Hi Keith-- > > I'm using a somewhat older force field (see below. The Lennard-Jones > potential is implemented differently.
Are you using a different version of Xplor-NIH than what we distribute? > For example S (radius = 1.05 angstroms) That seems rather short. > sigma is 3.7458 in this setup. After double checking the > formula given (sigma*2^(-5/6))I believe is wrong - it should include > division by 2 . So for S I get 3.7458(0.561)/2 = 1.05 angstroms. > > I pretty sure the sigma I should use is 4.635 to approximate r = 1.20 > angstroms. > In the nonbonded VDW potential used in Xplor-NIH, http://nmr.cit.nih.gov/xplor-nih/doc/2.34/xplor/node118.html (hopefully this is working by the time you look) the basic form of the potential is: E = sigma^12/R^12 - sigma^6 /R^6 where R is the interatomic distance, not radius (rather, it's 2*radius for interactions between two of the same type of atoms). The minimum in this energy is at R = sigma * 2^(1/6). Therefore, atomic radius = sigma * 2^(1/6) / 2 = sigma * 2^(-5/6) best regards-- Charles _______________________________________________ Xplor-nih mailing list Xplor-nih@cake.cit.nih.gov http://cake.cit.nih.gov/mailman/listinfo/xplor-nih
