Sorry for the forum/blog downtime today. Many things went wrong during davean’s heroic upgrade. (I blame the LHC.)
Feynman used to tell a story about a simple lawn-sprinkler physics problem. The nifty thing about the problem was that the answer was immediately obvious, but to some people it was immediately obvious one way and to some it was immediately obvious the other. (For the record, the answer to Feynman problem, which he never tells you in his book, was that the sprinkler doesn’t move at all. Moreover, he only brought it up to start an argument to act as a diversion while he seduced your mother in the other room.)
The airplane/treadmill problem is similar. It contains a basic ambiguity, and people resolve it one of a couple different ways. The tricky thing is, each group thinks the other is making a very simple physics mistake. So you get two groups each condescendingly explaining basic physics and math to the other. This is why, for example, the airplane/treadmill problem is a banned topic on the xkcd forums (along with argument about whether 0.999… = 1).
The problem is as follows:
Imagine a 747 is sitting on a conveyor belt, as wide and long as a runway. The conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?
The practical answer is “yes”. A 747’s engines produce a quarter of a million pounds of thrust. That is, each engine is powerful enough to launch a brachiosaurus straight up (see diagram). With that kind of force, no matter what’s happening to the treadmill and wheels, the plane is going to move forward and take off.
But there’s a problem. Let’s take a look at the statement “The conveyor belt is designed to exactly match the speed of the wheels”. What does that mean?
Well, as I see it, there are three possible interpretations. Let’s consider each one based on this diagram:
1. vB=vC: The belt always moves at the same speed as the bottom of the wheel. This is always true if the wheels aren’t sliding, and could simply describe a treadmill with no motor. I haven’t seen many people subscribe to this interpretation.
2. vC=vW: That is, if the axle is moving forward (relative to the ground, not the treadmill) at 5 m/s, the treadmill moves backward at 5 m/s. This is physically plausible. All it means is that the wheels will spin twice as fast as normal, but that won’t stop the plane from taking off. People who subscribe to this interpretation tend to assume the people who disagree with them think airplanes are powered by their wheels.
3. vC=vW+vB: What if we hook up a speedometer to the wheel, and make the treadmill spin backward as fast as the speedometer says the plane is going forward? Then the “speedometer speed” would be vW+vB — the relative speed of the wheel over the treadmill. This is, for example, how a car-on–a-treadmill setup would work. This is the assumption that most of the ‘stationary plane’ people subscribe to. The problem with this is that it’s an ill-defined system. For non-slip tires, vB=vC. So vC=vW+vC. If we make vW positive, there is no value vC can take to make the equation true. (For those stubbornly clinging to vestiges of reality, in a system where the treadmill responds via a PID controller, the result would be the treadmill quickly spinning up to infinity.) So, in this system, the plane cannot have a nonzero speed. (We’ll call this the “JetBlue” scenario.)
But if we push with the engines, what happens? The terms of the problem tell us that the plane cannot have a nonzero speed, but there’s no physical mechanism that would plausibly make this happen. The treadmill could spin the wheels, but the acceleration would destroy them before it stopped the plane. The problem is basically asking “what happens if you take a plane that can’t move and move it?” It might intrigue literary critics, but it’s a poor physics question.
So, people who go with interpretation #3 notice immediately that the plane cannot move and keep trying to condescendingly explain to the #2 crowd that nothing they say changes the basic facts of the problem. The #2 crowd is busy explaining to the #3 crowd that planes aren’t driven by their wheels. Of course, this being the internet, there’s also a #4 crowd loudly arguing that even if the plane was able to move, it couldn’t have been what hit the Pentagon.
All in all, it’s a lovely recipe for an internet argument, and it’s been had too many times. So let’s see if we can avoid that. I suggest posting stories about something that happened to you recently, and post nice things about other peoples’ stories. If you’re desperate to tell me that I’m wrong on the internet, don’t bother. I’ve snuck onto the plane into first class with the #5 crowd and we’re busy finding out how many cocktails they’ll serve while we’re waiting for the treadmill to start. God help us if, after the fourth round of drinks, someone brings up the two envelopes paradox.
xkcd 
Don’t confuse ground speed and air speed…
Liftoff can be as soon as 200kmh.
