For the given same mass (say it 400kgs), which one requires more power (in HP) between helicopter and airplane? Say that both are designed maximum (the required material are considered during design) to minimize the required power.

Edit: Question should be like this: Which one is requires more power to lift a helicopter (to make it hover) and to make a fixed wing airplane fly? Not to make them move at the same speed.

  • $\begingroup$ Actually it is flying at the same weight and speed. Please see answer below. $\endgroup$ – Robert DiGiovanni Dec 1 '18 at 8:50
  • $\begingroup$ Either you compare lifting and moving, or you compare the airplane nose up and hovering. The helicopter requires LESS power to hover. Why? (compare propeller dimensions and RPM to helicopter rotor). $\endgroup$ – Robert DiGiovanni Dec 1 '18 at 23:19

Airplanes fly by creating lift from their wings. This causes some drag, but good wings have lift/drag ratio's in the range 15-20. That means the lift-dependent drag is only 5% of the lift. Helicopters on the other hand generate lift directly from trust; there is no multiplier involved.

E.g. a 4000 kg plane will have a weight of 40.000 Newton, so the drag incurred will be 2000-3000 Newton. A 4000 kg helicopter will need to produce 40.000 Newton of lift just to hover.

Of course, planes and helicopters both have additional drag from forwards airspeed, and for planes this is obviously unavoidable to prevent stalls.

  • $\begingroup$ Dear @mstalters, is it possible you make calculation so will be clearer? A helicopter is lifting the load vertically, while airplane is lifting the load by dragging. $\endgroup$ – AirCraft Lover Dec 2 '18 at 6:32
  • $\begingroup$ Please see comments below. Thanks for the discussion. $\endgroup$ – Robert DiGiovanni Dec 2 '18 at 9:57
  • $\begingroup$ @AirCraftLover: Added. $\endgroup$ – MSalters Dec 2 '18 at 15:11

Let's look at this in an extremely simplified fashion.

An aircraft with mass $m_{ac}$ stays up in the air by pushing air downwards, or specifically, by giving a mass flow $\dot{m}_A$ [kg/s] of air a certain speed downwards $v_A$ [m/s]. This gives a momentum 'flow' $\dot{m}v$ [kg m/s²] which is the lift force $F_{lift}$ [N]

$$F_g = F_{lift}$$ $$m_{ac}g = \dot{m}_A\cdot v_A$$

The power required for this comes from having to give the air flow a kinetic energy flow $$P_{lift}=\dot{m}_A{v_A}^2$$

This is purely the power required for lift generation (power required to overcome induced drag, specifically). One can see that by making $\dot{m}_A$ arbitrarily large and $v_A$ arbitrarily small (while keeping their product constant), the power requirement can be made arbitrarily small. This can for example be done by making the wings or rotors longer so that they affect a larger air volume (and thus air mass), or by flying faster (so they move through more air, again increasing the mass flow).

However, this assumes perfect efficiency. In reality, wings will experience drag even if no lift is being generated, and the same goes for the fuselage. You will often find a minimum of total power required at some speed such that the induced drag is quite small but the friction drag is not quite as large. This goes for both fixed and rotary wing aircraft. These factors are a result of the aircraft's practical design, not of theoretical considerations.

So, there is no theoretical answer to this question. There is only a practical answer, which is that hovering in a helicopter is very inefficient and requires a lot of power (because it can only affect a small mass of air since it is not allowed to move), so given the constraints in your question (a hovering helo vs a fixed wing at 100kts), the fixed wing is probably more efficient in practice.

