What are your assumptions here? It makes a difference, which is sorta the not-very-well-stated point of Cecil’s column.
Ah. I asked a question about that in the other thread and never got an answer (probably because everyone else was terminally bored by that time). Thanks!
But you’re ignoring the force which would be applied because the wheels are rotationally accelerating. As long as the wheels have mass, and the treadmill can accelerate indefinitely, the treadmill will hold the plane in place. No friction is required (other than at the tire/treadmill interface, as Axiom points out).
Are you thinking about gyroscopic force?
I think I can summarise the two viewpoints as follows:
A) The question means:
The plane is on a treadmill roughly the size of the plane itself. The action of the treadmill is such that that *the plane does not move relative to the ground beneath the treadmill.
*
B) The question means:
The plane is on an enormous treadmill the same length as a runway. The action of the treadmill is such that the plane does move relative to the ground beneath the treadmill, and takes off in roughly the same time and space as if it were on solid ground.
And the outcomes are as follows:
A) Plane doesn’t take off. No motion relative to the air, no lift, despite what the engines may be doing.
**
B)** Plane does take off, the same way as normal planes take off.
To me, option B) just makes the thing into a kind of trick question, nothing to do with physics or aeronautics at all. It’s no better than “Think about words ending in -GRY. There are three words in the english language” or something like that.
When I heard the question, my mental image was A), and it seems to be the case for a lot of other people. So that’s the “right” answer, *if you interpret the question that way, *and the only way to get to Cecil’s answer is to realise that the wording of the question allows for a completely different interpretation with a mile-long treadmill and a plane zooming down it at regular plane speed.
Am I right?
heh. you said it.
there’s another very big difference between an airplane and a car with wings. the speedometer in a car is attached to the wheels. the “speedometer” in an airplane is called an airspeed indicator and is attached to tubes that stick out into the airstream and measure the speed the air is moving over the wings. oddly enough.
there are many speeds in this problem. let’s write them all down.
va = airspeed = speed of the airplane through the air.
vg = ground speed = speed of the airplane relative the ground.
vs = “speedometer” speed - assume our airplane is equipped with a speedometer = tire speed.
vb = belt speed = speed of the treadmill’s belt. the top of the belt travels at vb and the bottom of the belt travels at -vb.
and how do these speeds relate?
let’s assume for simplicity that there’s no wind.
va = vg
the speedometer speed will be the air/ground speed plus the belt speed.
vs = va + vb
now… the problem as stated doesn’t specify which speed the treadmill is using to set its belt speed.
group #1 assumes they mean the airspeed, va (=vg). which is really the only meaningful speed for an airplane. and they conclude that the airplane pretty much ignores the treadmill and takes off.
group #2 assumes they mean speedometer speed, vs. and there’s no difficulty with that so far. the problem then introduces a constraint: the belt speed is set to the speedometer speed. in equation form:
vb = vs
but from above we know that:
vs = va + vb
therefore:
va = 0
and the airplane doesn’t take off. it’s simple math, right?
well… the math is correct. but the conclusion isn’t.
the constraint can be restated like this: what has to happen in order for the belt speed to be equal to the ground speed?
the conclusion is the same:
va = 0
ie the airspeed is zero. but now the interpretation is different. this time around it’s clear we need to tie the airplane to the ground to set va=0. and the airplane doesn’t take off. oddly enough. note that once you set va=0, you can run the belt at any speed you want. it will always match the plane’s speedometer. however, if you let the plane move, ie va!=0, then you cannot set the belt speed to the speedometer speed no matter how fast you turn the belt.
group #2 runs into trouble when they draw their conclusions. the wrong conclusion is that the treadmill somehow keeps the plane from taking off. the right conclusion is that the only way to set the belt speed according to the stated problem is to prevent the plane from moving. note that in this case it doesn’t matter that it’s an airplane. a car, cat, person, author would be equally restrained.
No. Gyroscopic forces arise when you try to change the rotational direction of an object–gyroscopic forces are a function of the rotational velocity.
I’m talking about the inertial forces due to rotational acceleration of the wheels. If the wheels accelerate, there must be a pair of forces–a torque in other words–associated with the acceleration. This is from the rotational equivalent of Newton’s law: T = I[symbol]a[/symbol]. Torque equals moment of inertia times angular acceleration. If the treadmill accelerates, then the wheels accelerate, and a force is transmitted from the treadmill to the wheels, and from there into the plane. Large accelerations produce large accelerations; large enough, and the engine thrust is cancelled.
I go into more detail in this post in the older thread, if this quick explanation doesn’t make sense.
That seems like a pretty straightforward summary to me. The one additional issue that some people have is the question: In case A, is it physically possible for a treadmill to prevent the plane from moving? The answer being either yes or no, depending on your assumptions.
That the hypothetical treadmill is, in the practical world, a contradiction in terms. That you can’t generate enough thrust ‘backward’ just by rubbing the free-turning wheels faster and faster to make a difference.
Thought experiment - hold a hot wheels car (with free-spinning wheels) on a sander with a belt. Turn the sander on. Is it hard to keep in place? You might have to exert some downward force to keep the car on, so that there’s enough friction to make the wheels turn, but essentially no forward force is necessary.
Move the car forward. Hard? No, pretty easy.
