From: Victor (ovchinnv_at_gmx.com)
Date: Wed Jun 19 2013 - 11:29:11 CDT
Hi Wenhao,
Though what you say about the order being immaterial when psi0=0 looks
reasonable, I suggest you use the notation used by CHARMM.
In the documentation file io.doc (see, e.g.)
http://www.charmm.org/documentation/c37b1/io.html
at the end of the topology (RTF) section, there is a section that states:
...
for the PSF usage the first atom is the central atom,
and the last atom is the atom to be restained relative to
the axis defined by the middle pair of atoms.
also in
http://www.charmm.org/documentation/c37b1/parmfile.html
there is a section on impropers:
Improper dihedrals (E) IMPH, used for out of plane motions are
specified in the same fashion as dihedrals. The use of wildcards, X,
is also allowed in a number of variations. Multiple improper
dihedrals are not supported. Ordinarily, improper dihedrals are given
a multiplicity of 0, which imposes a harmonic restoring potential
instead of a cosine function. In this case, the central atom must be
either the first or the last atom in the definition, else the
resulting potential will be asymmetric and exhibit a discontinuity at
equilibrium! By convention, the first atom of an improper dihedral
(type A-X-X-B or A-B-C-D) should be the central atom. See io.doc for
more information.
--Victor
On 06/18/2013 07:30 PM, Wenhao Liu wrote:
> Hi namd people,
> I am a little confused about the impropers in NAMD. Is the order of
> the improper in parameter file will affect its psi0? Since I read that
> most of the impropers in charmm force field use 0 as their psi0, so if
> the order of the 4 atoms in improper do affect psi0, then is there
> rule or clear example tell me how to deal with this? For example, if I
> want to restrain four atoms, A, B, C and D to be in a plane. And these
> four atoms are bonding with each other like following:
> A B
> \ /
> C
> |
> D
> Can I use the order A B C D Kpsi 0 0.000(psi = 0.00) ? Thanks in
> advance!
>
> Wenhao
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