[gmx-users] TIP5P calculating the dummy positions
demon_lord_pratik at hotmail.com
Thu Oct 27 14:23:13 CEST 2011
First of all i really appreciate the help. I 've figured out the cross product part of the equations, but the |A+B| part i still have figured it out.
I've been trying to crack this bit but I still can't do it.
0.07 * cos (109.47/2) / | xOH1 + xOH2 | -> this bit is what i dont get below is what i've done
based on what you said xOH1 is the distance of O to H (bond lenght of OH being 0.09572, x-axis 0.07595, y-axis 0.058588) and the bond angel of HOH=104.52
putiing it into the equation:
0.07 * cos (109.47/2) / | xOH1 + xOH2 | ---(expansion)---> 0.07*cos (109.47/2) /(2*|OH|*cos(HOH))
plugging in the numbers:
0.07*cos (109.47/2) /(2*0.09572*cos(104.52)) = 0.040440/0.0479974 = -0.842546 (the answer i get)
-0.842546 is the answer i get how ever the answer i should be getting is -0.344908 . I've tried to look for solutions to this but i still don't understand it.
It would be very helpful if you could show me where i have gone wrong.
> Subject: Re: [gmx-users] TIP5P calculating the dummy positions
> From: richard.broadbent09 at imperial.ac.uk
> To: gmx-users at gromacs.org
> Date: Fri, 21 Oct 2011 11:40:04 +0100
> Dear Pratik,
> > I am trying to create the tip6p itp file. In order to do that, since
> > it is an overlap of the tip4p and tip5p model (visually)
> > I am trying to understand the a, b, and c values for the position of
> > the dummy charge in the tip5p models.
> > Below is the part of the script that is of my concern.
> > _________________________________________________________
> > [ dummies3 ]
> > ; The position of the dummy is computed as follows:
> > ;
> > ; The distance from OW to L is 0.07 nm, the geometry is tetrahedral
> > ; (109.47 deg)
> > ; Therefore, a = b = 0.07 * cos (109.47/2) / | xOH1 + xOH2 |
> xOH1 is the vector from O to H1, (not just the x component)
> | xOH1 + xOH2 | is normalisation factor as these vectors are not of unit
> > ; c = 0.07 * sin (109.47/2) / | xOH1 X xOH2 |
> again | xOH1 X xOH2 | is a normalisation factor as the cross product of
> the vector from O to H1 with the vector from O to H2 will not be a unit
> > ; =20
> > ;
> > ; Using | xOH1 X xOH2 | = | xOH1 | | xOH2 | sin (H1-O-H2)
> This is a standard vector identity the modulus of the cross product of
> two vectors is the product of the moduli times the sine of the angle
> between them:
> |V X U| = |V||U|sin(theta)
> Hope that's helpful
> > ; | xOH1 + xOH2 | = 2 | xOH1 | cos (H1-O-H2)
> > ; Dummy pos x4 = x1 + a*x21 + b*x31 + c*(x21 X x31)
> > ; Dummy from funct a b c
> > 4 1 2 3 4 -0.344908 -0.344908 -6.4437903493
> > 5 1 2 3 4 -0.344908 -0.344908 6.4437903493
> > _______________________________________________________
> > I do understand everything except the bolded bit.
> > so far i have understood that |xOH1| is the magnitude of OH on the
> > x-axis, but putting those values in i don't the the correct a, b, c.
> > I'm not good in vectors and i have had a look at the gromacs manual
> > (the 3out model ni figure 4.16).
> > I don't know where i am going wrong.
> > I would appreciate the help thanks.
> > Thanks,
> > Pratik Kaku
> > --
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