AMBER Archive (2005)Subject: Re: AMBER: how is truncated octahedron box calculated
From: Chris Moth (chris.moth_at_vanderbilt.edu)
Date: Thu Jul 07 2005  16:50:24 CDT
Here is the source code where the Volume of the bounding cube is multipled by
0.7698004 in tools.c:
dVolume = dMaxX * dMaxY * dMaxZ;
if ( bOct ) (Yes in this case)
dVolume *= 0.7698004; /* = sqrt(1  3*cost^2  2*cost^3),
where cost = 1/3 = cos(109.471) */
VP0(( " Volume: %5.3lf A^3 %s\n", dVolume, (bOct ? "(oct)" : "") ));
http://www.gromacs.org/documentation/reference_3.2/online/editconf.html
asserts this constant (.77) as well....
I don't have a quick reference on the trigonometry, unfortunately.
Quoting "Lwin, ThuZar" <ThuZar.Lwin_at_stjude.org>:
> Dear Amber Users,
>
> I am trying figure out how the volume of the truncated octahedron box is
> calculated. tleap printed out the following information:
> Scaling up box by a factor of 1.222563 to meet diagonal cut criterion
> Solute vdw bounding box: 49.588 45.810 35.593
> Total bounding box for atom centers: 78.930 78.930 78.930
> (box expansion for 'iso' is 27.7%)
> Solvent unit box: 18.774 18.774 18.774
> Volume: 255327.326 A^3 (oct)
> Total mass 134124.640 amu, Density 0.872 g/cc
> Added 6433 residues.
>
> The box information at the end of the restart file (produced from tleap) is
> 69.2212952 69.2212952 69.2212952 109.4712190 109.4712190 109.4712190
>
> Using 69.2212952 as the length of one side of the truncated octahedron and
> using the formula:
> (a^3)*8*(sqrt(2)), the volume I got is 3752529.786 A^3 which is significantly
> higher than 255327.326 A^3 tleap printed out. Can someone help me with how
> volume of truncated octahedron is calculated?
>
> Thank you very much,
> ThuZar
>
> 
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Chris Moth
Vanderbilt University
Email: christopher.moth_at_Vanderbilt.Edu

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