Now do the same experiment with the sander on various speeds. The force required stays relatively constant.
Cecil doesn’t seem to get much into large-scale experimentation on this sort of thing …
Aha! An epiphany! This is a job for …
Jamie and Adam!
(Of course, they’d also have to find some kind of excuse to blow up the airplane by the end of the experiment …)
Why doesn’t someone get a motorized treadmill, an RC airplane and a video camera and settle this once and for all?
In other words, a real-world treadmill won’t have near enough acceleration capability to transmit the necessary force to a plane? I’ll buy that.
Cecil’s done a thorough job explaining BR#1 and BR#2, although I think originally Cecil wasn’t aware of BR#2: he says
Actually, the question thus stated leads to either the BR#2 solution (if the forces work out) or to constant conveyor acceleration (if they don’t). The impossible bit is infinite acceleration, if the author meant “exactly and instantly match”, but BR#2 doesn’t require that to stop a plane.
Anyway, that’s not my point. I submit that the statement of the problem is supposed to be BR#2-like, because the problem solved by BR#1 is inherently less interesting:
You could walk, drive a car, ride a bike on this conveyor and actually get somewhere; you’d just get there half as fast. Given that, the contrast between plane and car is pretty mundane. Ooh, the plane’s more efficient because less work is done on the conveyor!
The whole problem statement would be a lot simpler replacing “powered conveyor” with “massless belt and rollers”, or even “frictionless surface”. Both seem to be what the problem’s originator was really trying to get at, but then I suppose the correct answer would be too obvious.
The more I think about this question, the more I think it’s got ludicrously out of hand.
I’m sure the question as originally posed equates to this:
*
Imagine a plane on a treadmill which operates such that, no matter how fast you run the engines, the plane stays stationary relative to the ground and never gets up any airspeed.*
It’s an interesting question whose solution some people will get wrong, but most logical people will grasp.
All other readings of the question are just nit-picking and sophistry and I’m not very interested in them.
This is actually not a bad idea. Is anyone here active on the Mythbusters message boards? I can only imagine it would cause as much furor there as it has here, and might be enough to trigger an investigation.
Bingo! It’s called a kite.
What you are not realizing is that the plane just not just sit in place on a treadmill and magically take off.
The treadmill needs to be as long as a normal runway. The plane makes its way down the treadmill from one end to the other, as it gains speed (relative to the ground and the air) before it takes off.
The treadmill doesn’t help a plane take off.
(This is all part of getting to BR #1.)
Real-world or not, I don’t think there is even an idealized version of this problem that would allow a treadmill to exert such a force (overcoming the jet engines) to the wheels of an airliner. bup’s beltsander is a good illustration of this. Offhand, I’d say Cecil pooted this. Why didn’t he at least vet it by Chronos?
Sure there is. Rotational acceleration of the wheels will transmit force from the treadmill to the plane fuselage. Cecil covers this (although admittedly not in a comprehensive way) at the end of the latest column. It does require continuous (and, I imagine, substantial) acceleration, so it’s unlikely something that could be realistically built–unless you’re talking tiny RC planes, maybe. bup’s beltsander isn’t really a good analogy, because it doesn’t include any element with rotational acceleration.
It’s also a plausible argument that wheel bearing friction will increase with velocity until the bearings themselves generate the required force. Again, not realistic, because they’re gonna melt into slag before that time, but plausible in a thought experiment. This is a more tenuous possibility, IMO.
I can probably hunt down some posts from the previous thread that goes into more detail here, if you like.
Cecil said that the idea behind BR#2 is so elegant that we can forget about real world. Well, I suppose it is elegant. After all I forgot about the force required to accelerate the wheel :o . However I don’t think it’s elegant enough to overcome the sheer folly of the mechanics.
I assumed a light plane with an engine thrust of 500 lb. The wheel was 18" diameter with a tire weighing 15 lb. on a 6" dia. hub weiging 5 lb.
Yes, a jumbo jet’s wheels have a lot more inertia than do a Cessna’s but the engines also put out a lot more thrust.
I think bup was addressing the idea that the frictional force between the treadmill and wheels does not increase as velocity/acceleration.
Anyway, at a certain point, the maximum would be complete abrasion of the wheel (as the bearings lock up completely) but aren’t we pretty sure that that is still not enough? If it were, why not just let it rip and sand it down to the belly to begin with? That’ll keep it grounded.
I don’t think Cecil got this one right.
That’s interesting.
Couldn’t agree with you more.
So. In the first second the belt would have to accelerate to 636 mph. Is this a umm… linear equation? Would it be 1272 mph in two seconds?
Not that it really matters much. In the first second of the ‘elegant’ BR#2, the wheel and plane would fail.
I understand the people that see this as more of a thought experiment, than an attempt to put it more into a real world (mis)application. We are obviously in two different camps.
‘Wheel speed’ (whateverintheheckthatissupposedtomean) vs. ‘plane speed’ relative to the ground.
![In order to accelerate the wheel enough to overcome a 500 lb. force the acceleration would be 1245 rad/sec[sup]2[/sup]. In one second the wheel’s periferal speed and thus the speed of the conveyer would be 636 mph.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=7e4fscd.jpg&.src=ph){kind=link}