Fly in Elevator: Does Compression Affect Hovering?

[lavoisier]

Suppose that a fly flies into an elevator and hovers in mid-air. The doors close and the elevator goes up.
In analogy with a similar example (link), I imagine that the fly will travel up, it won't fall to the elevator floor.
And I guess this is because the 'air' in the elevator, which is what the fly pushes down with its wings to create an upward force counteracting its weight, is carried up as well.
However, if I stand in the elevator, the force that is counteracting my weight is the reaction of the elevator floor, which is equal to my weight at the beginning, and temporarily increases as the elevator accelerates.
This is felt as some additional pressure under my feet. If I am carrying a heavy sack of potatoes, for a moment it will 'feel' heavier.
Is this correct so far?
Now, if this is true, doesn't it imply that the air in the elevator will undergo some compression, i.e. for a brief moment there will be 'less' air near the roof of the elevator than near the floor?
And if so, doesn't that affect the fly's ability to hover at a certain level, i.e. won't we see it go down for a brief moment as the elevator accelerates?

 

[sophiecentaur]

The air is being pushed up the same as your feet. The effect of the rising floor would be like an enormous loudspeaker, with a very low frequency pulse on it. The speed of sound in the air would be so high compared with the elevator speed that the displacement could be considered to be the same throughout the lift (a tiny fraction of the wavelength of the 'pulse).

 

[russ_watters]

Suppose that a fly flies into an elevator and hovers in mid-air. The doors close and the elevator goes up.
In analogy with a similar example (link), I imagine that the fly will travel up, it won't fall to the elevator floor.

And I guess this is because the 'air' in the elevator, which is what the fly pushes down with its wings to create an upward force counteracting its weight, is carried up as well.

The fly isn't floating, it is flying. It flaps its wings to counteract its weight. What just happened to its weight?

Now, if this is true, doesn't it imply that the air in the elevator will undergo some compression, i.e. for a brief moment there will be 'less' air near the roof of the elevator than near the floor?

Yes, but this is negligible for the acceleration of an elevator.

 

[rcgldr]

Now, if this is true, doesn't it imply that the air in the elevator will undergo some compression, i.e. for a brief moment there will be 'less' air near the roof of the elevator than near the floor?

Treating the elevator as a sealed container, this already happens when the elevator is not accelerating, due to the force from gravity. The pressure of the air at the top of the elevator will be less than the pressure at the bottom of the elevator, and this pressure gradient results in a net downforce exerted by the air onto the elevator equal to the weight of the air in the elevator.

And if so, doesn't that affect the fly's ability to hover at a certain level, i.e. won't we see it go down for a brief moment as the elevator accelerates?

As posted by russ_waters, what happens to the fly's weight if it accelerates upwards to match the upwards acceleration of the elevator?

 

[lavoisier]

Thank you all for your answers!
Unfortunately I did not manage to convince any fly to travel with me in the elevator, so this still remains a theoretical question for me.
I think the point about the weight is clear: the fly flaps its wings, i.e. it pushes some air down, to create a corresponding force pointing up that serves to counteract its own weight. That allows it not to fall down but instead hover in mid-air. A bit like a helicopter would do.
What is not clear to me is whether there is a difference between me and the fly when the elevator starts going up.
Surely an upward force is applied to the elevator, to cause it to go up in the gravitational field and to go from rest to a constant speed. Some work must be done for the potential and kinetic energy of the whole thing to increase, right?
I may be considered part of the elevator when I'm standing in it, so the upward force is applied to me as well.
The fly, on the other hand, is not 'touching' any part of the elevator: it is only touching the air in the elevator. Is the air going to apply a force on the fly in the same way as the elevator floor applies a force to my feet?
I know this is pretty basic stuff - it's obvious that my physics is extremely rusty, and some concepts were never clear to me in the first place

 

[SlowThinker]

Let's imagine a drone. The air flows through the rotors quite quickly, certainly much faster than an elevator. I believe the same applies to a fly.
Thus the speed of air has little effect on the lift generated.
Even if the elevator moved very fast, the drone needs to generate power to increase its altitude and thus potential energy in the Earth's gravity field. It needs to work harder in a rising elevator than in a stationary one. The exact amount depends on its actual size, rotor speed etc.

