Hi An!
I think I've figured it out. cpptraj density command gives you a linear density along the desired coordinate i.e. amu/A along the z axis, in your case. If you want to compute a volumetric density, then you have to divide the linear density (cpptraj output) by the area spanned by the other two dimensions (xy cross section, in your case).. That would result in a amu/A^3 density. Then you can apply the 1660x factor to get the kg/m^3.
Let me know if you could work this out!
regards!
Gera!
> Dear AMBER users,
> I have been looking into density profiles of a biphasic box primarily
> composed of water and dodecane. Specifically, analyses were performed on
> these solvent residues w.r.t. the z-axis. From my understanding, the
> density values for the mass density command are in amu/A^3. But I was not
> able to find anything that explicitly states this. Can anyone help confirm
> or correct me if I am wrong?
> Going off the basis that mass density is reported in amu/A^3, the
> conversion to kg/m^3 is by a factor of x1660. However, this conversion
> gives me values that are exceedingly high from what is expected. For
> example, the "bulk area" or maximum of the density profile for water should
> be ~1000kg/m^3. But the converted maximum values that I calculated are off
> by a magnitude of 1000. Is there something that I may be missing or not
> accounting for when converting the units of the reported values?
> P.S. I have attached my results for my water mass density as well as
> relevant input scripts. Please keep in mind that there are some acid in the
> water and, thus, the density may be slightly off from the 1000kg/m^3 but I
> am sure that it should not differ by x1000. Thank you for all of your
> considerations and advisement.
> Sincerely,
> *An T. Ta *
> Department of Chemistry | Ph.D. Candidate
> Coolbaugh Hall | Room 118
> Golden, CO 80401
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Received on Thu Mar 03 2022 - 14:30:02 PST
