[Chimera-users] Structure measurements

Elaine Meng meng at cgl.ucsf.edu
Sat Jun 23 09:18:19 PDT 2018


Hi Joe,
(A) One idea is to use “define axis,” e.g.

open 2mnr
define axis protein
<http://www.rbvi.ucsf.edu/chimera/docs/UsersGuide/midas/define.html#axis>

Reply Log gives:
axis name, ID, length, center, direction: axis: a1 63.317 ( 30.664,  11.322,  12.737) (-0.566, -0.439,  0.698)

(B) Alternatively, “measure inertia” to get principal axes, e.g.
measure inertia protein
<http://www.rbvi.ucsf.edu/chimera/docs/UsersGuide/midas/measure.html#inertia>

Reply Log gives:
Inertia axes for 2mnr 2698 atoms
	v1 = -0.565 -0.435  0.700   a = 33.352   r1 = 13.200
	v2 =  0.611  0.349  0.710   b = 22.381   r2 = 17.220
	v3 = -0.554  0.830  0.069   c = 19.243   r3 = 17.962
	center =   30.353   11.093   13.054
(a,b,c are half-diameters)

Axis is a little shorter than first principal axis, will look into that.

(C) For bounding box aligned along X,Y,Z see this previous post:
<http://plato.cgl.ucsf.edu/pipermail/chimera-users/2008-April/002540.html>

I just tried it to make sure it still works.  I put the lines in a file named box.py, then opened a PDB file, then opened box.py.  The box was shown in the graphics window and the dimensions reported in the Reply Log.  However, this is just a box aligned with X,Y,Z, so it’s not really the molecule dimensions (the molecule could be diagonal within the box).  Also, apparently it depends on what is displayed, so you should display all the atoms you want in the calculation and not the ones you don’t, like solvent.

I hope this helps,
Elaine
-----
Elaine C. Meng, Ph.D.                       
UCSF Chimera(X) team
Department of Pharmaceutical Chemistry
University of California, San Francisco

> On Jun 18, 2018, at 6:13 AM, Healey, Joe <J.R.J.Healey at warwick.ac.uk> wrote:
> 
> Hi Chimera Team,
> Is there a quick/automatic way to determine the largest dimensions of a model (maybe its minimal bounding box or similar?) without having to select 2 atoms specifically to draw the distance between?
> Thanks
> 
> Joe





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