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Fit a conic section without mirror image hyperbola

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x I'm tc
x I'm tc 2020 年 1 月 13 日
編集済み: Matt J 2020 年 1 月 14 日
I would like to fit conic sections to my data using something like the general form a.*x.^2+b.*x.*y+c.*y.^2+d.*x+e.*y+f. I have found a number of tools for this on MATLAB Central, and am currently using https://www.mathworks.com/matlabcentral/fileexchange/22683, which very often works brillianlty. One of the things I am curiuos to know is when my data are best-approximated by an ellipse and when they are best approximated by a hyperbola, so I don't want to force on or the other. When the data are ellipsiodal (or theoretically parabolic or circular, but that never happens in real life) I get exactly what I want. And when the data are hyperbolic, very often I get what I want, but sometimes the best-fitting hyperbola involves both sides of the hyperbola pair. I want to constrain the fit so that only one "mirror image" of a hyperbola can be used.
Here is an example: For the data below, I hope to get something like the blue hyperbola (just drawn by hand), but instead, I get the orange hyperbola pair as the best fit. I want to force the fit to use only one half of the pair. But I don't want to exclude the possibility of getting an ellipse as the best fit.
Picture1.png
XY = [0, -2.975; ...
2.5, -2.975; ...
5, -2.75; ...
7.5, -2.25; ...
10, -2.41667; ...
15, -2.25; ...
20, -1.75; ...
25, -1.9; ...
30, -1.675; ...
35, -1.575; ...
40, -1.625; ...
-2.5, -3; ...
-5, -3.25; ...
-7.5, -3.375; ...
-10, -3.08333; ...
-15, -2.85; ...
-20, -2.225; ...
-25, -2.0625; ...
-30, -1.375; ...
-35, -1.5625];
Any tips?

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Matt J
Matt J 2020 年 1 月 14 日
Is there any way to constrain the fit to have only the one hyperbola,
See my answer.
but still include the other forms?
It's probably not possible (or recomendable) with a single fit operation. The simplest solution would be to just force fit each of the different families of conic sections to your data and choose the conic type that gives the least fitting error.
x I'm tc
x I'm tc 2020 年 1 月 14 日
Unless I am mistaken--and I very well might be--the answer you provided only fits hyperbolas that "open" up or down, rather than to the left or right, which is quite possible. I think this is rather a hard problem, since the "standard form" of the conic section lends itself to solutions that have two values in Y for given points X and two values in X for given points, Y. It is also not immediately obvious how to determine which error is better, since we want an orthogonal regression value, not a minimization in either Y or X.
Matt J
Matt J 2020 年 1 月 14 日
It is also not immediately obvious how to determine which error is better, since we want an orthogonal regression value, not a minimization in either Y or X.
Then your question is really, "how do I do orthogonal regression for a 1-sided hyperbola"? If you have an orthogonal regression routine for ellipses, then once you have one for hyperbola, my answer applies.
the answer you provided only fits hyperbolas that "open" up or down, rather than to the left or right, which is quite possible.
No, it will fit an arbitrarily oriented hyperbola. In its current form it is not an orthogonal regression, though. Do you know for certain that the Taubin algorithm you are currently using from the File Exchange is orthogonal regression? I don't think it could be, because it claims to be non-iterative. I don't think it's possible to do orthogonal regression of conics non-iteratively. Are you sure you need a fit as robust as an orthogonal regression? It's a lot of extra effort...

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回答 (2 件)

Matt J
Matt J 2020 年 1 月 14 日
編集済み: Matt J 2020 年 1 月 14 日
This approach uses John's fminspleas FEX submission (Download). Although the fit is, technically, a hyperbola, it diverges to a piecewise linear fit here when you don't impose some lower bound on the inter-focal distance. Apparently, there is no curvature suggested by your data, so the curve fitting routine's iterations try to make the hyperbola as sharply pointed as possible.
lbound=0;
tmin=fminsearch(@(t)thetaCost(t,XY,lbound),0);
[fitError,xy0]=thetaCost(tmin,XY,lbound);
lbound=10;
tmin=fminsearch(@(t)thetaCost(t,XY,lbound),0);
[fitError,xy5]=thetaCost(tmin,XY,lbound);
plot(XY(:,1),XY(:,2) ,'o',xy0(:,1),xy0(:,2),'r.-',xy5(:,1),xy5(:,2),'k.-');
legend('Raw Data', 'Inter-Focal Distance >=0','Inter-Focal Distance >=10')
function [r,xy,p]=thetaCost(theta,XY,lbound)
R=[cosd(theta), -sind(theta); sind(theta),cosd(theta)];
xy=XY*R;
x=xy(:,1); y=xy(:,2);
[~,idx]=min(x);
p0=[x(idx),y(idx)];
flist={@(p,xd) sqrt(((xd-p(1))/p(2)).^2+1),1};
[p,AD]=fminspleas(flist,p0,x,y,[-inf,lbound/2]);
A=AD(1);D=AD(2);
x0=p(1);a=p(2);
yfit=@(x) A*sqrt(((x-x0)/a).^2+1)+D;
r=norm( yfit(x) - y,1);
if nargout>1
x=linspace(min(x),max(x),1000).';
xy=[x,yfit(x)]*R.';
p=[theta, A,x0,a,D];
end
end

  2 件のコメント

x I'm tc
x I'm tc 2020 年 1 月 14 日
Thank you very much, but if I am reading the code right, this will force a hyperbolic fit, but will not fit an ellipse, when that's better, which is what I need to do. I will post an update in a moment clarifying what I'm looking for.
Matt J
Matt J 2020 年 1 月 14 日
No clarification is needed. You must force fit your points with both an ellipse and a hyperbola and use the resulting fit error from each to decide which is better.

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John D'Errico
John D'Errico 2020 年 1 月 14 日
編集済み: John D'Errico 2020 年 1 月 14 日
This goes back to basic algebra. When does a conic section describe an ellipse versus a hyperbola?
Compute the discriminant, thus b^2 - 4*a*c. If it is negative, then the result is an ellipse, if positive, a hyperbola.
The point being, while the unknowns in the conic appear to be essentially linear, the choice of which specific form you get is a nonlinear thing. There are some special cases where you can get a circle, or a parabola.
Next, if you just want one of the branches? That enters in a nonlinear way too. You can probably think of it as a choice of whether the positive or negative branch of a sqrt is taken.
Effectively all of that makes the choices you want to be made in some automatic way not so trivial. The basic tools you will find to do this are not that intelligent, as to allow you to make those choices.
There are always things we want. Achieving our goals often takes work.

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x I'm tc
x I'm tc 2020 年 1 月 14 日
Thank you very much, but I know how to tell what I have, I just want to be able to constrain what I have. I will post an update.

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