[MITgcm-support] north-south or south-north gridding?

[Martin Losch]

Hi Jill,

the convention is this:
phiMin specifies the SOUTHERN boundary of the model domain (that is  
the lat-position of the southernmost v-velocity point). The NORTHERN  
boundary is then phiMin + sum(delY) and the latidute coordinates of  
the cell centers are something like Y(1) = phiMin + .5*delY(1), Y(2)  
= Y(1) + .5*delY(1) + .5*delY(2), etc. (have a look at  
ini_spherical_polar_grid.F. The variable names are different from  
what I used here. The grid layout is shown in GRID.h). So j=1  
corresponds to the southern edge of the domain and j=sNy to northern  
part of the domain.

Similarily thetaMin marks the western edge of the domain (default is  
0). This is usually not critical as a conventional lat-lon grid is  
independent of longitude, it's just convenient to have the correct  
lons for, say, the Ross Sea in the output.

If you specify the "wrong" phiMin, all that happens is the your lats  
are all shifted north (instead of -80 they start at -50 and go north  
from there). The bathymetry looks that same. It is just placed at a  
wrong latitude and the main effect is that your coriolis parameter  
(sin(2*om*lat)) is wrong (too far north, or even on the wrong  
hemisphere).
Have a look at grid.t001.nc and in particular YC and YG, fCori to see  
whether you got the grid you wanted.

Martin

On Sep 20, 2006, at 5:31 AM, jschwarz at awi-bremerhaven.de wrote:

> Hi all,
>
> It's been suggested to me by an experienced modeller that my grid  
> definition, which runs from north (-50) to south (-80) is  
> unconventional and will cause Bad Things to happen when i run  
> mitgcm.  However, it seems to me that mitgcm reads in the latitude  
> spacing just fine, and the bathymetry/initialisation and state...nc  
> files also look fine.  Can anyone tell me whether i need to  
> redefine my grid to run from south to north?
>
> Specifics:
> domain specified in /data/parm01 using
>    delXFile  and delYFile, where nx=81, x(1) ~ -50, grid is 1 x 1  
> degree isotropic
> phiMin = -50
>
> many thanks,
> jill.
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