I am trying to write a code to solve a simple problem as shown below. I believe what am I am doing is correct up until i begin the spline fitting method. Am i interpolating correctly?

Dd=[6 3 2 1.5 1.2 1.1 1.07 1.05 1.03 1.01];
c=[0.88 0.89 0.91 0.94 0.97 0.95 0.98 0.98 0.98 0.92];
a=[0.33 0.31 0.29 0.26 0.22 0.24 0.21 0.2 0.18 0.17];

ktoriginal=(c.*((0.5.*Dd)-0.5).^(-a)); %stress concentration factor for a stepped circular shaft
%based on table

%1)obtain an expression for c(Dd) annd a(Dd) with a 5th order polynomial

c5fitpoly = polyfit(Dd,c,5); %polyfit finds coefficents of a polynomial of degree 5 that fits a as a function of Dd
a5fitpoly = polyfit(Dd,a,5);

%next use polyval to determine the values for a and c using the coefficient
%vectors found for each

c5=polyval(c5fitpoly,Dd); %evalutes coefficent vector at each Dd
a5=polyval(a5fitpoly,Dd);

%calculating new kt values from fitted data
ktfitpoly5=(c5.*((0.5.*Dd)-0.5).^(-a5));

%comparing this kt with the original kt
polyerror=abs(ktoriginal-ktfitpoly5)';

%2)obtain an expression for c(Dd) annd a(Dd) with spline interpolation
cspline=interp1(c,Dd,'spline'); 
aspline=interp1(a,Dd,'spline');

ktspline=(cspline.*((0.5.*Dd)-0.5).^(-aspline));

splineerror=abs(ktoriginal-ktspline)';

disp('The error of kt using the polyfit method gives an error of:')
disp(polyerror)
disp('for each kt, whereas the error of kt using the spline method gives an error of:')
disp(splineerror)
disp('for each kt. In conclusion the first method is better to use as the errors are smaller.')

[Rik]

I don't see any obvious mistakes, so if it runs without error, I don't see why this would be wrong.

You could also interpolate to a finer grid, to see what happens between values, to get a better idea of what the difference between the two methods is.

[Caylyn MacDougall]

It does run without error, however my biggest concern i suppose is whether i should be wruting interp1(Dd,c,'spline') rather than interp1(c,Dd,'spline')

[Caylyn MacDougall]

also with what i currently have, I find that using polyfit is more accurate. however im pretty sure interpolation is actually more accurate

Answer 1

[Rik]

I just read the documentation page for interp1, and it turns out you are using this syntax:

vq = interp1(v,xq,method)

What you should be using is this:

vq = interp1(x,v,xq,method)

So for your code that means this:

%2)obtain fitted values for c(Dd) annd a(Dd) with spline interpolation
cspline=interp1(Dd,c,Dd,'spline'); 
aspline=interp1(Dd,a,Dd,'spline');

[madhan ravi]

Yes you are right

cc=linspace(c(1),c(end),1000);
aa=linspace(a(1),a(end),1000);
cspline=interp1(c,Dd,cc,'spline'); 
aspline=interp1(a,Dd,aa,'spline');

[Caylyn MacDougall]

Thank you guys! i mistakenly assumed using that syntax was valid, thanks for pointing out my mistake! i know get an error of zero using the spline method as expected.

[Caylyn MacDougall]

actually, would this syntax not just essentially be returning the same values for c and a as they are provided since the 'xi' we are interpolating at is the same original x vector provided, Dd. This seems redundant. Also am I correct in my statement that the spline fitting method is more accurate?

[Caylyn MacDougall]

@rik not @madhan. @madhan's answer may be more what im looking for, taking my last comment into context

[Rik]

It is indeed a bit redundant. The reason I wrote it like this is because you have a check at the end that depends on having the true value as well. If you don't, then you can indeed use a finer interpolation, as I suggested and as Madhan wrote.