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Re: [ccp4bb] Rotation axis |
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CCP4bb navigationCCP4bb <-- 1999 <-- November 1999 <-- 30 November 1999Subject: Re: Rotation axis From: Phil Evans pre {- at -} MRC-LMB {- dot -} CAM {- dot -} AC {- dot -} UK Date: 2008-08-07 My original posting on this question generated quite a few replies, for which many thanks. I found the most useful pointer to the required algorithm was the link to Mark Gerstein's site given by Ian, and I have now programmed this in a C++ class using Clipper routines, which I can make available if anyone wants it. There are many programs which will give the direction of the rotation axis, but defining a point lying on the axis, such that the minimal translation is just along the axis (ie a screw) is a bit more elusive. Thanks Phil On 29 Jul 2008, at 12:30, Ian Tickle wrote: > > Phil > > What I suggested works only if the point x is transformed onto itself, > i.e. there's no screw component. > > The general solution is here: > http://bioinfo.mbb.yale.edu/geometry/screw-axis/ > > There may be a neater way of deriving this in the general case using a > homogeneous matrix & co-ordinates but I haven't worked it out yet! > > Cheers > > -- IAn > >> -----Original Message----- >> From: owner-ccp4bb@jiscmail.ac.uk >> [mailto:owner-ccp4bb@jiscmail.ac.uk] On Behalf Of Phil Evans >> Sent: 29 July 2008 09:11 >> To: CCP4BB@JISCMAIL.AC.UK >> Subject: Rotation axis >> >> If I've go a superposition transformation (x' = Rx + t), as >> it happens >> from a superposition in ccp4mg, how do I get the position & >> direction >> of the rotation axis (to draw in a picture)? >> I know that any (orthonormal) transformation can be represented as a >> rotation about an axis + a screw translation along that axis >> >> I'm sure I've done this before ... >> >> thanks >> Phil >> >> > > CCP4bb navigationCCP4bb <-- 1999 <-- November 1999 <-- 30 November 1999 |
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