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I've been told that, generally speaking, rotor efficiency increase with rotor diameter. This is because the thrust generated by a helicopter rotor equals the mass of air moved times the delta V of the air moved, while the kinetic energy imparted to the air by the helicopter is proportional to the mass of the air times the square of the delta V. Therefore, since increasing the rotor area increases the mass of air moved, and since hovering requires a constant thrust force, increasing the rotor area will decrease the delta V of the air. And according to E = 1/2 m*v^2, doubling the mass and halving the delta V of the air will decrease the energy imparted to it.

From this, you can theorize that a helicopter with an infinitely large rotor diameter would require zero energy to remain hovering.

Now, assuming this is all correct, let's assume that the helicopter briefly accelerates upwards until it is moving vertically at 1 mph. Let's also assume that wind resistance is negligible. As it moves steadily upward, it will need the same force as when it was just hovering. And the above analysis suggests that the helicopter with the infinitely large rotor would not need to consume any energy to maintain the constant upward speed. But we know that a helicopter climbing into the sky is gaining potential energy, meaning that this conclusion of zero work done by the helicopter can't be correct.

So, what am I missing? Why does the analysis that suggests that neither a hovering helicopter nor a steadily rising helicopter with an infinitely large rotor need to consume any energy?

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    $\begingroup$ So we have an infinitely long rotor, hovering over an infinitely long/wide flat earth, powered by an engine that had infinite power to get those infinitely long blades moving in the first place, but perhaps no power required to keep them moving, so that all the air in the universe is being swept by these infinitely long blades, whose tip velocity would be infinite except that there ARE no rotor tips (tho much of the blade itself will be supersonic+ )... we're pretty far into the realm of theoretical / detached from reality here. VTC as not being about aviation, as defined in the Help Center. $\endgroup$
    – Ralph J
    Jul 18, 2021 at 4:21
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    $\begingroup$ @RalphJ Really? Helicopters and rotors are not about aviation? Voting to Leave Open, welcoming PrestonBecker to this site. Very good question. $\endgroup$
    – Koyovis
    Jul 18, 2021 at 8:38
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    $\begingroup$ @Jpe61 An infinite rotor creating finite lift will turn infinitely slowly. Don't look at the infinite case but several with increasingly large rotors and see where this is heading. The asymptotic case will be the solution. $\endgroup$ Jul 18, 2021 at 14:43
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    $\begingroup$ You see where this is going? Nowhere. No point in playing with infinities in cases like this. $\endgroup$
    – Jpe61
    Jul 18, 2021 at 15:51
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    $\begingroup$ @MichaelHall Aviation is ruled by physics, and Einstein has shown us how useful thought experiments can be. $\endgroup$
    – Koyovis
    Jul 19, 2021 at 2:17

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We don't need to look at the infinite case to see where the error in your thought experiment is. It will be sufficient to see what is different between a hovering rotor and an ascending one.

Basically, the angle of inflow into the rotor is different between the two and this will not change as diameter is changed. Without considering downwash and induced effects (which are zero in the infinite case), the inflow angle of the hovering rotor is zero, resulting in a lift vector which is exactly perpendicular in case of the infinite rotor in inviscid flow. This means lift is pointing straight up. Its creation does not use engine torque, as you correctly observed.

Now consider the ascending rotor: Here the inflow angle is somewhat positive, namely the angle defined by the inverse tangent of climb speed divided by rotational speed at the radius concerned. This inclined inflow angle will tilt the lift vector backwards (against the direction of rotation) such that the horizontal component of lift will cause a moment which acts against the direction of rotation. This moment needs to be overcome by added engine torque. And you need to adjust collective pitch in order to trim the helicopter for this climb.

Change the climb into a descent and you will see that the same mechanism will add rotational energy to your rotor. Now the inflow angle is from below, tilting the lift vector into the direction of rotation. This is how gyrocopters work!

If you make the rotor larger, the rotor extension will see a very shallow inflow angle because most of the angle change resulting from a climb speed occurs at the inner rotor. Thus, going to infinity will not change the picture.

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  • $\begingroup$ Regarding a hovering rotor requiring 0 torque (end of 2nd paragraph above)... is there any distinction between such a rotor spinning on a helo sitting on the ground, and the rotor spinning on (i.e. supporting) a helo hovering a few feet off the ground? $\endgroup$
    – Ralph J
    Jul 18, 2021 at 18:36
  • $\begingroup$ @RalphJ In real life yes because of ground effect. In this case with an infinite rotor no since downwash is infinitely small. $\endgroup$ Jul 18, 2021 at 20:15
  • $\begingroup$ @PeterKämpf Thanks for this answer. It is exactly what I was looking for. Where did you learn about inflow angle and everything else you discussed? Was if from school or is there a good book or webpage out there on the subject? I have a Mechanical engineering background myself and, while not having not done a lot of work with aerodynamics so far, am hoping to learn more about the topic. (not that I am trying to give you extra work, but if you have any sources off the top of your head, that would be cool) $\endgroup$ Jul 19, 2021 at 6:48
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    $\begingroup$ @PrestonBecker Oh boy, I am a voracious reader but cannot point to a single source. Things that interest me simply get stuck in my mind on first encounter. All that nonsense about sports or celebrities, on the other hand, is soon forgotten. At the root is the desire to drill down to the exact explanation for what is around us. And decades of reading and questioning the correctness of what I have read. Remember, especially on the Interwebs you will find a lot of silly and unscientific stuff. That might even include some of my answers here. $\endgroup$ Jul 19, 2021 at 7:52
  • $\begingroup$ The engine must always produce the kinetic energy of the accelerated air mass = ½$\dot m \cdot \Delta V^2$. Equals engine torque times rotational velocity. Engine torque cannot be zero. $\endgroup$
    – Koyovis
    Jul 19, 2021 at 9:16
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In order to hold a mass m at a fixed height h within a gravitational field of acceleration g, you don't need any energy. A pillar is enough...

