计算椭圆体（或WGS84坐标）上的点和线段之间的距离？

Question

我在地球表面（或者在椭球体上）的两个 WGS84 坐标之间有一条线段 AB，并且需要计算点 P 与线段 AB 上距离 P 最近的点之间的距离。我该怎么做？

非常感谢任何帮助。

问候， 约亨

Answer 1

我从来没有处理过 WGS84 坐标，对这种类型的数学也很生疏，但我会告诉你我想到的。我并没有真正回答这个问题，但这太多了，无法发表评论，而且它有伪代码。我只是觉得这个问题很有趣，想尝试一下。无论如何，这里...

想象一下以 P 为中心的所有球体的集合。

现在，这些球体中只有一个应该与 AB 恰好共享 1 个点。如果你有一个描述 AB 的方程，那么你应该能够找到以与 AB 恰好共享一个点为中心的球体。

可能有比反复试验更快的方法来做到这一点，但我想到的是二分搜索。这应该找到到 AB 的直线距离。如果你想要 P 和 AB 之间的距离，我会在代码后面介绍。
psudocode：

   Radius_lo = 0
   Radius_hi = minimum( distance(P, A), distance(P, B) )
   Radius_test = Raduis_hi // you may want to add a miniscule amount to this to keep from floating point inprecision from effectively rounding down which could lead to no actual intersection
   while true
      intersections = intersection_points( sphere( P, Radius_test), AB )
      if ( count(intersections) == 1 ) // or close enough.  With floating pt math you'll likely need to recognize close enough or it will never exit.
            return  Radius_test
      if ( count(intersections) == 2 )
            // Here you may do one of two things, both of which are binary searches.
            // 1. The first and simplest would be to divide average _test with _lo and try again
            Radius_hi = Radius_test
            Radius_test = (Radius_test + Radius_lo) / 2
            Continue
            // 2. The second would be to attempt to divide the segment fromed by the two intersection points, which is more complicated, but would almost always result in fewer times through the loop
            X = midpoint(intersection[0], intersection[1]) // midpoint would have to know how to find the midpoint of them along the elipse that they live on
            Radius_test = distance( P, X)
            Continue
     if ( count(intersections) == 0)
           // Gone too far.  If you had done (1) above, then do
           Radius_lo = Radius_test
           Radius_test = (Radius_test + Radius_hi) / 2
           Continue
           // If on the otherhand you had done (2) above then you shouldn't ever have this case
           error("0 intersections!")
     // Now, this should be for any intersection count other than 0, 1, or 2.  This shouldn't happen so
     error("bad intersection count")
  endloop // while true

只要椭圆线段 AB 比 AB 所在椭圆的任何其他部分更靠近 P，就可以找到 P 和 AB 之间的直线距离。如果这不是真的，那么测试这一点应该很容易，只需使用 A 或 B 中较接近的点作为最近的点即可。

好的，所以您实际上想要 P 和 AB 之间的表面距离。这更复杂，但你可以改变我的算法来处理它。