Here’s the kicker…the force my friend is applying is going to be a constant additive force. As soon as the treadmill increases speed to match the initial push, if my friend is applying constant force, it will always be causing the wheels to spin at a faster speed than the treadmill.
Picture a balanced system, where the wheels are spinning at 10mph and the treadmill is going at a speed that counteracts that. As soon as my friend applies some force…say that would cause the wheels to turn 1mph more…the treadmill matches that. But that 1mph force is constant…when the new system is balanced at 11mph, that 1mph force is still there, and the forward movement continues…the treadmill is playing a constant game of catch-up and will never get there…
The problem with the “thought experiment” explanation is that, as a thought
experiment, it is inordinately boring, because the question is reduced to “Is it possible to restrain a plane so that it cannot take off?”. The obvious answer is yes, whether it is via cables, a brick wall, bolting it to the runway, superglue, or a magical treadmill that creates friction but doesn’t destroy the plane. It’s not even marginally interesting or complicated. It’s not a thought experiment if the answer is so obvious you don’t need to think about it.
So the other part of the question is “will a plane with functioning wheels take off on a moving treadmill”, the answer is also obviously yes, although it may take a moment’s thought to see why.
What that means is we are left with a trick question, because it is not framed as either of these questions, but as some hybrid in between that causes confusion and controversy. In fact, the controversy is more interesting than the question.
While we’re on silly assumptions that answer the question:
a. The treadmill is bolted down.
b. The plane has fuel.
c. There is an atmosphere.
d. There is gravity.
e. The treadmill isn’t sitting on a second treadmill going the opposite direction.
f. The treadmill’s power source will outlast the plane’s.
g. The plane doesn’t turn 90 degrees and drive off the treadmill.
In a real-world control system, your position (or the plane’s position, whatever) would be constantly dithering back-and-forth as the control system seeks to balance the velocities. Just how quickly and accurately the control system is able to react is a function of a lot of things, but there’s certainly no reason to think that the treadmill would be playing catch-up all the time.
In addition, you are confusing forces and velocities. The treadmill control system will try to do something to oppose the additional force your friend applies. With an ideal system, the response would be quick, and your friend would feel increasing resistance as he pushed harder. (note this comment assumes not only the treadmill control system, but also high accelerations and wheel masses and things (post 106) that I’d rather not repeat in every single post, so I’ll just hope it’s understood.)
Well, actually I thought the other way around. The “original” problem in Cecil’s column (the one where the treadmill matches fuselage speed) seems boring to me, because the answer is obvious: if the plane (or car, or bicycle) is allowed to move forward, then of course it will move forward. The “second” phrasing (the one where the treadmill matches wheel speed) seems much more interesting, because the question is much more subtle: can force be transferred to a plane via a treadmill? The answer here isn’t obvious, and makes a much more interesting thought experiment.
Nevertheless, I understand what you’re saying: the phrasing leads to confusion over semantics, and the real “trick” to the question winds up having nothing to do with planes and aerodynamics, really. So I completely understand the sentiment.
There are several posts that imply that an accelerating belt could somehow keep the plane from moving. This is the first post I found this assumption, but there are several posts later, that also say the same thing.
(Emphasis mine.)
I have to correct the wording a bit. The frictional force that provides the torque for the wheels is not backwards, it is tangential to the wheel. This seems like splitting hairs, but I think the distinction is important. This tangential force only provides the torque, and the angular acceleration is a function of the torque. But this is all this force does, it does not in anyway have any impact on the plane. And it is certainly not applied to the center of mass of the wheels, that is what tangential means. Thus it can not “accelerate the center of the mass of the wheel”, to accelerate the center of mass, a force has to be applied to the center of mass. The only way the frictional force can have any effect on the plane is through the friction on the bearings of the wheel.
I also don’t understand what you mean by “simply model the system as a single wheel of the combined moment of inertia with the mass of the total system.” The whole system (plane + wheels) does not rotate, only the wheels rotate. The frictional force can only be coupled with the inertia and the mass of the wheels, not with the plane. Since the angular velocity (or acceleartion) of the wheels does not in anyway have an effect on the plane (except possibly through increased friction coefficient on the bearings of the wheels), the acceleration or the velocity of the belt have no effect on the plane.
I’m sorry, but this is completely incorrect. An unopposed force applied to an object will accelerate it linearly. If the force is offset from the center of mass, it will also accelerate the object rotationally. This post has a more complete argument.
Well, yes, if you build the conveyor belt runway with 20’ tall thistle bushes all over it, this should provide enough friction to keep the 747 in place so it won’t fly. That is, as long as you replace them as they get run over.
Even with theoretical frictionless bearings there is still the force generated by angular acceleration of the wheels to contend with. So, yes, I’m saying that even when your friend is generating an infinite amount of force to push you forward, the treadmill will be able to generate an infinite amount of force in the opposite direction to hold you in place. In theory.
I get the feeling that this whole discussion should have started off with, “Assume a spherical cow.”
