Answer:
One vorticity unit equals 0.00001 s-1 or 1 * 10 ^ -5 s ^ -1
Explanation:
The change in wind speed is 30-20 = 10 knots = 5 m / s
The change is the distance is 1,600 m.
The change in wind speed over distance = 5 / 1,600 s ^ -1 = 0.003125 s ^ -1
The above calculation is multiplied by 1000 to simplify the number. This will give a speed cutoff value of approximately 3 units. The following is an operational interpretation of the cutting speed.
0 to 3 weak, 4 to 5 moderate, 6 to 7 large, 8+ extreme
The shear vorticity is found similarly and has the same units. The shear vorticity is determined on a vertical axis. Here is an example problem:
At location A at 500 mb the wind speed is 30 knots, at location B at 500 mb the wind speed is 50 knots. The distance between the two places is 400 km. How many shear vorticity units occur between the two locations?
The change in wind speed is 50-30 = 20 knots = 10 m / s
The change in distance is 400,000 m.
The change is the wind speed over the distance = 10 / 400,000 = 0.000025 s ^ -1
One vorticity unit equals 0.00001 s-1 or 1 * 10 ^ -5 s ^ -1
This produces a shear vorticity value of 2.5 units.