 

[Drakkith]

The fly, on the other hand, is not 'touching' any part of the elevator: it is only touching the air in the elevator. Is the air going to apply a force on the fly in the same way as the elevator floor applies a force to my feet?

Yes it is. The same way that your torso isn't touching the elevator, but the force is transferred up through your feet and legs.

 

[russ_watters]

Thank you all for your answers!
Is the air going to apply a force on the fly in the same way as the elevator floor applies a force to my feet?

Not on its own, no - it's just air. The wings have to apply the force to the air, needed to keep the fly hovering.

 

[russ_watters]

Yes it is. The same way that your torso isn't touching the elevator, but the force is transferred up through your feet and legs.

I think that sounds more like the situation where you have a floating balloon instead of a flying fly...

 

[Drakkith]

I think that sounds more like the situation where you have a floating balloon instead of a flying fly...

How so?

 

[SlowThinker]

How so?

If the fly was massless, and it was pulling a massless string attached to the floor at a force equivalent to a fly's weight, then yes, in an elevator moving up, the massless fly would need to work just as hard as in a stationary elevator to maintain the pull.
But in the real scenario, the fly needs to pump some energy into its potential energy, so it must work harder.
A baloon is more similar to the massless scenario.

 

[Drakkith]

If the fly was massless, and it was pulling a massless string attached to the floor at a force equivalent to a fly's weight, then yes, in an elevator moving up, the massless fly would need to work just as hard as in a stationary elevator to maintain the pull.
But in the real scenario, the fly needs to pump some energy into its potential energy, so it must work harder.

The fly (and everything else in the elevator, including the air) is now moving upwards at a steady velocity. That means that the force exerted by the fly on the air is identical to the force exerted prior to the elevator accelerating up to speed. If the force exerted by the fly is identical to when the elevator was stationary, then no extra energy is being expended by the fly. The energy to move the elevator and all of its occupants upward against gravity is provided by the elevator's motor.

 

[SlowThinker]

The fly (and everything else in the elevator, including the air) is now moving upwards at a steady velocity. That means that the force exerted by the fly on the air is identical to the force exerted prior to the elevator accelerating up to speed. If the force exerted by the fly is identical to when the elevator was stationary, then no extra energy is being expended by the fly. The energy to move the elevator and all of its occupants upward against gravity is provided by the elevator's motor.

I'm afraid you're wrong. A helicopter that is rising at a steady pace generates as much force as a hovering helicopter (it is not accelerating upwards, only rising), but needs more power.
The fly in an elevator is a fairly tricky scenario and I guess it will remain unsolved until someone tries it with a drone, but I believe that my arguments are valid.

 

[rcgldr]

The fly (and everything else in the elevator, including the air) is now moving upwards at a steady velocity.

The original question asks what happens after the doors close and the elevator goes up. This would include an initial upwards acceleration of the elevator.

 

[Drakkith]

The original question asks what happens after the doors close and the elevator goes up. This would include an initial upwards acceleration of the elevator.

Yes I know. I was replying to SlowThinker's post, which I thought was about the non-accelerating portion of an elevator's ascent.

I'm afraid you're wrong. A helicopter that is rising at a steady pace generates as much force as a hovering helicopter (it is not accelerating upwards, only rising), but needs more power.

Of course. Just like the elevator motor needs to use power to perform work and raise the elevator and its occupants against gravity. Are you suggesting that the pilot of a helicopter, who is just sitting in a chair, is somehow performing work to raise himself against gravity?

The fly in an elevator is a fairly tricky scenario and I guess it will remain unsolved until someone tries it with a drone, but I believe that my arguments are valid.