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Not one argument based on Bernoulli’s principle. Shame.
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I understand everything you wrote, but I think you dismissed my position too easily.
Review the question: “…the conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction…” – that means that *by definition* the speed of the belt instantaneously matches the speed of the wheels.
Of the three definitions of belt speed you give, I am firmly in the vC=vW+vB camp. Regarding this definition, you write:
“The problem with this is that it’s an ill-defined system. For non-slip tires, vB=vC. So vC=vW+vC. If we make vW positive, there is no value vC can take to make the equation true.”
I disagree that this is an ill-defined system. It is a precisely defined system: vC=vW+vB period. I do agree that as long as the wheels do not slip that vB=vC, so therefore vC=vW+vC.
Your statement “If we make vW positive, there is no value vC can take to make the equation true” misses the point completely. Given the equation vC=vW+vC then vW can only be zero. You cannot make vW positive.
Would you ever write “given the equation 4=vW+4, if we make vW positive, there is no value 4 can take to make the equation true”? Of course not – you simply wouldn’t assume that vW could ever be positive.
Believe me – I understand the problem, and I understand everything that both sides have written about it. But as I read the problem, *by definition*, the plane has zero velocity. Zero velocity means it does not take off.
Of course, that just raises the thorny issues of the nature of friction in the bearings of the wheel, and how you can have a treadmill of infinite velocity. But those issues are external to the real question. The real question boils down to this: Can an airplane that by definition has zero foward velocity take off?
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(Admittedly tongue-in-cheek, not taking the problem seriously):
The speed the treadmill will eventually reach in order to maintain the same backward speed as a forward-speed engine will break the treadmill, and the loose belt will go galloping across the countryside at hundreds of miles per hour.
In the first second, the shockwave of breaking will cause said belt to ripple; that and the sudden pressure of the jet engine thrusting against the mass of air rushing towards it as the belt gallops backwards will cause the jet to perform an involuntary backflip.
So it will obviously get into the air, but I definitely wouldn’t want to be the one flying it when it did.
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This is one of those things designed to let n00bs and tr0lls drive rationally-minded people insane with their un(?)intentional ass-hattery. If you’re smart enough to figure out the correct answer, instead of trying to beat it into the heads of people with not enough headroom to encompass such stuff, just loftily shake your head in pity at them. They’ll never get it. Bonus points if you can correctly figure out which side of the understanding division you’re on.
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Not one argument based on Bernoulli’s principle. Shame.
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“Imagine a 747 is sitting on a conveyor belt, as wide and long as a runway. The conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?”
With all due respect to the Myth Busters, the problem stipulates that the wheels move the same speed as the conveyor belt.
In their experiment the wheels must be moving faster than the conveyor belt. In fact the speed of the conveyor plus the forward speed of the plane. So I don’t think they met the conditions of the problem.
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To those of you subscribe to the 3rd interpretation
The question does not stipulate that the tires will not slip thus VsubB does not necessarily = VsubC.
And if you want to add stipulations, then let me add one too: there is a strong gust of wind that blows across the wings providing sufficient lift to the aircraft that it breaks contact with the treadmill, and can take off.
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Besides if you strictly follow interpretation 3 then either the treadmill and the wheels are at rest, or they can speed up to infinity but the airplane doesn’t move. The first seems to be a physical impossibility and hence not in the domain of physics, perhaps philosophy but only a time wasting tangent. The second possibility is equally unpractical plus if it were possible, then as the speed continues to increase the treadmill would actually produce enough air current across the wings to produce lift in the aircraft that has 0 velocity relative to the ground.
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I believe that the problem is stated in a poor way. I would rather see this part of the problem defined as:
“The conveyor belt is designed to exactly match the AIRPLANE’S GROUND SPEED, moving in the opposite direction.”
Therefore the wheels, when moving, would move at twice the plane’s ground speed, and the additional friction (and power required to accelerate the spin of the wheels) would not be enough to prevent the plane from taking off.
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I must not be spending enough time online; I’m not aware of this controversy!
It sounds like the puzzles is describing an airplane which, from the viewpoint of a stationary observer, isn’t moving, since the treadmill cancels out any positional change relative to the countryside. But, from the viewpoint of someone aboard the plane, it is moving, expending fuel, producing thrust, and covering many meters on the treadmill “runway.” So, the puzzle asks, does the airplane take off or not?