But in order to increase that height by ∆h, you have to add energy. Exactly, m·g·∆h

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At hover an infinitely large rotor would move an infinite amount of air downwards at zero speed. If the helicopter is ascending though, it won’t see a zero vertical airspeed and so will need to input some energy in order to balance the increasing potential energy.

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    $\begingroup$ Rotating infinitely long rotors takes an infinite amount of energy. $\endgroup$
    – Jpe61
    Jul 18, 2021 at 11:15
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    $\begingroup$ And an infinite amount of material, and you’d need an infinite atmosphere to fly in. This is, I hazard, largely a thought experiment. Nevertheless in this fanciful scenario the amount of energy that would need to be expended in supporting the aircraft would be zero. $\endgroup$
    – Frog
    Jul 18, 2021 at 19:50
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In a loss free world, many perpetual mobile could exist. But unfortunately in the real world, we can be very certain that there never is such a thing as a free movement lunch. A hovering helicopter has the following losses:

  • Profile drag and induced drag, both quantified in this answer. An infinitely long rotor blade has zero induced drag, but unfortunately an infinite amount of profile drag.
  • Parasitic drag. From rotor/body interference, and from the vertical velocity when climbing.
  • Power loss from the tail rotor, or interference loss from the coax main rotors.

So in the hover, engine power is required for compensating losses in creating thrust = $ m \cdot \Delta V $. With $\dot m$ being the mass flow of air through the rotor in [kg/s] and $\Delta V $ the increase in velocity of the airflow through the rotor. So the equation is not dimensioned in W but in N, not in energy but in a static force. So indeed the whole thought experiment with infinitely long rotor blades is unnecessary. Whatever the length of the blades is, the engine power is necessary to overcome losses in creating hover thrust.

But we know that a helicopter climbing into the sky is gaining potential energy, meaning that this conclusion of zero work done by the helicopter can't be correct.

In a climb, the thrust is higher in order to overcome the vertical drag component. The extra energy required for the climb is equal to $F \cdot c$, with $F $ = thrust and $c$ = climb velocity. As everyone instinctively understands when they can hear their car engine revving up when running up a hill in cruise control.

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You have fallen into a common mathematical trap.

You note that "increasing the rotor area increases the mass of air moved." So an infinite rotor moves an infinite mass of air. Mathematically, 1/infinity = 0, or infinity = 1/0.

So your energy equation E = 1/2 m*v^2 becomes:

E = 1/2 (1/0)*0^2 = 1/2 0/0

But 0/0 is undefined (as is infinity*0), so E is undefined.

Thus, infinity is what is called a boundary condition, at which the mathematical model breaks down. This is in accord with reality, since no infinity has ever been observed in nature.

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    $\begingroup$ "This is in accord with reality..." Reality, the ultimate boundary condition! ;) $\endgroup$ Jul 18, 2021 at 16:31
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    $\begingroup$ @Guy Inchbald Well, since we are dealing with limits, we can use L'Hôpital's rule to determine what the value of the term is L'Hôpital's rule says lim(x->c) f(x)/g(x) = lim(x->c) f'(x)/g'(x) Therefore lim(x->0) 1/2 * (x^2)/x = lim(x->0) 1/2 * (2x)/1 = 0 Therefore, E = 1/2 1/x * x^2 as x goes to 0 gives that E = 0 $\endgroup$ Jul 19, 2021 at 6:02
  • $\begingroup$ @PrestonBecker You appear to have simply substituted x it for 0 and expected your equation to be valid. That is not how L'Hôpital's rule is applied. $\endgroup$ Jul 19, 2021 at 9:29
  • $\begingroup$ The accelerated air mass is infinitely large, the velocity approaches zero. There is no divide-by-zero in the equation. $\endgroup$
    – Koyovis
    Jul 19, 2021 at 10:45
  • $\begingroup$ @Koyovis That is quite beside the point. As my answer states, mathematically, 1/infinity = 0, or infinity = 1/0. Solving equations always involves such manipulations between equivalences, it is how maths works. $\endgroup$ Jul 19, 2021 at 11:55

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