I don't agree. I've ridden an elevator many times in my life. Once the initial acceleration is over, I've never had to exert more of an effort to remain standing than I had to when stationary. I just did an experiment in which I rode the elevator in the building I'm currently in from the ground floor to the top floor, a height of about 50 ft, in 12.8 seconds. That's roughly 9500 Joules spent over 12.8 seconds to get to the top floor, for a power of 742 Watts. 9.5 kJ is about 2.26 kCals, or 2.26 "food" calories. That's about the same energy required to run 120 feet.

I did not just run nearly half a football field worth of distance in 12.8 seconds. I didn't even break a sweat while standing in the elevator.

If you're thinking that the situation is different because the fly has to perform work on the air, then just think about the situations prior to and after the acceleration upwards. Both situations involve the fly hovering in air that is stationary with respect to the elevator. There is no acceleration of the general mass of air in either case. The forces exerted by the fly on the air are identical in both situations. If all of these things are the same, how could the fly be expending more energy in one situation than the other?

 

[SlowThinker]

Are you suggesting that the pilot of a helicopter, who is just sitting in a chair, is somehow performing work to raise himself against gravity?

No but he's not flapping his wings inside the helicopter either.
A fly sitting on the floor of the elevator does not perform any work, a flying one does.

I admit I'm starting to have doubts. Your arguments seem valid but I can't see a flaw in mine either.
I would be glad if someone could present sound reasoning either way.

 

[russ_watters]

How so?

When you say: "The same way that your torso isn't touching the elevator, but the force is transferred up through your feet and legs." it implies to me that the force the air provides to the fly changes on its own as the air accelerates with the elevator. Technically it does, but the amount is so small it only shows up as buoyancy, which can't keep a fly aloft. If the fly makes no adjustment, it accelerates toward the floor (in the elevator's reference frame).

 

[russ_watters]

If the fly was massless, and it was pulling a massless string attached to the floor at a force equivalent to a fly's weight, then yes, in an elevator moving up, the massless fly would need to work just as hard as in a stationary elevator to maintain the pull.

This doesn't make a lot of sense to me. String or not, the only scenario where the object doesn't change its elevation more than via a minor amount of pressure gradient in the air is if the weight equals the buoyancy: the balloon scenario.

A helicopter that is rising at a steady pace generates as much force as a hovering helicopter (it is not accelerating upwards, only rising), but needs more power.

That is because it is moving through the air as opposed to moving with the air.

The fly in an elevator is a fairly tricky scenario and I guess it will remain unsolved until someone tries it with a drone, but I believe that my arguments are valid.

It isn't a tricky scenario (and is a common and well documented question), but sure it would be fun if someone took video of a drone in an elevator. You can also try hovering a drone in a moving car . Balloons don't work as well, but if you have helium balloons they will move opposite the gravitational vector in the car (meaning of you turn right you will feel yourself pulled left, but the balloon will move right).

 

[russ_watters]

...someone tries it with a drone...

Ask and ye shall receive:

 

[Paul Woods]

You need to understand this: the fly will know only about its motion upwards due to the lift moving by its interaction with the air in the lift. The air 'knows' only about its motion upwards due to the lift moving by its interaction with the lift's floor. So only once the air has been affected by the lift's floor will the fly feel any effect, via the air's molecules' pressures on its wings. The rest follows from that, and in your case, the lift affects your feet without involving the air. (If you imagine a slow-time version of the situation, the lift moves before the air, which leaves the fly not moving until the message gets to it via the air.)

 

[russ_watters]

You need to understand this: the fly will know only about its motion upwards due to the lift moving by its interaction with the air in the lift. The air 'knows' only about its motion upwards due to the lift moving by its interaction with the lift's floor. So only once the air has been affected by the lift's floor will the fly feel any effect, via the air's molecules' pressures on its wings. The rest follows from that, and in your case, the lift affects your feet without involving the air. (If you imagine a slow-time version of the situation, the lift moves before the air, which leaves the fly not moving until the message gets to it via the air.)

Or if it can see where it is in the elevator. But yes, you are correct that initially all it might notice is that air is moving past it faster than it was before. It's probably very confusing!

