From: Giacomo Fiorin (giacomo.fiorin_at_gmail.com)
Date: Tue Jul 12 2011 - 23:19:31 CDT
Hi Robert, what is numerically speaking, the difference between what you get
and what you expect?
You could change the force constant to another physical unit, for as long as
the transformation is simply a rescaling, and not a change of variables,
such as taking the acos(), which has a non-constant derivative. This second
would require some math, before or after WHAM. So, if you're not OK with
using the cosine that "tilt" returns, you could try manipulating the code to
get the tilt angle. Beware though that its gradients are discontinuous
around 0° and 180°, though (as you can visualize easily): the atomic
gradients of its cosine that you're using, aren't.
Yes, as the manual "suggests", the 1/2 factor is included in the potential.
On Tue, Jul 12, 2011 at 5:48 PM, Robert Elder <rmelder_at_gmail.com> wrote:
> Hi all,
> I'm a little unclear about the units of the force constant (K) for the
> new 'tilt' collective variable. The units of the output for this
> colvar appear to be cos(angle in radians), such that I can extract the
> angle in degrees using acos(colvars output) * (180/pi).
> I have been using the raw output from the colvars module (i.e.
> cos(radians)) and the force constant in my colvars configuration file
> (no clue what the units are) for WHAM. However, it appears to me that
> using this force constant may be incorrect, based on a) the expected
> results of the simulations and b) some testing with lower force
> constants, which show behavior more closely resembling what I expect.
> So this made me wonder if I need to convert the K from colvars.conf to
> some other set of units where the value of K would be lower. Of
> course, I don't want to choose random numbers until I get desirable
> So, my questions (maybe Jerome or Giacomo can comment):
> 1. Do I need to convert the tilt force constant to some other set of
> units for WHAM? Or should I just continue to use the value as it is in
> my colvars.conf file?
> 2. Is the factor of 1/2 included in the colvars calculations? That is,
> is the potential K/2*(x-x0)^2 or simply K*(x-x0)^2? The manual
> suggests the 1/2 is included, but I'm still unclear.
> I can provide more details if necessary.
> Robert Elder
> University of Colorado at Boulder
> Chemical and Biological Engineering
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