How to interpolate a graph to find a point halfway from the max peak

Hi, I am wondering how to find a point halfway from the maximum point (thats pointing down the way) on the graph I created:
http://www.2shared.com/photo/pIak1Za9/graph1.html
It was made using the code
I = imread('f_000021.tif','TIFF'); imshow('f_000021.tif') x=[128 180]; y=[50 50]; improfile(I,x,y) plot(max)
There isn't many pixels plotted so I was wondering if I could pick a point in between a couple of pixels on that graph (half of the maximum value)?
Thanks, Graham

5 commentaires

It is not getting clear to me: Do you have the values of the curve, or just the image? What is the result you want to achieve? You search the "point halfway from maximum point" to which other point?
In the image I have selected a line across a beam and plotted the intensity of white to black which gave me the graph I uploaded. I would then like to find the edge of the beam from the graph which is somewhere along the big peak downwards. I was wondering if I could interpolate one half of the peak and find a point halfway down it. I hope this makes everything clearer, Thanks Graham
As far as I understand, the fact that you measure something in an image does not matter the question at all - correct? You have some values in a vector and search for the index, where the line crosses the half of the range - did I understand this correctly?
Yes I think so, Sorry its not so easy for me to convey what I am trying to do - my mathematics language is not so good!
No problem. Finding a solution is often a iterative process, which clears the question at first. If the question is absolutely and mathematically clear, the solution is usually found already.

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 Réponse acceptée

Do you search for something like this?
V = sin(0:0.01:pi/2);
maxV = max(V);
minV = min(V);
Range = maxV - minV;
HalfIndex = find(V > maxV - Range / 2, 1, 'first');
You have to adjust the signs and > or < according to your exact problem.

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