# [gmx-developers] Re: Force calculation

Chenyue Xing cxing at ucdavis.edu
Wed Oct 18 20:22:20 CEST 2006

```Hi Anton,

Thanks for the nice suggestions.
As for the first possibility, I've been investigating and printed out all
the zi, zj, but haven't got a clue yet.
As for the 2nd, the modified code still works in the same manner, that rsq =
z^2 and then take the sqrt, so a negative r should not be a problem.

If you have any other guess, please let me know. That will help a lot.

Thank you !
Chenyue

On 10/18/06, Anton Feenstra <feenstra at few.vu.nl> wrote:
>
> Chenyue Xing wrote:
> [...]
> > Thank you for your discussion and I hope I can provide more info soon so
> > that you might be able to give me more advices.
>
> There are two problems I can think of that may be bothering you.
> First is related to what David already mentioned: if you have an LJ-type
> interaction and your z becomes very small, the resulting force is HUGE.
> This can happen suddenly in your case when a particle enters the cutoff
> (as determined by the neighbourlist update), with relatively large x or
> y, but small z.
> Second is that r=z is not really equivalent to the 3D case, because your
> r can become negative (I haven't looked at the code, but I suspect
> something like z=zi-zj for particle i and j). In the 3D case, you have
> the squares and square-root. I can imagine that a negative r will not be
> present in the lookup table ;-)
>
>
> --
> Groetjes,
>
> Anton
>
> _____________ _______________________________________________________
> |             |                                                       |
> |  _   _  ___,| K. Anton Feenstra                                     |
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