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 The Poleward Acceleration of Meteorological Vortices* Due to Forced Precession

#### *(Hurricanes, Typhoons, Cyclones,Anticyclones, Northeasters and Tornadoes)

In a previous work a model of hurricanes was presented which explains the general path of hurricanes, including the phenomenon of recurvature. This seems to be the only explanation available for why hurricanes eventually head toward the northeast.

This model of course also applies to typhoons and cyclones, but it also applies to anticyclones, midlatitude cyclones such as northeasters and to tornadoes. The basis of the model is that the turning of the vortex with the Earth constitutes a forced precession which creates a global torque on the vortex that moves it toward the pole that has the same spin orientation as it does. The verticality of the vortex is enforced by the rising or falling of air in its eye. This is essential for the creation of the torque. Cyclones move toward the nearest pole; anticyclones appear to be moving toward the equator but that is only because the equator is on the way to the other pole.

A model of course is just a theory and while the model is consistent with the qualitative characteristics of meteorological vortices it is necessary to pursue some more quantitative verification.

The tangential velocities of cyclones are reasonably approximated as modified Rankine vortices; i.e., the tangential wind velocity v beyond the eye is inversely proportional to the distance r from the center of the vortex:

#### v = ζ/r

The parameter ζ will later be referred to as the Rankine vortex parameter.

Each band of a Rankine vortex has the same amount of angular momentum mrv so the total angular momentum of the vortex depends directly upon the physical dimensions of the vortex. However for a hurricane and other such vortex the limit of the physical dimension is ambiguous. What is presented below is an elegantly simple bit of analysis which shows that each band of a Rankine vortex is accelerated the same amount and thus the vortex moves as a unit and maintains its integrity.

Consider a meteorological vortex being made up of a continuum of rotating cylindrical shells. Let the radius of one such cylindrical shell be r and let h be its height and Δr its thickness. If ρ is the air density then the mass m of air in the cylindrical shell is given by:

#### m = (2πrΔr)ρ

Assume the tangential wind velocity v is given by the Ranking formula. In the following a symbol displayed in red represents a three dimensional vector. The same symbol not in red represents the magnitude of the corresponding vector.

The spin angular momentum of the rotating cylindrical shell is L. It points along the axis of the cylinder and is upward (positive) if the shell is turning in a counterclockwise direction when viewed from above. The Earth is turning in a counterclockwise direction at a rate denoted as Ω. The Earth's spin vector is then Ω. The vortex has another type of angular momentum having to due with its turning with the Earth which is dependent upon its distance from the rotation axis of the Earth. This will be called the rotational angular momentum of the vortex to distinguish it from the spin angular momentum. This will be important in the analysis later.

The magnitude of the spin angular momentum L of the cylindrical shell L = |L| is given by

#### L = mvr = (2πρrΔr)(ζ/r)r = 2πζrΔr

The dynamics of the vortex under forced precession is given by

#### dL/dt = T

where T is the global torque exerted on the vortex. The torque is equal to R×F where R is the vector from the center of the Earth to the vortex and F is the force on the vortex in a meridianal (poleward) direction and R is thus the radius of the Earth.

The magnitude of the spin angular momentum does not change so

#### dL/dt = L(dk/dt)

where k is a unit vector in the vertical direction.

The acceleration a of the cylindrical shell is given by

Therefore

#### a = (T/R)/m = (Ldk)/(Rm) a [(2πρζ)rΔr)(dk/dt)]/[R(2πρrΔr)]

This reduces to

###### a = (ζ/R)|dk/dt|

This is independent of the radius of the cylindrical shell. Thus all of the component shells of the vortex would be accelerated the same amount. This means that the vortex moves as a unit and its integrity is perserved as it moves.

It is determined elsewhere that the poleward component of the force on angular momentum is proportional to Ωcos(φ) so the acceleration of each component of the vortex is

###### a = ζΩcos(φ)/R

The Rankine vortex parameter ζ is the product of wind speed and radial distance at any point. Suppose for a hurricane 20° N the wind velocity is 50 m/s or 180 km/hr at a radial distance 20 km. The value of ζ is then 36000 km²/hr. This represents the order of magnitude for a category 3 hurricane.

The value of Ω 2π radians in 24 hours, or 0.2618 radians per hour. The value of R is about 6400 km. the value of cos(30°) is 0.15425. Therefore the poleward acceleration of the hurricane is

#### a = 3.6×104(0.2618)(0.94)/6400 = 1.38 (km/hr)/hr

In the absence of frictional drag in 24 hours the hurricane would be moving north at a speed of 33.22 km/hr. However soon the acceleration would be matched by the frictional drag of the gigantic blob of air (the hurricane) moving through other air. Hurricanes typically travel about 120 km/hr over the ocean. This speed would be acchieved within a period of about 87 hours (3 to 4 days). The lifespan of a hurricane is on the order of two weeks.

Because the acceleration depends upon the cosine of the latitude the model predicts that the travel speed would slow as the hurricane moves northward even without an increase in frictions.

## The Preservation of Rotational Angular Momentum

If φ is the latitude of an object on the Earth's surface its distance from Earth's axis is Rcos(φ). An object of mass M with an eastward velocity of V will then have a rotational angular momentum of MVRcos(φ). If the object moves to a new latitude φ' then the preservation of angular moment requires that

#### MV'Rcos(φ') = MVRcos(φ) which reduces to V'cos(φ') = Vcos(φ)

This can be restated in terms of the eastward accleration dV/dt as

#### (dV/dt)cos(φ) = Vsin(φ)(dφ/dt) or, equivalently dV/dt = Vtan(φ)(dφ/dt)

Thus a poleward velocity of a vortex translates into an eastward acceleration. This is the basis for the recurvature of the paths of meteorological vortices. The actual paths of such vortices are influenced by spatial pressure variations, terrain conditions and the frictional forces on a mass of air moving through other air.

Leut U be the northward (meridianal) velocity of the vortex. Since this meridianal velocity U is equal to R(dφ/dt) the above equation can be expressed as

#### dV/dt = Vtan(φ)(U/R)

The dynamic equations for the motion of a vortex in the absence of frictional forces are then

###### dU/dt = ζΩcos(φ)/R dV/dt = Vvtan(φ)/R

Or, in terms of latitude φ and longitude θ

###### d²φ/dt² = ζΩcos(φ)/R² and d²θ/dt² = R(dθ/dt)(dφ/dt)tan(φ)

(To be continued.)