[gmx-users] Planar group (again)

Yiannis Nicolis ioannis.nicolis at free.fr
Wed Apr 14 02:20:02 CEST 2004

Le 27 mars 04, à 09:18, David a écrit :

> and some more in the rings. For pure benzene you need six impropers
> around the ring and six to keep the hydrogens in the plane. You are

It's a two week old mail but I just saw it.

Let's say we have a benzene numbered C1 to C6 with the respective 
hydrogens numbered H1 to H6 in oplsaa.

Do you mean impropers of the
C1 C2 C3 C4 style around the ring?
C1 C3 C2 H2 style for hydrogens?

I looked at the PHE residue in the ffoplsaa.rtp and only the second set 
of 6 impropers is defined. And in the manual I find no tip for using 
impropers on 4 contiguous  atoms. So I guess I did not understood what 
you mean with "six impropers around the ring".

I tried a benzene with only 6 or all the 12 impropers. Both minimise 
towards the correct structure but of course with different energies. 
For a dynamics run that would make a difference in flexibility. And in 
addition the pdb2gmx also produces the RB dihedrals of the kind
C1 C2 C3 C4
H1 C1 C2 H2
H1 C1 C2 C3

with C1=0 and C2=-C0 which also gives a 180° periodicity (but with 
smaller min-max differences than the impropers). Are there no 
redundancies there? I assume both improper and RB are summed for the C1 
C2 C3 C4 for instance?

Thanks for any attempt to clarify a little bit this benzene ring plane 


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