ASK THE EXPERTS: Wind Speed Cubed

Kestral Wind Turbine

When I was in fifth grade, I was told that doubling wind speed quadrupled the force of the wind. However, I’ve recently learned that the wind’s power is proportional to the cube of its speed, rather than the square. The first equation in David Laino’s article on wind physics in HP161 says that kinetic energy equals 1/2 times mass times velocity squared, which raises two issues. First, the article doesn’t explain why the velocity is squared, which seems necessary for following the rest of the logic leading to the cube-of-velocity concept.

Second, the coefficient 1/2 presumably requires that specific units be used, such as kilograms and meters. Can you better explain these?

Malcom Drake • via email

What you learned in fifth grade was correct: The force of the wind is a function of the wind speed squared. However, force and pressure are static measures, whereas power and energy are dynamic measures. It is important not to confuse them. Static does not consider motion and dynamic does. Thus, it makes sense that the addition of motion—velocity—to a static value that already has a squared dependency on velocity yields a cubic dependency on velocity.

A presentation of the concepts in the article required accepting the laws of physics at some level. The details of the derivation of the kinetic energy equation were outside the scope of the article, but if we accept Newton’s second law of motion—that force equals mass times acceleration (F = ma), and the definition that energy (E) is the integration of force acting over a distance (s)—then we get this calculus equation:

E = ∫F × ds = ∫m × a × ds = m∫ (dv/dt) ×  ds = m∫ (ds/dt) ×  dv = m∫ (v) ×  dv,

which leads to 1/2mv2

If you are familiar with integral calculus, you’ll recognize that both the 1/2 and the squared terms are the result of integration, with the result that E = 1/2mv2. Thus the 1/2 term is not a coefficient for specific units—you are free to use any units you choose. If you use SI units (kg, m, and sec) you will get an answer for energy directly in watt-seconds.

The velocity-cubed function of kinetic energy is unique to continuous fluid flow. It also applies to water flow because the mass of the fluid passing through an area depends directly on how much fluid—and how fast that fluid—is flowing through the area. This is not true for the kinetic energy of a solid mass, such as a baseball. The baseball’s mass is a constant and thus its energy is only a function of the velocity squared. So take advantage of the power of the cubic by choosing wind power over baseball power!

David Laino • Cofounder, Endurance Wind Power

Comments (1)

MrEnergyMan's picture

Everything in theory is correct. What most people don't recognize is that there annomoiter tells them one windspeed in the generator is not producing the energy per manufacturers specifications. Most wind generators are tested in a controlled windtunnel environment. In reality most when turbines see winds that are disturbed which totally decreases the Energy produced. Site location of the turbine is critical for the end result of energy produced. If you're not sure on this read up on the various articles available to cite your w site location of the turbine is critical for the end result of energy produced. If you're not sure on this read up on the various articles available to cite your Wind turbine in the best location. If it's not the perfect spot then understand the energy production will be less than expected.

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