Re: [ferret_users] Two interpolation syntax questions

[Ansley Manke <ansley.b.manke@xxxxxxxx>]

Hi again,
I realized I missed one important point for your data.  If the time 
variable is in the form of hours and minutes HHMM then of course 
interpolating based on it isn't correct. Maybe this is what you have in 
mind with your own interpolation method. You'll get the right answer if 
you use the variables "day" and "time" to define a time axis, and put 
the lon and lat variables on that axis.  You can decide what units it 
should have and define things accordingly but for example,

 let hr = INT(time/100)
 let mn = time - hr*100
 let daytime = 24*day + hr + mn/60

 def axis/t/units=hours tax = daytime
 let tlon = reshape (lon, t[gt=tax])
 let tlat = reshape (lat, t[gt=tax])

 ! Define time point at which we want to interpolate
 !   day 20, hour 10, 30 minutes:
 let timepoint = 20*24 + 10 + 30/60

 let tt = t[gt=tax]
 let test = tlat* tt[t=@weq:`timepoint`]
 let nlat = test[L=1:48@sum]

-Ansley

Ansley Manke wrote:

> Hi Steve,
> Your second attempt is much more the preferred way to do this kind of 
> thing in Ferret. It uses the grid mindset of Ferret to work with the 
> variable as a whole and apply Ferret transformations or other 
> operations to do what you want. Either the @WEQ or @EVNT transform 
> will locate your point, and the variable you define
>
>  let test = lat*time[x=@weq:1030]
>  let nlat = test[x=1:48@sum]
>
>
> should be the latitude at the time point that you've specified, 
> computed by linear interpolation. When you define a variable
>
>   myvar = time[x=@weq:1030]
>
> the @WEQ result returns the weights for the points on either side of 
> the specified value.  It finds where the variable "time" crosses the 
> specified value and and gives the corresponding weights. There are no 
> other places where the variable crosses the value 1030, so the field 
> is missing everywhere else. Doing a calculation based on the value of 
> your variable "test" will give you the latitude just where time 
> crosses 1030, which is what you want.  If you want to ignore the 
> weights and do something else, you could for example define a new 
> variable
>
>   let mytimes = 1 + 0*time[x=@weq:1030]
>
> and the value would be 1 on each side of where TIME crosses 1030 and 
> missing elsewhere.
>
> Or, the output of time[x=@EVNT:1030] is 0 until the variable crosses 
> the specified value, then 1 until it crosses it again, (and would take 
> on the value 2 if it crossed again).
>
> (For the general question about IF statements, the thing to understand 
> is that the expression inside the IF needs to be a scalar or string, 
> that evaluates to a simple TRUE or 1, or FALSE or 0.  There is not an 
> AND clause but something like this could be done with nested IF's. It 
> is cumbersome, which is a good reason to use Ferret expressions like 
> the ones above instead.
>
>  repeat/i=1:48 (let di = day[i=`i`]; let ti = time[i=`i`]; if  `20 eq 
> di` then if `0325 gt ti` then if `0325 lt ti` then ...)
>
>
>
>
> Steve Guimond wrote:
>
> > Dear Ferreters:
> >   I have two separate, but related questions.  I have latitudes,
> > longitudes and times for the positions of several tropical cyclones 
> > and need to co-locate with some
> > satellite data.  Basically, just need to do a simple interpolation to 
> > find the position of the storm at satellite overpass.  So, the 
> > positions of the storm are at 20.50,53.50 at 0000Z and 21.34,52.10 at 
> > 0600Z and the satellite overpass is at 0325Z.    Data looks like:
> >
> > day     i=1:48
> > time    i=1:48
> > lat       i=1:48
> > lon      i=1:48
> >
> >   I wanted to find the appropriate "i" position in the array that 0325Z
> > was between and then apply my own interpolation calculations, but 
> > found this difficult.  In
> > fortran I would code:
> >
> > do i=1,48
> > if ( 0325 .gt. time(i) .and. 0325 .lt. time(i+1) ) then
> > complete interpolation
> > endif
> > enddo
> >
> > I tried this in ferret by saying:
> >
> > repeat/range=1:48:1/name=a (if 20 eq day[i=`a`] and 0325 gt time[i=`a`]
> > and 0325 lt time[i=`a`+1] then complete interpolation)
> >
> > I tried various sequences, but could not get the equivalent of my 
> > fortran code in ferret to work.      Additionally, I tried using the 
> > @weq transformation as:
> >
> > let test = lat*time[x=@weq:1030]      !trying a different time here
> > let nlat = test[x=1:48@sum]
> >
> > This worked for the first time in the array between 0000Z and 0600Z, but
> > did not produce a field of coefficients for the rest of the times 
> > that I needed.  This is
> > what I got from listing test and time together...
> >       time    test
> > 1/1: 600.    3.173
> > 2/2:1200.   8.027
> > 3/3:1800.   ....
> > 4/4:      0.   ....
> > 5/5: 600.     ....
> > 6/6:1200.   ....
> > etc.......
> >
> > why are there not coefficients for the rest of the times at 1030?
> >
> > Can anyone help to clear up (1) how to cycle through data element by
> > element like fortran for the statement above and (2) how to get 
> > coefficients for all the times and not just the first instance of 
> > 1030 like I showed above?
> >
> >
> > Many thanks...
> >
> > Steve Guimond
> >
> >
> >
> > *****************************************
> > Stephen R. Guimond
> > Graduate Research Assistant
> > Center for Ocean-Atmospheric Prediction Studies 
> > *****************************************
> >
> >
> >
> >