equilibrating water in sander

From: Elizabeth K. White <ekwhite_at_goober.cs.colorado.edu>
Date: Wed 20 Jun 2001 17:00:00 -0600 (MDT)

When I follow the excellent tutorial (Example 2, a more complicated
protein example), the sander run for constant P fails. It seems that
ntb=2 and ntp=2 are mutually incompatible:

  NTP ( 2) must NOT be 2.

      Non-isotropic scaling is currently NOT supported.

     Use NTP of 0 or 1

 *** input error(s)

Is this really true, or have I made a mistake? Can anyone suggest a nice
workaround? Can I get away with isotropic scaling if I make a cubic water
box?

Here is the input file:
Constant pressure simulation
&cntrl
   imin=0,
   nmropt=1,
   ntx=7, irest=1,
   scee=1.2, cut=9.0,
   ntt=1, tempi=300.0, temp0=300.0,
   ntb=2, ntp=2,
   ntc=2, tol=0.000001,
   ntpr=100, ntwr=99999,
   ntcm=1, nscm=1000, ndfmin=-3,
   ntr=1,
   nstlim=20000,
&end
&wt type='REST', istep1=0, istep2=1000, value1=1.0, value2=1.0, &end
&wt type='END', &end
LISTOUT=POUT
DISANG=RST
Group input for restrained atoms
 5.0
RES 1 99
END
END

Thanks, Elizabeth White
Received on Wed Jun 20 2001 - 16:00:00 PDT
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