# [gmx-users] Three-site model for acetonitrile

Christoph Freudenberger christoph.freudenberger at chemie.uni-ulm.de
Thu Feb 27 14:28:34 CET 2003

```Anton Feenstra wrote:
> christoph.freudenberger at chemie.uni-ulm.de wrote:
>
>>Zitat von David <spoel at xray.bmc.uu.se>:
>>
>>
>>>On Wed, 2003-02-26 at 17:46, Christoph Freudenberger wrote:
>>>
>>>>[ dummies2 ]
>>>>;  ai    aj    ak       funct   a
>>>>     3     1     2       1       0.2652197
>>>>     4     1     2       1       0.8203527
>>>>     5     1     2       1       1.2652197
>>>
>>>Shouldn't one of these numbers be negative, or is the dummy further out
>>>than the atom?
>>
>>no, that's fine D1 is ~0.2nm away from the com, D2 is ~0.06nm.
>
>
> Your (Christoph's) equations look fine, but the numbers in your
> topology don't make sense to me.
>
> First, your constraint distance between the masses is 0.263, while
> your Me-N distance (from a previous topology in your mail) was 0.262.
Thats maybe due to some typo in a previos version in the topology.
The correct Me-N-distance from the reference i cited is 0.263nm.
The distance R between D1(m1) and D2(m2) is in priciple free to choose.
only if you choose it shorter than r(Me-N) the quadratic equation
you find for the first unknown paramter to define becomes undefined:

M*R +- sqr(M^2R^2-4MIa)
r1 = -----------------------
2M

I got two solutions for r1. Calculating r2=r1-R will give you the
negative of the other solution. I have chosen the solution
that makes physical sense to me:

D1--Me-------+-C--D2---N
^
|
com
You have to consider that m(D1)<<m(D2). Hence the above config makes more
sense than the other mathematical solution:

Me-D2----+-C-------N-D1

> In other words, your masses are *further* out than your atoms, which
> cannot be true, since if you shift mass (from the C) outwards, your
> moments of inertia increase which must be balanced by shifting the
> two masses inwards with respect to the positions of the atoms.
>
> Also, since Me (15.035) and N (14.0067) are about the same mass, I'd
> expect the mass distribution over D1 and D2 to be much more symmetric.

The mass distribution over D1 and D2 has to represent the masses
of all three atoms and we have Me on the left and C and N on the
right side of the com... that is rather unsymmetric.

regards
--
Christoph Freudenberger
Abt. Organische Chemie I AK Siehl - Uni Ulm -Tel: ++49-731-502-2785

```