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

[Nimrod Micael]

Actually I think I fixed the problem but still didnt get the wind barb...

 zinterp(u,height,6)
 'u6km=interp'
 zinterp(v,height,6)
 'v6km=interp'
 'shear0006=1.94*mag(u6km-u10,v6km-v10)'
 'set gxout shaded'
 'set cmin 20'
 'd shear0006'


 'set gxout barb'
 'set cmin 20'
 'd shear0006'


What's going on?


On Mon, Sep 2, 2013 at 1:40 AM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:

> 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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>>>>
>>>>
>>>
>>>
>>> --
>>> Nimrod Micael
>>>
>>
>>
>>
>> --
>> Nimrod Micael
>>
>
>
>
> --
> Nimrod Micael
>



-- 
Nimrod Micael
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