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 
Of course the plane doesn’t take off- it burrows into the ground!
Are you guys idiots??
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I can’t belive there is still debate!
The plane flips over because of the treadmill/wheels.
then jet and treadmill push in the same direction, it goes flying,
hits the escape velocity of the earth, free satellite delivery method.
duh!
seriously though, the plane has to take off.
friction depends on forces going through the surface, not the relative speeds.
the jets push the plane forward, causing slight wind that steadily increases until there is enough lift to support the plane. THis is assuming the treadmill is long enough (read runway)
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degan manton
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Does no one realize that the plane can be moving faster than the treadmill, with the wheels skidding? Depending on the coefficient of friction of the treadmill with respect to the tires, all the pilot would have to do is force the wheels to skid by using the wheel brakes.
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I find the people seeking to end the problem by showing up with a rather arbitrary selection of factors that are considered or assumed as the only interpretation to be very disappointing. The result depends entirely on what you do and do not consider.
In the 12th grade physics class case is a frictionless treadmill with frictionless wheels can be treated very simply. Without friction the motion of the wheels or the treadmill can be ignored since their movement has no bearing on the net force on the airliner. The jet rolls down the runway and takes off as normal. This is a very boring case.
Throwing friction into the mix makes things more interesting. In a slightly more complex case the speed of the treadmill has a net drag on the aircraft. The net force on the jet is the combination of the jet thrust wanting the jet to move forward and the powered treadmill applying force backward through the frictional surfaces. An infinite-speed treadmill is hardly necessary as there is a steady state solution where the thrust of the jet engines is exactly matched by the drag of the treadmill. This is well below infinite velocity.
Getting more and more real is up to the patience and determination of the thinker. First of all one must recognize a zero lag speed compensator is impossible what with the parts being not co-located and the speed of light being only so much. Also there is the similar tale of moving the frictionless brick with your finger. At first it is at rest and you push it with your finger. You accelerate it from zero speed to some finite speed in effectively no time which requires infinite force, which is impossible but obviously doable. It is possible that examining this closer shows how one can “gain” forward displacement of the jet even if the treadmill is so good at speed matching.
You can also get into fun things like hydroplaning on the cushion of air building up in front of the tire as it rolls faster, the stress limits of the tire (if you’re a boring engineer), gyroscopic precession of the wheels, angular momentum effects, and relativistic effects if you want to be exotic.
However I have to comment on the following:
1. The Mythbusters “experiment” was laughably flawed.
2. “Running on a treadmill feels no wind” suggests that 747s run down the runway to takeoff?
3. “Jet engines push against air” is a very Neanderthal view of jet propulsion.
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My brother just reminded me to wave my credentials around as a licensed pilot and a physicist :p
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What a great problem!
Interpretation to conclude #3 by following reasoning:
Speed of wheel V(wheel) = R(wheel)*d(theta)/dt
Speed of conveyor V(belt) = dl(belt)/dt
(contrary to the less interesting definition of V(wheel)=V(plane) (in my opinion is an incorrect assignment) in which case of course it will take off as outlined previously)
define scenario #3
R(wheel)*d(theta)/dt = dl(belt)/dt
by the above definition of the problem would imply that there is a controller driving the limitless power belt to match the rotational speed of the wheels.
Now we outline contributing factors in the experiment:
Review of physics 101
Sum(F) = m*sum(A)
Force components considered:
F(air) = h*abs(V(plane))
F(wheel) = k*abs(v(wheel))
F(turbine) = nonzero constant (assumed constant at Max thrust) (generous as max thrust is attained with some V(plane) = nonzero
Next identify ambiguities in the scenario and solve for both.
Wheel rotational resistance yes/no
Next we model both systems F(wheel) = (k*dw(theta)/dt) where k=zero, nonzero
no wheel rotational resistance (k = 0)
Sum of forces are as follows (draw a picture if you want):
Sum(F) = F(thrust constant) – 0*dw/dt – h*(V(Plane))
F(thrust) = h*(V(Plane)) =F(Air Drag)
/* it becomes clear here that without an opposing force this plane will move by this defined fundamental relationship and knowing that air drag is zero at rest. However lets look into how the rest of the model is behaving.*/
V(Plane) = nonzero
V(Plane) = V(wheel) – V(belt)
// Substitute
nonZero = V(wheel) -V(Belt)
V(wheel) = V(Belt) //defined in problem
nonZero = v(wheel) – v(wheel)
nonZero = zero
belt speed control divides by zero, supplies 1/0 joules of energy to the belt and the plane still takes off leaving behind a fractured universe.
