[MITgcm-support] rSphere definition when using 'deepAtmosphere' option for ocean modelling - quick question

[Jean-Michel Campin]

Hi Flynn,

I will try to clarify how the deepAtmosphere (deepAtmosphere=TRUE) scaling is computed:

It really depends what vertical coordinate you are using:
1) using Z-cordinate: 
   scaling fct (i.e., distance from center of the sphere) is "rSphere+rF" (level interface)
   or "rSphere+rC" (level center), and since rF(1) = seaLev_Z
  (with default seaLev_Z value =0)), it means the "rSphere" value correspond to the top
  of the model (where the free-surface is).
2) using P-coordinate:
   scaling fct (i.e., distance from center of the sphere) is "rSphere+phiRef/gravity"
  which means that:
  a) for Ocean in P-coord, since at the top phiRef = seaLev_Z*gravity (but this time 
   the top is at rF(Nr+1)), the "rSphere" value correspond also to the top of the model
   (assuming we are using the default seaLev_Z=0 value).
  b) for Atmosphere in P-coord,
   since phiRef = seaLev_Z*gravity at the bottom (i.e., at rF(1)), the "rSphere" value 
   correspond to the bottom of the model.

Cheers,
Jean-Michel
   
On Mon, Feb 27, 2023 at 04:39:26PM +0000, Flynn Ames wrote:
> Dear MITgcm community,
> 
> Hello - I'm hoping to sanity check my interpretation of the 'rsphere' parameter in the MITgcm.
> 
> I've been interpreting rsphere as the distance to the top of the model domain (surface of the ocean).
> However, it was recently pointed out to me that rSphere could instead potentially refer to the distance to the bottom of the model domain (bottom of the ocean).
> 
> In the code, in the file 'set_grid_factors.F', it appears the computation of the deepAtmosphere adjustments to the grid cell size assume the radius is defined at the top of the domain (the adjustment factor appears to create a smaller cell for increasing Nr). So it seems my initial interpretation could be correct. Though I was wondering if someone else could confirm this?
> 
> My ocean model has the 'deepAtmosphere' option turned on (with an ocean depth comparable to the radius of the sphere), so this interpretation will have significant effects upon my results.
> 
> 
> Thanks for your time!
> Flynn
> 

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