Dear Amber users,
I am trying to understand periodic conditions but missing something;;;
I used following:(unit name is "sol")
solvateoct sol TIP3PBOX 8.0
Scaling up box by a factor of 1.284224 to meet diagonal cut criterion
Solute vdw bounding box: 10.101 8.342 6.437
Total bounding box for atom centers: 30.648 30.648 30.648
(box expansion for 'iso' is 20.5%)
Solvent unit
box:
18.774 18.774 18.774
Volume: 15849.965 A^3 (oct)
Total mass 7233.466 amu, Density 0.758 g/cc
Added 392 residues.
and used this also:
desc sol
UNIT name: MOL
Head atom: null
Tail atom: null
Periodic box: 27.40838, 27.40838, 27.40838
Contents:
R<MOL 1>
R<WAT 2>
........
............
R<WAT 393>
Volume: 15849.965 A^3 (oct) is volume of truncated octahedron (pleasecorrect me if I am wrong)
desc gives 27.40838, 27.40838, 27.40838 which I
think is representing dimension of truncated box of each side = 27.40838
but if it so then volume of truncated box = 11.313 x a x a xa = 11.313 x (27.40838)*3 = 232931.32
but this is not equal to 15849.965 A^3 (oct)
pls tell me where I am wrong ?
thanks
Jiomm
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Received on Thu Aug 20 2009 - 07:11:21 PDT
