计算立方贝塞尔长度的廉价方法 [英] Cheap way of calculating cubic bezier length

问题描述

一个立方贝塞尔长度
的解析解决方案似乎并不存在，但这并不意味着
编码的廉价解决方案不存在。便宜的我的意思是在50-100纳秒（或更少）的范围内。

有人知道这样的事吗？也许在两个类别：

1）更少的错误，如1％但更慢的代码。
2）更多错误，比如20％但更快？

我通过google扫描了一下，但没有找到任何看起来像是一个很好的解决方案的东西。只有像N线段上划分
和总和N sqrt - 对于更精确的速度太慢，
，对于2或3段可能太不准确。

有什么更好的？另一种选择是将弧长估计为和弦和控制网之间的平均值。在实践中：

  Bezier bezier = Bezier（p0，p1，p2，p3）; 
 
 chord =（p3-p0）.Length; 
 cont_net =（p0  -  p1）.Length +（p2  -  p1）.Length +（p3  -  p2）.Length; 
 
 app_arc_length =（cont_net + chord）/ 2; 
  

然后，您可以递归地将样条线段分成两段，并计算到收敛的弧长。我测试了自己，实际上它收敛得非常快。我从这个论坛。

An analytical solution for cubic bezier length seems not to exist, but it does not mean that coding a cheap solution does not exist. By cheap I mean something like in the range of 50-100 ns (or less).

Does someone know anything like that? Maybe in two categories:

1) less error like 1% but more slow code. 2) more error like 20% but faster?

I scanned through google a bit but it doesn't find anything which looks like a nice solution. Only something like divide on N line segments and sum the N sqrt - too slow for more precision, and probably too inaccurate for 2 or 3 segments.

Is there anything better?

解决方案

Another option is to estimate the arc length as the average between the chord and the control net. In practice:

Bezier bezier = Bezier (p0, p1, p2, p3);

chord = (p3-p0).Length;
cont_net = (p0 - p1).Length + (p2 - p1).Length + (p3 - p2).Length;

app_arc_length = (cont_net + chord) / 2;

You can then recursively split your spline segment into two segments and calculate the arc length up to convergence. I tested myself and it actually converges pretty fast. I got the idea from this forum.

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