[gradsusr] Meridional gradient?

[Jeffrey Duda]

Put the literal argument "lat" in place of "19.5" in the cdiff command for
dy and put "lon" for "30" in the cdiff for the dx definition.  There's
nothing wrong with setting your dimensions first, as that will only affect
the horizontal scope of the subsequent fields you define.

Jeff

On Sat, Jun 25, 2011 at 12:11 AM, sushant puranik
<sushantpuranik at gmail.com>wrote:

> 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
>>>>>
>>>>> _______________________________________________
>>>>> gradsusr mailing list
>>>>> gradsusr at gradsusr.org
>>>>> http://gradsusr.org/mailman/listinfo/gradsusr
>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> 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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>


-- 
Jeff Duda
Iowa State University
Meteorology Graduate Student
3134 Agronomy Hall
www.meteor.iastate.edu/~jdduda
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