Re: 2004102: vertical derivatives in GEMPAK?

David,

The quo(ldf(thta),ldf(pres)) would be giving

 d(thta) of layer / d(pres) of layer

That would be d(thta)/dp. The ldf(pres) would be a constant for
the calculation of 100mb given your 2 levels +/-50mb. 

The two mul() terms would be d/dx (omega_bar) and d/dx (thta_bar).
The assumption there is that using LAV() where the 2 levels are
+50 and -50 are providing a layer average, rather than just using
the quantity at the level, eg using lav(thta@100:200) instead of
thta@150. Since the dp is a layer, using the horizontal gradient
in the layer instead of at the midpoint level is probably for
consistency.

Steve Chiswell




On Mon, 2004-10-25 at 15:24, David Ovens wrote:
> On Mon, Oct 25, 2004 at 03:19:56PM -0600, Steve Chiswell wrote:
> > David,
> > 
> > LDF(s) will give you the layer difference where
> > you specify GLEVEL=level1:level2, and as
> > appropriate GVCORD=pres. This is the
> > equivalent of SUB(s1@level1,s2@level2).
> > If your vertical layer "delta-p" is reasonably small,
> > then you would be approaching D()/DP.
> > 
> > Steve Chiswell
> 
> Steve, 
> 
> Can you take a look at this PERL code snippet and see if you agree
> with it?  What I don't understand is the decision to use LAV() on
> OMEG and THTA; otherwise, it looks like it is giving an approximate
> D()/DP as you suggest, right?
> 
>       for( $level = 150; $level <= 950; $level += 50 ) {
>             $level1=$level-50;
>             $level2=$level+50;
> 
>             open TMP, ">/tmp/addfields$$";
>             print TMP <<END;
> gdfile = 9303${time}_eta.gem
> gdout =  9303${time}_eta.add.gem
> gdattim = $fhr
> glevel = $level1:$level2
> gvcord = pres
> gfunc = mul(mul(quo(ldf(thta),ldf(pres)),ddx(lav(omeg))),ddx(lav(thta)))
> grdnam = tiltt1\@$level
> r
> 
> gfunc = mul(mul(quo(ldf(thta),ldf(pres)),ddy(lav(omeg))),ddy(lav(thta)))
> grdnam = tiltt2\@$level
> r
> 
> END
> 
>             close TMP;
>             system "gddiag </tmp/addfields$$";
> 
> David