The thrust isn't directly coupled, but if there can't be any slip between the wheels and the treadmill, how does an object move forward relative to a given point? Any movement forward would represent an increase in speed, which would then be counteracted.

Again, the whole thing is moot becuase it's basaed on a logical fallacy; wheelspeed is irrelevant to airspeed, true (like taking off in a headwind), but the way the question is phrased implies an active counteraction of the forward motive force.

It's technically the same as having a jet dragster on the treadmill, no motive force applied to the wheels, right? The thrust starts moving the thing forward, then the treadmill counteracts by moving backward the same amount (again, assuming an active treadmill)...........the wheels on the treadmill don't decelerate the object in terms of wheelspeed, but they do effectively move it back in space.

Think of the speed link on the treadmill as being like a toothed gear, no skipping allowed......that notch on the gear is going hold the object on the treadmill in a given place, andthe the counteracting motive force (thrust) moves it forward, turning the gear, which now enages the next notch, is physically moved backwards. counteracting any gain of airspeed.


Again, with a non-active treadmill setup, the object would move forward as there doesn't have to be any motive force applied through the wheels to move the object. No question there, and I could even see where the 2x wheelspeed hypothesis comes from, though it does seem like an arbitrary number (it could be late and I'm just not processing right)


Anyways, I'll chalk it up to a question worded as to leave some gaps to be employed to engage in discussions by geeky sorts like us, and acknowledge that the Occam's Razor solution, ignoring hypotheticals, would be that the thrust against the air would move the object independient of groundspeed, and ignore friction coffecients and mechanical drag and counteracting movement because I'll hurt my brain if I go through the stuff again; I graduated college for a reason, I wanted to be done with hypotheticals like this, lol.......

 

Because how does it get there?

If it's matched perfectly at all times, that leaves no room for acceleration. Even in a tightly matched gearset, changes in velocity occur becuase a gear occupies more of the mesh space ahead of its rotational vector for acceleration, and behind the vector for deceleration.

In other words, if the wheels are at 0, and the treadmill's at 0, neither can ever move if they're perfectly matched, becuase as soon as the plane generates thrust and starts moving through the air, the wheels turn that amount, the treadmill turns the same amount int he oppsosite direction......the change in physical location is 0.

It can never move forward, and the wheels can never turn, becuase there is no point at which you could move one element if the other one is matched to it perfectly. Think of pushing two blocks together, with equal force on each side........same theroy. If it's the same force, it's not going to move form that center point, and if those two forces (in this case wheelspeeds) are defined as always being idenctical, there could never be a change. Like Nick said, any change in wheelspeed automatically goes to infinity, becuase there is no velocity other than infinity you could accelerate to given the wording of the problem.

Again, this is impossible in a real life situation, as everyting is a reaction to an action, not a counteraction that occurs perfectly simultaneously.

In real life, the plane moves forward, becuase the treadmill can't respond perfectly when ther'es no motive force applied to it to turn it. The 2x speed isn't valid, becuase the wheelspeed would effectively be infinity.

also, Nick forgot about Tinkerbell and the Tooth Fairy.

 

it will fly.

ive seen this on a few forums, and the general consesus on all of them was that it will fly.

 

One by one or all at the same time?

 

The Earth is flat, too, I've heard a bunch of people say that

If you ignore the logical fallacy inherent in the question, the plane takes off, becuase I see what the question is implying: does someone understand the simple fact that groundspeed is irrelevant in terms of a thrust source that's not pushing agaisnt the ground. That's clear.

The problem is that the way the question is stated, the plane might as well have its wheels bolted to the ground with no treadmill involved, because it can never accelerate past 0 mph.

 

I don't care, I'll take the smart blonde

 

sybir, I'm not sure where you're coming up with this BS, but it's certainly interesting, I'll give you that. Let's try and clear up some of it.



This is simply not correct. Assume the plane is stationary and the engines have just been turned on. Let's look at an instantaneous force balance in the x-direction of the plane. There is a large positive thrust force from the engine, and a small inertial load from that of the wheels opposing their instantaneous angular acceleration. At this instant there is a net positive force forward on the plane, so it's acceleration is positive.

Now let's look at an instantaneous instant after time 0 (time 0 + dt). The plane will have some infinitesimally small positive velocity relative to an inertial frame (stationary earth), and the magical treadmill underneath will have the same infinitesimal velocity in the reverse direction. Assuming the wheels do not slip relative to the treadmill, the velocity of the contact point between the wheel and the treadmill is moving at twice this infinitesimal velocity relative to the plane in the reverse direction. The presence of a positive wheel rotation rate will add some more forces on the plane (namely a rolling resistance), but this is still incredibly small compared to the engine thrust. Thus the acceleration of the plane is positive and the velocity will again increase.

Again as the velocity of the plane increases, that of the treadmill also instantaneously increases. At this point it should be clear that since the plane is moving forward with some velocity dV and the treadmill is moving backward with velocity -dV, the plane's wheels are spinning at a traced linear rate of 2dV. Whatever the linear acceleration rate of the plane, the wheels see twice that.

Let's try and decouple this mind bending scenario so it is more clear to understand. Forget about the treadmill for a second, and imagine just a single force balance on the plane, and how it differs from a plane taking off from a normal runway. Normally, the forces on a plane from the wheels are a rolling resistance and an inertial load from angularly accelerating the wheels. Rolling resistance force is ideally independent of speed, and the force from the inertial load of accelerating the wheel is proportional to the angular acceleration of the wheel. In the case of the treadmill, the only change in force on the plane through the wheels is that from double the angular acceleration (rolling resistance stays the same since the normal loading has not changed).

So we can simply transform this problem in to that of a plane taking off from a normal runway, but with wheels with twice the inertia (since the torque required to angularly accelerate a wheel is the acceleration rate times the inertia, twice the acceleration is equivalent to twice the inertia).

I don't think anybody here is going to debate that a plane that would normally take off just fine isn't also going to take off just fine with wheels that have twice the inertia.

Now back to more of your confusing incorrect analysis...

Again, completely incorrect. Nowhere in the problem does it say "a magical treadmill will move fast enough to cancel all of the forward thrust of the aircraft such as to prevent it from moving". In fact, such a statement would completely contradict the problem statement. The problem simply states that at whatever speed the plane moves forward, the ground below it will artificially move backward at that same rate. The ground can only exert a force on the plane through the wheels, and the only forces present at the wheels are rolling resistance and inertial load as described above.

I still don't see where you're getting this infinity stuff from. It's all pretty simple once you understand what's happening. There is indeed an instantaneous counteraction that's occurring. At the contact point between the wheel and the treadmill, there is an equal and opposite force on both the wheel and the treadmill. As described above, this force is small. And based on the problem statement, the ground is instantaneously accelerated at the same rate as the plane (in order to maintain exactly the same opposite velocity). There is nothing wrong with this assumption either. No infinite velocities or other magic are necessary to accomplish this.

I think what you are missing is simply how the forces are applied to the plane through the treadmill. If you are standing on the motionless ground far away from the runway, you'll see the plane moving to the left with some velocity, and the ground under it moving to the right with the same velocity, and the plane's wheels will rotate at a rate equivalent to the plane moving at twice its velocity on stationary ground.

Hopefully this is clear. I wasn't exactly sure what I was arguing against (your points don't make any sense), so I simply tried to re-explain it all.

 

MechEE

So you are saying the plane will take off?

[ ]yes [ ]no