Hi,
On our eORCA025 config, NEMO v4.0.1 with jperio == 4. The first two and last two columns (x dims) are the same due to eastwest cyclicity.
Then, my understanding is that each wet cell on the j == jpjglo
row has exactly one twin on the other side (here “other side” is with respect to grid index i
) of the North American continent, on the j == jpjglo  1
row. No other cells are duplicated, and no cell is there three times. At least that’s what comes out when searching for geographical duplicates in our mesh_mask files.
Is that correct with jperio == 4? Got confused because because if it is, then I don’t get this line of the CDFTOOLS seaice diagnostics:
CASE (4) ! ORCA025 type boundary
tmask(1:2,:)=0.
tmask(:,npjglo)=0.
tmask(npiglo/2+1:npiglo,npjglo1)= 0.
It’s meant to compute integrated seaice metrics (volume, extent etc.). The code above masks the duplicated cells to avoid double accounting: first the eastwest cyclicity duplicates, then the top row, and finally one half of the secondtotop row. I don’t get that last step  seems to me that it’s removing an extra halfrow… but I want to be sure that it’s not due to some inconsistencies between jperio
and our coordinate files.
Salut Charles
About North Pole Folding: you may find these sketches useful (thanks to Seb), Migrating from 4.0.x to 4.2.x — NEMO release4.2.0 documentation
You have a “Tpoint” folding, i.e. the symmetry at Tpoints concerns 3 lines along j ; the center (invariant) of the transform is located at the Tpoint located at (ip = npiglo/2 + 1 ; jp = npjglo1)
. Hence, row at j=npjglo1
is symmetric along i
around this point. So it makes sense counting points only once for half of this line.
This said, I would have rather written in the block above:
tmask(npiglo/2:npiglo,npjglo1)= 0.
,
e.g. kept the pivot as part of “nonduplicated” points.
@smasson, aka “NEMO NPF master”, would easily confirm this or not.
Jérôme
NB: Everything above is based on Fortran indexing
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Many thanks Jérôme, and thanks to the person (presumably the NEMO NPF master) who made these new sketches. So if I got it right:

jperio == 4
:
 every cell on the
j=npjglo1
line has a symmetric on the same line;
 every cell on the
j=npjglo
line has a symmetric on the j=npjglo2
line.

jperio == 6
: every cell on the j=npjglo
has a symmetric on the j=npjglo1
line, and that’s it.
You got it (there is aditionnal complexity for vector fields of course).
I wrote too fast. Of course, it should rather be;
tmask(npiglo/2+2:npiglo,npjglo1)= 0.
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