确定围绕对角线的边界矩形 [英] Determine a bounding rectangle around a diagonal line

问题描述

用户将在屏幕上定义一条线，该线在绘制时将具有给定的厚度(或宽度).

A user will define a line on screen which will have, when drawn, a given thickness (or width).

我现在需要能够确定围绕它的边界矩形的坐标.

I now need to be able to determine the coordinates of a bounding rectangle around this.

我有坐标A和B，以及线宽(W).

I have the coordinates A and B, along with the line thickness (W).

如何计算坐标A1，A2，B1和B2.

How can I calculate the coordinates A1, A2, B1 and B2.

我搜索了但找不到与已经问过的问题相对应的问题.

I searched but was unable to find a question corresponding to this already asked.

推荐答案

Dx= Xb - Xa
Dy= Yb - Ya
D= sqrt(Dx * Dx + Dy * Dy)
Dx= 0.5 * W * Dx / D
Dy= 0.5 * W * Dy / D

这将计算(Dx, Dy)在AB方向上长度为W/2的向量.那么(-Dy, Dx)是垂直向量.

This computes (Dx, Dy) a vector of length W/2 in the direction of AB. Then (-Dy, Dx) is the perpendicular vector.

Xmin = min(Xa, Xb) - abs(Dy) 
Xmax = max(Xa, Xb) + abs(Dy)
Ymin = min(Ya, Yb) - abs(Dx)
Ymax = max(Ya, Yb) + abs(Dx)

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更新:

我错误地回答了AABB.

I answered for the AABB by mistake.

对于笔触的四个角

Xa - Dy, Ya + Dx
Xa + Dy, Ya - Dx
Xb - Dy, Yb + Dx
Xb + Dy, Yb - Dx

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