Calculate work done by a hovering helicopter over time

Question

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.

Answer 1

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: $$ P = \pi A f L/\gamma $$

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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.

Answer 2

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.

Answer 3

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.

Answer 4

Allow me to offer an alternative answer to this question. Let $A$ represent the area swept out by the helicopter blades, $M$ is the mass of the helicopter, $m$ is the mass of the air that is pushed downward, and accelerated from an initial $v_i=0$ to some final $v_f=v$ in a time $\Delta t$ over a distance $\Delta z$. $\rho$ is the density of air. $m=\rho A \Delta z$.

Newton's 3rd law states that the force of the helicopter on the air is equal in magnitude to the force of the air on the helicopter, and that should support its weight $Mg$. The work done on the air in a time $\Delta t$ (i.e., the power $P$) is therefore $$P=F \frac{\Delta z}{\Delta t}=F \frac{v}{2}$$ where I've used the fact that $\Delta z/\Delta t=\bar{v}=v/2$, where $\bar{v}$ is the average speed of the air while it's being accelerated (I assume $a_\text{air}$ to be constant). The acceleration of the air must be $a_\text{air}=F/m_\text{air}=Mg/\rho A \Delta z = v/\Delta t$. This gives $$v=\sqrt{2Mg/\rho A}$$ and therefore $$P=\frac{1}{2}Mg\sqrt{\frac{2Mg}{\rho A}}$$ which simplifies to $$P = \frac{(Mg)^{3/2}}{\sqrt{2 \rho A}}.$$ This neglects energy lost to drag, but appears to behave at least plausibly with respect to the weight of the chopper, the density of air, and the sweep area of the blades.

Plugging in some reasonable numbers: $M\approx 2000$ kg, $A \approx 100 \text{m}^2$, $g=10 \;\text{m}/\text{s}^2$, $\rho \approx 1.3 \text{kg}/\text{m}^3$, we get $P\approx 175 \text{kW}$.