寻找分段二次贝塞尔曲线上的控制点

发布于 2024-11-10 11:43:06 字数 233 浏览 3 评论 0原文

我需要编写一个程序来生成并显示分段二次贝塞尔曲线，该曲线对每组数据点进行插值（我有一个包含数据点的 txt 文件）。曲线应具有连续的切线方向，每个数据点的切线方向是两个相邻弦方向的凸组合。

以上是我拥有的数据点，有谁知道我可以使用什么公式来计算控制点？

原文

I need to write a program to generate and display a piecewise quadratic Bezier curve that interpolates each set of data points (I have a txt file contains data points). The curve should have continuous tangent directions, the tangent direction at each data point being a convex combination of the two adjacent chord directions.

The above are the data points I have, does anyone know what formula I can use to calculate control points?

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少女净妖师 2024-11-17 11:43:06

您将需要使用三次贝塞尔曲线来很好地处理多个斜率变化，例如数据集中发生的情况。对于二次贝塞尔曲线，数据点之间只有一个控制点，因此每条曲线段大部分都位于连接线段的一侧。

很难解释，所以这里是数据（黑点）、二次控制点（红色）和曲线（蓝色）的快速草图。（假装曲线是平滑的！）

在此处输入图像描述

查看三次 Hermite 曲线作为通用解决方案。

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诗酒趁年少 2024-11-17 11:43:06

从这里： http://blog.mackerron.com/2011/ 01/01/javascript-cubic-splines/

要生成如下所示的插值曲线：

natural曲线
monotonic curve

您可以使用这个咖啡脚本类（它编译为 javascript）

class MonotonicCubicSpline

# by George MacKerron, mackerron.com

# adapted from: 
# http://sourceforge.net/mailarchive/forum.php?thread_name=
# EC90C5C6-C982-4F49-8D46-A64F270C5247%40gmail.com&forum_name=matplotlib-users
# (easier to read at http://old.nabble.com/%22Piecewise-Cubic-Hermite-Interpolating-
# Polynomial%22-in-python-td25204843.html)

# with help from:
# F N Fritsch & R E Carlson (1980) 'Monotone Piecewise Cubic Interpolation', 
#   SIAM Journal of Numerical Analysis 17(2), 238 - 246.
# http://en.wikipedia.org/wiki/Monotone_cubic_interpolation
# http://en.wikipedia.org/wiki/Cubic_Hermite_spline

constructor: (x, y) ->
  n = x.length
  delta = []; m = []; alpha = []; beta = []; dist = []; tau = []
  for i in [0...(n - 1)]
    delta[i] = (y[i + 1] - y[i]) / (x[i + 1] - x[i])
    m[i] = (delta[i - 1] + delta[i]) / 2 if i > 0
  m[0] = delta[0]
  m[n - 1] = delta[n - 2]
  to_fix = []
  for i in [0...(n - 1)]
    to_fix.push(i) if delta[i] == 0
  for i in to_fix
    m[i] = m[i + 1] = 0
  for i in [0...(n - 1)]
    alpha[i] = m[i] / delta[i]
    beta[i]  = m[i + 1] / delta[i] 
    dist[i]  = Math.pow(alpha[i], 2) + Math.pow(beta[i], 2)
    tau[i]   = 3 / Math.sqrt(dist[i])
  to_fix = []
  for i in [0...(n - 1)]
    to_fix.push(i) if dist[i] > 9
  for i in to_fix
    m[i]     = tau[i] * alpha[i] * delta[i]
    m[i + 1] = tau[i] * beta[i]  * delta[i]
  @x = x[0...n]  # copy
  @y = y[0...n]  # copy
  @m = m

interpolate: (x) ->
  for i in [(@x.length - 2)..0]
    break if @x[i] <= x
  h = @x[i + 1] - @x[i]
  t = (x - @x[i]) / h
  t2 = Math.pow(t, 2)
  t3 = Math.pow(t, 3)
  h00 =  2 * t3 - 3 * t2 + 1
  h10 =      t3 - 2 * t2 + t
  h01 = -2 * t3 + 3 * t2
  h11 =      t3  -    t2
  y = h00 * @y[i] + 
      h10 * h * @m[i] + 
      h01 * @y[i + 1] + 
      h11 * h * @m[i + 1]
  y

From here: http://blog.mackerron.com/2011/01/01/javascript-cubic-splines/

To produce interpolated curves like these:

natural curve
monotonic curve

You can use this coffee-script class (which compiles to javascript)

class MonotonicCubicSpline

# by George MacKerron, mackerron.com

# adapted from: 
# http://sourceforge.net/mailarchive/forum.php?thread_name=
# EC90C5C6-C982-4F49-8D46-A64F270C5247%40gmail.com&forum_name=matplotlib-users
# (easier to read at http://old.nabble.com/%22Piecewise-Cubic-Hermite-Interpolating-
# Polynomial%22-in-python-td25204843.html)

# with help from:
# F N Fritsch & R E Carlson (1980) 'Monotone Piecewise Cubic Interpolation', 
#   SIAM Journal of Numerical Analysis 17(2), 238 - 246.
# http://en.wikipedia.org/wiki/Monotone_cubic_interpolation
# http://en.wikipedia.org/wiki/Cubic_Hermite_spline

constructor: (x, y) ->
  n = x.length
  delta = []; m = []; alpha = []; beta = []; dist = []; tau = []
  for i in [0...(n - 1)]
    delta[i] = (y[i + 1] - y[i]) / (x[i + 1] - x[i])
    m[i] = (delta[i - 1] + delta[i]) / 2 if i > 0
  m[0] = delta[0]
  m[n - 1] = delta[n - 2]
  to_fix = []
  for i in [0...(n - 1)]
    to_fix.push(i) if delta[i] == 0
  for i in to_fix
    m[i] = m[i + 1] = 0
  for i in [0...(n - 1)]
    alpha[i] = m[i] / delta[i]
    beta[i]  = m[i + 1] / delta[i] 
    dist[i]  = Math.pow(alpha[i], 2) + Math.pow(beta[i], 2)
    tau[i]   = 3 / Math.sqrt(dist[i])
  to_fix = []
  for i in [0...(n - 1)]
    to_fix.push(i) if dist[i] > 9
  for i in to_fix
    m[i]     = tau[i] * alpha[i] * delta[i]
    m[i + 1] = tau[i] * beta[i]  * delta[i]
  @x = x[0...n]  # copy
  @y = y[0...n]  # copy
  @m = m

interpolate: (x) ->
  for i in [(@x.length - 2)..0]
    break if @x[i] <= x
  h = @x[i + 1] - @x[i]
  t = (x - @x[i]) / h
  t2 = Math.pow(t, 2)
  t3 = Math.pow(t, 3)
  h00 =  2 * t3 - 3 * t2 + 1
  h10 =      t3 - 2 * t2 + t
  h01 = -2 * t3 + 3 * t2
  h11 =      t3  -    t2
  y = h00 * @y[i] + 
      h10 * h * @m[i] + 
      h01 * @y[i + 1] + 
      h11 * h * @m[i + 1]
  y

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