[gradsusr] Meridional gradient?

Mon Jun 27 11:26:56 EDT 2011

Since dx is dependent on latitude, the contour plot for dx is correct.  The
largest values of dx are going to be at the equator where there is the
largest meridional distance between longitude lines.

For your dy plot, I guess we need to make sure you actually have a latitude
variable in your data.  Try just 'd lat'.  You should not get a constant
value.  If that doesn't work, send me either the .ctl file or the result of
ncdump -h if your data is coming from a netCDF file.

Jeff

On Mon, Jun 27, 2011 at 12:13 AM, sushant puranik
<sushantpuranik at gmail.com>wrote:

> Jeff
> I tried with making changes asper your suggestion.
> I got results as Constant field value =333533 for
> dy=6.37e6*cdiff(lat,y)*(3.1416/180) command
>
> and also tried dx=6.37e6*cos(lat*(3.1416/180))*cdiff(lon,x)*(3.1416/180) in
> this case it shows simply parallel lines on the plot.
> Results are attached with this mail.
>
> Thank you
>
> Sushant
>
> On Sat, Jun 25, 2011 at 8:58 PM, Jeffrey Duda <jdduda at iastate.edu> wrote:
>
>> 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>
>>>>>>
>>>>>> _______________________________________________
>>>>>> gradsusr mailing list
>>>>>> gradsusr at gradsusr.org
>>>>>> http://gradsusr.org/mailman/listinfo/gradsusr
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> _______________________________________________
>>>>> gradsusr mailing list
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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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-- 
Jeff Duda
Iowa State University
Meteorology Graduate Student
3134 Agronomy Hall
www.meteor.iastate.edu/~jdduda
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