creating periodic grid ...

[turuncu]

Hi,

I want to create a periodic grid in East-West boundary. The grid box is

Code: Select all

0,20N  --------- 360,20N
|                  |
|                  |
0,-20N --------- 360,-20N

So, it is like a belt over the earth equator. So, can i use ROMS to simulate such domain? My question is that do i need to add grid point in Eastern or Western side to create periodic grid? I mean that does the grid coordinates include the both 0 and 360 longitudes for the periodicity? I think that the projection must be Mercator. Am i correct? I never create a periodic grid and i just wonder that are there any trick for it?

Best Regards,

--ufuk

[mathieu]

First of all you have to realize that ROMS does not use the longitude/latitude internally. It uses pm, pn, Coriolis factor f and a few others. That is why you can make periodic grid at all because if longitude were used then you would have a problem with the ambiguity 0=360.

Hence you can actually do periodic grid as you want with periodicity along the longitude. For the latitude, you cannot do that with finite difference and so you have to let the north pole out.

In general when running ROMS in your grid domain (1:eta_rho, 1:xi_rho) only (2:eta_rho-1, 2:xi_rho-1) are active. If you choose NS_PERIODIC then

ZETA(1,:)=ZETA(eta_rho-1,:) and ZETA(eta_rho,:)=ZETA(2,:)

and similarly for other variables and the option EW_PERIODIC.

I would advise you to look at the code of easygrid, seagrid and others to see how grids are build and then build yourself the periodic grid, because there is no tools for periodic grids.

[kate]

For longitude, the psi points must be repeated, so the far east psi point is the same as the far west psi point. Then in rho, you need two overlapping edge points, such that i=0 is a copy of i=Lm and i=Lm+1 is a copy of i=Lm. There are Lm cells all around.

[turuncu]

Hi,

First of all, thanks for your help. From Kate's post,

i=0 is a copy of i=Lm and i=Lm+1 is a copy of i=Lm

are you sure about that because i expect like following, i=0 is a copy of i=Lm and i=1 is a copy of i=Lm+1

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0, 1 ......... Lm, Lm+1
               0 , 1 .............. Lm, Lm+1 

What do you think? Anyway, thanks for your help.

Regards,

--ufuk

[kate]

Of course you are right - just not typing what I meant. However, it's that i=Lm+1 is a copy of i=1 since i=1 is computed and i=Lm+1 is a boundary condition.