**SHOCK AND AWE OVER GAZA**— Jonathan Cook reports from the West Bank on How the Media and Human Rights Groups Cover for Israel’s War Crimes; Jeffrey St. Clair on Why Israel is Losing; Nick Alexandrov on Honduras Five Years After the Coup; Joshua Frank on California’s Water Crisis; Ismael Hossein-Zadeh on Finance Capital and Inequality; Kathy Deacon on The Center for the Whole Person; Kim Nicolini on the Aesthetics of Jim Jarmusch. PLUS: Mike Whitney on the Faltering Economic Recovery; Chris Floyd on Being Trapped in a Mad World; and Kristin Kolb on Cancer Without Melodrama.

# The Energy of a Hurricane

In Fidel Castro’s account of Hurricane Gustav’s impact on Cuba, he speculates about the energy such a storm can hold:

In all honesty, I daresay that the photos and film footage shown on national television on Sunday reminded me of the desolation I saw when I visited Hiroshima, victim of the first nuclear strike in August 1945. With good reason, it is said that hurricanes release an enormous amount of energy, equal, perhaps, to thousands of nuclear weapons like the ones used on the cities of Hiroshima and Nagasaki. It would be worthwhile for a Cuban physicist or mathematician to do the relevant calculations and make a comprehensible presentation.

Being Cuban (from New York) , and a physicist (when I was employed), I felt challenged to provide details. What follows are the results of a simple model.

I consider a hurricane to be a vortex with a local rotational speed that increases as the point of observation moves to smaller radius. The center of the hurricane is a zone of low pressure, while the exterior of the hurricane is a region of high pressure; air is drawn radially inward by this pressure gradient. Air drawn into the outer radius of the hurricane, Ro, rotates with angular (or circular) speed vo. Let us say this condition applies to a ring (inward along the radial direction) of thickness dr. This ring has a height L, from sea level up to the top of the storm (what satellite cameras photograph). The rotation of this large cylindrical shell carries a total angular momentum (the product of total air-mass in the shell times average rotational speed), which we will label Mr.

The air in the outermost shell moves inward to the next nested shell of thickness dr (these shells are arbitrary mathematical conceptions), which must also carry momentum Mr since there is no significant force to impede the flow (this is the law of conservation of angular momentum). However, since there is less volume in this next inner ring, and since the air is not significantly compressed as it moves, the circular velocity of the flow must increase (the radial velocity also increases, but is of lower magnitude). This effect continues so that the circular wind speed, vi, at the inner radius, Ri, is very high. The total number of rings in the storm is given by (Ro-Ri)/dr, or simply Ro/dr when Ro is much larger than Ri. Since each ring has the same momentum Mr, the total momentum of the storm is equal to the product M = Mr x (Ro-Ri)/dr, or to a reasonable approximation M = Mr x (Ro/dr). [x has been used as the multiplication sign.]

The circular speed v at any radius r can be expressed as the product v = w x r, where w is the rotational or circular frequency. This frequency is low at the outer edge of the hurricane and increases as the point of observation moves inward. From the model here, one can find that the circular frequency varies with radius as follows,

w = M/(2 pi rho Ro L r r).

[I have dropped the x for multiplication.] This quantity is the total momentum of the storm, M, divided by the product of 2, pi = 3.14159…, rho = density of air, Ro, L, and the square of the radius in question.

Hang on, we’re almost there.

Because M, 2, pi, rho, Ro and L are constants, the angular frequency can be written as w = A/(r r), where A is a constant equal to M/(2 pi rho Ro L). Notice that for any pair (r1, w1), where perhaps these specific values were measured at radius r1, that

A = r1 r1 w1,

the product of w1 times the square of r1. So, once we measure the wind speed at any given radius, we can infer the rotational frequency there (w1 = v1/r1), and then we can define constant A and use this in the formulas shown to get w and v at any radius.

Now, we find the kinetic energy of every layer (shell), which is essentially the product of its mass times the square of its average circular velocity, divided by 2. Then, we add up all the layers for the total energy of the storm. I do this with differential layers and form the double integral, radially and vertically, but we will skip over that discussion. The result for the total energy of the storm is

E = pi rho L A A LOGe(Ro/Ri)

in units of energy called joules (if you fell asleep, this is the answer).

