This answer says

A helicopter uses a LOT more fuel hovering than it does in forward flight.

Is this correct? Why?

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    $\begingroup$ Very related: aviation.stackexchange.com/questions/13397/… $\endgroup$ Commented Nov 12, 2017 at 1:00
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    $\begingroup$ Obviously it uses a LOT more fuel per mile if it's just hovering, duh! :) $\endgroup$
    – pipe
    Commented Nov 12, 2017 at 9:02
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    $\begingroup$ The simple answer to why it does is that the blades have less air to push on in hover. And moving less air faster takes more energy than moving more air slower. $\endgroup$ Commented Jan 24 at 1:33

7 Answers 7


Yes it is correct that helicopters use more fuel when hovering: the engine needs to apply more power to overcome drag. Here is a graph of the engine power required for different airspeeds, from J. Gordon Leishman, Principles Of Helicopter Aerodynamics:

enter image description here

The line for total power goes down between 0 - 70 kts with increasing airspeed, this is caused by the line for induced power: power required to overcome the induced drag of the helicopter blade. The total required engine power is the summation of:

  • Induced power. The power required to overcome the induced drag of lift creation, further detailed below. Propulsive power relates to the engine exhaust velocity now being helpful, and to the rise in induced power at higher speeds due to compressibility drag.
  • Profile power, required for the blade profile drag.
  • Parasitic power, for the drag caused by the air frame, rotor hub etc. Zero in the hover, very dominant at top speed. Helicopters have shapes that are way less aerodynamic than fixed wing planes, and this source of drag becomes very significant at higher speeds.
  • Tail rotor power. Up to 20% of the main rotor power both in the hover and at top speed, very low in the middle due to the helpful vertical tail. At top speed the torque of the main rotor is high and the tail rotor must do more work, unless the vertical tail can be adjusted.

from Wikipedia for induced drag

Induced power is dominant in the hover. Induced drag is caused by the backwards tilt of the lift vector: the higher the angle between blade and free stream, the more the vector is tilted backwards, which causes both loss of lift and increase of drag. The equation for lift L is:

$$ L = C_L \cdot \frac{1}{2} \cdot \rho \cdot V^2 \cdot S$$

and at a given altitude, the two variables here are $C_L$ (lift coefficient) and $V$ (airspeed at the blade). $C_L$ is an approximately linear function of angle of attack at the blade, so lift increases linearly with blade tilt-back and quadratically with increasing airspeed over the blade.

enter image description here

Above graph from Leishman shows the velocity distribution over the blades when hovering, and at airspeed. Quite a complicated situation - when hovering, the airspeed reaching the blade is only the rotational speed of the rotor, at forward speed the blade going forward has rotational speed plus airspeed.

The helicopter does not roll over and both forward blade and retreating blade deliver the same amount of lift, with the rearward going blade tilted back more than it was in the hover. But the forward going blade is tilted back a lot less: airspeed has a quadratic influence.

Note that the circle in the plot at fwd airspeed is not stalled flow, but reverse flow: the air streams in at the back of the blade. So drag is now negative, the airstream helps to propel the blade! However there is loss of lift in the reverse flow area.

Induced power reduces with airspeed at first according to the simple 1-D impulse consideration (more air mass through the disk), and later increases as the disk is increasingly tilted forward and must do more work to overcome losses from rotor profile drag, airframe parasitic drag, and compressibility drag.

There is also an interference effect of the downwash over the fuselage: in the hover the air streams straight down, while in forward flight the rotor wash is more aligned with the fuselage, catching more of a streamline shape. Parasitic drag is of course dominant at top speed, while offloading the rotor by using fixed wing surfaces reduces the induced power at high speeds - but from hover to moderate forward speeds it is purely the reduction in lift induced power that creates translational lift.

