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

[Nimrod Micael]

Oh my model height is in KM though. Hence why 6 KM and not 600 meters.


On Mon, Sep 2, 2013 at 1:39 AM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:

> 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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>>> gradsusr at gradsusr.org
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>>>
>>>
>>
>>
>> --
>> Nimrod Micael
>>
>
>
>
> --
> Nimrod Micael
>



-- 
Nimrod Micael
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