This is likely to be very simple, but...

How does one calculate work done by a hovering unmoving aircraft over time?

As in work in Joules.

In this scenario, to remain hovering the aircraft has to exert a force that counters gravity (which would be its mass times G, pointing in opposite direction of gravitational pull).

However, work is defined as newton per meter, and the aircraft does not move.

Meanwhile the battery charge or fuel is going to be expended, so the energy is spent.

The other question here: Conservation of energy for a hovering helicopter

is dealing with conservation energy and not calculating drain/work performed by hovering.

  • 1
    $\begingroup$ Think about what would happen in a vacuum. $\endgroup$ Commented Jul 20, 2019 at 21:28
  • 1
    $\begingroup$ @AaronStevens in a vacuum "A rocket is hovering above surface of the moon at fixed height...". <-- Same. Possible duplicate Closely related, but not a duplicate. As I want to calculate amount expended energy and the other question is not dealing with that. $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 21:45
  • $\begingroup$ I've given it a bit of thought and so far can think of only one situation when an object is hovering but no work is done. That would be magnetic levitation. In other scenarios: * Propellant hovering. * Properller-based hovering. * Electromagnetic Hovering. * Ion drive hovering. There is work done, but it seems to calculate it I'll need to go about it in roundabout way that involves secondary quantities. Namely for electromagnetic levitator I'd calculate it from voltage/current of the circuit, and propellant based seems to be related to volume of air. Isn't there a direct way $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 22:13
  • $\begingroup$ Possibly related. $\endgroup$
    – rob
    Commented Oct 28, 2022 at 13:43

3 Answers 3


The work done to keep the helicopter hovering is that generated by the motion of the rotor blades as they move through the air. The two key notions that go into the calculation of this work are lift and drag. Lift is the upwards-pointing force that balances gravity and keeps the helicopter hovering. Drag is the force pushing back horizontally against the rotors in the direction opposite to their motion. Lift is the "useful" force that helicopters and other aircraft need in order to fly, and drag is the unwanted force that is the price we have to pay to achieve lift.

In the hovering scenario, all the work goes into resisting the drag. If the drag force is equal to $D$, the work will be obtained by multiplying $D$ by the distance the rotor blades moved. We can encode this in the equation $$ P = \left(2\pi \frac{A}{2}\right) \times f \times D, $$ where \begin{align*} P &= \textrm{power (work per second) to keep the helicopter hovering} \\ A &= \textrm{length (radius) of the rotors} \\ f &= \textrm{frequency of rotation of the helicopter's rotors (in revolutions/second)}, \\ D &= \textrm{drag} \end{align*} (for obvious reasons, I find it easier to talk about power output in Watts rather than total energy in Joules). The meaning of the $2\pi A/2$ factor is that this is the circumference of a circle whose radius lies along the midpoint of the rotor blades - I'm assuming (which seems reasonable as an approximation) that drag is distributed uniformly along the length of the blades, so that would be the distance we would want to multiply by to get the work for a single revolution of the rotors.

Now, $A$ and $f$ are parameters whose values are easy to figure out (see the numerical example below), but how do we compute $D$? Well, recall that the helicopter is hovering, which means that the lift force $L$ is exactly equal to the weight of the helicopter: $$ L = \textrm{helicopter weight} = \textrm{(gravitational constant)} \times \textrm{(helicopter mass)}. $$ The last thing we need is the helicopter's lift to drag ratio, which will tell us how much drag will develop for a given amount of lift. This is a number that I think cannot be calculated from first principles, but must be measured (or approximated using numerical simulations). If the lift-to-drag ratio is $\gamma$, that simply means that $L$ and $D$ satisfy the relation $$ L = \gamma D. $$ This enables us to write down our final formula for the power output needed to keep the craft hovering: $$ P = \pi A f L/\gamma $$

A numerical example:

According to this Reddit thread, one of the most common helicopters used in civil aviation is the Robinson R22. From the specs on the Wikipedia page for this helicopter, its rotor length is $$A = 3.83 \ \textrm{m} $$ (they give the rotor diameter which is twice this value). The lift-to-drag ratio would vary depending on various complicated factors (see here), but according to the Wikipedia article on lift-to-drag ratios, a typical value for a helicopter is $$ \gamma = 4, $$ so I'll work with that.

Again referring to the helicopter specs on Wikipedia, the weight of the helicopter would be between 417 and 622 kilograms. Let's assume a value of 500 kilograms. The lift force is therefore equal to $$ L = 9.8 \times 500 = 4900 \textrm{ N}. $$ Since $\gamma=4$, the drag is therefore $$ D = 4900/4 = 1225 \textrm{ N}. $$ The final parameter is $f$, the number of revolutions per second. According to this source, a typical rate of rotor rotation for a helicopter is between 250 and 600 revolutions per minute. Let's assume 400 rpm, which translates to $$ f = 400/60 = 6.66 \textrm{ revolutions per second}. $$ We can now plug all these numerical parameters into the formula above to get the power output. The result is $$ P = 3.14159 \times 6.66 \times 3.83 \times 1225 = 98165 \textrm{ Watt} = 133.46 \textrm{ metric horsepower}, $$ i.e., the helicopter needs to do 98 kJ of work per second to maintain hovering.

Are these numbers reasonable? Well, according to the Wikipedia page for the R22, this helicopter uses a powerplant with a power output of 124 horsepower, which is less than my number (even with the unrealistic assumption of 100% engine efficiency). It looks like my numbers are a bit off, but it's not bad for an order-of-magnitude calculation.

