# How is “F2/sig(F2)” the data quality in SXRD?

When I recieve a dataset from single crystal X-ray diffraction, I can determine the quality from a graph in which F2/sig(F2) is displayed in dependence of the frame.

But I forgot the basics... I have found that the $R_{int}$ value is calculated as such: $$R_{int}=\frac{\Sigma|F^2-F_{mean}^2|}{\Sigma|F^2|}$$ Reference: http://www.canadiancrystallography.ca/cccw/files/CIF-file%20and%20validation.pdf
But what exactly is F2/sig(F2) now?

• Would "sig" be short for sigma $\sigma$ (i.e. the standard deviation). – Ben Norris Jan 23 '15 at 20:51
• That could be as well as the sum $\Sigma$. Maybe F2/sigma(F2) is part of the formula by which $R_{int}$ is defined. But would that make sense? – Crystal Lettuce Jan 23 '15 at 20:59

$R_{int}$ considers in this summation symmetry relevant reflections, e.g. in your equation the term $\Sigma|F^2-F_{mean}^2|$ takes an intensity of reflection and subtracts from this number the mean intensity of all its symmetry equivalents. If redundancy of your data collection is high (the case for all area detectors), then with this number you can tell how close in intensity are the equivalents, and thus how good is your crystal.
Another measure of quality is $R_{\sigma}$ $$R_{\sigma}=\frac{\Sigma[{\sigma}(F^2)]}{\Sigma|F^2|}$$ It tells you how close are the estimated standard deviation of equivalents (e.g.: were all equivalents collected with the same precision?)
$F^2/\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R_{int}$ at all. Indeed, to calculate $R_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\sigma(F^2)$, you have it as soon as you've collected a little bit of data.