

AMBER Archive (2009)Subject: RE: [AMBER] box volume not understood? From: Jio M (jiomm_at_yahoo.com)Date: Thu Aug 20 2009  13:26:41 CDT
Thanks Ross,
>Where did you get 11.313 from? Are you assuming 8xsqrt(2). This assumes that
>the side lengths are those of the square pyramid you can draw inside the
>truncated octahedron. This is not the case here. They are the lengths of the
>cube that encloses it (as far as I am aware). I am not sure what the exact
>equation is for determining this. But it should be possible to locate it in
>a good maths text book.
I got 11.313 from just a quick search on wiki :) it is 8xsqrt(2) only...
Actually I want to reduce the box volume ,after solvation, by using set
>I may be wrong here but I think this is reporting the box edges defining the
>'cube' that the truncated octahedron would fit into. Thus the volume of this
>cube is 27.40838^3 = 20589.704 A^3
as above clears that 27.40838 is dimension of cube, so in set command
Thanks and regards
Jiomm
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> I used following:(unit name is "sol")
>
> solvateoct sol TIP3PBOX 8.0
> Total bounding box for atom centers: 30.648 30.648 30.648
> Volume: 15849.965 A^3 (oct)
> Total mass 7233.466 amu, Density 0.758 g/cc
> Added 392 residues.
Something to note is that Leap deliberately sets the box size too large to
avoid large initial steric clashes at the edge of the box.
> desc sol
> Periodic box: 27.40838, 27.40838, 27.40838
> Volume: 15849.965 A^3 (oct) is volume of truncated octahedron
> (pleasecorrect me if I am wrong)
Yes I believe this is correct. Note Leap may report different sizes here
since in some cases it may be reporting the actual box size as will be
written to the inpcrd file and in others it would be reporting the distance
from furthest atoms on each edge.
> desc gives 27.40838, 27.40838, 27.40838 which I
> think is representing dimension of truncated box of each side =
> 27.40838
I may be wrong here but I think this is reporting the box edges defining the
'cube' that the truncated octahedron would fit into. Thus the volume of this
cube is 27.40838^3 = 20589.704 A^3. Which is larger than the 15849.965 it
reports for the box volume. As you would expect since a truncated octahedron
is effectively a cube with the corners cut off.
> 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)
Where did you get 11.313 from? Are you assuming 8xsqrt(2). This assumes that
the side lengths are those of the square pyramid you can draw inside the
truncated octahedron. This is not the case here. They are the lengths of the
cube that encloses it (as far as I am aware). I am not sure what the exact
equation is for determining this. But it should be possible to locate it in
a good maths text book.
 
