I see what you mean, so say if I am doing RDF, the values beyond 9A would be bogus?

Quoting "Justin A. Lemkul" <jalem...@vt.edu>:



nishap.pa...@utoronto.ca wrote:
Hello,

I wanted to know, if say my box is 30A (cubic) and if I set my cut-off lengths i.e. rvdw-switch = 0.8 and rvdw =0.9 , does that mean, the simulation will cut-0ff at 9A? Should I be changing that to 1.5nm (to get half the box cut-off)? I am a little confused about that.


Setting rvdw = 0.9 means that van der Waals interactions beyond 0.9 nm
are zero.   It has no effect on, for example, Coulombic interactions.

Most force fields have defined cutoffs that should be used in order to
be consistent with the original derivation of the parameter set, so it
is generally inadvisable to make ad hoc changes to the cutoffs in order
to meet some arbitrary criterion.  The minimum image convention
specifies that your longest cutoff must always be less than half the
smallest box vector.  If you set up a system with a cutoff equal to
exactly one half of a box vector, if that box vector decreases even
slightly (i.e., under the influence of pressure coupling), then you
will be calculating spurious forces.

-Justin

Thanks.

Nisha Patel



--
========================================

Justin A. Lemkul
Ph.D. Candidate
ICTAS Doctoral Scholar
MILES-IGERT Trainee
Department of Biochemistry
Virginia Tech
Blacksburg, VA
jalemkul[at]vt.edu | (540) 231-9080
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin

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Nisha Patel
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Leslie Dan Faculty of Pharmacy
Department of Pharmaceutical Sciences
144 College Street
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