[gradsusr] Meridional gradient?

[sushant puranik]

Hi Jeff
Can i use same command for longitudinal gradient computation by replacing
lat --> lon in the command
dy = Re*cdiff(lat,y)*(3.14159/180)

thanks

Sushant


On Tue, May 4, 2010 at 2:33 AM, Jeffrey Duda <jdduda at iastate.edu> wrote:

> Li,
> If you want to find a temperature gradient, why not just use cdiff?  Just
> realize that you'll have to use the following expression for dy (the
> incremental change in meridional distance between grid points, in meters):
>
> dy = Re*cdiff(lat,y)*(3.14159/180)
>
> where Re is the radius of the Earth (about 6.371e6 meters, I believe).
> This is to account for the projection Grads uses to display your data.  Hope
> this helps.
>
> Jeff Duda
>
>
> On Mon, May 3, 2010 at 3:20 PM, Li Dong <ldong at unm.edu> wrote:
>
>> Hello list,
>>
>> I need to calculate the meridional temperature gradient across 20 degree
>> latitude belt using global gridded data. Below is what I did:
>>
>> 'open ta.ctl'
>> 'set t 1'
>> 'set lev 1000'
>> 'set lon 0 360'
>> 'set lat -80 80'
>> 'define Tgrad=ta(lat+10)-ta(lat-10)'
>>
>> However, Grads complained "Cannot use an offset value with a varying
>> dimension". If I set 'lat' at a fixed value, such as 'set lat 40', the
>> above script did work, but the downside is that this only gives
>> temperature meridional gradient at the specified latitude instead of
>> over the entire latitude range. Any inputs would be highly appreciated!
>>
>> Best,
>>
>> Li
>>
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>
>
>
> --
> Jeff Duda
> Iowa State University
> Meteorology Graduate Student
> 3134 Agronomy Hall
> www.meteor.iastate.edu/~jdduda <http://www.meteor.iastate.edu/%7Ejdduda>
>
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