# Re: amber-developers: Truncated Octahedron box volume (fwd)

From: Wei Zhang <zweig.scripps.edu>
Date: Mon, 23 Jul 2007 17:07:36 -0500

Ilyas Yildirim wrote:

>amber-developers,
>
>I sent the following email to the amber mailing list but noone responded
>back to me. I will appreciate if someone from here can answer my question.
>Thanks.
>
>
>

Hi, the following is my understanding to this problem:

say we has a cubic, the length is a, then its volume is a*a*a.

say we set up a coordinate system on the cubic. set its origin to
cubic's center,
x, y, z axis parallel to cubic's axis. Then the cubic's equation should be

|x|<a/2, |y|<a/2, |z|<a/2

the trucated octahedron is defined a the intersection of the cubic and
another
shaped defined by equation |x|+|y|+|z| < 0.75*a;

Thus the volume of the trucated octahedron is a*a*a/2.

Meanwhile the length of the trucated octahedron is defined as:
a*sqrt(3/4)
(not sure why though).

Thus the volume of a trucated octahedron should be:

1/2 * sqrt(4/3)^3 * (l*l*l) = 4/(3*sqrt(3)) * (l*l*l) =
0.7698*(l*l*l)111

In your case, l =51.9046302, so the volume is 107645.6322,

which matches the output of amber perfectly.

Sincerely,

Wei Zhang
Received on Wed Jul 25 2007 - 06:07:24 PDT
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