With wheel rotational resistance (K = nonzero)
Sum of forces become
F(wheel) + F(Air Drag)= F(Thrust)
considering air drag again to be negligible (advantage to reach critical takeoff velocity), V(Air drag)=h*abs(V(Plane))=0
F(wheel) = F(Thrust)
The belt motor is defined to match the power output of the turbines (magnitude aside, we have defined it by the constraints of our problem)
Real world application to this last scenario (as it doesn’t divide by zero as the previous scenario did) will result in small accumulation of speed gained from the of the belt speed latency (attributed from belt speed controller response and belt power delivery response). Once max turbine output is achieved and our controller has responded to the speed input, we can conclude the following:
-Sum of forces horizontal forces = 0 (acceleration = 0, v= nonZero dependent on belt control system latency (assuming non zero belt response time))
-Conservation of Energy shows that virtually every Joule of energy applied as thrust to the airplane gets applied as heat to the wheel bearing friction and tire deformation friction (Now this would be fun to watch!)
-A rotational moment would be generated on the plane from these opposing forces with vertical displacement (engine height from ground). Given a close-to-theory belt assembly, this will result in the exceeding of the specified vertical loading of the front landing gear and would most likely identify the point where things fail and become exciting!
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Frederf is right: considering frictionless wheels is a boring case. Moreover, the text states that we are considering a 747. I haven’t checked a 747’s datasheet, but I might guess that they don’t have frictionless bearings and wheels without rolling resistance.
The conveyor belt, however, is in fact “designed to exactly match the speed of the wheels”: it is a (possibly rather expensive) thing that reaches the speed required for the friction of the wheels to match the thrust of the jet engines. Even if it is impossible to build something like this at our current level of technology.
–> The airplane is real, the conveyor belt is ideal.
Moreover, the “v_C=v_W” (#2) case is quite boring, because “All it means is that the wheels will spin twice as fast as normal” — which means that the plane would take of _even if_ it were powered by its wheels.
Sorry, Randall, you lose — twice 😉
P.S. I know that I will be unheard and buried among all those trolls, but making an unheard true and seemingly overconfident statement is more fun than making no statement at all
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As a member of the group of people in the aforementioned 12th grade physics class, we have dealt with friction and motion. Based on that, I did the calculations, and frictionless, it works. With friction, it still works as long as your coefficient of friction isn’t too high(has to be very inconceivably high). The jets push the plane forward, and friction works against it. THE VELOCITY OF THE TREADMILL HAS NO BEARING! Force is the determining factor in considering friction. But as I referenced above, the treadmill has to be long enough for the plane to accelerate(at a slightly reduced acceleration) to take-off speeds.
I think people are confused about the high speed=more friction. In air, this is true, because the force at which the air hits the object is higher when the object moves faster. In our theoretical treadmill, there is no push against air. Just the rollers on the end.
Side thought. What happens on an infinite speed treadmill when the surface turns around, switches directions from infinite one way to infinite the other way?
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Is the treadmill electric? Or is it one of those old school ones that requires the person – in the case the airplane – to physically move the belt?
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I put a VTOL jet on on the treadmill, say “Screw You Treadmil”, and take off regardless of fast my jet is or isn’t moving.
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I don’t have time to read all your witty patter….
1) Ice down a runway–effecting the same thing as the treadmill–and try to start, or stop the plane.
2) Exit the icy Patomac via emergency exits and inflatible slides, return to hotel and post on XKCD.COM: “Problem solved.”
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I don’t have time to spell check Potomac either. Sorry.
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It wasn’t clear to me which way you think.
icing the runway would make it almost impossible to stop the plane.
but since it starts with it’s engines it should take off okay.
and i don’t think inflatable slides would do much better than the boats they were in.
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It’s logic and not physics… but physics is just applied mathematics, and mathematics is just applied logic. (Where was the logician in comic 435?)
