# Relative efficiency of airplanes and helicopters, why is this argument wrong?

I'm trying to better understand why helicopters are less fuel efficient per unit distance than airplanes. One argument I keep seeing in other questions on the topic is that helicopters actively contribute energy toward both horizontal translation and maintaining vertical position, whereas airplanes get the vertical "for free" and only need to overcome drag to stay in the air. To me, this argument seems wrong, but if it isn't, I'd like to understand where my reasoning is leading me astray.

My understanding is that the force needed to fight gravity is the same in an airplane as it is in a helicopter, $$mg$$. The difference is that a helicopter contributes this force directly through its rotor whereas an airplane contributes this force through induced drag subtracted from the force in the horizontal provided by its engines as lift.

Why am I wrong?

• The only difference between an airplane wing and a helicopter blade is that the helicopter blade travels in circles. Oct 31, 2021 at 14:48

One argument I keep seeing in other questions on the topic is that helicopters actively contribute energy toward both horizontal translation and maintaining vertical position, whereas airplanes get the vertical "for free" and only need to overcome drag to stay in the air. To me, this argument seems wrong, but if it isn't, I'd like to understand where my reasoning is leading me astray.

I agree. I have always found there are holes in the reasoning that airplanes are more efficient than helicopters because an airplane wing only has to provide lift while helicopter rotor has to provide both lift and thrust. The reason is that lift production isn't free in an airplane; That's why airplanes need propellers to provide thrust to overcome form drag and induced drag; If lift production were free, you would be able to produce lift at zero degrees AOA with a symmetrical airfoil thus incurring no induced drag penalty, but of course we that's not the case.

This reasoning seems to boil down to two things that aren't often made explicit enough (if at all):

1. Lift production in an airplane isn't free but the forward motion used to get from point A to point B is. The distinction is that you can have forward motion without lift but you can't have lift without forward motion. The same motion that is required to produces lift to keep the plane aloft can also inherently be used to move it from point A to point B. Whereas they are separate things for a helicopter. This would affect distance efficiency, but not necessarily flight time efficiency.
2. Somewhat related to #1 is that in a helicopter, you have a lift/thrust producing motion in in the spinning rotor which encounters drag that must be overcome, and you have the forward translational motion of the aircraft which experiences a drag that must also be overcome. But this is ignores the fact that airplane propellers also have a rotational drag to overcome. It is this translation of thrust into lift on an airplane where comparisons become murky with this line of reasoning which is why I don't like it unless you have enough information to balance out all the numbers.

Below are more distinct, less murky reasons airplanes beat helicopters in both flight-time efficiency. Though they don't necessarily imply distance efficiency, when you combine them with #1 it certainly does.

1. An airplane's airframe is more streamlined and therefore producess less drag for translational motion than the helicopter airframe which reduces fuel consumption per distance traveled. And of course, as previously mentioned, this translational motion is also directly used for lift production as well.
2. Airplane wings are more aerodynamically efficient than helicopter rotor blades because they are much larger (what they might lack in aspect ratio compared to rotor blades is made up for in chord which affects the Reynolds number they operate at). The larger area allows an airplane wings to move a larger masses of air slowly in order to produce the same momentum $$mv$$ which results in lower energy usage via $$1/2mv^2$$. Which leads into...
3. On top of that, airplane wings have higher mass utilization because lift distributions are more evenly distributed from root to tip compared to helicopter rotors. This is because the entire wing moves through the air at more or less the same speed from root to tip (ignoring differences in span-wise flow) whereas spinning rotor blades have near zero airspeed at the root close to the center of rotation and produce a majority of their thrust somewhere past mid-span on the outer edges. That means you get more lift produced per mass invested in wing structure.

If you had unobtanium materials that were super-strong, super light (near-zero density in some cases), you could try and build a helicopter with an ENORMOUS rotor that traces out the same path as the wings of a circling airplane would, with the same chord. Obviously this is impossible in the real world since such a larger rotor would:

• experience incredible bending stresses
• be very heavy
• experience incredible centripetal stresses since you are swinging an enormous and heavy rotor
• require a gearbox that can handle the stresses involved in swinging such a rotor which means the gearbox is big and heavy if made of real materials

But such an imaginary helicopter made of unobtanium would have similar aerodynamics to a circling airplane made of real materials and thus have similar flight time efficiency assuming they ended up with similar weight. But even in forward motion this imaginary helicopter loses out because the translational drag of such a rotor is enormous and the motion of the rotor airfoil isn't simply moving towards the destination so you can't take advantage of it like the fixed wings on an airplane are.

The forces may be equal, but you need to compare fuel efficiency per unit distance, not the forces. If the helicopter soars in place, the force is still $$mg$$, but fuel efficiency per unit distance is zero. EDIT (10/24/21): The term "fuel efficiency per unit distance" used by the OP is incorrect, and I was wrong to use it. I should have said just "fuel efficiency", which is measured in MPG.

• Fuel efficiency per unit distance for something that stays in place is undefined, effectively infinite, as the denominator is zero.
– Nij
Oct 25, 2021 at 0:30
• @Nij : Then what do you have in the nominator? How do you define fuel efficiency? Oct 25, 2021 at 0:50
• It doesn't matter what you put in the denominator, if the value of that thing is zero, because division by zero is undefined: something that does not move has zero distance covered, so it will never be able to spend enough fuel to cover a unit distance. Your answer says that it does not need to spend any fuel to travel unit distance, which is the total opposite.
– Nij
Oct 25, 2021 at 1:23
• @Nij : So there is zero distance, that means fuel efficiency is zero. Maybe the the expression "fuel efficiency per unit distance" used by OP is unfortunate, but the fuel efficiency in terms of the distance is clearly zero. Oct 25, 2021 at 1:38
• You wrote yourself, "per unit distance". No movement = zero distance. If you meant fuel efficiency, that already incorporates distance travelled, and if you meant to correct their error, them do so. But as it stands, the answer is simply incorrect.
– Nij
Oct 25, 2021 at 3:58