David, why does the order matter, aside from consistency, as long as the
function is planar, and possibly the central atom is in position 2 or 3?
My impression is that if you randomly assigned atoms for each improper
and ran a long enough simulation, the ensembles would converge. (Bearing
in mind that impropers seem like a somewhat hacky way of enforcing
planarity in the first place.)
Regards,
Bill
On 3/24/16 4:23 AM, David A Case wrote:
>> 2.) Can anyone confirm that, for any atom types A, B, and D, an improper A
>> >B C D is equivalent to an improper D B C A and also B D C A, or any other
>> >permutation that leaves C in position 3?
> Above is wrong: the fact that the order of A,B and D matters is the origin of
> requiring them to be an alphabetical order. The only invariant thing you can
> do with a torsion (proper or improper) is to reverse it: ABCD == DCBA. All
> other permutations are different.
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Received on Thu Mar 24 2016 - 12:30:08 PDT
