AMBER Archive (2008)

Subject: AMBER: Fw: rmsd vs frame

From: Francesco Pietra (chiendarret_at_yahoo.com)
Date: Wed May 28 2008 - 18:19:11 CDT

Sorry, mail has left accidentally before being completed. I.e,
333334 x 4 = 1333336 + 195000 = 1528336 : 1000 = 1528.336. Should I take 1528 as the frame corresponding to the lowest_(potential)energy_frame on my scale ? Rounding off is just at the origin of my doubts.

Thanks

--- On Wed, 5/28/08, Francesco Pietra <chiendarret_at_yahoo.com> wrote:

> From: Francesco Pietra <chiendarret_at_yahoo.com>
> Subject: rmsd vs frame
> To: "Amber" <amber_at_scripps.edu>
> Date: Wednesday, May 28, 2008, 3:56 PM
> In order to plot rmsd vs frame with reference to the
> lowest_potential_energy frame, I know the latter from
> ptraj as NSTEP = 195000 of prod5.out, which is the last
> production run out of five runs).
>
> prod.in reads:
>
> &cntrl
> imin=0, irest=1, ntx=5,
> nstlim=333334, dt=0.0015,
> cut=10, ntb=2, ntp=1, taup=2.0,
> ntc=2, ntf=2,
> ntpr=1000, ntwx=1000,
> ntt=3, gamma_ln=2.0,
> temp0=300.0,
> /
>
> Is that correct 333334 x 4 = 1333336 + 195000 = 1528336 as
> the corresponding frame ?
>
> Having the lowest energy in the last production run, I
> assume to be at a preliminary stage of the production.
> Correct?
>
> Thanks
> francesco pietra

      
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