Re: Pressure Discrepancy

From: Richard Wood (rwoodphd_at_yahoo.com)
Date: Wed Mar 28 2007 - 22:33:48 CDT

Hi all, It seems to me that whether he's running NPT or NVT or bananas, he still needs to decrease the STARTING (or maximum, or whatever you want to call it) size of his box and let it fluctuate about that value. The name of the algorithm used isn't important, it's the concept that is. Not enough of us seem to get that sometimes. Good calculating, Richard Richard L. Wood, Ph. D. Computational Chemist Cockeysville, MD 21030 rwoodphd_at_yahoo.com ----- Original Message ---- From: Sterling Paramore <paramore_at_hec.utah.edu> To: Richard Wood <rwoodphd_at_yahoo.com> Cc: namd-l_at_ks.uiuc.edu Sent: Wednesday, March 28, 2007 11:19:36 PM Subject: Re: namd-l: Pressure Discrepancy But Morad's running in NPT, right? The algorithm is designed so that the volume is automatically adjusted to reproduce the desired pressure, on average. It could very well be the piston decay time giving rise to a very large autocorrelation time (so getting the average to be 1atm would require an enormously long simulation time). But before you go and run another long simulation, I really suggest that you do a quick calculation to determine the pressure autocorrelation function.... that'll tell you a lot. -Sterling On Mar 28, 2007, at 5:23 PM, Richard Wood wrote: Hi all, 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, 3 box length, 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 Richard L. Wood, Ph. D. Computational Chemist Cockeysville, MD 21030 rwoodphd_at_yahoo.com ----- 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. -Sterling 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 > > Thanks. > > Morad Alawneh* The fish are biting. Get more visitors on your site using Yahoo! Search Marketing. ____________________________________________________________________________________ Finding fabulous fares is fun. Let Yahoo! FareChase search your favorite travel sites to find flight and hotel bargains. http://farechase.yahoo.com/promo-generic-14795097

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