This application is just working very well and is very useful.
I am for the moment working on version 2011a and will soon update to 2013. Do you know if any update will be done in order to be compatible with this version ? (To avoid to loose my work...)
Anyway, great job !
There is a typo there on my part; the number should be 1.06447.... Here's why: The simplest expression for a Gaussian peak is y=exp(-x^2). This has a full-width at half maximum of 2 * sqrt(ln(2)=1.66561 and the area under the curve is sqrt(pi)= 1.77245. So the width/height ratio is 1.06447.... The peak area is proportional to both peak height and width. Therefore the area is 1.06447 * peak height * peak width. Hint: You can use Wolfram Alpha (http://www.wolframalpha.com) to do all the algebra for stuff like that.
Sorry about this really basic question, but could you please explain to me the reasoning behind calculating the area of the peak using 1.0646 * peak height * peak width? I'm not sure where the 1.0646 value comes from.