Hi Lewis,

The reason why it takes two particle indices as input is that MMFF94 uses rather complex combination rules for the van der Waals term.

In particular, EvdW between two particles /i/ and /j/ is given by the following equation:

where /R/_/ij/ is the distance between particles /i/ and /j/, while /R/_/ij/^* and /ε//_ij / are constants computed through these equations:

/B/ is 0 if one of the particles is a HBD, otherwise it is 0.2, while the expression for /γ/ is:

This is the reason why you need to pass two particles to get the values of /R/_/ij/^* and /ε//_ij /: you need to know if either particle is a HBD to compute /R/_/ij/^* , and /ε//_ij / depends on R_/ij/^* .

Finally, if /i/-/j/ is a HBD-HBA interaction, R_/ij/^* is further scaled by the factor DARAD = 0.8, while /ε//_ij / (which is computed using the unscaled R_/ij/^* ) is scaled by the factor DAEPS = 0.5.

More information is available on the MMFF94 paper by Thomas A. Halgren which describes the non-bonded terms:

Merck molecular force field. II. MMFF94 van der Waals and electrostatic parameters for intermolecular interactions <https://doi.org/10.1002/(SICI)1096-987X(199604)17:5/6%3C520::AID-JCC2%3E3.0.CO;2-W>

Cheers,
p.

On 03/07/2019 03:24, Lewis Martin wrote:
Hi all,
Can anyone please help explain what values are returned by GetMMFFVdWParams? It takes two indices as input, so is it an interaction term between the two? Or is it the well depth and minimum (i.e. epsilon and R)?

Example:
In:
m = Chem.MolFromSmiles('C1CCC1OC')
m2=Chem.AddHs(m)
AllChem.EmbedMolecule(m2)
AllChem.MMFFOptimizeMolecule(m2)
pyMP = AllChem.MMFFGetMoleculeProperties(m2)
pyFF = AllChem.MMFFGetMoleculeForceField(m2, pyMP)
pyMP.GetMMFFVdWParams(0,1)

Out:
(3.9377389919289634,
  0.06779699304291371,
  3.9377389919289634,
  0.06779699304291371)
The code at https://github.com/rdkit/rdkit/blob/master/Code/ForceField/Wrap/ForceField.cpp says it returns R_ij_starUnscaled, epsilonUnscaled, R_ij_star, and epsilon, but Im unclear on what the scaling is or why two indices are needed.
Thanks!
lewis



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