Plotting a faired curve through a data set that curves in on itself

I am trying to plot a curve that goes through all the points in my data set using a fit method. I have tried using interp1 with smoothing spline and makima, I have tried fit, and I have tried polyfit. In all cases it does not work. My data set is values of CM (pitching moment) vs CL (lift coefficient). The first 11 of my 14 data points are increasing CL and decreasing CM, but then from the 12th to 14th data points, my values of CL decrease whilst the values of CM also decrease (due to stall). In this way, the curve goes in on itself. My question is, is there a way to plot a single curve that goes through (or at least mostly) goes through all of the points even though CL is not strictly increasing over the entire interval?

Here is what the raw data looks like:

Any help would be greatly appreciated. Many thanks in advance.

[the cyclist]

Can you upload the data and/or the code you've tried? You can use the paper clip icon in the INSERT section of the toolbar.

I expect you'll have more success fitting CL as a function of CM, rather than vice versa (even though that is not the "right" functional relationship).

Is there some functional form you expect (in which case fitting a non-linear function makes sense, and maybe there is error in the observations), or you mostly just want a smooth curve with no theoretical basis?

[Jayden]

I am onsidering expressing CM as function of CL but when I tried it, it produced an even worse fit (very large oscillations in the curve). I expect the curve to be a polynomial, I am not quite sure of the order I expect to get, in theory I expect mostly linear, at least pre stall, but post stall I am not sure.

The code I have used is the following:

CL data

-0.117323468280501

0.109533077818298

0.349535877911274

0.578159011570390

0.799525368741050

0.989291914638142

1.15205588979230

1.23246920496028

1.29909262861280

1.35437407382001

1.35912940969856

1.34136836172825

1.29541307340786

1.11501423385178

CM data:

0.150258158337099

0.117407098855973

0.0884047743485015

0.0670114607977137

0.0517955492445810

0.0305383477518936

0.00355285590629767

-0.00978950167650419

-0.0257317663408145

-0.0378647559486718

-0.0873543217991825

-0.0904845481307666

-0.103361954196992

-0.184021637962439

CM_i2 = fit(Data_i2.CL,Data_i2.CM,'smoothingspline');

i2 = plot(CM_i2,'g-.');

i2.LineWidth=2;

When I run this I get this plot:

[Voss]

Here's how it looks if you fit CL as a function of CM, as the cyclist suggested:

CL = [-0.117323468280501
0.109533077818298
0.349535877911274
0.578159011570390
0.799525368741050
0.989291914638142
1.15205588979230
1.23246920496028
1.29909262861280
1.35437407382001
1.35912940969856
1.34136836172825
1.29541307340786
1.11501423385178];

CM = [0.150258158337099
0.117407098855973
0.0884047743485015
0.0670114607977137
0.0517955492445810
0.0305383477518936
0.00355285590629767
-0.00978950167650419
-0.0257317663408145
-0.0378647559486718
-0.0873543217991825
-0.0904845481307666
-0.103361954196992
-0.184021637962439];

F = fit(CM,CL,'smoothingspline');
CMi = linspace(min(CM),max(CM));
CLi = F(CMi);
plot(CLi,CMi,'g-.',LineWidth=2);
hold on
plot(CL,CM,'.k')
xlabel('C_L')
ylabel('C_M')

[Jayden]

Yes, this is exactly what I was trying to acomplish, thank you very much for this answer and the Cyclist for your advice as well. Can I ask just so I actually understand the logic, when you create the new vector CLi by feeding CMi into F, does F basically act like a function of CM? So F = fit(CM,CL, 'smoothingspline') creates a function of CM, called F, that can be evaluated for any value of CL?

Again thank you very much.

[Voss]

"when you create the new vector CLi by feeding CMi into F, does F basically act like a function of CM? So F = fit(CM,CL, 'smoothingspline') creates a function of CM, called F, that can be evaluated for any value of CL?"

Yes, except "F ... can be evaluated for any value of CM".

[John D'Errico]

As you have been told, you CANNOT model this as a function. You CANNOT use interp1 as you are trying to do, or to use fit directly. In all those cases, trying to model C_M as a function of C_L would result in something that is not a single valued function. That is, for C_L == 1.3, exactly which single choice for C_M should the model return? That is what a single valued function would do. And that is why your attempts with a spline, with a polynomial, with interp1, etc. all failed. Any such tool MUST presume a single valued function. For any give value of x, there would be EXACTLY one y.

