# Plane on Conveyor Belt **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [March 4, 2010, 1:24am UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224 "2010-03-04T01:24:22Z")\ **Posts on this page:** 16\ **Page:** 2
**Author:** ![Vorpal\_Blade](https://avatars.discourse-cdn.com/v4/letter/v/9de0a6/32.png) [@Vorpal\_Blade](https://boards.straightdope.com/u/Vorpal_Blade)\ **Post date:** [March 4, 2010, 3:32pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/21 "2010-03-04T15:32:14Z")
Noooooooooooooooooooooooooooooooooooooooooooooooo ---
**Author:** ![brewha](https://avatars.discourse-cdn.com/v4/letter/b/91b2a8/32.png) [@brewha](https://boards.straightdope.com/u/brewha)\ **Post date:** [March 4, 2010, 3:52pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/22 "2010-03-04T15:52:00Z")
The problem is poorly stated and the conditions are undefined. Depending on how you interpret the question, the plane will or will not take off. ^^^This is the correct answer. The only correct answer. ---
**Author:** ![Deeg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/deeg/32/2955_2.png) [@Deeg](https://boards.straightdope.com/u/Deeg)\ **Post date:** [March 4, 2010, 6:37pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/23 "2010-03-04T18:37:30Z")
> [@JXJohns](#): > > A plane can take off from water with floats… Hmmm…that brings up a possible experiment–would the doubters be convinced if a float-plane took off going up-river? ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [March 4, 2010, 7:01pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/24 "2010-03-04T19:01:28Z")
The problem with the original question is that in an effort to constrain the problem properly, it is unclearly worded and thus ambiguous. It doesn’t help that the question originated in Russian, so you are reading a translation - perhaps a poor translation. Here is the difficult part: > [@](#): > > This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). What does that mean? Interpretation 1: The conveyor is matched to the thrust output/airspeed indicator/whatever of the airplane such that it keeps the plane stationary with respect to the ground. Can the airplane take off? Answer 1: Of course not. Interpretation 2: The conveyor is matched to the airspeed indicator/speedometer/whatever in such a way that if the plane is trying to pull forward at 50 mph, the belt is moving exactly 50 mph backwards. Can the conveyor keep the airplane from taking off? Answer 2: No. In the first interpretation, you have defined the problem the same was as tying the tail of the plane to a wall. The plane cannot take off because it is not moving, thus not generating any airspeed, thus not generating lift. (We will ignore prop wash as negligible for most airplanes and irrelevant to the question.) The second interpretation is comparing how an airplane gets thrust vs how a car gets thrust. A car gets thrust from it’s wheels, so moving the ground backwards underneath the wheels directly counters the wheels pushing forward. An airplane gets thrust independent from the wheels - the wheels just provide support and reduce friction. So dragging the ground backwards just means the wheels have to spin faster, but the thrust comes from the engine pushing air, which means the plane can move forward. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [March 4, 2010, 9:45pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/25 "2010-03-04T21:45:36Z")
Got interrupted. **nilum** said: > [@](#): > > To further my point… imagine a plane in a vacuum with this same set up. Obviously there is no lift because there is no air flow. The plane can’t take off. But it’s not just the lack of air, but also the lack of acceleration AGAINST the air. > > You are accelerating the plane in this scenario, but you’re not accelerating it against the air current. Ever notice that a in a wind tunnel a model plane can be in a still state and attain lift once the wind is turned on? The acceleration isn’t what is important. It’s the air going around the foils at increased speed. Either you are confused, or you did not state clearly what you meant. Lift on airplane wings does not come from _acceleration_, it comes from _velocity_ - wind velocity across the wings. An airplane in a vacuum cannot get lift because there is no air, independent of any motion. An airplane with its fuselage tied to the ground such that it does not move forward cannot get lift because there is no airspeed. (I am assuming no wind for discussing the relevant issue.) However, the issue in question is, **can** a treadmill prevent an airplane from moving forward? If you _assume_ the answer is yes, then of course the airplane cannot take off. But if you _interpret_ the question differently, then that is the question being asked. And the answer to that question is “no”. > [@](#): > > Example a light craft achieves lift at 70Mph… and the platform moving against moves at 60 then the light craft has to achieve 130Mph roughly to achieve lift. > > If the moving platform was always the same speed as the craft, the plane could NEVER achieve lift. You are correct with your _ifs_, but here’s a question for you - how does the treadmill prevent the plane from achieving 130 mph? The force applied by the treadmill is through the wheels. At constant velocity, there is no force, the wheels spin. Meanwhile the plane thrust comes from the engines on the air, not related to the ground at all. Therefore, it would be like tieing the engine to a rope tied to some distant wall - the engine pulls on the rope, the rope pulls on the wall, the plane moves forward, and the wheels spin. Speed up the treadmill, the wheels spin faster, but the plane still moves forward. > [@](#): > > The propellers are really doing all the work, but the wheels are still an important part of the mechanism obviously. The wheels provide vertical structural support, and they serve to reduce friction between the ground and the fuselage. They do not add thrust. > [@](#): > > If this propeller plane starts at rest along with the moving platform, and accelerates with the platform accelerating at the same rate, but in an opposite direction the plane will appear stationary. This is an assumption on your part. How do you couple the moving of the platform to the thrust of the airplane? With a car on a treadmill, the car wheels push the belt, so if the belt is motorized at the same rate, then they cancel out. Thus a car does sit stationary with respect to the ground. But airplane wheels do not get thrust, they only spin based upon the difference in velocity between the plane structure and the runway. If the runway speeds up, the wheels speed up, without affecting the fuselage. Thus, the airplane still gets thrust from the engines pushing air, and still moves. > [@](#): > > I think the argument here is that the planes propulsion is from the propeller and not the wheels that in those cases the wheels would