[gradsusr] Meridional gradient?

[sushant puranik]

Thanks Jeff.
But i am not able to plot the output. It gives error with the latitudinal
command

ga-> define dy=6.37e6*cdiff(19.5,y)*(3.1416/180)
Error from CDIFF:  Specified dimension non varying
Operation Error:  Error from cdiff function
  Error ocurred at column 8
DEFINE error:  Invalid expression.

and also tried with the longitudinal command

ga-> define dx=6.37e6*cos(10*(3.1416/180))*cdiff(30,x)*(3.1416/180)
Error from CDIFF:  Specified dimension non varying
Operation Error:  Error from cdiff function
  Error ocurred at column 29
DEFINE error:  Invalid expression.
Segmentation fault

i think i am have to set dimensions first?

Thank you

Sushant

On Fri, Jun 24, 2011 at 10:14 PM, Jeffrey Duda <jdduda at iastate.edu> wrote:

> Sushant,
> Unfortunately, no.  Getting the longitudinal gradient is not as simple
> since the distance between longitude lines is a function of the latitude.
> However, the formula is
>
> dx = Re*cos(lat*(pi/180))*cdiff(lon,x)*(pi/180).
>
> The cos(lat*(pi/180)) term is the adjustment for dependence of longitude
> space on latitude.
>
> Jeff
>
>
> On Fri, Jun 24, 2011 at 12:55 AM, sushant puranik <
> sushantpuranik at gmail.com> wrote:
>
>> 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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>>>
>>>
>>
>>
>>
>>
>>
>>
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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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>


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