[phenixbb] Equations of transformation

Nathaniel Echols nechols at lbl.gov
Wed May 25 12:24:14 PDT 2011


FYI, for anyone inclined to write their own programs, there is a similar
Python function for this in CCTBX:

from mmtbx.refinement.fit_rotamers import rotate_point_around_axis
xyz_new = rotate_point_around_axis(axis_start, axis_end, xyz, angle_degrees)

I'm not posting the code here because Paul's version is clearer, but it's in
$PHENIX/cctbx_project/mmtbx/refinement/fit_rotamers.py.  (It wouldn't
surprise me if this duplicates a more general function in one of the
lower-level math modules, but I'm not sure where to look.)

-Nat

On Wed, May 25, 2011 at 12:11 PM, Paul Emsley <paul.emsley at bioch.ox.ac.uk>wrote:

> On 25/05/11 19:41, Hena Dutta wrote:
>
>>
>> Dear Members,
>>
>>
>>
>> I want to know the mathematical relations between 2 sets of co-ordinates,
>> one before and one after the rotation. I used COOT to rotate the side chain
>> (containing atoms N1, C2, N3,  C4, C5, C6, N6, N7 and C8) of Adenine by 180
>> degree about the glycosidic bond (C1’-N9). How the initial co-ordinates (Xi,
>> Yi, Zi) and final co-ordinates (Xf, Yf, Zf) are mathematically related? I
>> shall be very grateful if someone can tell me the equations.
>>
>>
>>
> I got it from the matrix from the Amore documentation.
>
> Coord_orth
> rotate_round_vector(direction, // vector of the bond about which we are
> rotating
>                                position,  // of the moving atom
>                                origin_shift, // typically the position of
> the B atom (e.g. N9)
>                                angle) {
>
>    unit_vec = direction.unit();
>
>   l = unit_vec[0];
>   m = unit_vec[1];
>   n = unit_vec[2];
>
>   ll = l*l;
>   mm = m*m;
>   nn = n*n;
>   cosk = cos(angle);
>   sink = sin(angle);
>   I_cosk = 1.0 - cosk;
>
>   // The Rotation matrix angle w about vector with direction cosines l,m,n.
>   //
>   // ( l**2+(m**2+n**2)cos k     lm(1-cos k)-nsin k        nl(1-cos k)+msin
> k   )
>   // ( lm(1-cos k)+nsin k        m**2+(l**2+n**2)cos k     mn(1-cos k)-lsin
> k   )
>   // ( nl(1-cos k)-msin k        mn(1-cos k)+lsin k
>  n*2+(l**2+m**2)cos k )
>   //
>   //  Thanks for that pointer EJD :).
>
>
>
>  Paul.
>
>
>
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