[gmx-users] TIP5P calculating the dummy positions

[Richard Broadbent]

Dear Pratik,

> Dear Richard,
>  
> 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))
this expansion is not quite right for the modulus of the sum of two
vectors,
| xOH1 + xOH2 | = sqrt(| xOH1 |^2 +| xOH2 |^2 +2 xOH1 [dot] xOH2 )
| xOH1 + xOH2 | = |xOH1|*sqrt(2+2*cos(HOH))

if you run that calculation through you should get the right result.

Regards,

Richard

>  
> 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.
>  
>  
> Thanks,
> Pratik Kaku
>  
> 
>  
> > 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
> > 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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