I must be missing the point of contention, because the answer seems pretty straightforward: No, it won’t take off. Takeoff speed is relative to the airflow under the wings, and not the turning of the wheels or the power level of the engines. Engines can drive an airplane forward, and wheels turn because they have to accommodate the change in the airplane’s physical position, but, in a 747, it’s the air under the wings that produces the actual lift. In the scenario described, the airspeed under the wings is (virtually) zero. Ergo, no takeoff.
We use the same principle for wind tunnel testing. By introducing lift under the wings through a turbine, an airplane will behave as if it were flying even though the wheels are not turning, and even if the engines are producing low or nil thrust.
If this is really such a big controversy, then there must be something obvious that I’m missing, but I can’t think of what it is. Clearly, therefore, everybody who doesn’t agree with me is a moran. =)
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Look at the earth as if it were a huge treadmill, rotating at aboutish 1000 mph on it’s axis. Obviously we are able to fly normally at this rate and have been for a hot minute. Now suppose someone were to speed up the rotation of the earth to 2000 mph. Would a plane taxiing/trying to take off to the west be able to fly?
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I’m with Josh on this one. Unless the treadmill is somehow dragging all the air in the room along with it (a wind tunnel maybe?), the plane won’t be going anywhere.
Frankly, I don’t even understand the premise of the debate otherwise.
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In my opinion, this plane will not take off!!!!
That’s because my interpretation of this problem, is the plane is at 0mph RELATIVE TO THE WIND.
The way i see this problem, is there is NO DIFFERENCE between a plane on a treadmill in a “stationary position” (relative to the wind, but moving on the treadmill) and a plane in “stationary position” on the ground. Can anyone else agree/disagree with me?
Imagine a plane hooked up to the ground with engines at 110%. Cut the cables/ropes and will it take off? I doubt it.
The same idea applies to the conveyor belt. That’s my opinion.
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but what if the plane had been swallowed by a goddamn snake that then ate its tail? Assuming that the snake is large enough that the force of gravity around its inner diameter is equal to that at the surface of Earth, and that the rate of rotation of the snake exactly matches the speed of the wheels, moving in the opposite direction. And that the snake’s digestive system is for some reason full of an earth-like atmosphere.
Okay it’s still basically the same question. But what if Samuel L Jackson was also in first-class?
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“Can an airplane that by definition has zero foward velocity take off?” forward velocity relative to the ground isn’t the thing that makes planes take off. Jet engines suck the air through the turbines resulting in the propulsion necessary to create lift.
The more interesting issue would be to actually creating this system since it would obviously be a far more efficient way of having planes take off (and also land – the space occupied by the conveyor belt would be at most half of a classic runway).
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I have a red pencil box
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A no flyer calculated this:
A Cessna 182 has a 230 horsepower engine. It also has a rolling resistance of about 50 pounds. Assuming an ideal prop, you would have to run the treadmill backward at a speed of 1,725 mph to keep it stationary at full power.
Which is an excellent tangible proof, but for the wrong camp.
A Cessna 182 will take off at about 70mph. So basically, the plane will have lifted off the conveyor belt long before the rolling friction force is high enough to hold it back at full engine power. At this point, the conveyor belt will be travelling at 70mph in the reverse direction.
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How can plane start without air flowing under the wings?
You are very good in maths, but you forgot the basic principles of flight
If the “take off speed” of the plane is 150 knots, it does’t mean, that wheels should reach that speed. It means, that AIR PASSING the plane must reach 150 knots
If we put plane on the belt an run – the speed of the air passing wings = zero, therefore plane will not take off.
If we do it in the air tunel (plane on the belt), and blow the air towards the plane (150 knots), plane will take off
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After few minutes I realised how stupid I was. Of course iw will take off!!! It will reach the take-off speed despite of the fact that beltis running in oposite direction
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can the airplane take off if the wheel brakes are constantly engaged?(assuming that there is no brake slippage)
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Thank you Randall for posting this. There is nothing that needs to be added to the original post in terms of the original problem. If anyone still feels the need to argue strongly for one side of the other I suggest you just reread the post and eventually that feeling should pass.
