[gmx-users] monoclinic super cell

Martyn Winn martyn.winn at stfc.ac.uk
Wed Feb 4 10:56:19 CET 2009


For a start, the 'P 1 21 1' on the CRYST1 line is a different spacegroup
from 'P 21/b'. The former implies Z=2.

In principle, it should be straightforward to generate it yourself by
applying the 4 symmetry operators to your starting coordinates. You then
need to know whether 'P 21/b' is 'P 1 1 21/b' or 'P 21/b 1 1'

Cheers
Martyn

On Tue, 2009-02-03 at 21:35 -0800, farzaneh fatahi wrote:
> Hi,
> I am simulating monoclinic hydroxyapatite CA10 (PO4)6 (OH)2.
> I have found a PDB of CA5 (PO4)3 (OH) in internet , whitch consits of 22 atoms.
> the monoclinic structure of HAP has however 88 atoms and space group (P 21/b)
> it means to generate the super cell of HAP i have to feed editconf with the 
> appropriate CRYST1 record in PDB file.  
>  
> CRYST1   9.42     18.86    6.88    90.00  120.00   90.00 P 1 21 1     4
>  
> according to CRYST1 record: because of  Z value and space group i should get a .gro
> file of the whol 88 atoms but editconf gives me only a .gro file with 22 atom.
> am i on the right way to generate the super cell with 88 atoms.
> I appreciate your help!
>  
>   
> 
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*                                                                     *
*               Dr. Martyn Winn                                       *
*                                                                     *
*   STFC Daresbury Laboratory, Daresbury, Warrington, WA4 4AD, U.K.   *
*   Tel: +44 1925 603455    E-mail: martyn.winn at stfc.ac.uk            *
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