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

[Nimrod Micael]

And it's fine to also use 10m wind for U since I wasn't able to retrieve
wind at 0 KM?


On Mon, Sep 2, 2013 at 1:08 PM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:

> Perfect. Just to verify I am doing the right procedure:
>
> 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'
>  'set clevs 20 30 40 50 60'
>  'set ccols 0 3 7 8 2 6'
>  'd shear0006'
>
>  'ushear= u6km-u10'
>  'vshear= v6km-v10'
>
>
>  'set gxout barb'
>  'set cmin 20'
>  'set arrlab off'
>  'set clopts -1 -1 0.14'
>  'set arrscl 5'
>  'd skip(ushear*1.94,40,40);skip(vshear*1.94,40,40)'
>
>
> Thanks Jeff you are awesome! :)
>
>
> On Mon, Sep 2, 2013 at 12:09 PM, Jeff Duda <jeffduda319 at gmail.com> wrote:
>
>> You're still only displaying a scalar quantity.  You need to display the
>> shear components separated by a ; to get the barbs to appear.
>>
>>
>> On Mon, Sep 2, 2013 at 10:39 AM, Nimrod Micael <nmicael at aggies.ncat.edu>wrote:
>>
>>> 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
>>>
>>> _______________________________________________
>>> gradsusr mailing list
>>> gradsusr at gradsusr.org
>>> http://gradsusr.org/mailman/listinfo/gradsusr
>>>
>>>
>>
>>
>> --
>> Jeff Duda
>> Graduate research assistant
>> University of Oklahoma School of Meteorology
>> Center for Analysis and Prediction of Storms
>>
>> _______________________________________________
>> gradsusr mailing list
>> gradsusr at gradsusr.org
>> http://gradsusr.org/mailman/listinfo/gradsusr
>>
>>
>
>
> --
> Nimrod Micael
>



-- 
Nimrod Micael
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