Fly in Elevator: Does Compression Affect Hovering?

[SlowThinker]

Is it the mass of air moved per unit time that changes? Do the rotors of a helicopter that wants to go up have to spin faster than the ones of a stationary one? But then shouldn't the upward force acting on the helicopter that goes up be larger than the one acting on the stationary one, and as they have the same weight, shouldn't the helicopter going up only be able to accelerate rather than move at a constant velocity?

The climbing helicopter sees air moving faster through its rotor, and accelerating air that is already moving is harder than air that moves slowly, because kinetic energy $E=\frac{1}{2}mv^2$ depends on $v^2$ rather than just $v$. So it has to expend more power, either by increasing speed, or angle of attack, or both.

After increasing the power, of course the helicopter accelerates upward. But as the climbing speed increases, the force decreases (as follows from $P=Fv$), until it only matches the helicopter's weight, and no more acceleration is possible. The climbing settles on a constant rate, until the power is changed again.

 

[rcgldr]

Ignoring issues that a climbing helicopter is operating with less of it's own induced wash (the induced wash from a hover takes some finite amount of time to setup into a steady flow), and ignoring the issue of drag related to the vertical velocity, then as already answered, the issue is the required energy is greater. Assume the increase in velocity for hover or stead climb is Δv. For the hover, assuming air's initial velocity is zero the increase in energy of the air = 1/2 m Δv^2. For a steady climb, the increase = 1/2 m (v₀+Δv)^2 - 1/2 m (v₀)^2 = m (v₀ Δv + 1/2 Δv^2). The climb versus hover difference is (m v₀ Δv) .

 

[russ_watters]

OK, I think I got it now, thank you both.

I watched the drone video; that was quite useful too.
So, if someone looks at this from outside, initially they will see the fly hovering at the same height compared to the 'ground'.
And it would continue to do so weren't it for the fact that the air in the elevator flows past it faster, and it probably sees that the floor is approaching, so it increases its power to go back to the same 'level' inside the elevator. At that point however it needn't keep the higher level of power, it will go back to the power it used before, I presume(?), as I believe @SlowThinker showed.
When the elevator decelerates and stops, I guess the force applied to the elevator is gradually decreased, so the sum becomes negative (if we consider positive a force pointing 'up'), and that's equivalent to applying a force that decreases the elevator's velocity. The air inside will also decelerate, the fly will feel it and see the ceiling approach, and decrease its power enough to avoid hitting the ceiling, but again, going back to the 'normal' level once the deceleration phase is over.

On the other hand, when the elevator and all its occupants are moving at a constant velocity, I suppose forces still apply, but their overall sum is zero (if we neglect attrition), because we only need to cancel out the weights of all the objects that are moving together.
If the sum of the forces applying to the elevator and its occupants were positive, shouldn't the system accelerate? And I would say this system doesn't, as its velocity is constant.

All of that is correct.

This actually makes me wonder what the difference would be between an object standing at constant height in a gravitational field, by cancelling its own weight somehow, and the same object moving up at constant velocity. Like the helicopter in an example made earlier.
Clearly the helicopter going up is doing more work than the one standing still, because its potential energy increases and its kinetic energy does not vary. What justifies this additional work, if it's true that the sum of the forces acting on the helicopter is zero in both cases? Is it the mass of air moved per unit time that changes? Do the rotors of a helicopter that wants to go up have to spin faster than the ones of a stationary one?

Yes.

But then shouldn't the upward force acting on the helicopter that goes up be larger than the one acting on the stationary one, and as they have the same weight, shouldn't the helicopter going up only be able to accelerate rather than move at a constant velocity?

No: the rotor has to spin faster to apply the same force because the air is moving past it faster.

There was that experiment, by Millikan I think, where oil droplets could be made to stand still in mid air by cancelling their weight by an electrostatic force.
Well, if you increased the electric field, wouldn't the drop accelerate up rather than move up at a constant velocity?

I'm not familiar with that experiment...

 

[jbriggs444]

In the Millikan oil drop experiment, one begins uses a mist of finely atomized oil drops. The mass of the drops is assessed by measuring their fall rate (terminal velocity) under gravity. A charge is then applied to the drops. The drops are small enough and the charge applied is small enough that the droplets will have a net charge that is a small integer multiple of an elementary charge. One then applies an upward electrical field with strength just enough to cancel gravity -- on some of the particles.

