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Thayer Watkins
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The General Paths of Tornadoes
Explained by the Polarward Force
Induced by their Forced Precession
with the Rotation of the Earth

Consider the general nature of the paths for tornadoes as shown in the following diagrams.

Reference: Jones, S. D. 2003. Spatial modeling of tornadoes. Unpublished Master’s Thesis, Department of Geography, University of South Carolina, Columbia, SC and B. J. Boruff, J. A. Easoz, S. D. Jones, H. R. Landry, J. D. Mitchem, and S. L. Cutter. 2003. "Tornado hazards in the United States," Climate Research 24:103-117.

Here is an enlargement for the tornado alley region of Oklahoma, Kansas and Nebraska.

Although there is considerable variation the general orientation of the paths to the northeast is evident. The purpose of this webpage is to explain why tornadoes in the Northern Hemisphere have a tendency to move to the northeast. It derives from a force that moves them north and any northward movement accelerates their motion to the east.

For an example of a tornado path in the Southern Hemisphere see Perth tornado of July 15, 1996. The path appears to be more east southeast but the general principle is confirmed. The wind speeds of this tornado were on the order of 100 km/hr, a relatively small value for a tornado.

Analysis

A tornado has spin angular momentum with respect to its center. It also has another angular momentum with respect to the axis of rotation of the Earth. A tornado is traveling with the Earth and maintaining roughly the same orientation with respect to the surface of the Earth as the Earth rotates. This creates a forced precession of its spin angular momentum. The forced precession creates a torque on the angular momentum which is manifested as a force pushing the tornado towards the pole.

This poleward force increases the poleward velocity of the tornado. As the tornado moves poleward the conservation of the terrestrial angular momentum requires an increase in its eastward velocity. The combination of the effects is a northeastern path in the Northern Hemisphere and a southeastern path in the Southern Hemisphere.

Tornadoes do occur in the Southern Hemisphere, notably in Australia, but not with the frequency that they occur in the United States. Tornadoes occur in Asia and Europe but again not with the frequency that they occur in the U.S. The topography and climate in the central U.S. make it exceptionally prone to tornadoes.

The Effect of a Poleward Movement
on an Object's Eastward Velocity
Relative to the Earth's Surface

The angular momentum per unit mass of any object on the rotating Earth is its eastward velocity times its distance from the axis of rotation. The distance from the Earth's axis of an object located at latitude angle φ is rcos(φ), r is the radius of the Earth. If u is the eastward velocity of an object relative to the Earth's surface and Ω is the angular velocity of Earth's rotation then the absolute velocity of an object is Ωrcos(φ)+u and its angular momentum per unit mass L is

L = rcos(φ)[Ωrcos(φ) + u]

Note that in all of this analysis the velocities being referred to are for the moment of the tornado center and not the circular wind velocities. The circular wind velocities can reach 300 miles per hour whereas the center moves at a more moderate speed in the neighborhood of 30 mph. This means that the eastward and northward velocities referred to below are approximately 20 mph.

Angular moment is conserved; i.e., remains constant when an object changes its latitude. (This is true only if the frictional forces of the interaction of the tornado with the Earth's surface and surrounding atmosphere are ignored.)

Therefore if φ0 is the initially latitude of a tornado and u0 its eastward velocity

L0 = rcos(φ0)[Ωrcos(φ0) + u0]

Thereafter L=L0 and hence

u = Ωr[cos²(φ0)/cos(φ) − cos(φ)] + [cos(φ0)/cos(φ)]u0

Thus if φ>φ0 then u will be great than u0.

The rate of change of eastward velocity with an increase in φ can be determined by putting the solution for u in the form

u(φ) = L/(rcos(φ)) − Ωrcos(φ)
and therefore
∂u/∂φ = −(L/(rcos²(φ))(−sin(φ)) − Ωr(−sin(φ))
which reduces to
∂u/∂φ = (L/(r)(tan(φ)/cos(φ)) + Ωrsin(φ)

Let y be the surface distance from the equator and thus y=rφ so

(du/dy)=(du/dφ)(dφ/dy)=(du/dφ)(1/r).
and hence
∂u/∂y = (L/(r²)(tan(φ)/cos(φ)) + Ωsin(φ)

Consider an object initially at 30°N with u=0. Its distance from the axis of rotation of the Earth is then approximately (6.38×106cos(30°)=5.525×106 meters. The angular velocity of the Earth in radians per second is 6.2832/(24*60*60)=7.27×10-5. Thus the absolute velocity of the object is (5.525×106)(7.27×10-5)=401.8 meters per second. Its angular momentum per unit mass is then (5.525×106)(401.8)=2.22×109 m²/s. This means that

∂u/∂y = (2.22×109/(6.38×69)²)(0.667) + (7.27×10-5)(0.5)
= 3.638×10-5 + 3.635×10-5 = 7.27×10-5 (m/s)/m.
or
7.27×10-2 (m/s)/km = 0.727 (m/s)/(10km).

This is roughly 2.6 miles per hour per 10 mile northern movement.

Meridianal Acceleration

Elsewhere an analysis is given that concludes that the acceleration of the latitude angle φ is given by

d²φ/dt² = γ/cos(φ)

where γ is a parameter that depends upon the spin angular momentum per unit mass, the terrestrial angular momentum per unit mass and the radius of the Earth. For tornadoes the lifespan is usually short enough that latitude does not change very much so the dynamics is that of constant poleward acceleration. Thus the poleward velocity v of the tornado increases driving an increase in the eastward velocity u. The combination is then a poleward-easterly motion for the tornado.

(To be continued.)


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