From: Richard Wood (rwoodphd_at_yahoo.com)
Date: Wed Mar 28 2007 - 18:23:22 CDT
I still say his box is too big, which is why his presure is too large (and negative!).
Here are some numbers from some MD simulations I did in which I varied the sides of a cube of water (512 water molecules) and calculated the average pressure, the goal of which was to find a box length that gave me an average pressure of 1 atm. If I plot the average pressure vs. box length, I get a straight line y = -2171.8x + 53499, so an average pressure of 1 atm would correspond to my water box being 24.633 A on a side.
volume, Å3 box length, Å Average pressure, atm.
14706.125 24.5 286.0323
14886.936 24.6 63.11659
15069.223 24.7 -155.146
15160.92188 24.75 -229.526
15286.41305 24.8181 -401.745
15366.93394 24.8616 -487.346
He needs to decrease his box size, since his average pressure is negative (see my data).
Richard L. Wood, Ph. D.
Cockeysville, MD 21030
----- Original Message ----
From: Sterling Paramore <paramore_at_hec.utah.edu>
To: Morad Alawneh <alawneh_at_chem.byu.edu>
Cc: namd-l <namd-l_at_ks.uiuc.edu>
Sent: Wednesday, March 28, 2007 5:12:06 PM
Subject: Re: namd-l: Pressure Discrepancy
Try calculating a pressure autocorrelation time. From this, you should
be able to estimate how many independent pressure observations you made
over the course of your 10ns simulation. Using that number, you can
calculate the standard deviation of the mean pressure. If the standard
deviation of the mean is larger than the average, you don't have enough
sampling; if not, then you've got another problem.
Morad Alawneh wrote:
> *The simulation was for 10 ns with the following SD:
> Pxx Pyy
> Pzz Pt P <P>
> average -105.7232 -119.0253 -16.4359 -112.3742
> -80.3948 -54.0773
> SD 516.3578 520.2953 578.8550 395.4912
> 362.1872 106.7834
> Morad Alawneh*
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