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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>
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Received on Mon Jul 06 2009 - 17:18:48 MDT
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