
        print("Object is East of Prime Meridian")
    } else {
        print("Object is at the Prime Meridian")
    }

    print("Vertical Distance: \(verDistance)")
    print("Horizontal Distance: \(horDistance)")
    print("Overall Distance: \(overallDistance)")

    //Test to see if vertical and horizontal distances are accurate compared to actual distance.
    print("Test: \(sqrt((pow(horDistance, 2.0)) + (pow(verDistance, 2.0))))")

}

谢谢!

Answer 1

你的代码没问题，但你的测试是错误的。

您错过了地球不是平坦，因此您正在考虑的直角三角形位于大地水准面上，并且斜边大于平方边和。

我建议您手动执行一些测试，以确保结果看起来真实，而不是深入研究曲面上的几何图形。

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小提示:如果这些点彼此非常接近，您的测试可能通过，因为在这种情况下，地球表面的曲率对计算的影响很小。