– MC I cannot say I am familiar with how planes tend to brake, but if the wheel was held so that it could not rotate, and there was no slippage between the wheels and whatever surface they are resting on, the plane would not move, though you would need some pretty ridiculous friction. Unless of course it was resting on a treadmill of infinite length that just so happenned to travel forwards at a speed exceeding the takeoff speed of the plane…..
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leblebi : Sounds like you need to read it again :).
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Im scared of flying its the treadmill for me haha
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im scared of flying its the treadmill for me im afraid lol
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what is important for the take-off of a plane is the build up of air pressure under it’s wings, and not the wheels….. hence, the plane’s engines will produce thrust, but there will be no lift built up below the plane’s wings…… so technically it shouldn’t be able to take off unless it slips off the treadmill…..
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Air speed is not tied to ground speed. And engines cause air flow over the wings even when the aircraft is not moving relative to the ground.
If you are endowing the treadmill with the magical ability to exert infinite thrust to infinite speed against infinitely durable wheels, to arrest the aircraft by the angular momentum of the wheels, regardless of any other factor, it’s fair enough to endow the engines with the magical ability to produce indefinite thrust, and thus be able to cause enough air flow to lift the aircraft from the treadmill.
If prohibit this, and eventually flat-out prohibit anything that would enable the aircraft to lift off, you’ve transformed the question into: If an aircraft is prevented from lifting off, can it lift off.
The real problem here, is that the original question provides insufficient data for a meaningful answer.
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A plane flies because of air flowing under its wings.
A 747’s engines cannot make the wall of air in front of said 747 come rushing towards it. They rather push the 747 into the wall of air in front of it. That, in turn, makes the air flow under the wings of the 747.
Premise 1: In order for a 747 to lift off, it must move forward.
Premise 2: Imagine a 747 in some contraption that prevents it from moving forward.
Question: will it lift off?
Honestly, people. You need to debate that?
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Imagine the plane is instead a helicopter.
let’s say that there is an assembly coming out of the cockpit parallel to the ground.
At the end of the assembly is a pair of wheels.
those wheels are up against a wall
the rotors are not spinning
If you engage the rotors to full power, will the helicopter take off?
Of Course
what if, instead, the wall is a vertical treadmill running at full speed, let’s say 100mph. Anyone who has ever seen a helicopter take off knows that it never rises at 100mph, so you already know the speed is more than enough.
Turn on the rotors
Does it take off?
OF COURSE
same problem, different axis.
THE PLANE WILL TAKE OFF!!
will the helicopter take off?
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The plane takes off. Simple. That treadmill cannot counteract the thousands of pounds of thrust from the planes engine(s).
If you have treadmill at home, you can do a small scale experiment yourself. Tie some weights to the front of a toy car. Turn your treadmill up to its max speed, and put the car on it with the weight hanging off the front end of the treadmill. Notice when you drop the weight, the car moves forward.
IT MOVES FORWARD!
Thus, it takes flight (if it was full scale).
I am an Aerospace Engineer btw.
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Peter, given case 3, if the treadmill does not counteract the engine thrust, it violates the problem specification in the first place, and only then the plane takes off, or not, one should never bother.
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There is a great website that explains this:
http://www.airplaneonatreadmill.com/
I highly recommend it.
It does not necessarily explain my thoughts on this though.
Imagine you stand in front of a 800,000 lb 747.
It starts up and runs full out and you try and hold it back. You would have to exert incredible force to hold it back.
You could cement the wheels in the ground and it would just rip them off.
You could try and hold it back with ropes, etc., and it would break them.
A FORCE against the plane is required to stop the plane in it tracks.
How much Force does a treadmill supply to stop the plane? None, really.
A person on a pair of skates will be held in place on a treadmill with almost no force.
No matter how much you speed up the treadmill the force to hold someone on the treadmill will be the same (basically nothing once you have overcome the initial inertia holding the person to the ground).
So essentially I could speed up a treadmill to 1000 mph and I would still be able to hold the person on the skateboard in place with basically on effort.
In this case the person is not even providing any forward momentum of their own (like a plane that is providing forward force).
So now ANY amount of forward force will move the person on the skateboard forward.
