Measuring Shear

[Bob Hartig]

From a recent post: "Shear could be better, but 35kts is good enough."

I notice that shear in the 40-50 knot range seems to put smiles on people's faces. What I'd like to know is, exactly what does that measurement mean?

I know that shear has to do with the potential of winds to create spin through differences in either speed or direction with height. It seems to me that directional shear would correlate to helicity, but helicity--at least SRH--is shown differently. So when you talk about 40kt shear, exactly what is that 40kt measurement indicating? If it's about speed shear, wouldn't you need to show a range of wind speeds varying with height in order to get a sense of how powerful the shear is? If it's about directional shear, I'm still lost as to how showing a single wind speed shows the potential for vorticity.

While I'm on it, what is the basic range of shear for tornadic storms? 30 to 50 knots? What would be considered minimal? What is extreme?

 

[Jeff Snyder]

The "Shear" you are referring to is likely 0-6km ("deep-layer") shear. Shear is a difference in wind with height (or on the horizontal, but we're talking about vertical shear here), so you need to define "shear" over some sort of distance. For supercells, many tend to use 0-6km / deep-layer / bulk shear in combination with CAPE (instability) to determine supercell potential.

How to find 0-6km shear -- take the vector difference between the surface wind (or 0-500m mean wind if you want) and the wind at 6km AGL. This is easiest to do on a hodograph (which itself seems to be an underused resource) -- Just make a line connecting the surface wind with the 6km AGL wind -- the length of that line (magnitude of the vector) is the 0-6km shear.

As you noted, supercells tend to requires 0-6km shear >30kts, but this is obviously not a rigid criterion. Typically (!!), the more instability, the less deep-layer shear that is needed for supercells. Most of the time, I tend to look for >40kts 0-6km shear for good supercell (instability aside). There are various publications available on the web comparing 0-6km shear for nontornadic and tornadic supercells.

 

[Bob Hartig]

Thanks, Jeff. I unstand--or think I understand--the basics of a hodograph, but I have a hard time reading all the numbers on the vector because some of them tend to crowd together.

In any event, I'm still not clear on how a single number (i.e. 40kt) gives an idea of the strength of shear. But maybe I'm getting a clue now. Is that number the difference in speed between winds at 0k and 6k? In other words, say the wind at the surface is 5kt and the wind at 6k is 40kt, so the difference would be 35kt. Is that difference of 35kt the 0-6k shear? (I realize, by the way, that this is a simplistic example that leaves out wind direction; I'm just using it for the sake of getting a general grasp of things.)

As I think of it, that brings up a related question. Say that the surface winds and 6k winds are nearly parallel; the shear would then be horizontal, right? Lots of tilt to any convection, but not much in the way of helicity. It would take winds changing direction with height to create a corkscrewing effect. So a strong shear reading in itself wouldn't necessarily indicate the potential for updrafts to rotate. Am I right? Or am I getting too involved for this discussion format?

 

[Jeff Snyder]

Bob,

I'm going to refer to a couple of sites that explain hodographs, as I think your question can be answered by a thorough understanding of hodographs. If winds between 0k and 6km are the same direction, then the difference in wind speeds is the 0-6km shear. For example, westerly surface winds at 10kt and westerly 6km AGL winds at 50kts yields 40kt deep-layer shear. If you have easterly sfc winds at 10kts and westerly 6km AGL winds, then the 0-6km deeplayer shear is 60kts. Much of the time, the winds at 0km and 6km aren't exactly the same direction, so simple adding/subtracting of wind speeds/velocities isn't going to work. You can split each wind (sfc and 6km) into their respective U and V components, and subtract from there (u: positive for westerly winds and negative for easterly winds; v: positive for southerly winds and negative for northerly winds). Since you don't really care what the components of the wind shear vector is, you can use a hodograph to quickly find the magnitude of the wind shear (length of the vector).

http://meted.ucar.edu/mesoprim/hodograf/frameset.htm
http://weather.cod.edu/sirvatka/hodo.html

Hodographs are an excellent resource that many don't really know about. It's easy to see on a hodograph that you can still have a UNIDIRECTIONAL shear profile even if you have a veering wind profile. Most people see se sfc winds at sw flow aloft and think supercells, but that profile can also be one of unidirectional shear (and thus less favoring of cyclonically-rotating supercells).

 

[Bob Hartig]

Ah! I get it! And what I don't get, I think the hodograph material you've included will help to clear up for me. Gracias!

One last question: When I'm reading shear and helicity maps, I'm assuming that the speed reading of the barbs gives me the strength of the shear, while the direction of the barbs tells me which way the shear is advecting. Is that right?