E is the product of five factors, the four leading ones being: pi, rho, L and the square of A; and the product of these four is multiplied by the mathematical function called the natural logarithm (LOGe) of the ratio Ro/Ri. The natural logarithm of 1 is equal to 0; LOGe(2.7182818…) = 1; LOGe(10) = 2.3026; LOGe(100) = 4.6052.

Let’s list parameters, choose values and find some numerical answers:

pi = 3.14159…

rho = 1.2 kilogram/(cubic meter)

[air at sea level]

L = 5000 meters (5 km)

[height]

A = 4,000,000 (meters squared)/seconds

[value chosen for rotational acceleration parameter]

Ro = 400,000 meters (400 km)

[outer radius]

Ri = 40,000 meters (40 km)

[inner radius].

Our choice for the value of A is consistent with:

v1 = 10 m/s,

r1 = 400,000 m (400 km) and

w1 = 0.000025 radians/s; and/or

v1 = 20 m/s,

r1 = 200,000 m (200 km) and

w1 = 0.0001 radians/s; and/or

v1 = 100 m/s,

r1 = 40,000 m (40 km) and

w1 = 0.0025 radians/s.

[There are 6.283 radians along a circle,

so 57.3 degrees/radian;

recall 360 degrees = 1 circle]

Note that 1 m/s = 2.24 mph (miles per hour);

1.61 km = 1 mile;

1 km = 0.62 mile.

For the values shown,

E = 6.944 x (10 to the 17th power) joules.

The energy released by the explosion of 1000 tons of TNT (a kiloton, abbreviated kt) is 4.182 x (10 to the 12th power) joules. So, E = 166,055 kt (or equivalently, 166.05 megatons). The atomic bomb exploded at Hiroshima on August 6, 1945 produced about 15 kt, so the model storm has the energy equivalent of 11,070 Hiroshima bombs. Most of the energy of a hurricane is dissipated as atmospheric turbulence and heating, and friction along the Earth’s surface, only a very tiny portion of it is absorbed by the structures built by humans.

Bear in mind that the energy of the hurricane is spread over a much larger volume than that of a nuclear explosion (so hurricane energy per unit volume is smaller), and it is released over a much longer period of time. But it is of awesome scale, and we are still as powerless before it as were our first ancestors four million years ago.

** MANUEL GARCIA, Jr./strong>. is a retired physicist; e-mail = mango@idiom.com**

**The Power of 11,000 Hiroshima Bombs The Energy of a Hurricane
**

# The Energy of a Hurricane

In Fidel Castro’s account of Hurricane Gustav’s impact on Cuba, he speculates about the energy such a storm can hold:

In all honesty, I daresay that the photos and film footage shown on national television on Sunday reminded me of the desolation I saw when I visited Hiroshima, victim of the first nuclear strike in August 1945. With good reason, it is said that hurricanes release an enormous amount of energy, equal, perhaps, to thousands of nuclear weapons like the ones used on the cities of Hiroshima and Nagasaki. It would be worthwhile for a Cuban physicist or mathematician to do the relevant calculations and make a comprehensible presentation.

Being Cuban (from New York) , and a physicist (when I was employed), I felt challenged to provide details. What follows are the results of a simple model.

I consider a hurricane to be a vortex with a local rotational speed that increases as the point of observation moves to smaller radius. The center of the hurricane is a zone of low pressure, while the exterior of the hurricane is a region of high pressure; air is drawn radially inward by this pressure gradient. Air drawn into the outer radius of the hurricane, Ro, rotates with angular (or circular) speed vo. Let us say this condition applies to a ring (inward along the radial direction) of thickness dr. This ring has a height L, from sea level up to the top of the storm (what satellite cameras photograph). The rotation of this large cylindrical shell carries a total angular momentum (the product of total air-mass in the shell times average rotational speed), which we will label Mr.

