[gmx-users] Tabulated potentials for dihedrals - regd

ramesh cheerla rameshgromacs at gmail.com
Wed Jun 6 20:33:59 CEST 2012


Dear Mark,

                    Thank you for your reply, According to my
understanding  functional form of dihedral function type 9 is same as
dihedral function type 1 i.e   k
(1 + cos(n
 (phi)- phi
s))  except the
difference that  function type 9 is used to handle the multiple potential
functions to be applied to the single dihedreal, Is my understanding
correct ?.
The functional form that I am using  Summn over 'n'  0.5*K [ ( Cos
 n(phi-phi0)] , Here 1<= n <= 3  is different from the  k
(1 + cos(n

(phi)- phi
s)). I am suspecting   can I use  function type 9 for my
function.
Can you please clarify my doubt.

Thank you in advance.
On Mon, Jun 4, 2012 at 12:55 PM, Mark Abraham <Mark.Abraham at anu.edu.au>wrote:

> On 4/06/2012 3:46 AM, ramesh cheerla wrote:
>
>> Dear Gromacs experts,
>>
>>                                  I am planing to use tabulated potentials
>> for dihedral functional form Summn over 'n'  0.5*K [ ( Cos  n(phi-phi0)] ,
>> Here 1<= n <= 3 . I have three different K values (i.e K1,K2,K3 ) one for
>> each 'n' value ( Here 'n' is multiplicity).
>>
>
> You do not need a table for this - type 9 will do the job easily, and for
> friendly values of phi0, there exists an equivalent form to the above that
> can be expressed as a Ryckaert-Belleman dihedral. See various places of
> manual chapters 4 and 5.
>
>
>   I have generated  table and given the table number in dihedrals
>> directive of the  ffbonded.itp file, According to my understanding one has
>> to supply K ( force constant ) value in addition to table number in
>> topology. In my case  I have three different  "K" values, which "k" value
>> do I have to take, Can I take sum of all  the three "k" values or can I
>> take  arbitrary value.
>>
>
> Your table would need to be the weighted sum of the three cosines, so
> apparently you have not yet generated such a table. But rather than buy
> trouble, go and do the easier and faster solutions above :-)
>
> Mark
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