Hi Ramiro,

Assuming your rings are nicely planar, and representing the ring as:

1-2-3
|     |
6-5-4

you can get the plane normal vector as the vector cross product from
(3)-(1) and (5)-(1).
Doing so for both rings gives you the two normal vectors. The angle
then follows from the dot product of the (normalized) normal vectors:

angle = acos(n1 . n2)

It becomes a bit more elaborate if the planes are not planar :)

Hope it helps,

Tsjerk

On Thu, Sep 23, 2010 at 10:53 AM, Ramiro Téllez Sanz
<urcind...@gmail.com> wrote:
>
>
> Hi everyone and thanks for reading this!
>
> I am interested in measuring the angle between aromatic ring planes.
> Is there any easy way/script to do it?
>
> One way that came to my mind is creating a pseudoatom representing the
> centroid for each ring (I already know how to do that), then drawing two
> lines perpendicularly to the planes from both centroids, and finally
> measuring the angle between the lines. Will that be possible? How could
> this be done?
>
> Is there any other way? I'm completely clueless. Any help will be
> greatly appreciated.
>
> Thanks in advance,
>
> Ramiro Tellez Sanz
> Dept. Physical Chemistry
> University of Almeria
> Spain
>
>
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-- 
Tsjerk A. Wassenaar, Ph.D.

post-doctoral researcher
Molecular Dynamics Group
Groningen Institute for Biomolecular Research and Biotechnology /
University of Groningen
The Netherlands

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