Even the slightest forward acceleration will accelerate the person on the skateboard (i.e. they have a leaf blower attached to their back end).
So no matter how fast the treadmill, the slightest attempt at acceleration will move the plane forward.
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Tommy, again, if the plain moves forward, the condition vC=vW+vB assumed in #3 is not met, so a plane moving forward is not a solution to #3.
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If you take a small ring-shaped object and push it along a flat surface, it will roll forward. Let’s say that the ring went at a speed of 1 m/s. Now let’s take that onto a small conveyor belt which is moving at a speed of 1
m/s in the opposite direction to the ring. Now if the ring is moving forward on an unmoving surface, it rolls forward fine. If this surface is moving backward (the conveyor belt) at the same speed as the ring is moving forward, then it is being carried backward at the same rate as it is moving forward, so it remains stationary.
Take this onto full scale, with the plane. The same principle applies – the plane is being carried backward by the surface it rests on at the same rate it is moving forward, so it doesn’t move.
The main dispute seems to be about the friction between the wheels and the conveyor belt, which is unspecified in the problem. If the tyres slip, there is less grip for the belt to carry the plane on. And this is when people start bringing in the origin of the thrust; thrust comes from the engines, not the wheels. And this is true. But planes are very, very heavy, and all this weight is resting on this conveyor belt. This creates more downforce, and thus more grip, more friction, so the tyres don’t slip. So for the plane to take off, it needs to be already in the air – to take off, it needs to take off. As long as the planes weight remains on the belt, it will never pickup enough speed to take off and remove this friction on the belt so that it can take off.
From my understanding of it, the simple answer is no. The plane is being carried backward at the same speed as it is traveling forward, so it remains stationary, so it cannot take off. Enlighten me to anything I have missed.
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Caerus: Yes, it does violate case 3. But case 3 has no basis in reality after all.
#3 assumes that vB = vC. But the very crux of the problem is that this is not true in general. Rather, vB = vC + vW (in the ideal case where there is no friction between the axes and the rest of the plane). Put in words, no matter how much you increase vC, it will only result in an equal increase of vB, while vW will remain unaffected. As vW is the speed that matters for take-off, the plane will do this despite the belt.
Arden: it’s not the friction between wheels and belt that matters, but the one between axis and bearing. This will of course not be zero in reality, but it will not be enough to stop the plane from moving. I.e., the plane is not being carried back at the same speed as it is traveling forward, only its wheels will spin backwards at a much faster speed than the plane is moving forward.
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Ulrich, indeed it’s not how much you increase vC that matters, it is _how fast_ you increase it, beacause of the wheels’ moment of inertia. sriracha had a perfect point on it earlier in the comments.
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Just want to add:
Everybody who says the plane wont move seems to have the idea that the wheels CAN NOT move faster then the belt gears if gears and wheels have same size (wheels and belt gears have same RPM), even though the wheels have CLOSE TO ZERO FRICTION TO IT’S AXIS on the plane’s “landing stuff” (I don’t know what it’s called, english isn’t my main language).
NO MATTER HOW FAST the belt goes, IF YOU SPIN IT UP while the plane is standing still, you will JUST BARELY move the plane. Now, when it’s been spun up – add the INCREDIBLY LARGE FORCE from the engines that are designed to lift TONS of metals into the air – will the plane really not lift?
It’s just crazy to assume that making the wheels spin “backwards” with the belt would somehow neutralize the force from the engine.
To say “oh but the wheels are not allowed to go faster then the belt” is crazy. Like said before, you can just as well ask “if the plane can’t move, will it move?” No, since you have a physical impossibility where most of the engine force vanishes into thin air. Literally.
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Huh, I guess the discussion of DDWFTTW (an entirely different problem) means that this is still ‘a thing’.
This is a horribly specified problem. I immediately found myself interpreting the problem as in camp #3, with the conclusion that of course the plane will take off.
The plane will take off because the only purpose of the landing gear is to reduce friction with the ground during takeoff and landing. Thrust gives rise to airspeed, airspeed enables flight.
The failure in the specification is that “in the opposite direction” doesn’t have a frame of reference. If that frame is the airfield, then as soon as the airplane starts to move, there’s no solution to vC=vW+vB. This is is the ‘treadmill goes to infinite speed’ problem.
