如何简化三次贝塞尔曲线？ [英] How to simplify cubic bezier curve?

问题描述

我有一个立方贝塞尔曲线包括多个段（左图）。它有一些粗糙的曲率，我需要使它更平滑，像正确的图像。这个问题有点像降噪，我该如何实现？

I have a Cubic Bezier curve comprises of a number of segments (left image). It has some rough curvature and I need to make it smoother like the right image. This problem is somewhat like "noise reduction", how do I achieve this?

有类似的线程这里，但输入是一组点并使用最小二乘拟合贝赛尔曲线，但在我的问题中输入已经是三次bezier。

There's similar thread here, but the input is a set of point and fitting a bezier curve on it using least square, but in my problem the input is already cubic bezier.

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在上面的图片上我不绘制段和控制点，但我希望你能得到想法。

推荐答案

你想保持相同数量的段吗？你需要保持Bezier段之间的连续性吗？你试图击中一定数量的细分，或只是保持在原曲线的一定公差内？

Do you want to keep the same number of segments? Do you need to keep continuity between Bezier segments? Are you trying to hit a certain number of segments, or just keep things within a certain tolerance of the original curve?

现在我假设你想减少贝塞尔曲线段的数量，您需要保持线段之间的G1连续性，并且您尝试在公差范围内平滑（仅从您的图像猜出）。

For now I'll assume that you want to reduce the number of Bezier segments, any you need to keep G1 continuity between the segments, and you're trying to smooth within a tolerance (just guessing from your image).

对于顶级算法，您将通过每个相邻的曲线对，并尝试将它们组合。重复此操作，直到两个相邻的曲线合并在公差范围之外。

For the top-level algorithm, you go through every adjacent pair of curves, and try to combine them. Repeat this until combining two adjacent curves would fall outside of your tolerance.

如何组合两个相邻的贝塞尔曲线？让我们假设它们是曲线P和Q，并且由于它们都是立方的，所以它们具有4个CV，每个具有P0，P1，P2，P3和Q0，Q1，Q2，Q3。我们还假设P3 == Q0。此外，我们会说输出曲线是R，由R0，R1，R2，R3组成。

How do you combine two adjacent Bezier curves? Let's assume that they are curves P and Q, and since they're both cubic they have 4 CVs each, P0, P1, P2, P3 and Q0, Q1, Q2, Q3. We'll also assume that P3 == Q0. Also, we'll say that the output curve is R, composed of R0, R1, R2, R3.

另一个非常重要的步骤 - 你需要分配t值对于您正在简化的较大曲线内的每个贝塞尔曲线段。因此，段0将从0..1开始，段1从1..2开始，段2从2..3开始等。

One other very important step - you need to assign t values for each Bezier curve segment within the larger curve that you're simplifying. So, segment 0 would go from 0..1, segment 1 goes from 1..2, segment 2 goes from 2..3, etc.

为了保持P与其邻居的连续性，以及Q与其邻居的连续性，您不能移动P0或Q3，由（P1-P0）和（Q2-Q3）形成的切向量必须保持在相同的方向。

If you want to maintain P's continuity with its neighbor, and Q's continuity with its neighbor, you can't move P0 or Q3, and the tangent vectors formed by (P1-P0) and (Q2-Q3) must stay in the same direction.. they can only be scaled.

由于你在R中只有4个CV，所以这两个比例因子是你唯一的自由度。我们称它们为kp和kq。

Since you only have 4 CVs in R, those two scale factors are the only degree of freedom that you have. We'll call them kp and kq.

R0=P0
R1=P0+kp*(P1-P0)
R2=Q3+kq*(Q2-Q3)
R3=Q3

如果两条曲线的结长度相等，则kp = 2和kq = 2。

If the knot length of the two curves are equal, than kp = 2 and kq = 2.

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