 

[Paul Woods]

Or if it can see where it is in the elevator. But yes, you are correct that initially all it might notice is that air is moving past it faster than it was before. It's probably very confusing!

Gravity equals acceleration. There is no difference. The acceleration for the fly here comes from its wings' interaction with the air's molecules. Einstein was amazing for realising stuff like this. :D (I just watched the last episode of 'Genius' and I'm woefully sad that it focussed so much more on his personal life.)

 

[SlowThinker]

This doesn't make a lot of sense to me. String or not, the only scenario where the object doesn't change its elevation more than via a minor amount of pressure gradient in the air is if the weight equals the buoyancy: the balloon scenario.

Clearly the fly will move down during the acceleration, just as your legs have to work harder. I was trying to figure out what is happening during the non-accelerated phase.

That is because it is moving through the air as opposed to moving with the air.

That indeed seems to be the difference. The rising helicopter works harder because it climbs through the air, not because its altitude is changing. This is where my mistake begins.

Since nobody has presented a clear and obvious argument so far, I'll do it myself, indeed using good old Albert's principle.
The inside of an elevator accelerating at 1g is the same, whether it is moving or not relative to some external observer. Thus, the inside of a climbing elevator is indistinguishable from a standing one.

 

[Drakkith]

When you say: "The same way that your torso isn't touching the elevator, but the force is transferred up through your feet and legs." it implies to me that the force the air provides to the fly changes on its own as the air accelerates with the elevator. Technically it does, but the amount is so small it only shows up as buoyancy, which can't keep a fly aloft. If the fly makes no adjustment, it accelerates toward the floor (in the elevator's reference frame).

I suppose I meant that the force keeping the air at a steady velocity is provided by the floor, and that the force of the air on the fly is the force that keeps it hovering. Thus I linked the force applied by the floor to the force on the fly as being transferred through the air. But I admit this entire situation is a confusing mess of terminology, context, and reference frames.

 

[Paul Woods]

Clearly the fly will move down during the acceleration, just as your legs have to work harder. I was trying to figure out what is happening during the non-accelerated phase.
That indeed seems to be the difference. The rising helicopter works harder because it climbs through the air, not because its altitude is changing. This is where my mistake begins.

Since nobody has presented a clear and obvious argument so far, I'll do it myself, indeed using good old Albert's principle.
The inside of an elevator accelerating at 1g is the same, whether it is moving or not relative to some external observer. Thus, the inside of a climbing elevator is indistinguishable from a standing one.

However, note that a lift accelerating constantly at 1g due to its movement caused by the cables *in the presence of Earth's gravitational field (of 1g)* will have an interior that seems as though it is either: accelerating in empty, distant space at 2g or; sitting still in a gravitational field of 2g. Those are both the same.

 

[SlowThinker]

However, note that a lift accelerating constantly at 1g due to its movement caused by the cables *in the presence of Earth's gravitational field (of 1g)* will have an interior that seems as though it is either: accelerating in empty, distant space at 2g or; sitting still in a gravitational field of 2g. Those are both the same.

I don't see how anyone could have doubts that the fly falls down during the acceleration and has to work harder to accelerate with the elevator. Your legs work harder during that time as well.
The only doubt I ever had was about the steady phase.

 

[lavoisier]

Thank you all for your replies!
So if I understand correctly, it's all down to inertia.
When the elevator accelerates, all objects that are inside it (including the air) will try to remain where they are, as seen from an external observer, and a force will have to apply to all of them to get them from 'still' to 'moving upwards at a constant speed' with the elevator.
But isn't the elevator doing the work needed to get the whole system up in the gravitational field? So why are the fly and I supposed to do some work too? Shouldn't the air push the fly up in the same way as the elevator floor pushes me up? And if there's a box on the elevator floor, is the box doing work too? How would that be possible?
Then when the elevator decelerates and eventually stops, the objects will want to continue to move upwards at a constant velocity. I don't feel that very much: I'm just slightly 'lighter' for a second; I guess a fraction of my body consumes my kinetic energy by going slightly up, then down again due to gravity, and the additional energy is then dissipated somehow. Surely though, if the elevator is going very fast and stops quite abruptly, I will probably jump up and be smashed against the ceiling, which is a different way to use that energy.
But what about the fly? Once it gets to a constant velocity, say 5 m/s, just hovering, not trying to fly up or down, and the air is just traveling with it, not pushing it either up or down, what impact will the elevator stopping have on it? Suppose that the fly is blindfolded (sorry about that) and doesn't know that the elevator is stopping - why isn't it continuing to go up? Is the air inside the elevator pushing down on it?

