Deep Layer Shear

[James Gustina]

So i have been doing forecasting for a year or two now and i've seen deep layer shear a lot but i haven't quite grasped what it is. I've seen "0-6Km vector" attached to it. So could someone explain exactly how to find and use Deep layer shear when it comes to svr forecasting?

 

[Chris Vagasky]

The 0-6 km shear can be used as a predictor to discriminate on storm type.

Supercells are generally found with values of 35-40 knots.

http://w1.spc.woc.noaa.gov/exper/mesoanalysis/help/help_shr6.html

 

[Bob Hartig]

To clarify, those values are a threshold, not a range. In other words, "35 to 40 knots and higher," not "between 35 and 50 knots."

 

[James Gustina]

Ah ok. Thanks guys. That helps a bunch.

 

[Jeff Snyder]

James,

When most operational meteorologists and chasers refer to "wind shear", we/they are referring to the vertical shear of the horizontal wind (or, how the horizontal wind changes with height). Technically, however, vertical wind shear actually has units of 1/s or s^(-1), since it's defined as d(Uh)/dz, where Uh is the horizontal wind vector. However, by appending the qualifier "0-6 km" to "shear", most folks (myself included) are actually referring to the magnitude of the vector difference between the wind at 6 km AGL and the wind at the surface (usually a mean wind between the surface and, say, 500 m). So, if the 6 km wind is from the west at 50 kts, and the surface wind is from the west at 20 kts, we'd say that the "0-6km shear" (again, actually "0-6km vector wind difference") is 30 kts. If the 6 km wind is westerly at 50 kts, and the surface wind is southerly at 20 kts, the "0-6km shear" is 54 kts.

Personally, I look for 0-6km wind "shear" >40 kts for sustained supercells, but that can drop to 30-35 kts if instability is very strong (e.g. 4000 j/kg CAPE). This is certainly just a "rule of thumb" for me, however, since there are many more pieces of the pie, if you will (including low-level storm-relative helicity, whether or not a boundary is nearby, etc.).

 

[James Gustina]

Thanks for that jeff. Yeah i understood helicity, storm relative inflow, etc. but this has really been a help with deep layer shear. Make it a lot easier on me trying to find a target area

 

[Paul Knightley]

Just remember that should a storm suddenly deviate by following an outflow boundary, or even back-build along a boundary, seemingly low deep-layer shear can, in a storm relative sense, become quite high!

 

[Darren Addy]

So, if the 6 km wind is from the west at 50 kts, and the surface wind is from the west at 20 kts, we'd say that the "0-6km shear" (again, actually "0-6km vector wind difference") is 30 kts. If the 6 km wind is westerly at 50 kts, and the surface wind is southerly at 20 kts, the "0-6km shear" is 53 kts.

Thanks for this post, Jeff. I get your first example where both vectors were from the west, but not your second (above). Can you "show your work" on how you calculate this with wind from different directions? TIA

 

[Chris Vagasky]

Thanks for this post, Jeff. I get your first example where both vectors were from the west, but not your second (above). Can you "show your work" on how you calculate this with wind from different directions? TIA

Darren,

His second example is a case of adding vectors. If you think back to high school geometry, to add vectors, you place the first vector (say, south at 20 knots) and from the tip of the first vector you place your second vector (west at 50 knots). Your new vector extends from the tail of the first vector to the tip of the second vector.

Understanding that my words aren't as good as my ability to draw it out...

 

[Darren Addy]

Ah! Excellent Chris. The fog clears.

 

[Jeff Snyder]

Darren -- no problem. Since it's easiest to see on a hodograph, I made up two hodographs that illustrate my post. To calculate the 0-6km shear vector magnitude for the second case, you can just use the pythagorean theorem since the surface and 6 km winds are normal to each other. The end result is about 54 kts.

EDIT: Chris did a good job of explaining the magnitude, but the direction of the vector is a bit off. For 0-6km shear, we want to subtract the sfc wind vector from the 6 km wind vector. Doing so is equivalent to adding -sfc wind vector (that is, adding the negative of the surface wind vector) to the 6 km wind vector. In Chris's post, he added the 6 km wind vector to the surface wind vector (not the negative of the surface wind vector), which results in an error in the direction of the vector. [Think of our first example of W sfc winds at 20 kts and W 6km winds at 50 kts... If we just add the two vector, we'd get 70 kts, when the magnitude of the 0-6km shear vector difference is actually 30 kts]. Because the sfc and 6 km winds in our second example are perpendicular to each other, the magnitude is the same whether we add or subtract.

 

[Darren Addy]

Thanks for the further explanation, Jeff.
I was going to say, the Pythagorean theorem only works for right triangles. So for other triangles (where the wind vectors are not 90 degrees from each other) one would need to use the Law of Sines or the Law of Cosines, eh?

Let's just say it has been a while since 10th grade Geometry.

 

[Jeff Snyder]

Thanks for the further explanation, Jeff.
I was going to say, the Pythagorean theorem only works for right triangles. So for other triangles (where the wind vectors are not 90 degrees from each other) one would need to use the Law of Sines or the Law of Cosines, eh?

Let's just say it has been a while since 10th grade Geometry.

It'd be easiest if you had a hodograph on a dry erase board. Then, you could just dry the vector between the tips of the sfc and 6 km winds, measure the length of the resulting vector, then line up that length with the hodograph. Since we're just looking of the magnitude, you can "move" the vector that represents the difference between the 6 km and sfc winds to the origin, and see how far the "tip" of the shear vector is from the origin. Then again, I'm not sure how many people have a dry erase board of blank hodographs... LOL.



If you want an estimate in the case that you run across a hodograph (or forecast hodograph) online (e.g. on TwisterData, WxCaster, etc), you can just take any object with a straight edge (a piece of paper works well), line up the edge along the tips of the 6 km and sfc winds, mark (on the paper) where the tips are, then move the mark for the sfc wind to the origin of the hodograph. Rotate the paper any way you want if it makes it easier, and find out how far from the origin the mark from the 6km wind is. I do this on a semi-regular basis.

 

[Chris Vagasky]

EDIT: Chris did a good job of explaining the magnitude, but the direction of the vector is a bit off. For 0-6km shear, we want to subtract the sfc wind vector from the 6 km wind vector. Doing so is equivalent to adding -sfc wind vector (that is, adding the negative of the surface wind vector) to the 6 km wind vector. In Chris's post, he added the 6 km wind vector to the surface wind vector (not the negative of the surface wind vector), which results in an error in the direction of the vector. [Think of our first example of W sfc winds at 20 kts and W 6km winds at 50 kts... If we just add the two vector, we'd get 70 kts, when the magnitude of the 0-6km shear vector difference is actually 30 kts]. Because the sfc and 6 km winds in our second example are perpendicular to each other, the magnitude is the same whether we add or subtract.

You're right Jeff, I wasn't thinking of it as Vector Wind Difference at the time...was writing the post as I was getting my A/C and kitchen cabinets fixed...