Hi, > > If we then require that the rms change in intensity be greater than the > average noise, then we can write down the requirement: > sigma(I_P) < rms(deltaI) > . . . > I_P/sigma(I_P) > 1.3*sqrt(MW/N_H)/fpp
one can actually require that abs(delta I) > 3 sigma(delta I) Using eq. 13 from Acta Cryst. D61, 1437–1448, and the same (sometimes (very) questionable assumptions) James used, the magic factor 1.3 inflates to 2.0 I once wrote a jiffy that allows one to simulate FOM's for various phasing + errors scenarios. I still have the code around, if someone is interested. Note that sometimes you cannot find sites, but with knowledge of the sites itself, phasing + density modification would result in interpretable maps (Acta Cryst. D60, 1085–1093). Also, the distribution of errors is rather important. I suspect that what matters in the end is that you have enough well-measured Bijvoet pairs to get the substructure. An empirical analysis possibly could shift the magic number back to 1.3 ;-) Cheers, P SAD Scientist ;-) > So, after doing these substitutions and rearranging, we get: > > sigma(I_P)/I_P < sqrt(2*N_H/(MW/14))*fpp/7 > sigma(I_P)/I_P < 0.756*sqrt(N_H/MW)*fpp > > I_P/sigma(I_P) > 1.3*sqrt(MW/N_H)/fpp > > There are some obvious approximations here. Probably the biggest is > assuming that fpp = F_H. In actual fact, anomalous differences "count > double" since fpp contributes both to F+ and F-. I think Peter Zwart > pointed this out earlier. There is also another sqrt(2) in the opposite > direction because sigma(delta-I) is the quadrature sum of two sigma(I_P). > It also matters if you are interested in the rms anomalous difference or > the mean absolute anomalous difference, as these are not the same thing. > Nonetheless, I think this last formula should be accurate to at worst a > factor of two. > In general, it is a good idea to have your signal be more than equal to > noise, so I consider this formula a limit to be avoided rather than a goal > to be met. The skill and expertise required to solve the structure > increases quite sharply as your I/sigma(I) approaches this limit, but you > can always double I/sigma(I) by merging data from four crystals. The latter > is a better strategy. > > -James Holton > MAD Scientist >
