Author 
Comment/Response 
Bill Simpson

10/26/12 9:53pm
In Response To 'Re: Re: Re: Re: Re: Re: Re: Re: Plotting nonli...'  Use your function and Sigma Plot's coefficients.
In[8]:= fy2[x_]=(b(Sqrt[b^22k^2 c x/s]))/(2k);
k = 2900000;
s = .84;
In[9]:= fy2[.79333]
Out[9]= 0.47222
In[10]:= fy2[10.313]
Out[34]= 1.32
Those seem to match your calculations.
Now a plot to see how good the fit is.
In[11]:= Show[ListPlot[data,Joined>True],
Plot[fy2[x],{x,0,10.313}],
PlotRange>All]
Out[11]= <plot snipped>
That fits the leading edge well, but nothing else.
Try FindFit with Sigma Plot's starting values.
In[12]:= k=.;s=.;
fit=FindFit[data,y2,{{k,2900000},{s,.84}},x]
Out[12]= {k> 0.21075,s> 0.184977}
Those are very odd results, how good is the fit?
In[13]:= k = 0.21074963872741664;
s = 0.18497695239903436;
fy2[x_] = b(Sqrt[b^22 k^2 c x/s]))/(2k);
Show[ListPlot[data, Joined>True],
Plot[fy2[x], {x, 0, 10.313}],
PlotRange>All]
Out[16]= <half way decent fit to your data snipped>
From this I think you might see why the first thing I always do is plot my fitted function over the data.
At this point I don't know what else to say. If all we have to go on is your student thought the value was supposed to be about a million then I am at a loss. If you could somehow scale your y2 down by 25% then it might not be a bad approximation, but I don't have any basis for doing that. If we can somehow track down and point out a likely calculation error anywhere then I'd be happy to try to pinpoint where that is happening.
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