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I was hanging upside down in the emergency room, listening to two doctors who seemed very far away (I’ve been to the emergency room a lot but even I didn’t know that hospital beds could be positioned so you were hanging upside down). My veins were too collapsed to get an IV or transfusion going, so they were trying to get all the blood to my jugular. One doc was very keen on drilling into my chest because she had “never got to do that before!!! And the guy who did it to himself on YouTube said it didn’t even hurt!!!” The other doc was giving me the classic “Push! Keep pushing!” instructions they scream at women giving birth, in the vain hope that my “bearing down!!!” would make my extrajugular veins cooperate with his cannula. I started to pass out. It had been made clear to me earlier that day that I was absolutely going to die if they couldn’t get some kind of fluid pumping through my body, as I was just about out of all my own type-O. It occurred to me that I might die alone listening to two cheeky residents argue about who got to stick me first. The possibility that I’d never get to make out with Debra Winger became freakishly real. I felt bad that I hadn’t called my parents earlier. I wished I had skateboarded more – been less afraid of being the only thirty-year-old woman trying to work the tranny at the park. I wished someone was there to whisper in Fermat’s theorem in my ear.
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Clarification: “Workin’ the tranny” is a California skate term from the 80s, referring to performing tricks on the section of the skate ramp that transitions from flat to vertical. In my experience it was never meant to be any kind of derogatory play on words, just shorthand for “transition”.
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I don’t think the plane will take off because there is no air movement,the wheels are moving the reactor are pushing,but if the plane isn’t moving through the air,the wings don’t make their job and u are a pure nature hater.
To make a plane take off you don’t have to make the wheel move you have to make the air around the wings moves,that’s what they do in “soufflerie”(sorry I don’t know the english terms).
Keep going your comics are cool.
From a french guy who get lost on the web.
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The answer to this is to look at what happens when the plane tries to take off. The plane is initially resting on the ground, so it requires the wheels to get moving so it can achieve lift. Unless the wind speed toward the plane (nose to tail) is equal to or greater than the speed required to achieve lift, it sits there making the treadmill go round really fast.
Which would be cool to watch.
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> Ok, but what if you put a helicopter on a giant turntable…
I’ve actually heard two variants of helicopter on a turntable. One is a direct analog: the turntable matches but opposes the speed of the blades. I hope everyone with some shred of common sense and understanding of how helicopters work sees that the helicopter would not lift off in this case.
The other, more interesting one is that the blades remain motionless, but the turntable turns in the same direction (and at the same speed) as the blades would turn. This is a little trickier, because some people think the blades do turn with the turntable, while others think the blades remain motionless relative to the earth (due to inertia). Theoretically, the answer is, yes, the blades do turn, but not as fast as the turntable (due to inertia), so the craft would probably not lift off.
But did anyone else notice the catch? It’s not that helicopters have two rotors – theoretically that would not be an issue, unless our particular helicopter is of the two-opposing-lift-rotors variety, and we’ll just assume it’s not (rear-rotor helicopters are more common). The actual problem is that when the helicopter starts spinning, the fuel (and any load it might be bearing) will shift away from the center of the turntable (which, presumably, would be aligned with the center of the helicopter’s blades), which in turn shifts the helicopter’s center of gravity out from under the helicopter’s rotor shaft. A Schweiser 300’s rotor (and thus, the turntable) can spin at 440 to 460 RPMs (and the Schweiser is a light helicopter) – even a slight inbalance would send the helicopter off the turntable by centrifugal force.
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Also:
>Moreover, the “v_C=v_W” (#2) case is quite boring, because “All it means is that the
>wheels will spin twice as fast as normal” — which means that the plane would take of
>_even if_ it were powered by its wheels.
You misread. “All it means is that the wheels will spin twice as fast as normal” is because a 747 is not powered by its wheels. It ceases to be true if the plane really was powered by its wheels, so your assertion is false.
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Alright, how about this:
On a theoretical conveyor belt that can quickly accelerate to infinity, and produces friction, a plane would still take off, because if there is a frictional force acting on the wheels, there is a frictional force acting on the air, moving it over the wings and thus producing lift. You can test this by turning a normal treadmill to its fastest setting and placing your hand just above the belt. The air gets pulled along as well.
On a frictionless conveyor belt, it would not matter either because the wheels would have no velocity at all, therefore the conveyor belt would not move.
captcha: vatore LIMITED
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>>a treadmill with no motor
>>I haven’t seen many people subscribe to this interpretation.
