At higher speeds, advancing helicopter blades will operate at velocities approaching mach 1, reducing efficiency. But what about lower speeds?

A helicopter is basically a wingless plane with a giant propeller. Why is it more efficient to move around using propellers to provide forward trust while using separate fixed wings to provide lift?

  • 2
    $\begingroup$ They don't fly so much as beat the air into submission... $\endgroup$
    – Ron Beyer
    Commented Aug 30, 2018 at 0:39
  • $\begingroup$ ...by using money ;-). $\endgroup$ Commented Aug 30, 2018 at 1:30

4 Answers 4


The answer lies in the difference between energy and momentum. Both have to be maintained, but energy goes up as the square of velocity, while momentum goes up linearly. To impart a given momentum, you can either move a lot of air slowly, or a little air quickly -- but as you move the air more quickly, it requires disproportionately more energy.

In other words, the bigger the wing, the more efficient (from an energy/momentum perspective, anyway).

A rotor's size is limited, both because its fast rotation means it needs to be light (and thus can't be very reinforced, etc) and because the larger its radius, the faster the end spins. That means it has to touch less air, but move it more -- which requires more energy.

There's more detail at this question on Physics StackExchange (disclaimer: I wrote that question).

  • $\begingroup$ Thanks for the link to your question! It's basically what I was asking, but better elaborated. I like the third answer to your question. $\endgroup$ Commented Aug 30, 2018 at 16:29
  • $\begingroup$ I don't see why the end has to spin faster as the radius is increased. In fact, it seems to me like the opposite should be true. As the radius is increased, the wings of the rotor are longer, so they don't need to move through the air as fast to provide the same lift. $\endgroup$ Commented Aug 30, 2018 at 16:30
  • $\begingroup$ @StephaneBersier That's getting a bit too far from my expertise. :-) But I wonder if the problem is that if you did that, then the inner parts of the rotor would be moving too slowly. There are basically a lot of "radius-squared" tradeoffs to be made. Beyond that, just the physical aspect of making such a wing that's both very long and very lightweight (it does have to move in a lot of different ways, so you want to keep its inertia down) is hard. $\endgroup$
    – yshavit
    Commented Aug 30, 2018 at 19:48

A helicopter is not exactly wingless - the rotor is the helicopter's wings.

A wing's drag is proportional to $W\cdot(L/D)$, where $W$ is weight and $L/D$ is lift/drag. The energy spent overcoming this drag each second is $V_{wing}\cdot W\cdot(L/D)$, where $V_{wing}$ is the wing's airspeed.

The amount of fuel required to travel distance $S$ is then $SFOC\cdot S/V_{aircraft}\cdot V_{wing}\cdot W\cdot(L/D)$, where $V_{aircraft}$ is the aircraft's ground speed, where $SFOC$ is specific fuel consumption.
For a fixed wing, $V_{wing}$ and $V_{aircraft}$, ignoring wind, are the same. In a rotary wing (helicopter), $V_{wing}$ is higher than $V_{aircraft}$ - the wing travels a longer path through the air, with rotation.

For this reason, a helicopter will always be less efficient than a fixed wing aircraft with the same lift/drag ratio. If the ratio is 3x, that's 3 times more fuel for the same distance. This is of course a vast simplification, ignoring all the details and just describing the principle.

To put it in even simpler terms, a helicopter is like a plane with its wings flying a path full of loops.

  • $\begingroup$ A nice and well-founded answer... $\endgroup$
    – xxavier
    Commented Aug 30, 2018 at 8:39
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    $\begingroup$ I feel like your answer is missing something, as it doesn't apply the same math to the propeller of the plane, which is moving even faster than the rotor of a helicopter... sadly I don't have more time to think about it for now. $\endgroup$ Commented Aug 30, 2018 at 16:35
  • $\begingroup$ @StephaneBersier A propeller's thrust is much smaller than that of a rotor, just ~1/10 of the aircraft's weight. The wings then turn that thrust into 10 times more lift, to keep the plane aloft. Since propellers are fairly small, their drag is also relatively small. Notable fact: helicopters use the most fuel per minute in hover; going faster takes less fuel, because the rotor no longer needs to provide entirely static thrust. $\endgroup$
    – Therac
    Commented Aug 30, 2018 at 18:20

the wing on a small plane moves through the air at around 100MPH, whereas the main rotor blade on a small helo moves through the air at around 400MPH. the blade is smaller than the wing to be sure, but the drag forces acting at 400MPH are still big, and soak up a lot of horsepower.

Try this comparison: imagine a Cessna 150 with two people in it at cruise conditions and compare it with a Robison R22 with two people in it, same altitude, etc. Look at their respective fuel burn rates.

  • $\begingroup$ What about the speed of the propeller blades on the Cessna 150? I'd imagine it's above 180 m/s. $\endgroup$ Commented Aug 30, 2018 at 16:38
  • $\begingroup$ yes, but the prop diameter on a C150 is quite small compared to the main rotor of an R22. The Cessna cruises at about 110mph on 100HP and burns between 5 and 5.6 GPH. The R22 requires 150HP and burns between 7.5 and 10 GPH. $\endgroup$ Commented Aug 30, 2018 at 16:45

As pointed out in this answer:

  • More interference drag
  • A circular shape of the rotor disk - it can be considered to be a wing in forward flight, but a circular wing is much less efficient than a long slender wing.

enter image description here

The picture from J. Gordon Leishman, Principles of Helicopter Aerodynamics, illustrates the complexity of the multitude of aerodynamic drags that can be found in a helicopter, and that decrease efficiency.


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