AMBER: how is truncated octahedron box calculated

From: Lwin, ThuZar <>
Date: Thu, 7 Jul 2005 15:38:05 -0500

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,

The AMBER Mail Reflector
To post, send mail to
To unsubscribe, send "unsubscribe amber" to
Received on Thu Jul 07 2005 - 21:53:01 PDT
Custom Search