RE: [AMBER] IDIVF and PN

From: Ross Walker <ross.rosswalker.co.uk>
Date: Fri, 17 Jul 2009 04:40:11 +0100

Hi Bill,

> 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
Ross


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Received on Fri Jul 17 2009 - 01:08:28 PDT
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