I must apologize, you are correct. Reading your post thoroughly helped me found the flaw in my reasoning (I accidentally made an assumption that isn’t true). The wording of some of the posts got me quite confused, since they seemed to imply something rather different than the actual mechanism. I had to go through the calculations to actually see what happens in the case of an accelerating belt.
Thinking further along: Now, if I got my reasoning correct, in the case of an accelerating belt, the frictional force opposes the force from the accelerating belt, so the frictional force actually helps the plane to move forwards.
Holy cow, someone actually changed their mind in this thread. Now I’m all tingly inside.
Assume a) the plane’s engines are thrusting forward, b) the belt surface is accelerating rearwards, and c) the forces in the plane fuselage cancel, so the fuselage does not accelerate.
In this case the belt is driving the wheel, so the force from the belt on the wheel points rearward (the force from the wheel on the belt points forward, of course). Since the wheel is not translating at all (although it is rotationally accelerating) the force from the plane on the wheel points forward to counteract the force from the belt. Thus the force from the wheel on the fuselage points rearward, counteracting the thrust. The total of the forces on the plane fuselage add to zero, and the total of the forces on the wheel add to zero (but create a torque which accelerates the wheel).
Or, in other words, the rearward force from the accelerating belt is transmitted to the plane’s fuselage.
It’s been too long since I’ve last thought about mechanics. Clearly, I should just stick to my own field. Anyway, I agree that an accelerating belt applies a force on the wheels.This is getting a bit off-topic now, but I’m slightly confused about what happens in different scenarios I’m trying to think about. Actually, it is the rotational motion and acceleration and how they relate to the translational motion and acceleration that I’m trying to understand. Maybe I’m making a wrong assumption about the causality in some place, or maybe my equations are wrong, or maybe I just don’t understand rotational friction. Please point out any mistakes I’ve made in the following (and if you could use equations to describe what happens, it would actually be easier for me to understand).
(I need vectors now, so let’s define by x,y and z the unit vectors to the directions of the principal axes and by axb the cross product between vectors a and b.)
Let’s assume that we have a wheel in free space. Then applying any force F (tangential or not) to the wheel would result in movement in the direction of the force by F = ma, and if the force is offset from the center of mass, it will also result in rotational motion that is defined by the right-hand rule, i.e. close your right hand with the thumb pointing upwards, then with your thumb pointing to the direction of the torque t, the wheel rotates to the direction your fingers curl (counter-clockwise). In other words, t = rxF, where r is the vector pointing from the center of mass to the point where the force F is applied. That’s clear.
Now, assume that we have a wheel (alone) rolling on the xy-plane (above it, to be precise) in the direction of the positive x-axis at an initial velocity v = vx, so that the axis of rotation is parallel to the y-axis. Eventually, the wheel will slow down and stop due the force exerted on it by the plane (let’s forget air drag for now). This means that the angular velocity of the wheel also decreases. The force exerted by the plane on the wheel is tangential, and it thus results in a torque ty = (-r)zxFx** = -rFy on the wheel, where r is the radius of the wheel, and Fx is the force the plane exerts on the wheels. The torque is related to the angular acceleration alpha by t = I***alpha**, where I is the inertia of the wheel. For the angular velocity to decrease, the angular acceleration must point towards the negative y-axis, by the right-hand rule. Is this so far correct? By this logic, the force of the plane on the wheels should counterintuitively point towards the positive x-axis to decrease the angular velocity. But the frictional force (which slows down the wheel) should point towards the negative x-axis, otherwise the wheel does not slow down. Or am I making a wrong assumption here? What is the connection between the force exerted by the plane on the wheels and the frictional force that slows down the wheel?
Let’s again consider the plane on an accelerating belt. The engines of the plane provide a force F*_e* on the the plane that is towards the positive x-axis. The acceleration of the belt is a = -ax, which is related to the angular acceleration alpha of the wheels by a = alphaxr, and r = -rz. From the point of view of the wheel, the point r is accelerating towards the negative x-axis, so the angular acceleration alpha points towards the positive y-axis, resulting in a torque t = Ialpha**. This torque can then be associated with a force by ty** = (-r)zx(-F_b)x, so the force of the belt on the wheels F_b points towards the negative x-axis. The force of the belt on the wheels thus pushes the plane backwards, counteracting the force of the engines F_e, and at the same time makes the wheels spin furiously. Again, by 2. there should be a force F_a* exerted on the wheels by the belt that should oppose the angular acceleration of the wheels by F*_b*. How does it affect the force F*_b*?
Frictional force is essentially a reactive force, right? So if you take a tire, or a block of rubber, and slide it whichever way across the plane (“plane” = “flat surface” here, not “thing with wings”) the frictional force is simply a vector pointing in whatever direction opposes the force. Clearly the frictional force could also be zero: a tire just sitting there has only z-direction forces on it (weight and reaction) and there’s no frictional force.