Wait what, you mean this isn’t the most obvious way of doing it? ;_;
Whatever happened to the KISS method? The less effort the better.
I also would have to agree with this previously posted comment:
“Any chance of a Pterosaurs For Stability inspired xkcd T-shirt? I love that drawing!”
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1-The wings will be ripped off the plane.
2-The plane will stall without taking off.
Simple experiment. Step on a treadmill, open your arms wide and start running. Have two friends try to pull you forward by holding your arms. They sure can pull you forward if the treadmill itself is moving at a constant speed. But since it is matching the speed of your running feet no matter what, your feet will be at the same spot. Meanwhile your friends (the engines) are only limited by the friction the treadmill imposes on you (plus air friction), which they can easily overcome. So the tension will make your arms bend forward.
With a plane, that will rip off the wings. Or if the wings can resist the tension, it will create a moment of force making the body of the plane stand up on its nose.
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i was glad that someone brought up the ‘helicopter on a turntable’ problem, but was surprised that nobody brought up the ‘hovercraft-on-a-vacuum-cleaner’. it’s been sadly overlooked.
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Only possible solutions:
#6: The engines are off. V_w is zero, and the plane will not take off.
#7: The conveyor itself will have to move (as implied by #3). Then:
V_r + V_c = V_w + V_b
=> The plane will take off.
#8: The airflow towards the airplane has to match the suction of the engines.
=> The plane will take off.
Conclusion: whether or not the plane takes off depends *only* on whether the engines are on or off. The wheels simply don’t matter
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Oh, so the plane doesn’t even stay on the conveyor belt, because it’s wheels will always move twice as fast. Or if the conveyor is the size of a runway, it will still take off because it will still roll down the belt.
but
What if we put a piece of cheese in front of the plane?
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yea…the sprinkler does move. If you goto the 4th floor near the Edgerton center at MIT they actually used to have a demonstration system that showed you it moved, I don’t know if it is still there.
Also, to be totally honest I think there is something really fishy about the way you explained the airplane problem…it shouldn’t be impossible to go into a reference frame where the plane remains stationary, which somehow you have argued you can’t….or maybe I misinterpreted your explanation. I think the key idea is that the velocity of the axle is the only velocity that really defines the behavior of the wheels, so it’s kinda pointless to talk about any of the other points on the wheel. The wheels twice as fast thing is really I believe a misinterpretation of the problem.
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You can’t place cheese in front unless you’re willing to sacrifice your chips as well. It’s demanded by the Laws of Physics.
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Just wanted to point out that Amazon’s DRM-Free mp3 store is 100% free, legal, drm-free, awesome, drm-free… did i mention it’s completely drm-free?
mp3.amazon.com
Pricing rivals iTunes, and although not every hot new thing released in the past year is available, it’s got tons of awesome music in general. I love the old punk rock titles they have 🙂
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I just deiced my freezer.
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This is actually pretty simple: Because the aeroplane is being propelled by the engines, not the wheels, it’s still going to be propelled forwards. The wheels are merely a way of minimising friction between the aeroplane and the ground.
The wheels and treadmill will accelerate perpetually, but the aeroplane will continue to accelerate and eventually achieve enough lift to take off.
If the engine drove the wheels, then the problem would be unsolvable, but they’re not.
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@DavidG
At first i was going to disagree with you on the airplane/treadmill problem. But then i thought about it (I know, an American using his brain.. what a crazy thought), the wheels have nothing to do with the plane taking off, all they do is allow the plane to roll on the runway. All the forward force is provided by the engines. So like JP said, the wheels would just spin twice as fast.
End result: Plane takes off.
The cheese would only anger the airplane, causing the passengers on-board to demand the in-flight movie be Gigli.
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As our good friend Zeno would have said: “in any one instant of time, for the plane to be moving it must either move to where it is, or it must move to where it is not. It cannot move to where it is not, because this is a single instant, and it cannot move to where it is because it is already there. In other words, in any instant of time there is no motion occurring, because an instant is a snapshot. Therefore, if it cannot move in a single instant it cannot move in any instant, making any motion impossible.”
Conclusion: The plane, treadmill, and everyone and everything else cannot move.