  • $\begingroup$ We agree on F=ma and vectors. What is interesting is that with equally opposing forces there is no movement. The rocket on the test stand is producing Power, the test stand (hopefully) is strong enough to hold it. Power equation = ForcexSpeed may be correct with the caveat "with no opposing force". In reality, friction, drag or any opposing force can bring velocity to steady state. $\endgroup$ – Robert DiGiovanni Dec 3 '18 at 14:54
  • $\begingroup$ And BTW, the solution is to compare the Hughes 269 with the Cessna 150 like @John K and you will see the the difference. A Cessna 150 flies with around 1/4 of its weight in thrust, the helicopter requires 100% of its weight in thrust but the (higher aspect) rotor is more EFFICIENT than the prop, ok? $\endgroup$ – Robert DiGiovanni Dec 3 '18 at 15:02
  • $\begingroup$ @RobertDiGiovanni In steady state, the engine still does work, but the drag does equal work but opposite in sign. Anyway, if we can agree that power is not the same as thrust (which we apparently cannot, despite the discrepancy in units), we can see that the Hughes comes with a 134kW engine for a loaded weight of 703kg, while the Cessna comes with a 75kW engine with a loaded weight of 726kg (numbers from Wikipedia) $\endgroup$ – Sanchises Dec 3 '18 at 15:09
  • $\begingroup$ Getting somewhere here. So the helo generates 4x the thrust with 2x the power. Now let's put a very large prop on an airplane (V-22 Osprey) ? $\endgroup$ – Robert DiGiovanni Dec 3 '18 at 15:31
  • $\begingroup$ @RobertDiGiovanni Indeed, a helo is more efficient at converting power to thrust, thanks to the large blades. An airplane can get away with a less efficient (less large, so smaller landing gear required) propeller because it has the highly efficient wings. You'll see that early airplanes had almost comically large propellers since the engines were so small; to this day, you see that turboprops tend to be high-wing (e.g. ATR-72, Dash-8) to accommodate larger props. $\endgroup$ – Sanchises Dec 3 '18 at 16:14

Helicopter rotors need to provide enough thrust to compensate the weight of the helicopter: $T_H = W$.

Fixed wing aeroplanes need to provide enough thrust to overcome drag, while the wing lift compensates for weight. As @MSalters correctly states, the wing provides much more lift than drag, plus there is a fuselage & tail as well.

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Torenbeek Synthesis of Subsonic Airplane Design gives some L/W ratios of complete aeroplanes, a medium size turboprop such as the F-27 is listed as having an L/D of 13.8 during takeoff, since aspect ratio A = 12. So this fixed wing aeroplane would have to provide 1/14 of the thrust of a helicopter of the same T/O weight: $T_F = W/14$

Simple impulse theory gives the following relation between thrust T and power P:

$$ T = C_T \cdot \rho A {(\Omega R)}^2$$ $$ P = C_P \cdot \rho A {(\Omega R)}^3$$ And therefore at constant disk area and tip speed: $$C_P = \frac{{C_T}^{3/2}}{\sqrt{2}} => P_F = (P_H/14)^{3/2} = 0.1$$

Tip speeds $(\Omega R)$ of propellers and helicopter rotors are comparable, disk area of a fixed wing is lower. So the required power at constant weight for a fixed wing is at least an order of magnitude lower than for a helicopter.


If you mean for a 400kg helicopter and 400kg fixed wing airplane to move at 100kt, it's generally going to be the helicopter that requires more power since the whole egg beating thrashing mess is a lot less efficient at converting energy to forward speed.

Of course you can make the airplane draggy enough that it can require more power than the helicopter to go 100kt if you want, and there are plenty of those, but I assume we're talking about optimized craft here.

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    $\begingroup$ I mean, which one is requires more power to lift a helicopter and to make a fixed wing airplane fly? I agree with you, if the intention is to make it move the same speed, of course helicopter requires more power. $\endgroup$ – AirCraft Lover Dec 1 '18 at 2:06
  • $\begingroup$ Well, a Hughes 269 weighs 1550 lbs and has 180 hp and is not exactly a vertical rocket with that, and a Cessna 150 weighs 1500 lbs and flies just fine on 100 hp, so there you go. $\endgroup$ – John K Dec 1 '18 at 5:08

the helicopter's "wings" (its main rotor blades) are moving through the air at something like 400 MPH at their tips even while the helo itself is sitting in ground effect and not traveling through the air at all. This requires work, and to fly forward, the engine has to overcome the main rotor drag at the same time it has to overcome the fuselage drag. So: more drag for the helo, less for the Cessna 150.


