Fitting plot and data to an equation

Question

If I have the following data:

data={{2, 66.7635}, {Log[300]/Log[10], 69.9679}, {Log[600]/Log[10], 
  71.54}, {3, 72.2428}, {-2.30103, 54.0023}, {-(Log[60]/Log[10]), 
  55.1941}, {-(Log[20]/Log[10]), 56.0038}, {-1, 
  56.9497}, {-(Log[6]/Log[10]), 57.305}, {-(Log[10/3]/Log[10]), 
  57.7213}, {-(Log[2]/Log[10]), 58.2489}, {-2.30103, 
  54.0367}, {-(Log[60]/Log[10]), 55.1157}, {-(Log[20]/Log[10]), 
  56.1704}, {-1, 56.7117}, {-(Log[6]/Log[10]), 
  57.2506}, {-(Log[10/3]/Log[10]), 57.7097}, {-(Log[2]/Log[10]), 
  58.1068}}

Which looks like this plotted:

I have two questions:

1) How can I fit and plot the fit of this data based on the following equation?

: where Tf'_ref=57.2506 , q_ref=0.166667 and c1 and c2 are the fitting parameters. Also, notice that data is Tf' vs Log q in the equation. "Log" in the equation refers to based 10 logarithm and not to natural logarithm.

2) How can I find the values of c1 and c2 which are the fitting parameters.

The fitting (orange line) is supposed to look like this (done in excel):

with fitting parameters c1=9.43903909249581 and c2=23.1214485816819

EDIT: I tried using NonLinearFitModel like this: Table[{NonlinearModelFit[data, Logqref - ((c1*(data[[i, 2]] - Tfref))/(c2*(data[[i, 2]] - Tfref))), {{c1, 8.6}, {c2, 17.2}}, x]; }, {i, 1, 11}] but this does not work. The reason I tried this is because data[[i, 2]] represents Tf' in the equation. Here Logref=Log10[0.16667]

EDIT2: When I do it in Excel c1=9.43903909249581 and c2=23.1214485816819. When I put this parameters into the equation they seem to fit the equation very well as well. Since the parameters found by @Bob Hanlon seems to fit the data very well as well, 3) How can I know what is the best parameters to fit and how could mathematica also find the parameters I found in excel, which seems to have a lower sum of errors squared?

Answer 1

Clear["Global`*"]

data = {{2, 66.7635}, {Log[300]/Log[10], 69.9679}, {Log[600]/Log[10], 
    71.54}, {3, 72.2428}, {-2.30103, 54.0023}, {-(Log[60]/Log[10]), 
    55.1941}, {-(Log[20]/Log[10]), 56.0038}, {-1, 
    56.9497}, {-(Log[6]/Log[10]), 57.305}, {-(Log[10/3]/Log[10]), 
    57.7213}, {-(Log[2]/Log[10]), 58.2489}, {-2.30103, 
    54.0367}, {-(Log[60]/Log[10]), 55.1157}, {-(Log[20]/Log[10]), 
    56.1704}, {-1, 56.7117}, {-(Log[6]/Log[10]), 
    57.2506}, {-(Log[10/3]/Log[10]), 57.7097}, {-(Log[2]/Log[10]), 
    58.1068}};

EDIT:

eqn = ((log10q - Log10[qref]) == c1*(Tfp - Tfpref)/(c2 + (Tfp - Tfpref)));

Solve the equation for Tfp to obtain the model

model = Tfp /. Solve[eqn, Tfp][[1]] // FullSimplify

(* Tfpref + c2 (-1 + (c1 Log[10])/((c1 - log10q) Log[10] + Log[qref])) *)

const = {Tfpref -> 57.2506, qref -> 0.166667};

Substituting the constants into the model

model2 = model /. (const // Rationalize) // FullSimplify

(* 286253/5000 + 
 c2 (-1 + (c1 Log[10])/((-6 + c1 - log10q) Log[10] + Log[166667])) *)

Using NonlinearModelFit with a constraint on the parameters

nlm = NonlinearModelFit[data,
     {model2, c1 > 5, c2 > 5}, {c1, c2}, log10q];

param = nlm["BestFitParameters"]

(* {c1 -> 9.29797, c2 -> 22.629} *)

nlm["ParameterTable"] // Quiet

confidInterval = Thread[{c1, c2} -> Around@*Interval /@
    nlm["ParameterConfidenceIntervals"]]

(* {c1 -> Around[9.297969154154664, 1.0182007406816869`], 
    c2 -> Around[22.62894607083057, 3.89519370692895]} *)


Plot[nlm[log10q],
 {log10q, Min[data[[All, 1]]], Max[data[[All, 1]]]},
 Epilog -> {Red, AbsolutePointSize[4], Point[data]},
 Frame -> True,
 Axes -> False,
 GridLines -> Automatic,
 GridLinesStyle -> Lighter[Gray, .8],
 FrameLabel -> (Style[#, 14, Bold] & /@
    {HoldForm@Log10["q (K/s)"], 
     Row[{Subscript[T, f'], " (", Degree, " C)"}]})]

Answer 2

Your example is a Table of nonlinear fits rather than applying NonlinearModelFit to all the data at once.

Two issues in using NonlinearModelFit in your case:

1) your equation gives Log[q] in terms of Tf whereas NonlinearModelFit needs an equation for the output, in this case Tf, in terms of the input, i.e., Log[q].

2) the Tf values in your data extend below the reference temperature, so unless C2 is large enough, the fit could divide by zero somewhere in the range of temperatures in your data.

For issue 1), you could either reverse the order of your data, fitting Log[q] in terms of Tf, or rewrite your equation to give Tf in terms of Log[q]. For issue 2), NonlinearModelFit may require specifying initial guesses for C1 and C2.

Here's how to use NonlinearModelFit by rewriting your equation, using logq to refer to Log[q]:

expr = Simplify[Tf /. Solve[
   PowerExpand@Log[q/qRef] == (C1 (Tf - TfRef))/(
    C2 + (Tf - TfRef)), Tf][[1]] /. 
Log[q] -> logq /. {TfRef -> 57.2506, qRef -> 0.166667}]

To fit to your data, I use initial guesses of 10 for the parameters:

fit = NonlinearModelFit[data, expr, {{C1, 10}, {C2, 10}}, logq]

To get fit parameters:

fit["BestFitParameters"]

and to plot:

Show[ListPlot[data], Plot[fit[logq], {logq, -3, 3}]]

Comparing the fit with the data isn't as close as the Excel plot you provide. Is it possible the reference q value is for log base 10, i.e., Log10[q] in your equation?