Re: CCL:A little geometrical problem



 That is about right. The condition No. 2) leads to inifinitely many
 solutions.... The points of possible CB forms a circle. By the way, the
 only anlges that satifsy this condition are: N-CA-C=180 !
 Daquan
 On Fri, 1 Mar 2002, Ingo Brunberg wrote:
 > Obviously this problem is not as trivial as it seems. Though I can
 > think of a rather complicated solution, I just want to point out, that
 > there is most likely no single solution. The point you are searching
 > is located on the intersection of two cones with the angles N-CA-CB
 > and C-CA-CB originating from CA. So there are three different cases:
 >
 > 1) N-CA-CB + C-CA-CB < N-CA-C: no solution
 > 2) N-CA-CB + C-CA-CB = N-CA-C: one solution
 > 3) N-CA-CB + C-CA-CB > N-CA-C: two solutions
 >
 > Seems like the problem leads to a quadratic equation.
 >
 > Regards,
 >
 > Ingo
 >
 > > A little geometrical problem :
 > >
 > > Taken a residue in a PDB file, how can I calculate the Cartesian
 > > coordinates of CB when I know the coordinates of N, CA and C, the
 angle
 > > N-CA-CB, C-CA-CB and the bond length CA-CB ?
 > >
 > > It seems to be trivial but I am unable to find a solution !
 > >
 > > I am sure that somebody can help me.
 > >
 > > In advance thank you
 > >
 > > LOGEAN Antoine
 >
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