如何创建一个三次Bezier曲线时，在3D给予N个点？

问题描述

所以，我需要找出控制点会是一个三次Bezier曲线时，只需要知道曲线上的点，这些点可以位于3D。这将是理想的，如果我能做到这一点的任何数量的曲线上的点。大部分我发现只能用2D的交易，或只为4分。

So I need to find out where the control points would be for a cubic bezier curve when only knowing points on the curve, the points can lie in 3D. It would be ideal if I could do this for any number of points on the curve. Most of what I have found deals only with 2D, or only for 4 points.

推荐答案

让我看看，如果我理解你：你想要一个插值贝塞尔曲线，经历一个给定的点集P0 P1，...
但绘制贝塞尔曲线，像函数

Let me see if I understand you:you want an interpolating Bezier curve,going through a given set of points P0 P1 ...
but drawn as Bezier curves, with a function like

bezier4( nstep, Pj, Cj, Dj, Pj+1 )  -- control points Cj, Dj

也就是说，你要导出两个贝塞尔控制点CJ，DJ每件PJ - PJ + 1

That is, you want to derive two Bezier control points Cj, Djfor each piece Pj -- Pj+1 ?

之一导出这样的控制点的方法是使用所述伯恩斯坦多项式基

One way of deriving such control points is to use the Bernstein polynomial basis

b0(t) = (1-t)^3
b1(t) = 3 (1-t)^2 t,
b2(t) = 3 (1-t) t^2
b3(t) = t^3

bezier4(t) = b0(t) P0  +  b1(t) C0  +  b2(t) D0  +  b3(t) P1
= P0 at t=0, tangent --> C0
= P1 at t=1,  tangent <-- D0

和查找或得到插值又名卡特莫尔-Rom样条即经过P-1 P0 P1 P2：

and look up or derive the interpolating aka Catmull-Rom splinethat goes through P-1 P0 P1 P2:

b0(t) P0
+ b1(t) (P0 + (P1 - P-1) / 6)
+ b2(t) (P1 - (P2 - P0) / 6)
+ b3(t) P1
= P0 at t=0, P1 at t=1

我们想bezier4（吨），以完全相同曲线作为CatmullRom（吨），所以：

We want bezier4(t) to be exactly the same curve as CatmullRom(t), so:

C0 = P0 + (P1 - P-1) / 6
D0 = P1 - (P2 - P0) / 6

由于N点P0 P1 ......（在2D转3D ... anyd），带他们4的时间;每个4，该公式为您提供了2个控制点CJ，DJ为

Given N points P0 P1 ... (in 2d 3d ... anyd), take them 4 at a time;for each 4, that formula gives you 2 control points Cj, Dj for

bezier4( nstep, Pj, Cj, Dj, Pj+1 )

这是否有道理，是不是你想要的？
（对于一个赏金，我会凑齐一些Python / numpy的在一起。）

Does this make sense, is it what you want ?
(For a bounty, I'd cobble some Python / numpy together.)

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