Re: gradient computation

From: Mateus Teixeira <mateus.teixeira_at_nyahnyahspammersnyahnyah>
Date: Mon, 6 Jul 2009 20:24:14 -0300

Thank Dennis,

I want K/m.

Best regards,

Mateus

2009/7/6 Dennis Shea <shea_at_ucar.edu>

> What units do you want on the gradient?
>
> If K/deg_of_latitude
> dT_dy = center_finite_diff( T( lon|:, lat|: ), lat, False, 0 )
>
> If K/m
> y = lat*6.37e6*acos(-1.)/180.
> dT_dy = center_finite_diff( T( lon|:, lat|: ), y, False, 0 )
>
> ------
> Note: if it is a global gaussian grid, use
> http://www.ncl.ucar.edu/Document/Functions/Built-in/gradsg.shtml
>
> Mateus Teixeira wrote:
>
>> Dear NCL users,
>>
>> I have a temperature field that is not equally spaced in y, but it is in
>> x. I'm not sure how to compute its gradient in y direction. I'm wondering if
>> I can apply this
>>
>> dphi = center_finite_diff( lat, 1., False, 0 ) * 6.37e6 * acos( -1. )/180.
>> dT_dy = center_finite_diff( T( lon|:, lat|: ), dphi, False, 0 )
>>
>> or I just use
>>
>> dT_dy = center_finite_diff( T( lon|:, lat|: ), lat * acos(-1.)/180.,
>> False, 0 ) / 6.37e6
>>
>> Thanks.
>>
>>
>> Best regards,
>>
>> --
>> Mateus da Silva Teixeira
>>
>> Registered Linux User #466740
>>
>>
>> ------------------------------------------------------------------------
>>
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>

-- 
Mateus da Silva Teixeira
Registered Linux User #466740

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Received on Mon Jul 06 2009 - 17:24:14 MDT

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