From: Jacqueline Schmidt (jacqueline_schmidt_1983_at_yahoo.com)
Date: Fri Mar 15 2013 - 12:35:52 CDT
I just learnt the units of force printed by NAMD are PN not kcal/mol/Ang. Converting the units, I actually get something close to -0.6..I appreciate your comments in any case
This amount is just the amount of work done for pulling the peptide for ~0.1 or less angstrom away from the surface. I expect the overall work or free energy ends up to less than 10-20 kcal/mol...
From: Aron Broom <broomsday_at_gmail.com>
To: Jacqueline Schmidt <jacqueline_schmidt_1983_at_yahoo.com>
Cc: namd <namd-l_at_ks.uiuc.edu>
Sent: Friday, March 15, 2013 1:10 PM
Subject: Re: namd-l: smd force is too high
I'm not really very familiar with SMD, but have you tried a smaller pulling velocity? I can imagine that if the pulling velocity is too fast to allow the system to be in a constant pseudo-equilibrium, you might appear to have a steeper energy barrier to pulling the protein because you have slowly relaxing orthogonal degrees of freedom that aren't being given time to relax. That is, you're crossing the energy barriers associated with those slowly relaxing degrees of freedom using a path that isn't the lowest energy one. Again, I'm not terribly familiar with SMD, but this kind of problem is fairly general to any biased MD and it will generally present itself as the energy barrier appearing to be much greater than you expected.
The above being said, is the value you are getting really that extreme? I suppose -0.6 kcal/mol/Ang is effectively the gradient for the binding energy? If so, what would you be expecting for the actual binding energy over the distance you want to pull? It seems like a gradient of this magnitude would only yield a binding energy of at most a few kcal/mol or am I confused?
On Fri, Mar 15, 2013 at 12:25 PM, Jacqueline Schmidt <jacqueline_schmidt_1983_at_yahoo.com> wrote:
Dear NAMD users,
>I have a question regarding the smd simulation using NAMD. I am pulling off a peptide from a surface (surface is fixed) using smd with force constant of 6 kcal/mol/Ang^2 and pulling velocity of5x10^-6 Ang/fs. My problem is the forces that are printed in the output are unreasonably highe!
>e.g. at step 100, the center of mass of peptide has been moved ~0.1 Ang in z-direction. The smd force printed in the output is fz~-43 and I assume the units are kcal/mol/Ang
>but this is so high! Doing a rough calculation for force Fz=k(vt-z+z0)= 6(5*10^-6 -0.1) ~-0.6 kcal/mol/Ang
>can anyone please help me with this?
>I will appreciate it
Aron Broom M.Sc
Department of Chemistry
University of Waterloo
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