将2d样条函数f（t）变换为f（x） [英] Transform 2d spline function f(t) into f(x)

问题描述

因此，我有一组特殊情况的三次样条曲线，其2d控制点总是会形成一条永远不会在 x 轴中穿过的曲线。也就是说，曲线看起来像是一个简单的多项式函数，使得 y = （ x ）。我想要高效地沿着样条线创建一个 y 坐标的数组，这些坐标对应于运行样条线段长度的均匀间隔的x坐标。

我想要高效地找到样条线上的y坐标，例如 x = 0.0， x = 0.1， x = 0.2等，或以另一种方式接近，有效地转换 f _x，y 我现在正在使用一个4x4常量（）风格的函数到 f （ x ）函数中。矩阵和四个二维控制点来描述样条，使用Hermite或Catmull-Rom样条的矩阵常量，并将它们插入从0到1的 t 三次函数。 b
$ b

给定矩阵和控制点，在x轴上获得这些y值的最好方法是什么？

编辑：I应该加上一个足够好的近似值来绘制就足够了。 你的问题表明你要均匀分布空格 x 坐标，并且大致的解决方案都是正确的。所以我建议使用以下算法：

• 决定你想要的网格点，例如每个0.1 x 单位。


• 以 l = 0和 r = 1开头。 li>
  
• Compute f _x （ l ）和 / i>（ ）并考虑由这些端点表示的区间。
  
  
  
  • 如果间隔足够小并且只包含一个网格点，则使用中心参数 t =（ >
    
  • 如果网格点中至少有一个网格点，则将其作为一个单元素列表返回。该间隔使用（l + r + r）/ 2作为分割点将其分成两部分，并将两个计算结果列表连接在一起。
  
  
  • 如果间隔中没有网格点，则跳过计算的当前分支，返回空列表。
  ul>
  
  

  这将放大网格点，在每一步中平分参数空间，并为所有网格点提供合适的参数。

  
  

  So I've got a special case set of cubic splines, whose 2d control points will always result in a curve that will never cross itself in the x axis. That is, the curves look like they could be a simple polynomial function such that y=f(x). I want to efficiently create an array of y coordinates along the spline that correspond to evenly-spaced x coordinates running the length of the spline segment.

  I want to efficiently find the y coordinates along the spline where, for instance, x=0.0, x=0.1, x=0.2, etc., or approached another way, effectively transform the f_x,y(t) style function into an f(x) function.

  I'm currently using a 4x4 constant matrix and four 2d control points to describe the spline, using matrix constants for either Hermite or Catmull-Rom splines, and plugging them into a cubic function of t going from 0 to 1.

  Given the matrix and the control points, what's the best way to obtain these y values over the x axis?

  EDIT: I should add that an approximation good enough to draw is sufficient.

  解决方案

  Your question states that you want evenly spaces x coordinates, and approximate solutions are all right. So I propose the following algorithm:

  • Decide on the grid points you want, e.g. one every 0.1 x units.
  • Start with l = 0 and r = 1.
  • Compute f_x(l) and f_x(r) and consider the interval denoted by these endpoints.
    • If the interval is sufficiently small and contains exactly one grid point, use the central parameter t=(l+r)/2 as a good approximation for this grid point, and return that as a one-element list.
    • If there is at least one grid point in that interval, split it in two using (l+r)/2 as the splitting point, and concatenate the resulting lists from both computations.
    • If there is no grid point in the interval, then skip the current branch of the computation, returning an empty list.

  This will zoom in on the grid points, bisecting the parameter space in each step, and will come up with suitable parameters for all your grid points.

  这篇关于将2d样条函数f（t）变换为f（x）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！