I like Paul's example of 2g vs 1g. This seems to imply that if the elevator is completely opaque, i.e. I can't see outside, I will have the same 'experience' if the elevator is actually going up OR if it stays still and a remarkably massive body suddenly appears below it.
In this context I understand the need for additional work by myself and by the fly: our weight doubles all of a sudden, so all the mechanisms that our bodies use to maintain our positions will have to work harder. Especially the fly, because the force that keeps it hovering is proportional to the mass of air it pushes down per unit time, I guess.
But when the massive body disappears again I won't be lighter for a second or jump up, will I, because I wasn't moving in the first place...?

Sorry to bring up so many doubts; there's clearly some fundamental concept that still escapes me...

 

[SlowThinker]

There are 3 different things going on during acceleration:
1. A box on the floor feels increased force, but is not doing any work. The floor performs work on the box.
2. A human standing on the floor has to expend more energy, because muscles aren't 100% efficient. You need energy just to stand, while the Eiffel tower doesn't, and that's the physically relevant example.
3. The fly is accelerating air above and pushes it down. I derived the following formula myself so I'm not sure if it's correct but anyway, if there were no turbulence and other losses etc., then power $P$ expended by the fly (or any helicopter or airplane) is approximately $P=F v$, where $F$ is the force generated, and $v$ is the speed with which air goes down relative to the fly (average of above and below the fly). So if the air starts to go up ($v$ decreases), and the fly uses as much power as before, the force increases, and the fly starts to accelerate up, until the relative air speed returns to the previous value, where the force matches weight.
So the blindfolded fly doesn't move with the rest of the elevator as a rigid body, but it will eventually move with it.

 

[russ_watters]

Thank you all for your replies!
But isn't the elevator doing the work needed to get the whole system up in the gravitational field?

Yes.

So why are the fly and I supposed to do some work too? Shouldn't the air push the fly up in the same way as the elevator floor pushes me up? And if there's a box on the elevator floor, is the box doing work too? How would that be possible?

They do. All of these are subsets or subdivisions of the total work done on the elevator. Dividing the system up into pieces doesn't change what is going on for the system as a whole.

Consider if you had 3 books stacked on your hand, each weighing 1N. The total force on your hand is 3N, the force the top book exerts on the second is 1N and the force the second exerts on the 3rd is 2N. Is that a total of 6N? No, it's a total of 3.

But what about the fly?

The fly really isn't any different than a person, you're just letting the fact that it is using its wings to hover confuse you. Treat it the same as a person standing still and the answers should come easier.

Once it gets to a constant velocity, say 5 m/s, just hovering, not trying to fly up or down, and the air is just traveling with it, not pushing it either up or down, what impact will the elevator stopping have on it?

The same as the impact on the person.

Suppose that the fly is blindfolded (sorry about that) and doesn't know that the elevator is stopping - why isn't it continuing to go up?

It would, until it noticed the wind and realized it was no longer "stationary" with respect to the elevator.
[quote Is the air inside the elevator pushing down on it? [/quote]
No, like the person, the air is pushing the fly up. Always.

But when the massive body disappears again I won't be lighter for a second or jump up, will I, because I wasn't moving in the first place...?

Right. The massive object would have to briefly re-appear above you for it to be equivalent to the elevator.

 

[lavoisier]

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.
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? 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?

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?

Maybe I will try to do some further reading on this...

 

[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.