Calculate a curve out of points and the curvature radius

I have a curve with a variable curvature radius and i want to calculate a function tha defines that radius.

x and y are the coordinates of some points in the x an y axis, and with that ponts i want to calculate a curve fitting them with splines and the calculate de curvature radius of that curve.

I dont know why theres is an error in this function.

Does anyone know how to solve it?

Thanks

function [r] = rtrans2(x,y)
%UNTITLED3 Summary of this function goes here
%   Detailed explanation goes here


s=csapi(x,y);

r=((diff(s,2)).^2)/(1+diff(s,1).^2);

end

[Sam Chak]

@Jon Bilbao, Can you run a test case to show us the generated error message in red?

x = 1;
y = 1;
rtrans2(x, y)

Error using chckxywp
There should be at least 2 data sites.

Error in csapi>csapi1 (line 90)
[xi,yi,sizeval] = chckxywp(x,y);

Error in csapi (line 75)
pp = csapi1(x,y);

Error in solution>rtrans2 (line 8)
s = csapi(x,y);

function [r] = rtrans2(x,y)
    %UNTITLED3 Summary of this function goes here
    %   Detailed explanation goes here
    s = csapi(x,y);
    r = ((diff(s,2)).^2)/(1+diff(s,1).^2);
end

[Jon]

It looks like you are using some additional functions (csapi, csap1,chckxywp) that are not included in my MATLAB distribution. Since this is where the errors are occuring you would need to provide these functions (attach with paperclip) to allow further debugging

[Sam Chak]

Hi @Jon

It appears that csapi() is a built-in function. I think x and y are coordinates (not scalars).

https://www.mathworks.com/help/curvefit/csapi.html

But a closer look shows that the supplied formula for the curvature is inaccurate.

https://en.wikipedia.org/wiki/Curvature#Graph_of_a_function

[Jon]

Hi @Sam Chak thanks for clarifying that, I don't have the curve fitting toolbox, which explains why I did not see the csapi function in my MATLAB. As I don't have that toolbox, I will not be able to help too much with this one.

[Jon Bilbao]

@Sam Chak yes you are right, thanks, the formula of the curvature radius is wrong.

[John D'Errico]

@Jon Bilbao - You could not use diff anyway on a spline. You could use fnder, but again, you don't have the curvefitting toolbox. Worth having, IMHO. But differentiating a spline is not that difficult. So you could differentiate a curve created by spline instead of csapi. Most important is to know if an interpolating spline is a good fit to your data. Is there (significant) noise in the data? How much noise? If there is, then computing an interpolating spline fit, and then hoping to find the radius of curvature from that will be a waste of time. And since we don't seee any data, it is difficult to know. But remember that computing a radius of curvature from an interpolating spline will be a highly noisy thing to do. It will greatly amplify any tiny noise in your data.

[Jon]

@John D'Errico I think the original poster, @Jon Bilbao, has the curve fitting toolbox. It was me @Jon responding, who said they didn't have the curve fitting toolbox. A little confusing as we both have the same first name

[Jon Bilbao]

The problem is solved thanks , with fnder the function worked, the function is the next:

function [r] = rtdef(x,y,Lc)

xi = 0:0.1:Lc;

pp = csapi(x, y);

yi = ppval(pp, xi);

yp = fnval(fnder(pp), xi);
ypp = fnval(fnder(pp,2), xi);

r = ((1 + yp.^2).^(3/2)) ./ abs(ypp);
end

[Sam Chak]

@Jon Bilbao, good to see that you've found the solution. I'd advise you to move your solution in the comment to the Answer section.

Answer 1

[Jon Bilbao]

Solved:

function [r] = rtdef(x,y,Lc)
xi = 0:0.1:Lc;
pp = csapi(x, y);
yi = ppval(pp, xi);
yp = fnval(fnder(pp), xi);
ypp = fnval(fnder(pp,2), xi);
r = ((1 + yp.^2).^(3/2)) ./ abs(ypp);
end