检查是否在同一圆周重叠两段/相交 [英] check if two segments on the same circle overlap / intersect

问题描述

由于同一个圆的两个圆弧段：A = [A1，A2]和B = [B1，B2]，有：

Given two circle segments of the same circle: A=[a1, a2] and B=[b1, b2], with:

• 在A1，A2，B1，B2在-INF和+ INF之间的度值
• 在A1中= A2; B1＆LT; = B2
• A2-A1中= 360; B2-B1＆LT; = 360

我如何才能知道，如果这两个圆弧段重叠？ （即，如果它们相交或触摸中的至少一个点）

How can I find out if these two circle segments overlap? (i.E. if they intersect or touch in at least one point)

例如：

A=[  -45°,    45°]; B=[   10°,   20°] ==> overlap
A=[  -45°,    45°]; B=[   90°,  180°] ==> no overlap
A=[  -45°,    45°]; B=[  180°,  360°] ==> overlap
A=[ -405°,  -315°]; B=[  180°,  360°] ==> overlap
A=[-3600°, -3601°]; B=[ 3601°, 3602°] ==> overlap (touching counts as overlap)
A=[ 3600°,  3601°]; B=[-3601°,-3602°] ==> overlap (touching counts as overlap)
A=[    -1°,    1°]; B=[ 3602°, 3603°] ==> no overlap 

这看起来像一个看似简单的问题，但我不能环绕它我的头。 我公司目前有涉及分割每个段成两个，如果它穿过0°的解决方案，一个基本的想法，但我不知道这涵盖了所有的情况，我想知道是否有一个优雅的公式。

This looks like a deceptively simple problem but I cannot wrap my head around it. I currently have a basic idea for a solution which involves splitting each segment into two if it crosses 0°, but I am not sure if that covers all cases, and I was wondering if there is an elegant formula.

推荐答案

由于@admaoldak提到的，首先规格化度：

As @admaoldak mentioned, normalize the degrees first:

a1_norm = a1 % 360
a2_norm = a2 % 360
b1_norm = b1 % 360
b2_norm = b2 % 360

现在，以检查是否B1内（A1，A2），

Now to check if b1 is within (a1,a2),

def intersect(b, as, ae
    Intersect = False
    If as > ae:
        if b >= as or b <= ae:
            return True
    Else:
        if b>=as and b<=ae:
            return True
    return False

最后的答案是：

Final answer is:

intersect(b1_norm,a1_norm,a2_norm)||intersect(b2_norm,a1_norm,a2_norm)||
intersect(a1_norm,b1_norm,b2_norm)||intersect(a2_norm,b1_norm,b2_norm)

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