[AMBER] Pairwise vs Sander with imin=5

From: kurisaki <kurisaki.ncube.human.nagoya-u.ac.jp>
Date: Thu, 20 Nov 2014 11:01:08 +0900

Dear amber developers and uses,

I'm now using "pairwise" in cpptraj to
Calculate inter-residue energy.

Assuming residues A and B,
I calculated energy three times, i.e., for A, B and A+B.
Finally, I got inter-residue energy as Ene(A+B) - {Ene(A) + Ene(B)}.

Now, I compaired the value with that obtained from sander with imin=5, and
idecomp = 4.
It is "almost" a half of the "pairwise" value.
I am assuming that "sander value" is divided by two
To assign inter-residue energy for each residue.

Is this observation right?

Even so, I found the difference between sander and pairwise by ca 1 kcal/mol,
(for pairwise and sander, -96.6274 and -94.492 (2 x -47.246) [kcal/mol],
Although I used the same value for non-bonded cutoff of value, 99 [A] for each

Considering that each calculation is executed under the vacuum.
This inconsistensy seems strange.
Is there any additional difference between sander and pairwise?

Thank you for your support.

Best regards,

                                    Ikuo KURISAKI


I believe "idecomp =3 or 4" does not make difference
When inter-residue energy is calculated.

AMBER mailing list
Received on Wed Nov 19 2014 - 18:30:03 PST
Custom Search