blocks|key|2447805|text|由于你所描述的原因，使用2D角度是一件痛苦的事情，所以最好使用矢量数学，因为它是旋转不变的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2447806|定义2D叉积A+%5E+B+=+Ax+*+By+-+Ay+*+Bx。如果A相对于B顺时针旋转，则该值为正，反之亦然。|offset|length|style|CODE|2447807|逻辑是：|2447808|2447809|Compute+C+=+(last_pt+-+center)+%5E+(first_pt+-+center)|unordered-list-item|2447810|If+C+=+0，圆弧要么是闭合的，要么是180度(忘记了这个名称)|2447811|如果是C+>+0，则圆弧必须是(i)顺时针且锐利/钝，或者是(Ii)E215逆时针且反射|BOLD|2447812|如果C+<+0，则相反的情况适用于|2447813|2447814|伪码：|2447815|int+arc_type(Point+first,+Point+last,+Point+center,+bool+clockwise)
{
++++//+cross-product
++++float+C+=+(last.x+-+center.x)+*+(first.y+-+center.y)
++++++++++++-+(last.y+-+center.y)+*+(first.x+-+center.x);

++++if+(Math.abs(C)+<+/*+small+epsilon+*/)
++++++++return+0;++++++++++++++++++++++//+180-degree

++++return+((C+>+0)+%5E+clockwise)+?+1+++//+reflex
+++++++++++++++++++++++++++++++++:+-1;+//+acute+/+obtuse
}|code-block|syntax|javascript|2447816|请注意，如果您不知道圆弧是顺时针还是逆时针，您可以在相邻的点上使用相同的叉积方法。您需要确保点的顺序是一致的-如果不一致，您可以再次使用叉积，按(相对)角度对它们进行排序。|2447817|entityMap^0|0|6|P|Y|1|12|1|0|0|0|8|18|0|3|5|0|3|5|F|3|U|E|0|0|H|2|5|0|0|0|3|0|0|0|2E|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@$D|18|E|19|F|G]|$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|1G|8|@$D|1H|E|1I|F|G]]|9|@]|A|$]]|$1|N|3|O|5|M|7|1J|8|@$D|1K|E|1L|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|M|7|1M|8|@$D|1N|E|1O|F|G]|$D|1P|E|1Q|F|R]|$D|1R|E|1S|F|R]]|9|@]|A|$]]|$1|S|3|T|5|M|7|1T|8|@$D|1U|E|1V|F|R]|$D|1W|E|1X|F|G]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1Z|8|@$D|20|E|21|F|R]]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|22|8|@]|9|@]|A|$10|11]]|$1|12|3|13|5|6|7|23|8|@$D|24|E|25|F|R]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|26|8|@]|9|@]|A|$]]]|15|$]]

Working with 2D angles is a pain for the reason you described, so it's better to work with vector math instead, which is rotationally invariant.

Define the 2D cross-product, <code>A ^ B = Ax * By - Ay * Bx</code>. This is positive if <code>A</code> is clockwise rotated relative to <code>B</code>, and vice versa.

The logic:

<ul>
<li>Compute <code>C = (last_pt - center) ^ (first_pt - center)</code></li>
<li>If <code>C = 0</code>, the arc is either closed or 180-degree (forgot the name for this)</li>
<li>If <code>C &gt; 0</code>, the arc must either be (i) clockwise and acute/obtuse or (ii) anti-clockwise and reflex</li>
<li>If <code>C &lt; 0</code>, the opposite applies</li>
</ul>

Pseudocode:

<pre><code>int arc_type(Point first, Point last, Point center, bool clockwise)
{
 // cross-product
 float C = (last.x - center.x) * (first.y - center.y)
 - (last.y - center.y) * (first.x - center.x);

 if (Math.abs(C) &lt; /* small epsilon */)
 return 0; // 180-degree

 return ((C &gt; 0) ^ clockwise) ? 1 // reflex
 : -1; // acute / obtuse
}
</code></pre>

Note that if you don't have prior knowledge of whether the arc is clockwise or anti-, you can use the same cross-product method on adjacent points. You need to ensure that the order of the points is consistent - if not you can, again using the cross-product, sort them by (relative) angle.

I have this seemingly simple but very confusing problem.

Given I have a set of vertices (x1,y1), (x2,y2), (x3,y3)...... representing an arc. The points can either be clockwise or counter clockwise, but are all similarly ordered.
And I know the center of the arc (xc,yc).

How can I tell if the arc subtends an acute/obtuse or reflex angle?

One obvious solution is to take the difference of atan2((last_pt)-(center)) and atan2((first_pt)-(center))). But if the arc goes through the point where PI become -PI, this method breaks down.

Also, since the arc points are derived from a rather noisy pixelated picture, the vertices are not exactly smooth. 

<a href="https://i.stack.imgur.com/3f4Jt.png" rel="nofollow noreferrer">Picture of a acute and reflex arc</a>

I cant wrap my brain around solving this problem. 

Thanks for your help!

Determine if an arc represents an reflex or acute angle

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个看似简单但却非常令人困惑的问题。假设我有一组顶点(x1，y1)，(x2，y2)，(x3，y3)......表示一条弧线。这些点既可以是顺时针的，也可以是逆时针的，但都是类似的顺序。我知道圆弧的中心(xc，yc)。如何判断圆弧是否包含锐角/钝角或反射角？一个明显的解决方案是取中心((Last_pt)-(中心))和...

确定圆弧代表的是反射角度还是锐角-腾讯云开发者社区-腾讯云

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