Re: CCL:A little geometrical problem
- From: Daquan Gao <dgao ^at^ chem.iupui.edu>
- Subject: Re: CCL:A little geometrical problem
- Date: Thu, 28 Feb 2002 19:49:45 -0500
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 !
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.
> > 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
> > 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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