San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Recurvature of the Paths of Cyclones Due to Their Forced Precession with the Rotation of the Earth

Although the path of a cyclone (hurricane or typhoon) may be erratic there ususally comes a point at which the path that was westerly moves in the direction of the nearest pole and then turns polar-easterly (northeastly in the Northern Hemisphere and southeasterly in the Southern Hemisphere). Below is a striking illustration of the phenomenon for a cyclone in the Bay of Bengal in 1991. The explanation for recurvature lies in the fact that cyclones (and anticyclones) possess angular momenta with respect to their spin axes. This will be referred to as their spin angular momenta. They also possess angular momenta with respect to the axis of rotation of the Earth, which will be called their terrestrial angular momenta. The turning of the spin angular momentum vector with the rotation of the Earth is a forced precession which creates a torque on the cyclone that accelerates it toward the nearest pole. As the cyclone moves toward its pole it gets closer to the axis of rotation of the Earth. The preservation of the terrestrian angular momentum then requires it to increase its velocity toward the east.

Below is the pattern for hurricanes (tropical cyclones in the Atlantic region). A similar phenomenon occurs for anticyclones in which they are accelerated toward the equator. Preservation of terrestrial angular momentum causes them to decelerate, but anticyclones are not strong enough to last to the point of recurvature.

## The Global Torque Produced by Forced Precession

In the analysis which follows vectors are displayed in red. Let L be the magnitude of the spin angular momentum of a cyclone. Vectorially angular momentum is L k, where k is the local unit vertical vector. When the cyclone moves to a different location the vertical unit vector is different so there is a vectoral change in angular momentum even though the magnitude does not change.

Let Ω be the angular velocity vector for the Earth's rotation. By a standard theorem in vector analysis

#### dk/dt = Ω×k

Newton's Second Law in terms of angular momentum is:

#### dL/dt = T

where T is the torque vector which in this case is r×F, where r is the radial vector from the center of the Earth and can be represented as rk and F is a force perpendicular to r.

Since the magnitude of the angular momentum vector is constant

Thus

#### T = rk×F = But also T = L(Ω×k) so rk×F = L(Ω×k) which can be rearranged to -r(F×k) = L(Ω×k) or (L(Ω+rF)×k = 0

This last equation implies that

#### LΩ + rF = λkand hence rF = -LΩ + λk

This means that the horizontal component of F is the same as the horizontal component of -(L/r)Ω; i.e., Fj = -(L/r)Ωcos(φ), where φ is the latitude.

Thus if the cyclone is to be carried along at the same latitude there must be a force of -(LΩ/r)cos(φ) applied to it. In the absence of such a force the cyclone will move poleward subject to a force equal to (LΩ/r)cos(φ). This is analogous to the matter of centrifugal force on a body moving in a circular orbit. In order to keep the body moving in the circular orbit a centripetal force equal to the centrifugal force must be applied. In the absence of this centripetal force the body flies away as though subject to a radial force equal to the centrifugal force.

Normally a cyclone does not follow the Earth's rotation exactly and so the effective rotation rate is somewhat different from that of the Earth. If u is the west-to-east velocity of the cyclone and R=rcos(φ) is the distance to the Earth's axis (where φ is the latitude angle) then the effective rotation rate of the cyclone with respect to the Earth's axis is:

#### Ω' = Ω + u/R = Ω + u/(rcos(φ)) and thus the horizontal force on the cyclone is (L/r)(Ω')cos(φ)

Let v be the equator-to-pole velocity of the cyclone and m its mass. The acceleration on the cyclone toward the pole is then:

#### dv/dt = (L/m)(1/r)Ω')cos(φ)

At this point it is necessary to take into account the other angular momentum associated with a cyclone. The angular momnentum considered above is that due to the spin of the cyclone. There is also the terrestrial angular momentum of the mass of the cyclone moving around the Earth's axis. This terrestrial angular momentum per unit mass is equal to the velocity of the cyclone with respect an inertial coordinate system times the distance to the axis of rotation. The absolute velocity of the cyclone is:

#### ΩR + u so if m is the mass of the cyclone then the magnitude M of the terrestrial angular momentum is M =m(ΩR+u)R = m(Ω + u/R)R² = m(Ω')R² and hence Ω' = (M/m)/R²

Thus, ignoring local pressure differences and friction drags on the cyclone, its dynamics is given by the two equations

#### dv/dt = (L/m)(1/r)Ω'cos(φ) Ω' = (M/m)/R² which together imply dv/dt = (L/m)(M/m)(1/r)cos(φ)/R²which reduces to dv/dt = (L/m)(M/m)/(r³cos(φ))

It is notable that the spin angular momenta enter into the equations only as angular momenta per unit mass. Thus bigger or smaller cyclones should follow the same general path.

