AMBER Archive (2009)
Subject: RE: [AMBER] IDIVF and PN
From: Ross Walker (ross_at_rosswalker.co.uk)
Date: Thu Jul 16 2009 - 22:40:11 CDT
> If the author didn't specify the IDIVF factor in dihedral parameter in
> his paper, so should I assume it equal to one?
Yes. Note IDIVF is generally only used for wild cards. E.g X-CA-CA-X. When
you explicitly define all 4 atoms it should generally be set to 1.
> Also, in Amber file format, under dihedral section, it states that:
> "The negative value of pn is used only for identifying the existence of
> the next term and only the absolute value of PN is kept." Sorry, what
> does "absolute value is kept" mean?
It means: ABS(-X) = X. I.e. just the value is kept, the - sign is discarded.
So essentially this means if, for a given set of 4 atoms you have one
dihedral then the last value should be +ve. If you have 2 or more then the
first N-1 dihedrals for this specific torsion should have -ve PN values and
the last one should have a +ve value.
> For example, in parm99
> HC-CT-CM-CM 1 0.38 180.0 -3. Junmei et
> al, 1999
> HC-CT-CM-CM 1 1.15 0.0 1. Junmei et
> al, 1999
> what is the PN here for this dihedral bond?
So this says for the dihedral HC-CT-CM-CM there are 2 torsion terms in use.
A 3 fold one and a 1 fold one. In this case the last entry has a +ve value
for PN (1 here) while the earlier one (-3) has a -ve value.
All the best
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