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

[Nimrod Micael]

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
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