But v=r|dφ/dt| and u=Rdθ/dt=rcos(φ)dθ/dt where θ is longitude.

Consider now the determination of the west-to-east velocity u. When a cyclone moves towards its hemispheric pole its distance from the Earth's axis is decreased so that in order to conserve terrestrial angular momentum the velocity u must increase. For M constant then

#### Ω' = Ω + u/R = (M/m)/R² and thus u = (M/m)/R − ΩRwhich reduces to u = (M/m)/(rcos(φ)) − Ωrcos(φ)

Thus while the initial value u0 might be negative a movement toward the nearest pole may increase u to a positive level. That is to say a cyclone initially moving west may eventually start moving east. Recurvature is at the point where u=0.

The dynamic equations of a cyclone's path can be reduced to ones involving only latitude and longitude as variables and the magnitudes of spin and terrestrian angular momenta per unit mass as parameters.

These equations would be

#### d²φ/dt² = (1/r)(dv/dt) = (L/m)(M/m)/(r4cos(φ)) and dθ/dt = u/R = (M/m)/(rcos(φ))² − Ω

To simplify the presenttion let [(L/m)(M/m)/r4]=γ so the differential equation to be solved is

#### d²φ/dt² = γ/cos(φ)

The solution to this differential equation can be obtained by multiplying both sides of the equation by (dφ/dt) and integrating from 0 to t.

#### ½[(dφ/dt)²(t) - (dφ/dt)²(0)] = γ[ln(tan(φ/2 + π/4) − ln(tan(φ0/2 + π/4)]

To simplify matters let ζ=φ/2+π/4 and thus dζ/dt=½(dφ/dt) so dφ/dt=2(dζ/dt). The previous equation then becomes

#### 2[(dζ/dt)² − (dζ/dt)²(0)] = γ*ln[tan(ζ)/tan(ζ0)] where dζ/dt(0) is the value of dζ/dt at t=0 and thus dζ/dt = {(dζ/dt(0))² + γ*ln[tan(ζ)/tan(ζ0)]}½

For the special case in which (dφ/dt(0))=0 and hence (dζ/dt(0))=0 the above equation reduces to

#### dζ/dt = γ½*{ln[tan(ζ)/tan(ζ0)]}½

In principle the solution of the differential equation for ζ is just

#### ∫0t dζ/g(ζ) = t where g(ζ) = {(dζ/dt(0))² + γ*ln[tan(ζ)/tan(ζ0)]}½

That of course is easier said than done. Nevertheless a solution for ζ(t) would give a solution for φ(t) which may be substituted into the equation for (dθ/dt) to obtain the θ as a function of time.

The parameter γ may be greatly simplified. First M=m(ΩR+u0)R. Thus (M/m)=ΩR²+u0R and since u0 is small compared with ΩR, (M/m) is approximately equal to ΩR². The spin angular momentum L is the integral of ρh(2πz)zV(z)dz where z is the distance from the cyclone center, h is the height of the cyclone, V(z) is the tangential wind speed, and ρ is the mass density of the air. The mass of the cyclone on the other hand is the integral of ρh(2πz)zdz so (L/m) is in the nature of a weighted average wind speed. Let us call this Vav. This likely to be directly related to the maximum wind speed such as a fraction of the maximum. Let us say (L/m) is equal to Vav. Thus

#### γ = (ΩR²)(Vav)/r4which can be reduced to γ = (Ωcos²(φ0)(Vav)/r2

The scale of γ is easily determined. The radius of the Earth is about 6.4×106 meters. The value of Ω is 2π/(24*60*60)=7.3×10-5 radians per sec. The value of cos²(φ) for φ=30° is 0.75. If the average wind speed in a hurricane is 50 meters per second then

#### γ = (7.3×10-5)(0.75)(50)(2.44×10-14)and thus γ = 6.7×10-17 m-1s-2and γ½ = 8.2×10-8 m-1/2s-1.

While the above equations explain the general path of a cyclone the short term behavior of a cyclone can be very erratic. It can be influenced by local topography and regional pressure patterns. But ultimately the polar-easterly path prevails. The situation is like the trajectory of a dry leaf falling from a tree. The leaf may be buffeted to the right and to left; it may even rise a bit; but ultimately the inexorable force of gravity will prevail. For cyclones the inexorable force due to their forced precession with the rotation of the Earth will ultimately prevail if the they last long enough.