我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了，所以我看到的例子让我很困惑，我很遗憾地承认。
当前回答
我制作了一个交互式Desmos图来演示如何实现这一点:
https://www.desmos.com/calculator/kswrm8ddum
红点是A点，绿点是B点，C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。
编辑:
有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标，AB是单位x轴，单位y轴垂直于AB。
其他回答
请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html
按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。
几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。
特征c++版本的3D线段和点
// Return minimum distance between line segment: head--->tail and point
double MinimumDistance(Eigen::Vector3d head, Eigen::Vector3d tail,Eigen::Vector3d point)
{
double l2 = std::pow((head - tail).norm(),2);
if(l2 ==0.0) return (head - point).norm();// head == tail case
// Consider the line extending the segment, parameterized as head + t (tail - point).
// We find projection of point onto the line.
// It falls where t = [(point-head) . (tail-head)] / |tail-head|^2
// We clamp t from [0,1] to handle points outside the segment head--->tail.
double t = max(0,min(1,(point-head).dot(tail-head)/l2));
Eigen::Vector3d projection = head + t*(tail-head);
return (point - projection).norm();
}
%Matlab solution by Tim from Cody
function ans=distP2S(x0,y0,x1,y1,x2,y2)
% Point is x0,y0
z=complex(x0-x1,y0-y1);
complex(x2-x1,y2-y1);
abs(z-ans*min(1,max(0,real(z/ans))));
WPF版本:
public class LineSegment
{
private readonly Vector _offset;
private readonly Vector _vector;
public LineSegment(Point start, Point end)
{
_offset = (Vector)start;
_vector = (Vector)(end - _offset);
}
public double DistanceTo(Point pt)
{
var v = (Vector)pt - _offset;
// first, find a projection point on the segment in parametric form (0..1)
var p = (v * _vector) / _vector.LengthSquared;
// and limit it so it lays inside the segment
p = Math.Min(Math.Max(p, 0), 1);
// now, find the distance from that point to our point
return (_vector * p - v).Length;
}
}
我需要一个Godot (GDscript)的实现，所以我写了一个基于grumdrig接受的答案:
func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
# Return minimum distance between line segment vw and point p
var l2: float = (v - w).length_squared() # i.e. |w-v|^2 - avoid a sqrt
if l2 == 0.0:
return p.distance_to(v) # v == w case
# Consider the line extending the segment, parameterized as v + t (w - v).
# We find projection of point p onto the line.
# It falls where t = [(p-v) . (w-v)] / |w-v|^2
# We clamp t from [0,1] to handle points outside the segment vw.
var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
var projection: Vector2 = v + t * (w - v) # Projection falls on the segment
return p.distance_to(projection)