enter image description here

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    $\begingroup$ Choppers are so technical ! Awesome. $\endgroup$
    – Fattie
    Commented Nov 12, 2017 at 15:52
  • $\begingroup$ This is only for going forwards though right? Does the same effect apply if the chopper is going sideways or backwards? $\endgroup$ Commented Nov 13, 2017 at 6:45
  • $\begingroup$ Yes it does, or if there is sidewind when it goes up and hovers in place. When there is sidewind the pilot needs to apply pedal to adjust anti-torque. With a tailwind the vertical stabiliser will make it hard to maintain heading. $\endgroup$
    – Koyovis
    Commented Nov 13, 2017 at 7:30
  • $\begingroup$ Great detail here - thx for the speed/power curves. $\endgroup$ Commented Nov 14, 2017 at 0:55
  • $\begingroup$ Regarding induced power, and this statement: "the higher the angle between blade and free stream, the more the vector is tilted backwards, which causes both loss of lift and increase of drag ". Should it instead be: "the higher the angle between the rotor plane and the effective relative airflow, ..."? Reasoning: 1. The effective lift vector direction doesn't change with the blade angle (although magnitude likely does). 2. The resultant wind vector is what's ultimately used to calculate lift/drag directions. 3. The portion of the lift in the rotor plane is what's felt as "drag" on the rotor. $\endgroup$
    – Philip
    Commented Aug 31, 2023 at 17:22

Yes, it is correct, if the helicopter doesnā€™t fly too fast. A helicopter will produce the necessary lift most efficiently at a moderate forward speed.

In a hover all the airflow which is available for lift creation must be generated by the rotation of the main rotor. This means that a small amount of air must be accelerated by a lot. If the helicopter adds forward speed, it can achieve a higher mass flow through the rotor, and now less acceleration of air is needed to achieve the same lift. This improves the efficiency of lift creation. If the helicopter goes faster than its speed for maximum rate of climb, aerodynamic drag grows too high and reduces efficiency again.

At high speed, the tips of the advancing blades might reach transsonic speeds, which produces a noticeable drag increase, and the inner part of the retreating blade will see very little airspeed, and to still produce lift, the whole blade will pitch to a high angle of attack, causing the inner part to stall, which again produces a noticeable drag increase. There is a sweet spot between hover and fast speed where the required power reaches a minimum.


Yeah, I'm not a physics student, but I work on Black Hawks. If you conceptualise a helicopter as just a main rotor disc producing lift, then Peter Kampf's answer about mass-flow through the rotor disc is the greatest factor. (Remember the disc is tilted forward as the helicopter moves forward). However, your question actually asked why do they burn less fuel: well, thousands of little design features on the airframe each help save precious pounds of fuel in forward flight. (You might want to do a Google image search to look at while you read this.)

The Black Hawk has a cambered vertical fin which unloads the tail rotor above 60kts, and this torque is redirected into the main rotor. It has a variable stabilator which changes angle with fwd airspeed (= changing main rotor downwash angle) in order to provide lift, further unloading the main rotor. The tail rotor is canted at an angle and spins backwards into the main rotor wash, again to unload the main rotor, freeing up more power for forward speed. It has on-flight computers and a mixer unit which flattens out the airframe in flight, so that it doesn't present a flat cabin roof into the airstream at high forward airspeeds. The flatter you can keep the disc into the relative airflow, the smaller the pitching angles of the blades, and the less parasitic drag from the rotor disc.

The main rotor blade tips are swept backwards to delay the onset of transonic tip drag as the advancing blade sees higher relative airspeeds in forward flight. Other helicopters have airframe fairings that generate lift off the cabin body in forward flight. All of these aerodynamic savings are present in forward flight, but not in the hover. And lastly, your turbine engine air inlets will benefit from some ram-air effect in forward flight, which means burning less fuel for the same torque. Every helicopter in the world uses some or all of these features to save fuel in flight, and if you compare generations of helicopters (Bell 47, Bell UH-1, Bell 412, Black Hawk), you can see these features gradually develop.

There are other considerations when a helicopter is hovering just off the ground, but I've tried to list just some of the ways helicopters are designed to save fuel in flight. Hope some of this helps.

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    $\begingroup$ talk about a real-world answer! awesome $\endgroup$
    – Fattie
    Commented Nov 12, 2017 at 15:53
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    $\begingroup$ "There are other considerations when a helicopter is hovering just off the ground": I suppose it uses less fuel due to the surface effect (basically, an air cushion emerges which increases lift)? That would imply that if you need to hover for extended periods of time, do so close to the ground. $\endgroup$ Commented Nov 13, 2017 at 10:52

The concept is known as "translational lift". When moving in forward flight, a helicopter's rotor disc acts a lot like an airplane's wing - it has a significant lift-to-drag ratio. The required thrust to maintain level flight is reduced by that ratio, and therefore necessary engine power and fuel flow are also reduced. In hover, the engine+rotor system has to supply thrust fully equal to the weight of the helicopter.