General rule of thumb with heavier than air craft is the bigger, slower lifting surface with less interference is more EFFICIENT at converting available power into lifting and/or propelling force.

The helicopter blade has greater efficiency in creating thrust compared with a propeller (see V-22 Osprey), but the wing is far more efficient at creating lift, therefore the Cessna 150 requires less POWER to create the same amount of total FORCE (lift and thrust).

You will notice this works for monoplane vs biplane (less interference), propeller vs fan (much less interference), hi aspect glider vs low aspect speed plane (more efficient wing), 2 bladed helicopter vs 8 bladed helicopter (less interference), bird vs airplane (slower "propeller")

Thrust Force = Power x Efficiency

You will notice birds propel and lift more like helicopters. Yet they are MORE efficient than the Cessna because their PROPELLERS are proportionally larger and slower. But we do not see 400 kg birds or 800,000 lb Cessnas because NET thrust is what actually moves the aircraft.

A less EFFICIENT jet has much more power, so much more thrust is available, albeit with more fuel consumption per pound of thrust produced.

Finally, force required to "make the aircraft hover" and "make the air plane fly" really is comparable! One must simply realize in both cases it is FORCE vs DRAG. So, for direct comparison, they both need to be at flying speed.


For the benefit of @ MSalters and @ AirCraft Lover an examination of units is in order

F = ma kg m/seconds squared

Now the equation for Power = F x velocity = kg m/seconds squared x m/second

This is an unfortunate corruption of James Watts original weight lifting work with horses (origin of "horsepower") and should read:

Power = massxgravity + massxacceleration to reach speed and Power = massxgravity at constant speed F=ma=weight being lifted. Force vectors are ADDITIVE. Air drag and pulley friction are negligible. (Watt knew to take only the first meter)

Notice distance has NOTHING to do with POWER! (I'm sure football players struggling at the line of scrimmage (Thrust vs Drag Superbowl) appreciate this).

Let's fast forward to the THRUST curve of an Estes C6-0 model rocket and viola! That is what James Watt's horse is doing!

So we can think in terms of thrust, lift, drag, and gravity as ADDITIVE forces (much easier), and better yet, break them down into vertical and horizontal components!

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    $\begingroup$ That's some handwaving physics there. Force x Speed = Power. $\endgroup$ – MSalters Dec 1 '18 at 23:46
  • $\begingroup$ @MSalters Power opposes drag. Efficiency produced is really thrust to drag (or lift to drag) ratio. Could rearrange to thrust or lift force/power=Efficiency. No hand waving here. It is the definition of power that includes speed (as d/t) that may need examination. It takes power (as force/time) to hold up an object (against gravity/time) even if there is no motion. I have written about James Watt and horsepower and concluded P= weight x d/t should be weight x v/t to accelerate weight to constant speed and simply P=weight at constant speed for horse. I wonder what the horse thinks. $\endgroup$ – Robert DiGiovanni Dec 2 '18 at 1:04
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    $\begingroup$ I was being friendly there. The equation doesn't even agree in its dimensions. Force is in Newtons, Power in Joules per second, and efficiency is a dimensionless fraction. And no, it doesn't take power to hold up objects against gravity. Unpowered tables do it al the time. $\endgroup$ – MSalters Dec 2 '18 at 1:09
  • $\begingroup$ Dear @Robert-DiGivani, is it possible you make to a real calculation? Say, what would be minimum blades and power required to lift a helicopter vertically, and so does with an airplane. From there, that would be clear which one is requires more power. $\endgroup$ – AirCraft Lover Dec 2 '18 at 6:31
  • $\begingroup$ Dear @Robert-DiGivani, from your equation, force unit is Newton (N), and power unit is Newton-meter/second (N.m/s). Almost there, but however still need correction. $\endgroup$ – AirCraft Lover Dec 2 '18 at 6:36

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