I am sorry for my carelessness on the definition of Euler angles. I am just
thinking of an Euler angle equivalent. Sorry for the confusion I have made.


On Fri, Sep 5, 2014 at 3:34 PM, Eleanor Dodson <[email protected]>
wrote:

> I don't think Eulerian angles are defined for a non-orthogonal axis
> system? ?
>
> They are defined relative to perpendicular axes X Y Z
> e.g.
> Rotate coordinates by gamma about Z, beta about Y', alpha about Z".
>
>
> Eleanor
>
>
>
>
>
> On 5 September 2014 16:27, Chen Zhao <[email protected]> wrote:
>
>> Thank you Eleanor for your reply. I am actually considering how to
>> describe a pseudo-NCS with an arbitrary rotational and translational
>> relationship. I don't have to do this but I am just curious. It is more
>> straightforward if I say how the two molecules are related by a rotation
>> around unit cell axis than around orthogonal coordinate axis, which does
>> not have an absolute physical meaning.
>>
>> The command line output after coot superpose prints out the rotational
>> and translational matrices for both the orthogonal and fractional
>> coordinate system.
>>
>> For using coordconv, my concern is that if I deal with a low-symmetry
>> unit cell, which is not orthogonal by itself, the Euler angles for the
>> fractional coordinate system and the orthogonal coordinate system should be
>> different. If I just feed some numbers into coordconv, will it consider
>> them as orthogonal coordinates?
>>
>> Thank you,
>> Chen
>>
>> On Fri, Sep 5, 2014 at 6:24 AM, Eleanor Dodson <[email protected]
>> > wrote:
>>
>>> Rotation matrices are rarely specified in a fractional coordinate
>>> system?  The criteria for checking such a matrix is "Is the determinant 1?"
>>> and this only holds for orthogonal matrices.
>>>
>>>
>>>
>>> I guess the way I would do this though.
>>>
>>> You presumably have two sets of fractional coordinates, before and after
>>> rotation?
>>>
>>> There is a ccp4 program - coordconv which will read the fractional
>>> coordinates and generate pdb format with the convention ncode = 1 (You may
>>> need to fudge the fractional format I suppose..)
>>>
>>> You can then use superpose to match the two sets of coordinates and the
>>> output will tell you the Eulerian angles used for the rotation!
>>>
>>> Lots of ways to kill cats!
>>>   Eleanor
>>>
>>>
>>>
>>>
>>>
>>> On 4 September 2014 21:21, Phil Jeffrey <[email protected]> wrote:
>>>
>>>> The orthogonal/fractional matrix is outlined here:
>>>> http://www.iucr.org/__data/assets/pdf_file/0009/7011/19_
>>>> 06_cowtan_coordinate_frames.pdf
>>>>
>>>> Sorry to say I apparently ditched my old Fortran o2f and f2o programs
>>>> to do that.
>>>>
>>>> Bear in mind, however, that orthogonal has no fixed orientation with
>>>> respect to fractional - for most space groups "ncode 1" is often used but
>>>> for primitive monoclinic "ncode 3" is sometimes used, and I think the
>>>> matrix shown in Kevin Cowtan's document above corresponds to "ncode 1".
>>>>
>>>> Phil Jeffrey
>>>> Princeton
>>>>
>>>>
>>>> On 9/4/14 3:55 PM, Chen Zhao wrote:
>>>>
>>>>> I am sorry, just to clarify, the fractional coordinate matrix I
>>>>> referred
>>>>> to is a rotational matrix in the fractional coordinate system.
>>>>>
>>>>>
>>>>> On Thu, Sep 4, 2014 at 3:52 PM, Chen Zhao <[email protected]
>>>>> <mailto:[email protected]>> wrote:
>>>>>
>>>>>     Hi all,
>>>>>
>>>>>     I am just curious whether there are some tools extracting the Euler
>>>>>     angles from a fractional coordinate matrix. I have no luck
>>>>> searching
>>>>>     it online.
>>>>>
>>>>>     Alternatively, I found the analytical solution for the Euler angles
>>>>>     from an orthogonal coordinate matrix. So in the worst case, my
>>>>>     problem reduces to calculating the transformation matrix between
>>>>> the
>>>>>     fractional and orthogonal coordinate system. I feel a little bit at
>>>>>     a loss because it is 6 years since I last studied linear algebra.
>>>>>     How can I calculate this for a specific unit cell?
>>>>>
>>>>>     Thanks a lot in advance!
>>>>>
>>>>>     Sincerely,
>>>>>     Chen
>>>>>
>>>>>
>>>>>
>>>
>>
>

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