If we’re talking about a tire rolling along a flat surface, the deceleration you see wouldn’t be coming from friction in the sense you’re thinking, it would be coming from tire flex. (Disclamer: tire dynamics and rolling resistance are somewhat far afield from my area of expertise, so… details may be fuzzy) As the tire rolls forward, the sidewalls on the leading edge of the tire contact patch are compressed, and the sidewalls on the trailing edge spring back (imagine rolling over a field of tiny springs). That creates a torque on the tire that will angularly decelerate it. Since we’re assuming no slip, the tire will also decelerate linearly, and the horizontal (x-direction) friction force will be whatever’s necessary to complete the force balance.
I’m not sure what you’re asking here: the force exerted on the wheels by the belt is Fb, which, since it’s off-center, causes the angular acceleration of the wheels. A second force Fa between the belt and the wheels isn’t necessary.
Ok. If the force is 100% through the center of mass the wheel doesn’t spin and accelerates to the left in response to the force.
If the force is 100% tangential the wheel both spins and translates. What fraction of the force results in spin and what fraction results in translation?
I’m just going to try to sum this up as best I can see it.
The problem most people seem to be having is the fact that traditionally, treadmills are just forces that stop movement by moving backward at roughly the same rate. This is because basically anything that is on a treadmill moves by pushing against the ground, which is how the treadmill works. We don’t get things that move (roughly) independently of the ground being put on treadmills all that frequently.
Okay, but the plane. Broken down, you have the following forces:
Treadmill moving backwards.
Engines moving the plane forwards.
Let’s say the speed of the two is roughly equal.
So they cancel out, right?
WRONG. Since the engines push against the AIR not the GROUND, the moving conveyor belt doesn’t work against them except through the minimal friction of the wheels. See again the example of putting on roller skates, getting on a treadmill, and pulling yourself forward by a rope attached to the wall. The rope represents a generic force not affected by the treadmill, much like the forward push of the plane’s mode of thrust.
I’m just curious… Are there still those who say that, given the scenario of the **axel **of the conveyer mounted airplane making forward movement in relation to stationary ground, that the airplane will NOT take off?
So you’re saying that the force of the belt turning backwards is powering the wheels on the plane to push it forward? I hope you’re not saying that. There is no force coming from the wheel. The ONLY thing accelerating the wheels are the engines. What is torquing? You’d have a TINY bit of torque from the tire to the wheel. That’s pushing the plane forward?
The ground (in this case, represented by a treadmill) is totally and completely irrelevant with regard to whether an airplane will fly. An airplane flies relative to the wind, not the ground.
One can fly backwards in a small airplane. If the stall speed is thirty-five knots, and the relative wind is fifty knots, I can point the airplane into the wind, and even though I’m flying forward at oh, forty knots, I’m moving backward at ten knots relative to the ground.
What makes an airplane fly? The difference in air pressure between the top and bottom of the wing. As the wing moves through the air, the shape of the airfoil dictates the differential between airflow above and below the wing. The curved surface of the wing is designed so air moves farther over the top than the bottom. This creates a lower pressure above the wing (faster movement == lower pressure), which in turn creates lift (ask Bournoulli if you don’t believe me).
The amount of lift contributed by the airflow generated by the propeller is minimal; not enough to lift the entire airplane off the ground (for jets whose jetwash never touches the wing, the effect is zero). Why? Because it would have to overcome two factors: 1) drag - the opposite of thrust, this is the friction caused by the airframe’s motion through the air; 2) weight - the opposite of lift, the propwash would have to generate enough negative pressure to equal the weight of the plane plus extra to generate forward flight.
A plane could care less about the ground (see my last post about flying backwards). If the relative wind is moving at less than the stall speed of the wing, then the plane will not fly (excluding the cushion caused by ground effect).
Therefore, I conclude that if the plane is not moving in relation to the wind surrounding the wing, then the plane will not fly. Even a jet spinning its wheels at 500 kts will never lift off because the air around the wing is stationary. There is no lift.
For any plane to get off the ground, the relative wind (to the wing, remember?) must be moving faster than the wing’s stall speed.
Well, I think that’s a bit misleading to think of it in that way. The force F produces a torque Fr around the center of mass, right? But the wheel still accelerates translationally at same acceleration it would if there was only a force F and no torque. And the wheel accelerates rotationally the same as if there were a total torque of Fr, with no total translational force. So all the force is used to accelerate the wheel linearly, and all the force (torque) is used to accelerate it rotationally.
However, you could recast the question as an energy question. Clearly the force is adding kinetic energy to the wheel at some rate, and the tangential force adds more energy than the center force (since it produces additional rotation). So it’s perfectly valid to ask what proportion of the added energy is going into translational motion, and what proportion is going into rotational motion. You can calculate that proportion by comparing the translational kinetic energy (1/2mv[sup]2[/sup]) to the rotational kinetic energy (1/2I[symbol]a[/symbol][sup]2[/sup]) Does that make sense?
Nope, not saying that at all. This particular scenario is where the plane fuselage isn’t moving at all (where the treadmill matches the “wheel speed” [the last paragraph of Ceceil’s column] as opposed to the fuselage speed [the original question in the column]). In that case (with certain assumptions), the force from the treadmill balances the force from the engines by accelerating the wheels.