UPWELLING test case

[schen]

hi all,
I'm a new user and trying to understand the UPWELLING test case first.
In the UPWELLING case, Lm is 41 and Mm is 80. But in the output file
the bathymetry (h) is 43x82. So, there is one extra point at the boundary (?).
I know the depth proflie is set up in analytical.F as below
///////////////////////////////////////////////////////////////////////////////////////////////
DO j=JstrR,JendR
IF (j.le.Mm(ng)/2) THEN
val1=REAL(j,r8)
ELSE
val1=REAL(Mm(ng)+1-j,r8)
END IF
val2=MIN(depth,84.5_r8+66.526_r8*TANH((val1-10.0_r8)/7.0_r8))
DO i=IstrR,IendR
h(i,j)=val2
END DO
END DO
//////////////////////////////////////////////////////////////////////////////////////////////

it looks like JstrR is 1, JendR is 80, IstrR is 1, and IendR is 41.

My question is the above code only gives depth for the interior points
(80 pints in ETA-direction).
Where can you set the depth for two extra boundary points in ETA-direction?

Or my understanding is totally wrong ? Thanks a lot

[kate]

There is always an extra rho point outside the box defined by Istr to Iend, Jstr to Jend. Where IstrR lies relative to Istr depends on if the domain is periodic or not (assuming it isn't tiled). In a periodic direction IstrR=Istr, with the value of the points outside being set by the code that makes the domain periodic. Otherwise IstrR is zero when Istr is one.

The upwelling case is a periodic channel, periodic in the i direction. The reason Lm is 41 and not 40 is historic from a different method of handling the periodic join, but it doesn't really matter since there should be no long-channel variation in the results. Lm could just as well be 5 or 10.

In this case, IstrR = 1, JstrR = 0, IendR = 41, JendR = 81. OK? I put some charts here:
viewtopic.php?t=106