Problems with @IIN integration

[Glenn Carver <Glenn.Carver@xxxxxxxxxxxxxxxx>]

I'd appreciate some help from someone experienced with the @IIN (indefinite
integral) function in Ferret. I'm having a real tough time getting it to
work correctly with my irregularly spaced z coord, but also there seems to
be some inconsistency in the way ferret calcs. spacing (ie ZBOX) which is
not helping.

Essentially I'm trying to compute half-level pressures from isentropic
(theta) levels using:  dp = -g.sigma.dtheta. In principle @IIN is precisely
what I want. In Fortran the code to do this is dead easy:

   ph(1) = 0.0     ! n.b. real :: ph(nlev+1), sigma(nlev) and ph(1) is
always 0.
   do k = 1, nlev
   ph(k+1) = ph(k) + g * sigma(k)*dtheta(k)
   end do

The problem in Ferret centres around getting the integral to work
correctly. In Ferret, SIGMA is on the model full levels which in my case
look like:
yes? list z[g=sigma], sigma
 Column  1: Z is Z (axis THETA)
 Column  2: SIGMA is Isentropic density (kg/(m2.K))
                  Z     SIGMA
344   / 1:  343.9950  16.97323
373.2 / 2:  373.1750  11.09886
404.8 / 3:  404.8350   7.45681
439.2 / 4:  439.1750   5.10337

I first thought something like  LET A = SIGMA[Z=330:5400@IIN] might work,
but it doesn't because if the integral ranges are specified in world coords
then it weights the first box by the fraction of the box in the range. So
whereas I want  ((373.2-344)/2 - 330)*16.9   I get ((373.2-344)/2 -
344)*16.9.

So that's no good. Next I thought I'd try putting sigma on the half level
grid which is defined as thetahalf = 0.5 (theta(k) + theta(k+1) ), and
integrate on this z axis. But now another problem crops up. Here's the
theta half levels (THDOTH happens to be variable on the theta half level
axis).

yes? list z[g=thdoth], zbox[g=thdoth]
 Column  1: Z is Z (axis THETAHALF)
 Column  2: ZBOX is ZBOX (axis THETAHALF)
                  Z      ZBOX
330   / 1:  330.0000  27.98999
358   / 2:  357.9900  29.17999
388.4 / 3:  388.3600  31.66000
421.3 / 4:  421.3100  34.34001

ZBOX is what's going to be used to weight the sigma values when the
integral is done. Look at the value for k=1. You'll see it's z(2) - z(1).
Ok there. Now, look at zbox for k=2. It's not z(3)-z(2), instead it's the
distance between the the midpoints either side of this node. That is,
zbox(2) = (0.5(z(2)-z(1)) - 0.5(z(3)-z(2)). This seems inconsistent because
it means zbox(1) is overlapping with zbox(2). Or rather then endpoint zbox
values are forward differences whereas the remainder are centred
differences.

This doesn't help me much when all I want is sigma(2)*(z(2)-z(1)).  It
would be nice if Ferret treated [z=@iin] and [k=@iin] differently so that
[z=@iin] was as now but [k=@iin] took the cell gridbox to be from the nodes
rather than midpoints. ie. zbox(2) = z(3)-(2). But maybe that's too subtle
and prone to confusion. Either way there doesn't seem anyway of using @IIN.

At this point, I'm not sure how to proceed. I think I need to go back to
the first approach and put sigma on a grid in which the z axis has been
extended at either end, but I've not figured out how to do that yet and I'm
not sure it would work anyway. Another way might be to do it by hand,
compute the differences the way I want and then use RSUM to do the
integration?

If anyone can help or has done something similar with @IIN I'd appreciate
hearing from them.

Many thanks,

              Glenn


----
Dr. Glenn Carver, Senior Research Associate,
Centre for Atmospheric Science, Chemistry Dept., Cambridge University, UK
mailto:Glenn.Carver@atm.ch.cam.ac.uk   http://www.atm.ch.cam.ac.uk/~glenn/

              "I never think of the future, it comes soon enough"
                   - Albert Einstein