如何从离散样本计算曲率半径? [英] How do I calculate radius of curvature from discrete samples?

问题描述

我从平面曲线中获得了一系列 (x,y) 样本，这些样本取自实际测量值，因此可能有点嘈杂且时间间隔不均匀.

x = -2.51509 -2.38485 -1.88485 -1.38485 -0.88485 -0.38485 0.11515 0.61515 1.11515 1.61515 ...y = -48.902 -48.917 -48.955 -48.981 -49.001 -49.014 -49.015 -49.010 -49.001 -48.974 ...

如果我绘制整个系列，它看起来像一个漂亮的椭圆形，但如果我仔细观察，这条线看起来有点摇摆不定，这可能是噪音.

我将如何提取基础椭圆曲率半径的估计值?

任何编程语言都可以！

解决方案

Roger Stafford 在这里给出了一些 MATLAB 代码:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/152405

为了实现这个功能，我有点冗长:

#给定一个负载点，用x,y坐标，我们可以估计半径# 通过使用最小二乘法将圆拟合到它们的曲率.函数 [r,a,b]=radiusofcurv(x,y)# 将点转换为质心坐标mx = 平均值(x)；我的 = 平均值(y)；X = x - mx;Y = y - 我的；dx2 = 平均值(X.^2);dy2 = 平均值(Y.^2);# 建立线性方程求导并求解RHS=(X.^2-dx2+Y.^2-dy2)/2；M=[X,Y];t = M\RHS;# t 是圆的中心 [a0;b0]a0 = t(1);b0 = t(2);# 从中我们可以得到半径r = sqrt(dx2+dy2+a0^2+b0^2);# 返回给定坐标系a = a0 + mx;b = b0 + 我的；端功能

它似乎对我的目的很有效，尽管它给出了非常奇怪的答案，例如共线点.但是，如果它们来自一个不错的曲线并增加了一点噪音，那么工作就大功告成了.

它应该很容易适应其他语言，但请注意，\ 是 MATLAB/Octave 的使用伪逆求解"函数，因此您需要一个可以计算伪逆的线性代数库来复制它.

I've got a series of (x,y) samples from a plane curve, taken from real measurements, so presumably a bit noisy and not evenly spaced in time.

x = -2.51509  -2.38485  -1.88485  -1.38485  -0.88485  -0.38485   0.11515 0.61515   1.11515   1.61515 ...

y =  -48.902  -48.917  -48.955  -48.981  -49.001  -49.014  -49.015  -49.010 -49.001  -48.974 ...

If I plot the whole series, it looks like a nice oval, but if I look closely, the line looks a bit wiggly, which is presumably the noise.

How would I go about extracting an estimate of the radius of curvature of the underlying oval?

Any programming language would be fine!

解决方案

Roger Stafford gave some MATLAB code here:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/152405

Which I verbosed a bit to make this function:

# given a load of points, with x,y coordinates, we can estimate the radius
# of curvature by fitting a circle to them using least squares.
function [r,a,b]=radiusofcurv(x,y)
  # translate the points to the centre of mass coordinates
  mx = mean(x);
  my = mean(y);
  X = x - mx; Y = y - my; 

  dx2 = mean(X.^2);
  dy2 = mean(Y.^2);

  # Set up linear equation for derivative and solve
  RHS=(X.^2-dx2+Y.^2-dy2)/2; 
  M=[X,Y];
  t = M\RHS;

  # t is the centre of the circle [a0;b0]
  a0 = t(1); b0 = t(2);

  # from which we can get the radius
  r = sqrt(dx2+dy2+a0^2+b0^2); 

  # return to given coordinate system
  a = a0 + mx;
  b = b0 + my; 

endfunction

It seems to work quite well for my purposes, although it gives very strange answers for e.g. collinear points. But if they're from a nice curve with a bit of noise added, job pretty much done.

It should adapt pretty easily to other languages, but note that \ is MATLAB/Octave's 'solve using pseudo-inverse' function, so you'll need a linear algebra library that can calculate the pseudo inverse to replicate it.

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