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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?

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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.

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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.

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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.

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