Jump to content
  • Welcome to American Weather

    Register now to gain access to all of our features. Once registered and logged in, you will be able to contribute to this site by submitting your own content or replying to existing content. You'll be able to customize your profile, receive reputation points as a reward for submitting content, while also communicating with other members via your own private inbox, plus much more! This message will be removed once you have signed in.

Sign in to follow this  
It's Always Sunny

Bulk Shear Formula

Recommended Posts

Have been having a hard time finding this online so figured I'd ask here. I am looking for a bulk shear equation to subtract two wind vectors with their wind directions included as well. An example would be:

 

What is the bulk shear between a 15kt wind out of the SW (220 degrees) at 5000 ft and a 40kt wind out of the W (270 degrees) at 8000ft?

 

Thanks in advance.

Share this post


Link to post
Share on other sites

I believe you use the law of cosines.

c^2 = a^2 + b^2 - 2ab*cos(gamma)

So

a=15, b=40, and gamma=270-220

Then

sqrt(15^2 + 40^2 - (2*15*40*cos(50))) = sqrt(225 + 1600 - 1200*0.64) = sqrt(1057) = 32.5 kts

You can play around with special cases such as gamma = 0 or gamma = 180 to see that the shear would reduce to the trivial sums (40 - 15) = 25 and (40 + 15) = 55 respectively.

 

Share this post


Link to post
Share on other sites
18 minutes ago, bdgwx said:

I believe you use the law of cosines.

c^2 = a^2 + b^2 - 2ab*cos(gamma)

So

a=15, b=40, and gamma=270-220

Then

sqrt(15^2 + 40^2 - (2*15*40*cos(50))) = sqrt(225 + 1600 - 1200*0.64) = sqrt(1057) = 32.5 kts

You can play around with special cases such as gamma = 0 or gamma = 180 to see that the shear would reduce to the trivial sums (40 - 15) = 25 and (40 + 15) = 55 respectively.

 

Thanks for this. Would elevation difference make a difference so in this case it would be 3000ft.

Share this post


Link to post
Share on other sites

I don't think the elevation makes a difference since normally you specify which layer the bulk shear is covering anyway. So in this case you'd just say the bulk shear in the layer 5000-8000ft is 32.5 kts.

I have seen some literature actually divide by the depth to get the units as s^-1 as a way of expressing the shear in relation to the depth. So in that case you'd just divide by the depth after converting 32.5 kts to m/s and 3000 ft to m first. Obviously the units are the same as vorticity once you do the division so this is one way of quantifying the horizontal vorticity of that layer. The RAOB sounding program actually does show shear in this manner.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

  • Recently Browsing   0 members

    No registered users viewing this page.

×