AMBER Archive (2009)

Subject: [AMBER] box volume not understood?

From: Jio M (jiomm_at_yahoo.com)
Date: Thu Aug 20 2009 - 08:54:52 CDT

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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