I've never performed the experiment myself, but I understand that one looks for the various field strengths where some fraction of the oil drops are suspended, motionless. One then infers the size of an elementary charge by looking for a periodicity in the field strengths that work -- that periodicity corresponds to a single elementary charge. You may or may not find an oil drop with a single excess electron.

 

[lavoisier]

OK, I see... very interesting, thanks.
I was trying to write a differential equation to check what effect applying a constant power would have on the helicopter's vertical velocity, but I reasoned that power and delta v can't be chosen independently, and I don't know how to relate them. The difference in kinetic energy does not appear to help.

So I found this article: https://en.wikipedia.org/wiki/Thrust but so far it raised more questions than it answered (sorry, I keep doing that).

Intuitively I think I got the concept, but I can't put it into figures. That's OK. I wasn't planning to quit the day job yet.

 

[SlowThinker]

So I found this article: https://en.wikipedia.org/wiki/Thrust but so far it raised more questions than it answered (sorry, I keep doing that).

The article is somewhat misleading, they use a formula for zero intake speed and then use is as correct, but obviously, if there is no air going in, the engine has no air to burn fuel. Same with helicopter, you can't have air at zero speed above the propeller, because then there couldn't be any air going down below it. (This doesn't apply to a rocket engine because air/gas is "created" inside).

I'll try to repeat my derivation, others are welcome to check it... I'm not an expert on this, just using common sense.
v...speed of air going into rotor
v+$\Delta$v...speed of air going out
S...area of the rotor
$\rho$...density of air going in
F...force (lift) generated
P...power used by the rotor
m...mass of air passing through the rotor during an arbitrary time T
a...acceleration of air
s...length that air spends inside propeller

$$m=\rho S v T$$
$$F=m a=m \Delta v/T=\rho SvT\Delta v/T=\rho Sv\Delta v$$
$$s=\frac{1}{2}aT^2+vT=\frac{1}{2}\frac{\Delta v}{T}T^2+vT=\frac{1}{2}\Delta v T+v T$$
$$P=F s/T=F(\frac{1}{2}\Delta v T+v T)/T=F(v+\frac{1}{2}\Delta v)$$
Speed $v$ that minimizes power:
$$P=F(v+\frac{1}{2}\Delta v)=F(v+\frac{1}{2}F/(\rho S v))$$
$$0=P'=1-\frac{1}{2}F/(\rho Sv^2)$$
$$2\rho Sv^2=F$$
$$v=\sqrt{\frac{F}{2\rho S}}$$
$$P=F(\sqrt{\frac{F}{2\rho S}}+\frac{1}{2}\frac{F}{\rho S \sqrt{\frac{F}{2\rho S}}})=F^{3/2}(\sqrt{\frac{1}{2\rho S}}+\frac{1}{2}\sqrt{\frac{2}{\rho S}})=\sqrt{2}\frac{F^{3/2}}{\sqrt{\rho S}}$$
My constant factor is $\sqrt{2}$ instead of Wiki's $1/2$, not sure if I made a mistake or what. But then, they are doing something different.

 

[rcgldr]

Same with helicopter, you can't have air at zero speed above the propeller, because then there couldn't be any air going down below it.

The idealized case is that you have air at zero speed at some point above the rotor. The air accelerates and decreases in pressure until it reaches the plane of the rotor, where there is a pressure jump (from below ambient to above ambient) and little change in speed. The air continues to accelerate until it's pressure returns to ambient, and it's velocity at that point is called the "exit" velocity. The velocity at the rotor would be 1/2 the exit velocity. Nasa has an article about this for propellers:

https://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

 

[SlowThinker]

Nasa has an article about this for propellers:
https://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

The good thing is that I seem to have the same result for F as they do (their $v_p$ is my $v+\Delta v/2$).
I find the explanations of wing or propeller based on pressure needlessly complicated. It's really just pushing air down or behind that generates lift or thrust.

 

[rcgldr]

I find the explanations of wing or propeller based on pressure needlessly complicated. It's really just pushing air down or behind that generates lift or thrust.

or "pulling" air down or behind from above or from in front ...