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

[Martin Losch]

Hi Jill and Mark,

On Sep 21, 2006, at 12:35 AM, Mark Hadfield wrote:

> jschwarz at awi-bremerhaven.de wrote:
>> Thanks for clarifying the convention - why do all you modellers  
>> think upside down?  What is that?!
>>
> Yeah, Jill, like I've noticed all the maps on your wall have south  
> at the top :-)
>> Like i said (  (o;  )  the grid i define is read in by mitgcm as i  
>> intended - the co-ordinates that are listed in the run_output.txt  
>> go from -50 to -80, the area i see in grid.nc covers the  
>> geographic region i intended.
> Just to clarify for the group, it's a spherical polar grid and the  
> upside-down-ness has been achieved using negative values in the  
> delY file.

Yes, I was afraid of that. This is part of the part of the code of  
ini_spherical_polar_grid.F where the input delX/Y are used to compute  
the model dXG, dYG, which --together with derived quantities, such as  
dxC/F/V/U dyC/F/V/U-- are essentially used for all calculations.
>           lat = 0.5*(yGloc(I,J)+yGloc(I+1,J))
>           dlon = delX( iGl(I,bi) )
>           dlat = delY( jGl(J,bj) )
>           dXG(I,J,bi,bj) = rSphere*COS(deg2rad*lat)*dlon*deg2rad
>           if (dXG(I,J,bi,bj).LT.1.) dXG(I,J,bi,bj)=0.
>           dYG(I,J,bi,bj) = rSphere*dlat*deg2rad
If your delY is negative, then so is dyG, right? Although you may not  
have immediate problem (e.g, when you evaluate something like (vVel 
(i,j+1)-vVel(i,j))/dyG(i,j) both numerator and denominator change  
sign and you end up with the correct sign), there may be places in  
the code where abs(dyG) is used, which could kill you. Personally, I  
have never tried this, and I don't think that there exists any  
experience with this.

I would not take the risk of trying this with a full blown simulation  
that takes a while to integrate and to interpret. Why not try it with  
a small test domain, but with all the "physics" (parameterizations)  
that you indend to use.

>>   Of course i can rewrite all my input files to run south to  
>> north, but i'm interested to know :-  will there be any  
>> repercussions in, for example, the direction the currents are  
>> calculated or anything bizarre that i might not be able to detect,  
>> or is MITgcm clever enough to calculate everything the correct way  
>> round regardless of whether j=0 is north or south of j=end  (given  
>> that it's got the right lat/lon so that the coriolis force is OK) ?
> Aside from the interesting question of whether MITgcm is clever  
> enough, there's the question of whether Jill's colleagues are  
> clever enough to cope with all the conceptual head-standing  
> required to understand the results.

Well, you could always look at them through a mirror that you place  
at the top of the figure (o:

Martin
>
> -- 
> Mark Hadfield          "Kei puwaha te tai nei, Hoea tahi tatou"
> m.hadfield at niwa.co.nz
> National Institute for Water and Atmospheric Research (NIWA)
>
>
>
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