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CALCULATION OF EARTH, SHEAR
AND CURVATURE VORTICITY

METEOROLOGIST JEFF HABY

Vorticity is displayed on the synoptic scale forecast models. One source is at UNISYS:

http://weather.unisys.com/nam/500.php

Notice the units of vorticity given on the chart are (10^-5/s). In English, this is 10 to the negative 5 with units of seconds to the negative 1. This model is showing absolute vorticity, which is earth, shear and curvature vorticity combined together. Notice the scale on the bottom shows values of vorticity ranging from the negative single digits to more than positive 20. These are known as UNITS of vorticity. A unit of vorticity is simply the whole number left after dropping the 10^-5/s.

To determine the value of vorticity at a fixed location, all three values of vorticity must be found independently then added together. We will go over the calculation of each term one by one.

The first term is earth vorticity. All that is needed is the earth's latitude to get this value. The formula for earth vorticity is: f =2*(omega)*SIN(THETA); where omega is the earth's angular momentum and THETA is the latitude. The value of omega for the earth is 7.292*10^-5 rad/s. This number comes from the fact that there are 2*PI radians in a complete circle (360 degrees). If the number omega is multiplied by the number of seconds in a day, the number comes up to 2*PI, which is equal to 360 degrees. 7.292*10^-5 rad/s times 86,400 seconds = 6.3 = 2*PI. PI is equal to 3.1416 Since the SIN of 0 is 0, earth vorticity is 0 at the equator. Since the SIN of 90 is 1, earth vorticity (earth spin) is equal to the value of earth's angular momentum at the pole. The earth vorticity (also called Coriolis), increases when moving from the equator toward the pole. The value of earth vorticity at 45 degrees north is equal to F=2*(omega)*SIN(45) = 10.312*10^-5 rad/s. Since 10^-5/s is equal to one unit of vorticity, the earth vorticity in this case accounts for a positive 10.312 units of earth vorticity. Earth vorticity is always positive except at the equator where it is 0.

The second term is shear vorticity. This is the change in wind speed over distance at 500 mbs. If the shear produces a cyclonic (counterclockwise in Northern Hemisphere) spin of air, the term is positive. If the shear produces an anti-cyclonic spin of air, the term is negative. The easiest way to determine the direction of spin is to put a short line segment on the chart. Next, determine how the wind flow will spin the line segment over time. If it spins cyclonically, then positive shear vorticity is occurring. Suppose the change in wind speed is 20 m/s over a distance of 250 km and a cyclonic spin ensues, the resulting shear vorticity is 20/250,000 = 8*10^-5/s. This produces 8 units of shear vorticity.

The third term is curvature vorticity. This value is determined by the turning ratio of the air over time. The curvature that is calculated is circular. The stronger the rate of directional turning, the higher the curvature vorticity value will be. If the turning is cyclonic, the value will be positive. If the turning is anti-cyclonic, the value will be negative. As an example, suppose an air parcel spins 10 degrees is 1 hour. 1 hour is 3,600 seconds. A circle is made up of 360 degrees or 2*PI radians. PI is equal to (3.14). 1 degree is equal to 0.01745 radians (found by dividing 2*PI by 360). The numerical value for a 10 degree cyclonic spin is = 10*0.01745 = 0.1745. Therefore, the value of curvature vorticity is 0.1745/3,600 = 4.8*10^-5/s. This produces about 5 units of positive curvature vorticity.

Here is an example problem for finding absolute vorticity.

An air parcel is located at 35 degrees north. The parcel spins cyclonically at 15 degrees per hour due to curvature. The wind shear produces an anti-cyclonic spin. The change is wind speed over distance is 10 m/s over 300 km. What is the value of absolute vorticity at this point?

As stated earlier, absolute vorticity is the addition of earth, shear and curvature vorticity. The earth vorticity term will always be positive. In this example, curvature vorticity is positive and shear vorticity is negative. The next step is the find the units of vorticity for each term.

Earth vorticity = 2*omega*SIN(35) = 8.37*10^-5/s = +8.37 units
Curvature vorticity = (15*0.01745)/3,600 = 7.27*10^-5/s = +7.27 units
Shear vorticity = - 10/300,000 = - 3.33*10^-5/s = -3.33 units

Absolute vorticity = 8.37 + 7.27 - 3.33 = +12.31 UNITS