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

[Nimrod Micael]

oh u10 and v10 is
      u10  1  0  U at 10 M (m s-1)
      v10  1  0  V at 10 M (m s-1)
respectively.


On Mon, Sep 2, 2013 at 11:29 AM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:

> 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
>>>>>
>>>>> _______________________________________________
>>>>> gradsusr mailing list
>>>>> gradsusr at gradsusr.org
>>>>> http://gradsusr.org/mailman/listinfo/gradsusr
>>>>>
>>>>>
>>>>
>>>>
>>>> --
>>>> Nimrod Micael
>>>>
>>>
>>>
>>>
>>> --
>>> Nimrod Micael
>>>
>>
>>
>>
>> --
>> Nimrod Micael
>>
>
>
>
> --
> Nimrod Micael
>



-- 
Nimrod Micael
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