The air in the outermost shell moves inward to the next nested shell of thickness dr (these shells are arbitrary mathematical conceptions), which must also carry momentum Mr since there is no significant force to impede the flow (this is the law of conservation of angular momentum). However, since there is less volume in this next inner ring, and since the air is not significantly compressed as it moves, the circular velocity of the flow must increase (the radial velocity also increases, but is of lower magnitude). This effect continues so that the circular wind speed, vi, at the inner radius, Ri, is very high. The total number of rings in the storm is given by (Ro-Ri)/dr, or simply Ro/dr when Ro is much larger than Ri. Since each ring has the same momentum Mr, the total momentum of the storm is equal to the product M = Mr x (Ro-Ri)/dr, or to a reasonable approximation M = Mr x (Ro/dr). [x has been used as the multiplication sign.]

The circular speed v at any radius r can be expressed as the product v = w x r, where w is the rotational or circular frequency. This frequency is low at the outer edge of the hurricane and increases as the point of observation moves inward. From the model here, one can find that the circular frequency varies with radius as follows,

w = M/(2 pi rho Ro L r r).

[I have dropped the x for multiplication.] This quantity is the total momentum of the storm, M, divided by the product of 2, pi = 3.14159…, rho = density of air, Ro, L, and the square of the radius in question.

Hang on, we’re almost there.

Because M, 2, pi, rho, Ro and L are constants, the angular frequency can be written as w = A/(r r), where A is a constant equal to M/(2 pi rho Ro L). Notice that for any pair (r1, w1), where perhaps these specific values were measured at radius r1, that

A = r1 r1 w1,

the product of w1 times the square of r1. So, once we measure the wind speed at any given radius, we can infer the rotational frequency there (w1 = v1/r1), and then we can define constant A and use this in the formulas shown to get w and v at any radius.

Now, we find the kinetic energy of every layer (shell), which is essentially the product of its mass times the square of its average circular velocity, divided by 2. Then, we add up all the layers for the total energy of the storm. I do this with differential layers and form the double integral, radially and vertically, but we will skip over that discussion. The result for the total energy of the storm is

E = pi rho L A A LOGe(Ro/Ri)

in units of energy called joules (if you fell asleep, this is the answer).

E is the product of five factors, the four leading ones being: pi, rho, L and the square of A; and the product of these four is multiplied by the mathematical function called the natural logarithm (LOGe) of the ratio Ro/Ri. The natural logarithm of 1 is equal to 0; LOGe(2.7182818…) = 1; LOGe(10) = 2.3026; LOGe(100) = 4.6052.

Let’s list parameters, choose values and find some numerical answers:

pi = 3.14159…

rho = 1.2 kilogram/(cubic meter)

[air at sea level]

L = 5000 meters (5 km)

[height]

A = 4,000,000 (meters squared)/seconds

[value chosen for rotational acceleration parameter]

Ro = 400,000 meters (400 km)

[outer radius]

Ri = 40,000 meters (40 km)

[inner radius].

Our choice for the value of A is consistent with:

v1 = 10 m/s,

r1 = 400,000 m (400 km) and

w1 = 0.000025 radians/s; and/or

v1 = 20 m/s,

r1 = 200,000 m (200 km) and

w1 = 0.0001 radians/s; and/or

v1 = 100 m/s,

r1 = 40,000 m (40 km) and

w1 = 0.0025 radians/s.

[There are 6.283 radians along a circle,

so 57.3 degrees/radian;

recall 360 degrees = 1 circle]

Note that 1 m/s = 2.24 mph (miles per hour);

1.61 km = 1 mile;

1 km = 0.62 mile.

For the values shown,

E = 6.944 x (10 to the 17th power) joules.

The energy released by the explosion of 1000 tons of TNT (a kiloton, abbreviated kt) is 4.182 x (10 to the 12th power) joules. So, E = 166,055 kt (or equivalently, 166.05 megatons). The atomic bomb exploded at Hiroshima on August 6, 1945 produced about 15 kt, so the model storm has the energy equivalent of 11,070 Hiroshima bombs. Most of the energy of a hurricane is dissipated as atmospheric turbulence and heating, and friction along the Earth’s surface, only a very tiny portion of it is absorbed by the structures built by humans.

Bear in mind that the energy of the hurricane is spread over a much larger volume than that of a nuclear explosion (so hurricane energy per unit volume is smaller), and it is released over a much longer period of time. But it is of awesome scale, and we are still as powerless before it as were our first ancestors four million years ago.

** MANUEL GARCIA, Jr./strong>. is a retired physicist; e-mail = mango@idiom.com**