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    $\begingroup$ Thanks very much for the simplest answer here that provides a basic understanding good enough for us laymen to quench our curiosity. šŸ¤˜ $\endgroup$ Commented May 5, 2022 at 3:56
  • $\begingroup$ You're welcome @JoelMellon, thanks for the kind comment. $\endgroup$ Commented May 14, 2022 at 19:38

When in a hover, the air has more time to setup into an induced wash from further upwards that translates into higher down flow speed by the time the induced wash reaches the plane of the rotor. When in translational flight, the rotor is continuously moving into clean air, so the down flow speed by the time the air reaches the plane of the rotor is less than that of a hover. Power equals force times speed, in this case consider the power output to the air. In both cases, the force is the same (equal to the weight of the helicopter), but in a hover, the down wash speed through the plane of the rotor is greater than during translational flight, so the required power in a hover is greater than in translational flight, until the translational drag becomes an issue.

Another issue is tip vortices. In a hover, these can get quite large, again due to all the time for the vortices to set up and the rotor tips moving into the vortices induced by the other rotor tip(s). In translational flight, the vortices are "washed" off by the relative horizontal wind, reducing the size of the tip vortices.


Another point to consider is whether helicopter has supplemental wings. Rather famous example is Mi-24 family of attack helicopters, where weapon pylons works as wings.

"At high speed, the wings provide considerable lift (up to a quarter of total lift)."

At high altitudes with full load the recommended lift-off procedure is to gain horizontal speed so wings pick up some lift.

  • $\begingroup$ Damn, a 25% weight reduction is no joke. $\endgroup$ Commented May 5, 2022 at 3:57

If gravity were the only force acting on an aircraft, then at each moment in time the aircraft would be gaining certain amount of downward momentum. So to maintain altitude, the aircraft must transfer that momentum to some other mass (i.e. air). That is, thereā€™s going to be some air that starts with zero velocity (in the simplest case) and ends up with some downward velocity. Since momentum is mass times velocity, the velocity that the air has to be accelerated to will be inversely proportional to the mass of air accelerated:

$velocity = momentum/mass$

However, the energy of that air is: $Ā½mvĀ²$

When we substitute velocity into that equation, we get:

$energy = mass \cdot (momentumĀ²/2 \cdot massĀ²)$

One power of mass in denominator cancels out the mass in numerator, and we get:

$energy = momentumĀ²/2 \cdot mass$

Thus, doubling the amount of air accelerated downward halves the energy required. When an airplane is traveling at high velocity, a large amount of air is coming into contact with its wings, meaning that it does not have to expend much energy to generate lift (of course, the faster it travels, the more drag it experiences, giving a lift-drag trade-off). A helicopter experiences something similar: when it is traveling horizontally, it naturally moves into new air. When it hovers, thereā€™s less air to accelerate downward, and what air there is has to be pulled towards the rotor by the rotorā€™s own effort.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Ralph J
    Commented Dec 30, 2022 at 15:16
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    $\begingroup$ @AdityaSharma Is it really a good idea to extensively re-write someone elseā€™s answer? $\endgroup$
    – Koyovis
    Commented Jan 5, 2023 at 18:09
  • $\begingroup$ @Koyovis No its not. I have only corrected the mathematical angle of the answer, while fully conveying the facts as intended by the original answerer. $\endgroup$ Commented Jan 5, 2023 at 20:41
  • $\begingroup$ @AdityaSharma I see a lot of phrases that the original poster did not write, yet their name is under it. Helping the author by changing the equations into Mathjax would be a good idea, correcting spelling mistakes etc. But extensive addition of phrases is not done on this site $\endgroup$
    – Koyovis
    Commented Jan 5, 2023 at 23:15
  • $\begingroup$ @Koyovis I apologise for that. Long story short: Sophit asked me to fix the math, I did so - but quickly realised that the changes made to the answer were drastic. I still submitted my edit proposal and against all odds it got approved. Now, I still believe that the message of the answer is still presented as originally intended by the answerer, and that all of those added phrases are only there to support the new fixed math. If you believe that my edits have been too drastic, let me know if you recommend me to revert to an older edit with only the Mathjax addition. $\endgroup$ Commented Jan 5, 2023 at 23:27

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