[cctbxbb] Attitude of rotation, Python to C++

[Ralf Grosse-Kunstleve]

Hi Graeme,

Ouch... this is an unfortunate mixup. I derived the C++ code from cgkit
many years ago without checking. I'm changing it now to be compatible with
http://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula
and the Python code in scitbx.matrix.
cctbx rev. 14644, labelit rev. 1923.
Thanks a lot for pointing out the inconsistency.

Ralf


On Thu, Jan 12, 2012 at 4:03 AM, <Graeme.Winter at diamond.ac.uk> wrote:

> Hi Folks,
>
> Just started moving code from Python to C++ - and seeing a 40x speedup
> which is nice - however I noticed something odd.
>
> scitbx.matrix.col.rotate(axis, angle)
>
> has the reversed attitude to
>
> scitbx::vec3<double>.rotate(scitbx::vec3<double>, double angle)
>
> viz:
>
> this python code
>
>
>            s = (self._ub * hkl).rotate(self._axis, angle)
>            q = (s + self._s0).normalize()
>
>            # check if diffracted ray parallel to detector face
>
>            q_dot_n = q.dot(detector_normal)
>
>            if q_dot_n == 0:
>                continue
>
>            r = (q * distance / q_dot_n) - self._detector_origin
>
>            x = r.dot(self._detector_fast)
>            y = r.dot(self._detector_slow)
>
>            if x < self._limits[0] or y < self._limits[2]:
>                continue
>            if x > self._limits[1] or y > self._limits[3]:
>                continue
>
>            observed_reflection_positions.append((hkl, x, y, angle))
>
>
> gives the same result as
>
>
>  scitbx::vec3<double> s, q, r;
>  double x, y, q_dot_n;
>
>  /* this is rather weird - the attitude of rotation is back to front in
>     the c++ code */
>
>  s = (ub * hkl).rotate(axis, - angle);
>  q = (s + s0).normalize();
>  q_dot_n = q * normal;
>
>  if (q_dot_n == 0) return false;
>
>  r = (q * distance / q_dot_n) - origin;
>
>  x = r * fast;
>  y = r * slow;
>
>  if (x < limits[0]) return false;
>  if (y < limits[2]) return false;
>  if (x > limits[1]) return false;
>  if (y > limits[3]) return false;
>
>  prediction[0] = x;
>  prediction[1] = y;
>
>
> where I had to use -angle. Code is linked into same routines - used
> identically. With -angle I get numerically identical results (which is nice)
>
> Anyone got any clues on this one?
>
> Thanks,
>
> Graeme
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