Re: [AMBER] Interplane angles using vector and vectormath in cpptraj

From: Daniel Roe <>
Date: Tue, 30 Oct 2018 08:53:35 -0400


What version of cpptraj are you using? Using the PDB and input you
gave me I get a value of 20.6452 deg (see attached output). One thing
I often find useful when working with the 'vector' command is writing
the resulting vectors into vector "trajectories" so I can visualize
them, e.g.:

writedata vecs.mol2 name1 name2 vectraj trajfmt mol2

This will write vector data from vector data sets name1 and name2 to a
Mol2 file named trajfmt, which can then be visualized using e.g. VMD
alongside the original trajectory. The name1 and name2 vector data
sets must of course exist and be populated (so this should be done
after a trajectory Run). An example of this is in the attached cpptraj

On Mon, Oct 29, 2018 at 6:35 PM Matthew Turner <> wrote:
> Dear AMBER users,
> I'm looking to calculate the angle between the planes of aromatic residues over a trajectory using cpptraj.
> I've looked through the manual and the mailing list (particularly noting and found the commands needed:
> vector name1 corrplane <mask1>
> vector name2 corrplane <mask2>
> vectormath vec1 name1 vec2 name2 dotangle out name1_name2_angle.dat
> This calculates the vectors normal to the planes defined by <mask1> and <mask2> and then does the inverse cos of the dot product of those two vectors.
> However, the results are not what I expected. I have selected a single frame pdb file (attached, zipped) and was looking at the interplane angle between His13 and His14 as a test case. In this frame, the residues sit nearly parallel - max 20 degrees out of parallel, by eye.
> Using masks :13.CG,ND1,CD2,CE1,NE2 and :14.CG,ND1,CD2,CE1,NE2 to select the aromatic rings of His (rather than the whole residue) and running the above commands gives a dotangle value of 127.3200 degrees.
> Firstly, this is clearly not correct, so where have I gone wrong? And secondly, shouldn't the angle between planes should be between 0 and 90 degrees (as simply explained here:, or am I missing something? Calculating the dot product of name1 and name2 above gives -0.6063. If I was to take the modulus of that, arccos(0.6063) = 52 degrees, still not the near zero value I expected.
> Any help appreciated!
> Kind regards,
> Matthew Turner
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Received on Tue Oct 30 2018 - 06:00:02 PDT
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