Yes -- Because of the form of the model, this can be done by constraining b. A notebook is attached. However, this is no longer the best fit of the model to the data -- it is the best fit subject to the constraint.

Best, David

Attachments:

Hi Syed, In the attached notebook, I made two changes. I deleted the 10th data element, which is clearly an outlier. I then added a scaling parameter for x. Without the scaling parameter, the model is not a good model for the data.

nlm3 = NonlinearModelFit[data2, 
  b ((0.1262/((1.025 - Cos[k x Degree])*Cos[k x Degree]))^3), {b, k}, 
  x]
Show[ListPlot[data2], Plot[nlm3[x], {x, 0, 30}], PlotRange -> All]

In[16]:= nlm3["AdjustedRSquared"]

Out[16]= 0.986928

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Hi David,

I managed to get a reasonable fit with the experimental data and the equation for a set of data at a particular condition. I did not use a scaling factor as I am using the equation directly from the literature, to see how my data fits.

Is it possible to set a constraint so that the value of y ? 100 when x = 0, when solving for b.

The nb is attached for your reference.

Attachments:

Hi David,

Thanks for that. The plot looks more like the experimental data in terms of shape.

However, the r squared comes out very low at 0.66 and the model plot is much lower than the exp data. Any idea why that might be?

Thanks,

Hi Jofre,

Thanks for your reply.

When I plot the data it does not look anything like the experimental data with the given b value, and the r squared comes out very low (0.13). For this reason I assumed something is wrong.

The nb file is attached, so you can see how the graph looks.

Much appreciated for your help.

Attachments: