[gradsusr] Question about why the wind vectors barbs not showing up for the 0-6 KM Shear difference

[Nimrod Micael]

The interpolation is not working for me.

I did this:

'umid=zinterp(u,height,6)'
 'vmid=zinterp(v,height,6)'
 'ulow="zinterp(u,height,0)'
 'vlow="zinterp(v,height,0)'
 'verticalshear =(umid-ulow)+(vmid-vlow)'
 'bulkshear=sqrt(verticalshear)'


Still not getting anything after many tries.


Thanks!


On Sun, Sep 1, 2013 at 11:35 PM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:

> Ohhhhh @Jeff...I see what you are saying...there's an angle to the
> direction of the winds...you cant assume that both levels are in the same
> direction. That makes perfect sense. Very excellent explanation by the way.
>
>
>
>
> On Sun, Sep 1, 2013 at 10:52 PM, Jeff Duda <jeffduda319 at gmail.com> wrote:
>
>> I suggest you have a look through this page (
>> http://mathworld.wolfram.com/VectorAddition.html) and the related page
>> linked under "see also" called "vector difference".  You can add/subtract
>> two vectors by adding/subtracting the corresponding *components* (i.e.,
>> the u and v components), and making a vector of the resulting
>> sum/difference.  What you seem to be stuck on is taking the difference of
>> the *magnitude* of the vectors rather than the difference of the
>> components.
>>
>> I'll use this example to help illustrate this.  Suppose you had a
>> westerly wind at 500 mb of 50 kts, and an easterly wind of 50 kts at the
>> surface (unlikely except for possibly within a severe thunderstorm, but
>> just bear with me).  There is no difference in the wind speed between 500
>> mb and the sfc, but there is a difference in the direction.  Clearly there
>> is shear, but only if you look at the component form of the wind.  The u-
>> and v-components of the 500 mb wind in this example are u = 50, v = 0
>> (kts), whereas at the surface, the components are u = -50, v = 0 (kts).
>>  Therefore, the shear is given by (u500-usfc)*i* + (v500-vsfc)*j* = (50
>> - -50)*i* + (0-0)*j* = 100*i* + 0*j*.  The shear vector is straight out
>> of the west here.  The magnitude is given by sqrt(ushear^2 + vshear^2),
>> which is sqrt(100^2 + 0^2) = 100 kts of shear.
>>
>> Jeff
>>
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>>
>>
>
>
> --
> Nimrod Micael
>



-- 
Nimrod Micael
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