'ave' zonal average PROBLEMS!

[Sestak, Dr. Michael]

I don't know about the example in the documentation, but if you have
one-degree lat/lon grids, your two expressions are NOT equivalent.  The
first averages the data for all longitudes 0 to 360, the second averages
grid points 1 to 180, which for one-degree data would only be longitudes 0
to 179.

> -----Original Message-----
> From: GRADSUSR at LIST.CINECA.IT
> [mailto:GRADSUSR at LIST.CINECA.IT]On Behalf
> Of Diane Stokes
> Sent: Friday, March 24, 2006 8:24 AM
> To: GRADSUSR at LIST.CINECA.IT
> Subject: Re: 'ave' zonal average PROBLEMS!
>
>
> Hi, Thomas.
>
> The "-b" option works as advertised for me (using v1.9b4).
>
> If you are still having this problem, posting your data
> descriptor file
> might get a better response. (The output from 'q config' and your
> platform info sometimes helps, too.  I don't know if it will
> in this case).
>
>     Diane
>
> Thomas Spengler wrote:
> > Dear all,
> >
> > I am using the function 'ave' to calculate zonal means of different
> > variables.
> >
> > If you average over a latitude circle, following the
> documentation, you
> > should do:
> >
> > ave(z,lon=0,lon=360,-b)   (1)
> >
> > right?
> > The '-b' option enabling the weighting of the longitude '0' which is
> > included twice in the zonal average here.
> >
> > Expression (1), following the GrADS documentation, should
> be identical to
> >
> > ave(z,x=1,x=180)   (2)
> >
> > right?
> >
> > The problem ist though, that this is not the case! I get
> deviations of
> > the two results reaching 350 m^2 s^-2.
> > Does someone have the solution for calculating real,
> consistent zonal
> > averages with GrADS?
> > Or am I doing something wrong or didn't understand the documentation
> > properly?
> >
> > Thanks in advance for your help and advice!
> >
> > Cheers,
> > Thomas
> >
>
> --
> Diane Stokes
> Environmental Modeling Center
> National Weather Service/NOAA
>