What can you do? You can use tools like cscvn. It allows a non-single valued function. Curves by that tool can follow any set of points you give it. The result will NOT be a simple function you can evaluate however, because again, if you want to evaluate a function, it must be single valued!

The other choice is to swap x and y, as has been suggested. In this case, swapping the relationship results in a single valued function that can be fit. (That need not always be the case. for example with points around a circle.) Once you do that on your data however, the result is a single valued relationship, x(y). Here that would be C_L(C_M). But it cannot predict in the other direction. How could it do that? If all you want to do is plot a curve from that data, you could do this:

CL = [-0.117323468280501
0.109533077818298
0.349535877911274
0.578159011570390
0.799525368741050
0.989291914638142
1.15205588979230
1.23246920496028
1.29909262861280
1.35437407382001
1.35912940969856
1.34136836172825
1.29541307340786
1.11501423385178];
CM = [0.150258158337099
0.117407098855973
0.0884047743485015
0.0670114607977137
0.0517955492445810
0.0305383477518936
0.00355285590629767
-0.00978950167650419
-0.0257317663408145
-0.0378647559486718
-0.0873543217991825
-0.0904845481307666
-0.103361954196992
-0.184021637962439];
spl = cscvn([CL,CM]')

spl = struct with fields:
      form: 'pp'
    breaks: [0 0.4788 0.9705 1.4496 1.9207 2.3577 2.7639 3.0494 3.3111 3.5490 3.7720 3.9063 4.1247 4.5693]
     coefs: [26x4 double]
    pieces: 13
     order: 4
       dim: 2

fnplt(spl)
hold on
plot(CL,CM,'ro')
xlabel 'C_L'
ylabel 'C_M'
hold off

It is a little bumpy on the right, but that is due to it trying to interpolate those points exactly. It needed to make some sharp turns. The curve generated by @Voss using a smoothing spline is nicer (smoother) looking, but it does not interpolate the data. You could also do this, as a pure interpolant:

spl2 = spline(CM,CL);
plot(CL,CM,'ro')
CMint = linspace(min(CM),max(CM));
hold on
plot(fnval(spl2,CMint),CMint,'-b')
xlabel 'C_L'
ylabel 'C_M'
hold off

So you need to make a choice. If all you want to do here is plot a smooth curve through the data, then there are at least a few options, depending on how smooth of a curve you want. Do you want a curve that follows the data exactly? Thus, an interpolant? Do you want a smoother curve, that comes close to the data?

[Jayden]

Thank you very much John for your insight. I see that trying to fit a curve to non-monotonic data sets is not as straightforward as I initially hoped! Can I ask what the benefit of using interpolation as opposed to smoothing spline is? As I understand it you can obtain R squared values and rmse from fit so is it just that interpolation can be applied to data set outside of original range of points? Sorry if my understanding of this is flawed.

Another approach I considered was trying to express CL and CM both as a function of some other parameter, say t, to allow me to plot(x,y,t) but I was unsure of how to find the t parameter given a set of data X, Y. Would this also be a valid approach?

Answer 1

[Mathieu NOE]

hello

experimental airfoil data can exibit some randomness especially at high angle of attack (stall)

we usually prefer to smooth the data rather to go exactly through the points (IMHO)

so a very simple code that interpolates and smooth the data :

CL = [-0.117323468280501
0.109533077818298
0.349535877911274
0.578159011570390
0.799525368741050
0.989291914638142
1.15205588979230
1.23246920496028
1.29909262861280
1.35437407382001
1.35912940969856
1.34136836172825
1.29541307340786
1.11501423385178];

CM = [0.150258158337099
0.117407098855973
0.0884047743485015
0.0670114607977137
0.0517955492445810
0.0305383477518936
0.00355285590629767
-0.00978950167650419
-0.0257317663408145
-0.0378647559486718
-0.0873543217991825
-0.0904845481307666
-0.103361954196992
-0.184021637962439];

% interpolate 
CM2 = linspace(min(CM),max(CM),100);
CL2 = interp1(CM,CL,CM2,'linear');
% smooth
 N = 20;
CL2s = smoothdata(CL2,'lowess',N);
 
plot(CL,CM,'*-');
hold on 
plot(CL2s,CM2,'-');
xlabel 'C_L'
ylabel 'C_M'
hold off

[Mathieu NOE]

hello again

problem solved ?

please accept the answer that best matches your expectations - it rewards people who have worked on your problem