(quoted from wikipedia, with the word arrow replaced by plane)
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I’d say the plane doesn’t take off because it’ll never start moving. But then again there’s a very high chance I’m wrong.
I think the pilot should just turn around and go the other way, and take advantage of the movement of the treadmill.
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PS (sorry for the double post)
Thor…I need to do that, everything in my freezer is really frosty.
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So, someone was close to what will really happen– someone said the wings would get ripped off. That might happen, but only if the plane were built like a cracker. It’s really irrelevant what the wheels do if the treadmill will counter their actions. However, because the thrust provided by the engines is a linear matter and has absolutely no relation to the treadmill, the plane will quite definitely move forward, at slightly less speed than if it had functioanl wheels. A physicist will then tell you that the motion will generate lift and allow the plane to take off. HOWEVER, an engineer will tell you that the wheels are being moved forward without actually rolling forward. The resulting friction would be a strong enough force to snap off the wheels, rather than rip off the plane’s wings. Anyone who’s seen a plane crash-land knows that when the plane comes in too fast the wheels snap off, rather than the wings ripping off. The engineer will then go on to tell you that the plane will not in fact take off, because the friction caused by the plane’s fuselage dragging on the runway will rip open the fuselage. The heat generated by the friction (which will be immense) will then ignite the now-exposed fuel tanks, thereby destroying the plane and conclusively answering the question: No, the plane will most definitely not take off, but not for the reasons you would think.
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I think that that’s the first post on this topic I’ve read that has addressed all viewpoints and rationales intelligently. Well done.
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@Ben Jones
Consider this:
…999 = x
…990 = 10x
subtract the two equations above:
9 = -9x
x = -1
Thus proving that …999 = -1
This isn’t true in the reals, as real numbers must terminate to the left (and if you translate it to sums of series, the series is divergent). However, it is true for the 10-adic numbers (which aren’t allowed more than a finite number of decimal places to the right). The 10-adic numbers aren’t a field, nor even an integral domain, as it’s possible to get two non-zero 10-adics that multiply to give zero (exercise for the reader). However, if the base is a prime number, you can obtain a perfectly nice field of characteristic zero for each one. These are known as the p-adics.
With real numbers, numbers are considered small if they have zero to the left of the decimal point and many zeroes to the right of it; the more, the smaller. In the p-adics, having more zeroes to the _left_ of the decimal point makes numbers smaller, so in the 5-adics, 400000 is smaller than 400, and (strangely) smaller than 400001.
You can also consider p-adics to be “base 1/p” with the digits flipped horizontally round the “decimal point”.
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The plane will not take off, because as the wheels spin at infinite speed (c), they have infinite mass. The plane thrusters are working fine and will rotate the body of the plane into the very fast-moving runway.
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The whole issue is the wording of the problem.
If the problem is meant to convey that the plane has no forward motion relative to the air around it, then it cannot attain any airspeed, and then cannot take off.
If the problem is meant to convey anything other than the plane is somehow managing to stay completely stationary relative to the air around it, then of course it will eventually take off.
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I doubt anyone wants to read another analysis of this… but I’m posting it anyway.
Problem Statement:
“Imagine a plane is sitting on a massive conveyor belt, as wide and as long as a runway. The conveyer belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?”
Ok… I think I know what various groups use as their various reasons, but here’s my thinking…
“The conveyer belt is designed to exactly match the speed of the wheels”
This phrase is very important. Now, a rotating object (like a wheel or belt) has two speed components. It has translation and it has rotation. The conveyor belt, of course, has a net translation of zero. It’s stationary. There’s also a translation component for the point on the belt that touches the tire. This, of course, is not zero if the belt is spinning. We do know that, unless the tire is skidding, the translation of the point on the tire touching the belt must equal the translation of the point on the belt touching the tire. Also, the translation of the wheel axle must be equivalent to the translation of the plane.
Ok… thing is, this is a bad question. We can take the quoted phrase to mean one of three things.
Option 1 – The net translation of the wheel equals the net translation of the conveyor belt. This basically means that the plane isn’t allowed to move forward.
Option 2 – The rotational speed of the belt equals the rotational speed of the tire. This is also a bad option, since we don’t know anything about the radii at work and so a smaller tire on a big runway would move backwards and a big tire on a small runway would move forwards.
