问题描述
我需要找到一个盒子和一个圆之间的最近距离,但是,我意识到这可以分解为一条线段和一个圆之间的最近距离.
I need to find the closest distance between a box and a circle, however, I realize this can be broken up into the closest distance between a line segment and a circle.
- 我有两点point1_x、point1_y和point2_x、point2_y<的线段/代码>
- 我有一个 circle,中心为 circle_x、circle_y 和 radius radius
- I have a line segment of two points point1_x,point1_y and point2_x, point2_y
- I have a circle with center circle_x, circle_y and radius radius
是否有一个可以开箱即用的 python 库,如果没有,有人可以提供一个函数来支持它吗?
Is there a python library that will support this out of the box and if not can somebody present a function to do so?
(我相信我必须在圆上找到与直线相同斜率的切点?)
(I belive I have to locate the tangent point on the circle with the same slope as the line?)
推荐答案
有一种方法可以找到圆到矩形的最近距离(此处为轴方向).
矩形边将平面分成 9 块.我们可以找到包含圆心的部分(中心、左上、左等),并计算所需的距离.矩形ABCD和圆心E:
There exists a method to find the closest distance from circle to rectangle (axis-oriented here).
Rectangle sides divide plane into 9 pieces. We can find what piece (central, left-top, left etc) contains circle center, and calculate needed distance. Rectangle ABCD and circle center E:
德尔福代码:
//returns closest distance from circle to rectangle //0 if intersection or inclusion occurs function CircleRectDistance(CX, CY, CR: Integer; RR: TRect): Double; var wh, hh, dx, dy, t, SquaredDist: Double; begin SquaredDist := 0; //halfwidth and halfheight wh := 0.5 * (RR.Right - RR.Left); hh := 0.5 * (RR.Bottom - RR.Top); //distances to rectangle center dx := CX - 0.5 * (RR.Left + RR.Right); dy := CY - 0.5 * (RR.Top + RR.Bottom); //rectangle sides divide plane to 9 parts, t := dx + wh; if t < 0 then SquaredDist := t * t else begin t := dx - wh; if t > 0 then SquaredDist := t * t end; t := dy + hh; if t < 0 then SquaredDist := SquaredDist + t * t else begin t := dy - hh; if t > 0 then SquaredDist := SquaredDist + t * t end; if SquaredDist < CR * CR then Result := 0 else Result := Sqrt(SquaredDist)- CR; end;