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#25808 - 11/19/10 12:12 PM Cell Basis Vectors
apredeus Offline
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Registered: 01/20/10
Posts: 147
dear all,

how can one get default cell basis vectors used by CRYSTAL facility to set up a crystal of particular type?

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#25810 - 11/19/10 12:49 PM Re: Cell Basis Vectors [Re: apredeus]
rmv Offline

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source/image/crystal.src
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#25846 - 11/22/10 02:32 PM Re: Cell Basis Vectors [Re: rmv]
apredeus Offline
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Registered: 01/20/10
Posts: 147
I looked at it and still is somewhat confused.

I'm interested in cell basis vectors for truncated octahedron.

it seems like they are defined as follows (if I understood the meaning of XTLABC array correctly)

A = (a, -a/5, -a/5)
B = (-a/5, a, -a/5)
C = (-a/5, -a/5, a)

these vectors do have correct angles between them, but at the same time they are not of appropriate length. I mean, aren't cell basis vectors supposed to be of length a?

If anyone can clarify this matter I'd greatly appreciate it.

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#25858 - 11/24/10 11:48 PM Re: Cell Basis Vectors [Re: apredeus]
apredeus Offline
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Registered: 01/20/10
Posts: 147
ok, apparently the vectors are normalized, I just had hard time finding it in the code.

so the set would be

A = (5b, -b, -b)
B = (-b, 5b, -b)
C = (-b, -b, 5b)

where b = 5a/3*sqrt(3)

These are the vectors that can be used in NAMD/Gromacs in order to obtain the crystal of the same orientation as in CHARMM.

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#33308 - 01/20/14 11:53 AM Re: Cell Basis Vectors [Re: apredeus]
cwbluesky2009 Offline
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Registered: 05/25/09
Posts: 2
there is an error in the formula of b,
it should be b = a/(3*sqrt(3)), where a is the vector length.

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#37505 - 05/16/19 03:43 PM Re: Cell Basis Vectors [Re: cwbluesky2009]
Antoniel Gomes Offline
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Registered: 09/13/18
Posts: 8
Loc: São Paulo, BR
Could anyone please tell me where is this information in Charmm's source?
I'm trying to do the same (convert the crystal parameters to simulate in NAMD) using the RHDO lattice.
If someone could help me how to define the Cell Basis Vector, I would appreciate.

And, Yes. I've checked an octahedral example and the formula b = a/(3*sqrt(3)) seems correct.

Best regards,


Edited by Antoniel Gomes (05/16/19 03:46 PM)

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