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A helicopter just circulates air in a hover and maintains a stable altitude. So, how much energy is used to do this? Using the standard equation $KE = \frac12 mv^2$; then the kinetic energy used would be $0.5$ times the mass of air displaced down each second ($m$), times the vertical velocity of this air squared ($v^2$). Is this correct?

I couldn't find this in any engineering textbook. Estimates (guesses) from fuel burn is helpful, but are very inaccurate; as they rely on estimates (assumptions) of how efficient the fuel conversion to mechanical energy is, how efficient the motor is at using this energy, and then how efficient the rotors are at displacing air down. Which is too many assumptions.

Qmechanic
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Nick Landell
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1 Answers1

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Using the standard equation $KE = \frac12 mv^2$; then the kinetic energy used would be $0.5$ times the mass of air displaced down each second ($m$), times the vertical velocity of this air squared ($v^2$). Is this correct?

No, unless you consider heat to be a form of kinetic energy. Air is a massively complicated, turbulent fluid, and the air the helicopter blades are interacting with isn’t pushed down as a single monolithic body moving at a fixed speed $v$, but instead gets thrown around in all sorts of directions, creates vortices, and much of it ends up dissipating as heat and sound waves, or moving in directions that don’t generate lift. So unless you’re willing to go down to the molecular level and add up the kinetic energy of lots of air molecules bouncing around in all directions (good luck with that), no, your suggested method of calculating the energy required for hovering is not correct.

As others said in the comments, the correct way to calculate the energy dissipation of a hovering helicopter is explained here.

GenlyAi
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