Re: [AMBER] Questions about reading the right LJ_Acoef values in *.prmtop file

From: Carlos Simmerling via AMBER <amber.ambermd.org>
Date: Sat, 26 Nov 2022 05:50:08 -0500

These values are much easier to read (and change) using parmed.

On Sat, Nov 26, 2022, 2:04 AM ChenXin via AMBER <amber.ambermd.org> wrote:

> Hello everyone,
> It says in the Appendix of Amber Reference Manual that "The index for two
> atoms i and j into the LENNARD JONES ACOEF and LENNARD JONES BCOEF arrays
> is calculated as index = NONBONDED PARM INDEX [NTYPES × (ATOM TYPE INDEX(i)
> − 1) + ATOM TYPE INDEX(j)]." I wonder that the range of index number for
> each section, which should affect the read values.
> For example, we want to obtain the LJ_ACOEF value between the first atom
> and the 15th atom for the a.prmtop file with 4731 atoms and 14 atom types.
>
> %FLAG ATOM_TYPE_INDEX
>
> %FORMAT(10I8)
>
> 1 2 2 2 3 4 3 5 5
> 3
> 6 6 7 3 6 6 6 8 9
> 1
> 2 3 6 3 5 3 5 5 5
> 3
> .......
>
>
> 1
> %FLAG NONBONDED_PARM_INDEX
>
> %FORMAT(10I8)
>
> 1 2 4 7 11 16 22 29 37
> 46
> 56 67 79 92 2 3 5 8 12
> 17
> 23 30 38 47 57 68 80 93 4
> 5
> 6 9 13 18 24 31 39 48 58
> 69
>
> ......
>
> %FLAG LENNARD_JONES_ACOEF
>
> %FORMAT(5E16.8)
>
> 9.44293233E+05 2.12601181E+03 1.39982777E-01 9.95480466E+05
> 2.56678134E+03
> 1.04308023E+06 2.01791425E+04 9.14716912E+00 2.27401052E+04
> 2.01823541E+02
> 8.96776989E+04 1.07193646E+02 9.71708117E+04 1.41077189E+03
> 7.51607703E+03
> 6.20665997E+04 5.94667300E+01 6.78771368E+04 8.79040886E+02
> 4.98586848E+03
> 3.25969625E+03 2.01562190E+06 5.97860700E+03 2.09861767E+06
> 4.93469320E+04
> 2.02461849E+05 1.42791446E+05 4.19430400E+06 8.82619071E+05
> 2.27577561E+03
> 9.24822270E+05 2.01619733E+04 8.61541883E+04 6.01816484E+04
> 1.86068943E+06
> 8.19971662E+05 6.06829342E+05 1.02595236E+03 6.47841731E+05
> 1.12780457E+04
> ......
> If the index number for each section ranges from 1, we can know that the
> ATOM_TYPE_INDEX for the first atom (i=1) and the 15th atom (j=15) is 1 and
> 6, respectively.
>
> Then according to the equation, the pointer for NONBONDED_PARM_INDEX is
> calculated as 14*(1-1)+6=6.According to the pointer, we can read the 6th
> number in NONBONDED_PARM_INDEX section, which is 16. And the corresponding
> LJ Acoef in the LENNARD_JONES_ACOEF should be the 16th number,
> 6.20665997E+04.
>
> However, if the index number for each section ranges from 0, though the
> ATOM_TYPE_INDEX for the first atom (i=0) and the 15th atom (j=14) is still
> 1 and 6,and the pointer calculated for the NONBONDED_PARM_INDEX is still 6,
> the pointer for LENNARD_JONES_ACOEF section read from NONBONDED_PARM_INDEX
> section should be the 7th number, 22. And the corresponding LJ Acoef should
> be the 23th number, 5.97860700E+03.
>
> The started number (0 and 1) can obviously affect the resulted pointer and
> subsequently affect the reading of LJ_Acoef.
>
> I would wonder which is the correct way.
>
> To make it clear, the a.prmtop with specific sections is provided as the
> attachment.
>
> Looking forward to the replys and thanks a lot.
>
>
> All the best,
> Chanyan_______________________________________________
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>
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Received on Sat Nov 26 2022 - 03:00:03 PST
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