  • $\begingroup$ I gotta ask something. Let's assume that both forces (gravity and lift) operate in opposite directions to affect object's kinetic energy. I.e. gravity decreases it, and lift increases it (let's represent it as a vector value). If only one force is present, in one second it would make the object reach kinetic energy of m*v^2/2 = 500 * 9,8 * 9,8 / 2 = 96040 Joyles. Since the value has been reached in one second, that is 96040 Watts exactly, which is 128.8 horsepower. Is this line of reasoning incorrect? $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 23:42
  • $\begingroup$ Yes, it is incorrect. It’s a cute coincidence (related to your arbitrary choice of seconds as the unit for measuring time) that this gives a number close to what I got, but this number doesn’t mean anything more beyond the fact that it’s the kinetic energy that would be gained by a 500 kilogram object after falling for one second from an initial position of rest. $\endgroup$
    – GenlyAi
    Commented Jul 20, 2019 at 23:57
  • $\begingroup$ Also, if your reasoning really said something about the work needed for hovering then Robinson R22 helicopters would be unable to maintain hovering flight at a weight of 500 kilograms (since their powerplant only has 124 horsepower), and yet clearly they can (maximum take off weight is 622 kg). $\endgroup$
    – GenlyAi
    Commented Jul 21, 2019 at 0:00
  • $\begingroup$ Ah, I see. With t =/= 1 kinetic energy follows parabolic curve, and numbers don't match anymore. So, basically... in all cases, if nothing is moving, there's no heat and no electric energy consumed, even if there's force, there's no work being done? $\endgroup$
    – SigTerm
    Commented Jul 21, 2019 at 0:08
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    $\begingroup$ @SigTerm correct. Work is force times distance. If nothing is moving then the distance is zero, and therefore the work is also zero. $\endgroup$
    – GenlyAi
    Commented Jul 21, 2019 at 0:20

However, work is defined as newton per meter, and the aircraft does not move.

Since the helicopter does not move, obviously there is no work being done on the helicopter. A force has to be applied over a distance to do work.

How does one calculate work done by a hovering unmoving aircraft over time?

However, the helicopter blades do work on the surrounding air. Can we determine this work? Well, neglecting friction in the rotation mechanisms or other energy losses, we can reason that the work done on the air is equal to the work done by the engine to rotate the blades. Therefore, if you know the torque $\tau$, the angular frequency $\omega$, and the time $t$ you are calculating the work over, then $W=\tau\omega t$.

This is therefore a pretty general method for similar scenarios like this one. You just multiply your power output by the time interval over which you are calculating the work over.

  • $\begingroup$ However, in this scenario, isn't amount of rotational torque on the properller zero? As you, once again, apply enough torque to keep the winch rotating at constant speed. The torque applied by your engine is countered by torque applied by air friction. So the propeller keeps rotating, fuel/battery charge is being drained, and the amount of work being done is zero. As there's no torque. $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 22:56
  • $\begingroup$ @SigTerm Edited $\endgroup$ Commented Jul 20, 2019 at 22:59
  • $\begingroup$ This one is going through intermediate quantity/value (blade air resistance, rpm, torque). Can the intermediate value/quantity be removed and is there a solution in general case? It feels like there should be one. Consider alternative approaches for hovering: Electromagnetic repulsion, Rocket propellant, Compressed air/water propellant, Electrostatic Propulsion and so on. $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 23:05
  • $\begingroup$ @SigTerm It is the work done on the air. The only other way would be to track all of the air. What I discuss in my answer is just the power output of the engine multiplied by time. It's a pretty direct method IMO. $\endgroup$ Commented Jul 20, 2019 at 23:10
  • $\begingroup$ The reason why I call it indirect is because instead of using "How much work it is to counter force F for t seconds", it relies on specific details of the situation. Namely presence of air. This splits the calculation into two-step approach. So I wonder if the problem can be approached from perspective of momentum (N * s), or kinetic energy instead. (J). For example, if we represent object kinetic energy as 3-component vector, and define gravity as a force that continuously drains the energy, while propellant as something that continuously increases it. $\endgroup$
    – SigTerm
    Commented Jul 20, 2019 at 23:21

I think the accepted answer used the wrong lift-to-drag ratio $\gamma = 4$ .

$\gamma = 4$ (From Wikipedia) is the helicopter lift-to-drag ratio (as a whole), not the blades' lift-to-drag ratio. The blades' lift-to-drag ratio should be much higher, since it has very high aspect ratio.

If we use $\gamma = 20$ which is reasonable, the power required for hovering is $26.69$ HP (100% efficiency), if we use 60% hover efficiency, transmission efficiency 80%, it will use $55.6$ HP engine output. ICE engine is most efficient at ~50% load, so the 50-60 hp number is probably pretty close.

  • $\begingroup$ That is the additional power required because of the drag of the blades. It doesn't lift the helicopter. $\endgroup$ Commented Oct 27, 2022 at 5:00
  • $\begingroup$ @FlatterMann when helicopter is hovering, almost all engine output is to overcome drag, the rest is friction. I edited my answer to include efficiency, you can see it's pretty close to real-world. $\endgroup$
    – mikewen
    Commented Oct 28, 2022 at 13:32
  • $\begingroup$ @mikewan That is simply not so. Most of the power of the helicopter engine is used to move air to produce the lift. The theory is both trivial and well established. You should be able to find it in many textbooks about aerospace engineering. $\endgroup$ Commented Oct 28, 2022 at 15:32
  • $\begingroup$ @FlatterMann Please prove it use the example above. You will see the real-world numbers. $\endgroup$
    – mikewen
    Commented Oct 29, 2022 at 18:49
  • $\begingroup$ I usually don't care to invalidate people's random numbers. Like I said, you can find the theory in the books. $\endgroup$ Commented Oct 29, 2022 at 21:40

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