How to fit the data points into a closed curve?

Hello everyone, I’d like to ask you a question. Now I have a series of data points and want to fit these points into a closed curve. But the effect is not good, I would be very grateful if anyone can help me. The data required for closed curve fitting is shown below.

[KSSV]

What did you try? Did you try with polyfit?

[John D'Errico]

Please don't use an answer just to make a comment. I've moved your answer into this comment.

[John D'Errico]

What about that data do you want to be a closed curve? No, you cannot use interpolation using tools like interp1 to fit the black curve.

Answer 1

[Stephan]

This will not solve your problem finally, but it shows a possible way to tackle this. The idea here is to use a cardiod curve and find parameters that try to minimize the difference between the single data points:

data = readmatrix('DataFit.xlsx');
a = [72.0839, 1.3643, 338.485, 289.4648, -21];
[x,y] = calcCurve(a);

figure
scatter(data(:,1),data(:,2))
hold on
scatter(x,y)
hold off

function [x,y] = calcCurve(a)
    theta = linspace(-pi,pi,1187);
    r = @(theta) a(1).*(a(2)+cos(theta));
    rho = r(theta);
    [x,y] = pol2cart(theta,rho);
    x = -x';
    y = -y';
    xy = sortrows([x,y],1);
    x = xy(:,1) + a(3);
    y = xy(:,2) + a(4);
end

The parameters for a were found by optimization techniques - maybe this is a local optimum and we could do better. But the point is, that you have to think about your mathematical problem and analyze what you have there. After that it makes sense to start coding:

Answer 2

[Wesley]

The curve you fit out is a bit larger than the original image. I am thinking of a fitting method that is more appropriate to the original curve. The method described in this article feels very feasible. But I don't know how to implement it with programs. The article is as follows:

Answer 3

[Wesley]

For convenience, we use this data for closed curve fitting.