[ncl-talk] R: NCL uv2vr_cfd function on Python

[Giorgio Graffino]

Thank you Dennis,
I will try to use that, or maybe I will find an other quicker solution.

Giorgio



----Messaggio originale----
Da: shea at ucar.edu
Data: 30-set-2015 17.52
A: "Giorgio Graffino"<giorgio.graffino at alice.it>
Ogg: Re: [ncl-talk] R: NCL uv2vr_cfd function on Python

Attached..

Not sure why you want to do python. Mimicking the NCL function could
be cumbersome.
Also, 'for/do' loops in any interpreted language can be slow.

Good Luck

On Wed, Sep 30, 2015 at 9:49 AM, Dennis Shea <shea at ucar.edu> wrote:
> The underlying code used by NCL is a f77 subroutine that handles the
> boundaries (half widths). Also, it allows both the lat and lon arrays
> to be unequally spaced. ie
>
>     dlat_a = lat(2)-lat(0)   , dlat_b = lat(10)-lat(8)
>
>     dlat_a does not need to be the same as dlat_b
>
> Allowing the above, makes the code a bit more involved than your simple script.
>
> The f77 code will be set to you
>
>
>
>
>
>
>
> On Tue, Sep 29, 2015 at 8:29 AM, Giorgio Graffino
> <giorgio.graffino at alice.it> wrote:
>> Ok, I think I've found the problem: dx2 and dy2 are distance in meters, not
>> degrees.
>> Since u and v are given in latitude/longitude, there must be an implicit
>> conversion inside the uv2vr_cfd function.
>>
>> Anyone knows how this conversion is performed?
>>
>> Giorgio
>>
>>
>> ----Messaggio originale----
>> Da: giorgio.graffino at alice.it
>> Data: 29-set-2015 11.31
>> A: <ncl-talk at ucar.edu>
>> Ogg: [ncl-talk] NCL uv2vr_cfd function on Python
>>
>>
>> Dear NCL Users,
>> I'm trying to adapt an NCL script I wrote to compute water vertical speed on
>> Python.
>>
>> In particular, I'm finding some difficulties on "translating" the NCL
>> built-in function uv2vr_cfd into Python; I used that function to compute the
>> vertical component of wind stress. On the web page
>> (https://www.ncl.ucar.edu/Document/Functions/Built-in/uv2vr_cfd.shtml) it's
>> stated that
>>
>> According to H.B. Bluestein [Synoptic-Dynamic Meteorology in Midlatitudes,
>> 1992, Oxford Univ. Press p113-114], let D represent the partial derivative,
>> a the radius of the earth, phi the latitude and dx2/dy2 the appropriate
>> longitudinal and latitudinal spacing, respectively. Then, letting j be the
>> latitude y-subscript, and i be the longitude x-subscript:
>>
>>     rv = Dv/Dx - Du/Dy + (u/a)*tan(phi)
>>
>>     rv(j,i) = (v(j,i+1)-v(j,i-1))/dx2(j)
>>               - (u(j+1,i)-u(j-1,i))/dy2(j)
>>               + (u(j,i)/a)*tan(phi(j))
>>
>> Now, I tried to write that function on Python in this way
>>
>> def Curl(u,v,nlat,nlon):
>>
>> rv = np.empty(shape=(nlat,nlon))
>>
>> rv.fill(np.nan)
>>
>> M0 = np.arange(1,nlat-1) # M0 = [1,...,nlat-1]
>>
>> N0 = np.arange(1,nlon-1) # N0 = [1,...,nlon-1]
>>
>> for m in M0:
>>
>> for n in N0:
>>
>> rv[m,n] = (v[m,n+1]-v[m,n-1])/(lons[n+1]-lons[n-1]) - \
>>
>> (u[m+1,n]-u[m-1,n])/(lats[m+1]-lats[m-1]) + \
>>
>> (u[m,n]/radius)*tans[m]
>>
>> return rv
>>
>>
>> but, even if the pattern is similar, there are four orders of magnitude
>> between the values. I attach results using NCL (MedDailyINDEX.png file) and
>> Python (WindStressCurl.png) for the same day.
>>
>> Could it be possible to know how exactly rv is computed inside uv2vr_cfd
>> function? Because I think there is something missing in my implementation...
>>
>> Thanks for your attention.
>>
>> Regards,
>> Giorgio
>>
>>
>>
>>
>>
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