[Wien] How to generate the C19 structure (alpha-Sm)

[Claudio Cazorla Silva]

> Hello,
>
> I think that (0 0 2/9) are coordinates in hexagonal setting. You want 
> them in rhombohedral setting, so try transforming them to (2/9 2/9 2/9) 
> - this is a special position generating 2 atoms in the rhombohedral 
> cell, i.e. 6 atoms in the hexagonal cell. Together with the atom at the 
> zero point, this delivers a total of 3 atoms per rhombohedral cell or 9 
> atoms per hexagonal cell. Is that the number of atoms that you expect 
> per cell?
> With this, I get each atom twelve-fold coordinated with an interatomic 
> distance of 3 Å when using a = 3 Å and c = 22 Å as approximate values 
> for the lattice parameters. I am not too familiar with the type of 
> structure you are investigating, but as it is derived from ccp- and 
> hcp-structures, I think that this should be ok.
>
> Best regards
> Michael
>
>   
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Dear Michael,
thank very much for your very useful advices and comments.
Certainly, the coordinates I sent in my previous message are in 
hexagonal setting so the basis vectors are
the ones to transform to rhombohedral. With your advice, (0,0,2/9) --> 
(2/9,2/9,2/9),
the group symmetry detected by 'sgroup' is the correct (166 R_3m), and 
in fact the rhombohedral
cell must contain 3 atoms.  Also each atom must be twelve-fold coordinated.
Best wishes,

Claudio Cazorla Silva