[Chimera-users] angles after match

[Tom Goddard]

Hi Julien,

  In Chimera 1.4 daily builds since January 7 there is a command called 
"measure" that can report the relative orientation of two models:

    measure rotation #0 #1

prints in the reply log (menu Favorites / Reply Log)

  Matrix rotation and translation
     0.22612416   0.31146762  -0.92296034  15.49596646
     0.00745874   0.94692067   0.32138080  -1.29273361
     0.97406993  -0.07955609   0.21179848 -13.27353380
  Axis  -0.20428607  -0.96657822  -0.15489914
  Axis point  16.21434627   0.00000000   3.22513422
  Rotation angle (degrees)  78.90585699
  Shift along axis   0.13997704

giving the 3 by 4 transformation (3 by 3 rotation matrix plus 
translation in last column) or in a different format the rotation axis 
vector, a point on the rotation axis, a rotation angle and a shift along 
the axis.  These numbers refer to the orientation of model #1 using the 
coordinate system of model #0.

  The command also makes a graphical depiction of the rotation axis.  
That becomes a new model which can be closed with Model Panel.  I added 
a new optional parameter to "measure rotation" today to not show the 
graphical depiction: "measure rotation #0 #1 showAxis false".

  We do not use or provide Euler angles in Chimera.  As you note there 
are problems with conventions and degeneracy with Euler angles.

    Tom


julien(GWDG) wrote:
> Dear all,
>
> thank you for the very nice software. I continuously like it.
>
> I have the following problem.
>
> For different molecules, I would like to know what are the 
> transformation that link them.
> I can do for example
> match #0 #1
> matrixget -
>
>  From this matrix I have the rotation information and the translation 
> information.
> I would like to extract the angles. For that I need to know what 
> convention you used.
> I guess those are Euler angles (?), but I need to know what is the order 
> of rotation xyz, xzy...
> Finally how do you get ride of the gimbal lock problem if you use Euler 
> angles.
>
> Thank you for your constant help.
> (Sorry if this question is a duplicate)
>
> many greetings
>
> julien
>
>
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