[gmx-users] TIP5P calculating the dummy positions

[Richard Broadbent]

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
length
> ; 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
vector.
> ; =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

Richard

> ; | 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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