Re: [AMBER] ncorr in rotdif command seems redundant to me. Please help.

From: case <>
Date: Tue, 3 Nov 2015 20:49:28 -0500

On Tue, Nov 03, 2015, Jose Borreguero wrote:
> I am using the difrot command for the first time. I have read the manual
> and I can't understand the purpose of "ncorr" that is not accomplished with
> ti and tf.
> My trajectory has 10,000 frames and dt=0.001ns.
> If I set ti=1ns and tf=2ns, my understanding is that I am calculating the
> correlation function between frames 1000 and 2000. What's the point of
> ncorr then?

ti and tf are the relative times between snapshots: i.e. how much the vector
moved in time t (ti<t<tf). There is generally no reason to ever have ti be
anything other than the default of 0. A reasonable rule of thumb is that tf
should be 10 to 50 times shorter than the length of the trajectory.

The time that corresponds to ncorr frames is less useful: it just has
to be longer than tf. (You will calculating the time correlation from zero
to ncorr*dt, but only using the part of that between ti and tf to estimate
the rotational diffusion constant). Leaving ncorr at its default is fine.
(You could save a small amount of time by setting ncorr to be tf/dt, where
"dt" is the time between snapshots.)

Dan: (a) calling ti the "initial time for calculating integral" is pretty
vague. (b) I think(?) that the description of ncorr in the documentation
is wrong: tf just needs to be less than (not "much less" that) ncorr*dt.
I'll take a stab at updating the documentation. (c) have ncorr input as a
frame number, but ti and tf as ns seems confusing.


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Received on Tue Nov 03 2015 - 18:00:03 PST
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