Open vvoelz opened 9 years ago
I don't really think it has a meaningful unit, unfortunately. If you convert each dihedral angle into a complex number (via the Euler form exp(i \phi)
), then the distance corresponds to the L2 norm on a vector of these complex numbers.
That, unfortunately, doesn't really have a nice unit associated with it.
Actually that’s exactly the info I need. I have N=15 angles, and you’re telling me that distances are computed as L2 norms. So, if the reported distance is d = ( \sum_{i=1}^{N} ||\delta z_i ||^2 )^{1/2}, then the "average" length of a complex difference vector is z = d N^{-1/2}. Thus, we can use the law of cosines, z^2 = (1)^2 + (1)^2 - 2(1)(1)\cos(\phi) = 2(1 - \cos(\phi) ) to back-calculate \phi, the "average" dihedral angle that separates any two dihedral clusters.
In my specific case; d = 1.8, so \phi = \cos^{-1}( 1 - d^2/(2N) ) = 26.8 degrees
Is this reasoning sound, or am I missing something?
Suppose I perform dihedral clustering and MSMBuilder reports that the cluster radius is 1.8. What are the units here? I know that the the angles phi_i get projected 2-fold onto x1_i = cos(phi_I) and x2_i = sin(phi_i). What is the "rmsd" that gets reported?