Welcome to American Weather
George BM

How to calculate DCAPE, ESRH, EBWD

4 posts in this topic

I need help figuring out how to calculate Dcape, "Effective"SRH (or SRH in general), and "Effective"BWD. 

 

Share this post


Link to post
Share on other sites
On 3/23/2017 at 4:01 PM, George BM said:

I need help figuring out how to calculate Dcape, "Effective"SRH (or SRH in general), and "Effective"BWD. 

DCAPE

SRH

As for Effective BWD ----> I assume this is meant to be bulk wind difference? I never actually learned how to do so.

But I believe instead of taking it say from 0-6km for bulk shear it takes it through the whole thunderstorm up to the EL. but feel there is more to it than just that.

hodofig18.GIF

Share this post


Link to post
Share on other sites
8 hours ago, so_whats_happening said:

DCAPE

SRH

As for Effective BWD ----> I assume this is meant to be bulk wind difference? I never actually learned how to do so.

But I believe instead of taking it say from 0-6km for bulk shear it takes it through the whole thunderstorm up to the EL. but feel there is more to it than just that.

hodofig18.GIF

Under SRH on the 17th slide how do you get the values for average U and average V shear? (-13 , 24.5) and (2 , 19.5)?

Share this post


Link to post
Share on other sites
On 3/25/2017 at 10:13 AM, George BM said:

Under SRH on the 17th slide how do you get the values for average U and average V shear? (-13 , 24.5) and (2 , 19.5)?

For finding average U and V you want to average the two layers. So Surface had (0,10) and 850 had (10,17) you just add the U together and divide by 2 since there are only 2 and do the same with V and should get (5,13.5).

 

They did calculations for figuring out C vector, storm motion vector, within the hodograph. If you then subtract the C from U and V average you should get those values for each Usr and Vsr in sfc to 850 and then again 850 to 700mb. C is determined through the drawing of the hodograph and is approximated if being hand drawn. 

They didnt show how they got C so that doesnt really help seeing it through PPT but they mention it just underneath I believe what they found as their mean wind for storm motion. Now remember that 300 degree and 22.5 is a vector that should be broken down into its components just as your answer was given to you.

That should do the trick.

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

  • Recently Browsing   0 members

    No registered users viewing this page.