Re: Urey-Bradley term

From: Fangqiang Zhu (fzhu_at_ks.uiuc.edu)
Date: Wed Nov 19 2003 - 17:39:23 CST

Hi, Cosmin,

I have a workaround that's not optimal but should work...

My solution is that you run namd twice. The first run is a normal one,
outputting
the total angle energy, and in the mean while recording the trajectory in a dcd
file. In the second run, you simply read in the dcd and ask namd to output
energies.
But this time you use a different parameter file, in which you change all of the

k_theta values to 0. In this way, namd still outputs the total angle energy, but

since the first term is 0, this energy is actually the Urey-Bradley energy for
the frame.

Good luck,
Zhu

Cosmin ROMAN wrote:

> Fangqiang,
>
> I am trying to build a tool that finds charmm force field parameters for
> carbon nanotubes. It is based on running namd in a loop with a dft program.
> CNTs in particular are very difficult to fit and I need all charmm's
> parameters to do the job. Namd computes internally the Urey-Bradley terms and
> it is a waste of information not to display it. I did a Tcl script but it's
> too slow, and I won't have the time to do it in C.
>
> Can namd developers help me in this matter, and build a patch for namd-2.5
> that simply displays the ANGLE term split in ANGLE and UREY-BRADLEY terms. I
> would have done it myself but to be honest I never was able to compile namd
> and that is because of the cvs system.
>
> Thank you,
> Kind Regards,
> Cosmin.
>
> On Wednesday 19 November 2003 01:24, Fangqiang Zhu wrote:
> > Unfortunately, NAMD can't display the breakdown of the angle energy. In
> > principle you could calculate it from the trajectory, but it will be very
> > complicated
> > if you want that energy for the whole system...
> >
> > Zhu
> >
> > Cosmin ROMAN wrote:
> > > Hello,
> > >
> > > Can namd display the Urey-Bradley contribution to the ANGLE energy? As it
> > > is, it only displays the sum, i.e. Ktheta(Theta - Theta0)**2 + Kub(S -
> > > S0)**2, and I would really like to have them separately.
> > >
> > > Thanks,
> > > Cosmin.

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