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From: Jung-Hsing Lin <jlin_at_chemcca51.ucsd.edu>

Date: Tue 10 Apr 2001 13:43:52 -0700 (PDT)

Hi! Artem,

I think your 1.5783 Ang is sigma, the vdW radius is 2^(1/6)*sigma.

So, 2^(1/6)*1.5783 = 1.7682 you will obtain the parameter given in AMBER

data base.

CN1_ij CN2_ij sigma_ij 12 sigma_ij 6

--------- - -------- = 4 epsilon_ij * ( (----------) + ( --------- ) )

r_ij^12 r_ij^6 r_ij r_ij

sigma_ij = sigma_i + sigma_j

R_ij* = 2^(1/6)*sigma_ij

R_i* = 1/2 (R_ii)

I hope this helps,

Jung-Hsin

----------------------------------------------------------------------------

Jung-Hsin Lin, Dr. rer. nat.

Howard Hughes Medical Institute and

Department of Chemistry & Biochemistry TEL: +1 858 534 0956

University of California, San Diego FAX: +1 858 534 7042

9500 Gilman Drive, MC 0365 WWW: http://mccammon.ucsd.edu/~jlin/

La Jolla, CA 92093-0365 U.S.A.

----------------------------------------------------------------------------

On Tue, 10 Apr 2001, Thomas Huber wrote:

*> Hi Artem,
*

*>
*

*> if you compare different papers about vdW radii, then you have to look for
*

*> the particular definition of the radius. Sometimes they are defined as the
*

*> half of the distance between identical atoms at minimum interaction
*

*> energy, and as I recall (unfortunately I don't have the time to look it
*

*> up) as the half of the distance at zero (!) interaction energy, which
*

*> would correspond to the hard spere radius.
*

*> Regardless of the actual definition, each paper should describe the
*

*> analytical term to calculate the interaction energy, so if you have a
*

*> parameter set and an equation, just fill in the values into the formula
*

*> and compare it with the other paper. After some rearrangements, you'll
*

*> probably find the parameters of being equivalent.
*

*>
*

*> Thomas
*

*>
*

*> -----------------------------------------------------------------------------
*

*> Dr.Thomas Huber University of Arizona
*

*> Tel.: (520) 621-2537 Department of Chemistry
*

*> FAX: (520) 621-8407 1306 E. University Blvd.
*

*> email thuber_at_physik.tu-muenchen.de Tucson, Arizona 85721-0041
*

*>
*

*> On Tue, 10 Apr 2001, artem wrote:
*

*>
*

*> > Dear Amber users,
*

*> >
*

*> > I've noticed that Van der Waals radius for TIP3P in Amber (parm94.dat)
*

*> > is different from original TIP3P model reported in J. Chem. Phys., 1983,
*

*> > 79, 926-935. Amber has 1.7683 angs and the original has 1.5753 angs. I
*

*> > can assume that Amber's radius comes from some additional
*

*> > parametrisation. Does anybody have references about the justificationof
*

*> > Amber's TIP3P Van der Waals radius? What radius should i use if i wantto
*

*> > switch on SPC model?
*

*> >
*

*> > thanks a lot
*

*> >
*

*> > Artem Mamonov
*

*> > Chemistry Department
*

*> > University of Pittsburgh
*

*> >
*

*> >
*

*>
*

*>
*

Received on Tue Apr 10 2001 - 13:43:52 PDT

Date: Tue 10 Apr 2001 13:43:52 -0700 (PDT)

Hi! Artem,

I think your 1.5783 Ang is sigma, the vdW radius is 2^(1/6)*sigma.

So, 2^(1/6)*1.5783 = 1.7682 you will obtain the parameter given in AMBER

data base.

CN1_ij CN2_ij sigma_ij 12 sigma_ij 6

--------- - -------- = 4 epsilon_ij * ( (----------) + ( --------- ) )

r_ij^12 r_ij^6 r_ij r_ij

sigma_ij = sigma_i + sigma_j

R_ij* = 2^(1/6)*sigma_ij

R_i* = 1/2 (R_ii)

I hope this helps,

Jung-Hsin

----------------------------------------------------------------------------

Jung-Hsin Lin, Dr. rer. nat.

Howard Hughes Medical Institute and

Department of Chemistry & Biochemistry TEL: +1 858 534 0956

University of California, San Diego FAX: +1 858 534 7042

9500 Gilman Drive, MC 0365 WWW: http://mccammon.ucsd.edu/~jlin/

La Jolla, CA 92093-0365 U.S.A.

----------------------------------------------------------------------------

On Tue, 10 Apr 2001, Thomas Huber wrote:

Received on Tue Apr 10 2001 - 13:43:52 PDT

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