Option 3 – The point on the wheel contacting the belt has a translational speed equal to the point on the belt contacting the wheel. Problem with this statement is that it only ensures no skidding. Think about it… the same thing can be said about a regular runway. Unless the tire is skidding, those points will always have the same speed.
So… where does that leave us? Well….
Option 1 – The plane can’t take off because it can’t move
Option 2 – Not enough information
Option 3 – Since planes can take off on regular runways, they can obviously take off on this as well.
Now… let’s be clear… if the question had said “WILL the plane take off” instead of “can”, we’d have an issue. Here’s why…. if you assume Option 3 was the intended option (as I do), then any case where the wheel isn’t skidding is a possible case. You could imagine, then, a case where the conveyor belt is designed to accelerate its rotational speed. In this case, the acceleration of the belt would induce a backwards force on the plane that could theoretically counteract the thrust of the jets. Now, obviously, the available friction force will be insufficient to resist the thrust force in basically every real world case, but we’re talking hypotheticals right?
Moreover, if this were the case, then both the translation of the contact point and the net translation would match, which actually best aligns with the statement that the belt must “exactly match the speed of the wheels”, because it matches in both translational forms.
In fact, if you wanted to go further and consider giant wheels, you could even say (still within the bounds of the problem) that the translational speed of the contact point, the net translational speeds, AND the rotational speeds would all be exactly the same except opposite by having the wheels be as big around as the runway belt. In this case, all three interpretations would be satisified and the plane could be unable to take off.
Except… consider any scenario with the accelerating belt inducing a backwards force that’s countering the thrust. After some unknown amount of time, that belt would be moving really really fast and still accelerating. Air directly against the belt would form a boundary layer in which the air against the belt is moving the same speed as the belt and the air further away is stationary. Eventually, this boundary layer would probably grow so large that the spinning belt itself would thrust enough air past the plane’s wings to generate lift.
This is assuming, of course, that the plane’s wings aren’t sufficiently flat that the effect of this flow over the wings would be counteracted by the pressure differential due to the air speed differential through the boundary layer (which would cause a downward suction). Faster air under the wings than above them (which is what the boundary layer would cause) would result in lower pressure under the wing, but this could be overridden by a sharp angle on the wing.
So can the plane take off? Sure. As long as the treadmill isn’t accelerating. Can the parameters of the problem be satisifed and somehow disallow takeoff of the plane? In a hypothetical world, absolutely.
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The grass is always greener in the other envelope, on the other plane, on the other treadmill.
::: claps with one hand then exits :::
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Amen, Erika! I’ve been wondering where the logician went in comic 435 for a long time now! Possibly years! My theory is that she just recently switched from Aristotelian to Boolean logic, and is now exploring the murky depths of the empty set (thus, unavailable for comment).
Also, it pains me and slightly depresses me that this comment thread has, even after our good sir Randall’s exhortations to the contrary, devolved into circular discussion of the original problem and its various mutations. In the end, the answer is that (whether this is a logical or a physical problem) neither the plane nor the treadmill exists outside of the internat, and since those particulars cannot be proven to exist outside of the internat, it may well be that the general ideas which they manifest cannot be proven to exist outside of the internat either (especially if you’re a nominalist).
As someone who has actually taken the time to read through many of the posts (find me something better to do at work; I dare you), I am going to move for an immediate referendum on the idea of changing the primary object of consideration here from the problem originally outlined in the blag post to the newer and much more intellectually challenging problem outlined in the comment thread by our good sir RM.
On the subject of that second thought problem, a friend of mine pointed out that we have yet to prove at least one of its underlying axioms: that goats have pockets. I figure we should assume the converse and try for a reductio d’ absurdum. In that manner, we should at least be able to prove whether such a goat as whose form you have assumed (grammar check?) can determine the denomination and number of a wad of soaking wet dollar bills.
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The Two Envelopes Paradox seems like a watered-down version of the Monty Hall Problem.
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??? ??? ???? ??????? ??? ???? )
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i believe in number 2 haha. you are totally awesome! i love the comics… i just started checking out the forums. man what you do is a laugh a minute, even tho i’m into the liberal arts, i like math and science… none of my not so smart frends understand your jokes, but i laff at all of them… they think i’m nuts because to them it doesn’t make sense. thanks for putting something other than porn on the internet 🙂
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