So, “in the opposite direction” must be in the frame of the aircraft. Since vC=vW+vB when all are zero, the only solution to the problem as stated is that the treadmill would move *with* the plane to keep the wheels from turning at all. The plane takes off because the landing gear isn’t part of the system that makes it fly. The treadmill just saves wear and tear on the landing gear. 🙂
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You got Feynman wrong. The sprinkler does rotate exactly as it should (it does not stay still), which is slowly forward into the fluid.
Here is a nice demo with air: http://www.youtube.com/watch?v=xQrkiH0U1Yg
Here is a write up in Wiki: http://en.wikipedia.org/wiki/Feynman_sprinkler
As for the aeroplane, it would indeed standstill — momentarily. The treadmill would exert a force on the wheels equal and opposite to that of the plane’s thrust. This becomes a Torque = I alpha problem, with the torque accelerating the wheels. In a few seconds (I estimate 17), the conveyour belt and the wheels would need to be traveling at escape velocity. Actually, the belt would get torn up by the centripetal forces that occur at the rollers.
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If the tangential velocity of the wheels is met instantaneously with the surface of the treadmill, then it will not take off.
Obviously this is hypothetical because the treadmill/wheels tangential velocity will approach infinity relatively quickly.
This is the ONLY answer if the question stipulates that “the tangential velocity of the wheel instantaneously matches that of the treadmill.”
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Sorry, forgot to mention that the wheels may not slip either. (otherwise imagine the airplane on a frictionless sheet of ice…it will take off)
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Hah, people are still talking here.
The wheels don’t matter. The plane could have pontoons, or skis, or rubber feet on a linoleum runway. They’re only there to reduce friction between the plane and the ground, or water, or snow.. whatever.
If the treadmill is spinning 5 million mph one way, and the plane is producing the thrust it normally needs to move at its take-off speed of 100 mph the other way, the wheels will be spinning at 5.0001 million mph and the plane will take off.
Or the landing gear will fail and the plane will nosedive onto the ground.
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This is simple. If you generalise, it quickly becomes obvious:
The treadmill spins backwards at an undefined, fast speed. This causes an amount of friction in the wheels that does NOT increase with velocity. In a perfect frictionless wheel, the plane would stand still with no engines. In the real world, a tiny amount of the power of the engines will be required to hold this, regardless of how quickly the wheels spin. Additionally, the wheels will quickly begin to slide from the ridiculous forces acting on them.
As soon as the engines are turned up even the slightest bit above that threshold, the plane will begin to accelerate forwards. Additionally, the movement of the treadmill will have no effect on the distance required to make the plane take off, in either way (unless it was exceptionally high friction, in which case it might increase it very slightly.
In summary:
NO, it is not possible to prevent a plane from taking off with a treadmill underneath it.
YES, it is possible for the pilot to work together with the treadmill operator to hold it still by keeping the engines turned all the way down (admittedly, this may be below the levels that the engine can maintain on some engines).
NO, it is not possible to take off a plane in a shorter distance by using this method (although similar things do work, such as attaching a clamp directly to the plain and pulling it forwards, a la aircraft carriers).
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A plane flies because of air flowing under its wings.
A 747’s engines cannot make the wall of air in front of said 747 come rushing towards it. They rather push the 747 into the wall of air in front of it. That, in turn, makes the air flow under the wings of the 747.
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Similar problem:
A car is driving on the interstate (let’s say at 70 mph). It approaches a car transport trailer truck with its ramp down (let’s say it’s driving 65 mph). What does the driver of the car need to do in order to drive onto the transport truck? Is it possible?
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First of all: You are my God and you Sir, are great.
Second, you are wrong about Feynman’s sprinkler, although barely measurable.
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This obviously all comes down to interpretation how you read the problem. I read it how situation 3 is stated. It doesnt matter that that forward force is due to the plane engines because the convey belt needs to apply an opposite but equal force on the wheels. Unless your assuming the wheels are frictionless and weightless then the conveyor belt cannot apply this force. But that’s never a case and the combined friction at the pin and the angular acceleration of a body with mass will cause 1 quarter of a million pounds of thrust therefore making the plane do nothing but want to spin and maybe nosedive into the ground.
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