Why do you want to quote "sigma" level anyway? It's more or less meaningless for the reasons you give. Stick to e/A^3
</flame> Phil On 22 Dec 2010, at 22:02, Dale Tronrud wrote: > Hi, > > I have a crystal structure at 3A resolution with six copies in the > asu. When I average the map over the ncs I find that the original > 2Fo-Fc style map has a sigma of 1.5 at 0.3 e/A^3. When I adjust the > contour level of the averaged map to match, by eye, the level of the > unaveraged map I find them equivalent at a sigma of 2.6 at 0.28 e/A^3. > These results imply that the "sigma" level of the original map was > 0.2 e/A^3 and the averaged map was 0.11 e/A^3. > > The "sigma" of a 2Fo-Fc style map is not an estimate of uncertainty, > of course, because nearly everything in the map is signal. It is > just a measure of the variability of the signal, i.e. the rms. With > averaging the signal should be preserved and the noise reduced, but > the noise of a 2Fo-Fc map is small compared to the signal. How is > it that the rms of my averaged map drops to half of the unaveraged > value and yet the electron density looks about the same when contoured > at the same e/A^3 (0.3 vrs 0.28)? > > I guess the real question is, how does Coot calculate the "sigma" > of an averaged map? You can't calculate the rms over the asymmetric > unit because the asymmetric unit is many millions of unit cells in > size (and hugely variable depending on small changes in the ncs > operators). > > The problem at hand is that I want to quote the sigma level I > insisted upon when creating water molecules and think it will sound > weird if I say I used a value of 12, which I did. The numbers just > don't seem right to me so I'd like a little reassurance. > > Dale Tronrud