just magically ignore the moving platform beneath it. But it’s not the case. Even though the propeller is causing the forward acceleration, it’s still the wheels which are in contact with the ground who are responsible for the acceleration. What? No. The wheels will spin, because that is what wheels do. On an airplane, they free-spin (if the brakes are off and they are well lubricated and yadda yadda yadda). If the engines are off and the belt starts to move from zero, inertia will work to keep the plane still with respect to the ground while the belt works to move the plane backwards. Depending upon the internal friction, you might get zero plane movement or partial plane movement or full plane movement. In the case of the airplane engines pulling, the wheels don’t “magically ignore” the moving platform, they function as wheels function. The surface of the wheel in contact with the moving platform surface moves backwards, the wheel spins, and the contact point on the wheels changes. If the fuselage of the plane were hung from a crane such that the wheels made contact but the fuselage could not move with respect to the ground, then the treadmill could run at any speed and the wheels would spin. If you take away the crane and instead apply thrust on air from the airplane engines, the wheels can still spin. But if the wheels free spin, then they speed up to accept whatever velocity difference is required. Try this - take an inline skate. Gently roll one wheel by hand. How fast does it spin? As fast as it has to. Now give that wheel a solid spin, so it goes very fast. How fast does it go? as fast as it has to. That is the bearing in the wheel doing its job. The same thing applies to the airplane wheels - they have a free-spinning bearing, and so will spin as fast as they have to. > [@](#): > > Water isn’t the same surface as say concrete or rubber. So moot point. Not a good example at all. You could theoretically have a take off without pontoons as well. Not sure what you think you are saying. Water is a different surface than rubber or concrete, sure, but it still provides friction. Seaplanes are designed to do two things: float when not under power, and glide across the water when under power to limit the friction. That is exactly the function of wheels on an airplane - limit the friction. > [@](#): > > Exactly… they are just floats. They don’t serve the same purpose as wheels on the ground. What kind of seaplane are you envisioning? [This](http://www.flyingthefish.com/images/seaplane_cape_town.jpg) seaplane uses the fuselage as the skid. The pontoons under the wings mostly work as float supports while the plane is stationary. Once under thrust, the plane hydroplanes on the belly and those pontoons are not touching the water. [This seaplane](http://www.aopa.org/images/whatsnew/newsitems/2005/050429seaplane.jpg) uses the two pontoons just like a normal plane uses its landing gear. They are the floats while stationary (structural support), but also the hydroplaning surface that reduces friction for take off. So on this plane, yes, the pontoons are precisely like the wheels. > [@](#): > > Lets add a lubricant to the moving platform that makes the friction almost 0. > > Now accelerate the plane and the platform at the same rate, but in opposite directions. > > The plane will stand still. Accelerate them with respect to what? Understand: when a car drives or a person walks on a treadmill, the treadmill belt is accelerated with respect to the ground, but the car or person is accelerated _with respect to the belt_. This is because the thrust comes from pushing on the belt. An airplane does not get thrust from pushing on the belt. When the airplane accelerates, it is also with respect to the ground. Both the belt and the airplane fuselage are speeding up with respect to the ground. Ergo, the interface between the two is going twice as fast. **nilum** > [@](#): > > It kind of makes sense and it kind of doesn’t. > > Someone needs to test it out with a real jet. Not a remote control plane, but a real airliner. Mythbusters did. Or you could get a treadmill, some inline skates, and a rope. Tie the rope to the wall in front of the treadmill, wear the skates and stand on the treadmill, hold onto the rope, and turn on the belt to some speed (5 mph or whatever). Then pull on the rope and see if you move forward or if the belt keeps you in the same place. **Frylock** said: > [@](#): > > I know Mythbusters did it, but to my recollection not everyone was happy with the way they interpreted the situation. I haven’t seen the episode myself. The Mythbusters interpreted the question as “Can the treamill keep the plane from moving forward”, so they demonstrated that a treadmill cannot keep the plane from moving forward. Well, they did by accerating at the same profile to the same set speed. They did it with a model plane on a treadmill for concept demonstration, they did it with a model on a sheet of paper for scale test, and then they did it with an ultralight airplane and a giant sheet of canvas. The airplane always took off. Some people interpret the question as “If the treadmill keeps the plane from moving forward, can the plane take off?” They did not try to replicate that, because that could just as easily be replicated by keeping the brakes on, the wheel chocks in place, and a chain to the fuselage. **nilum** said: > [@](#): > > But I still say that there is enough friction between a commercial jet and the ground to keep it on the ground due to gravity or what not against the surface of a moving platform. Why? On what basis do you make that assertion? You agree the whole point of wheels is to reduce friction, correct? So if the wheels are doing their job, how does the friction magically increase? > [@](#): > > If say the plane was on something like a chassis dynamometer… could it still take off? Would it jump off the dynamometer or would it stay stationary. This once again highlights the difference in interpretation of the question. If you _assume_ the airplane cannot move with respect to the ground, the dynamometer makes sense, because there is no forward motion. But if you think the question is whether or not the moving surface can keep the fuselage from moving, then you need a longer surface so the wheels have somewhere to go. On the dynamometer, depending upon how deep the groove an how low the wheels sit in the groove, the airplane would either be stuck because it would function like wheel chocks, or the plane would roll over the front wheel drum and then take off. So that is a worse design, because it is dependent upon how the thing is built, not what principle is in work. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [March 4, 2010, 11:19pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/26 "2010-03-04T23:19:28Z")
> [@Irishman](#): > > Some people interpret the question as “If the treadmill keeps the plane from moving forward, can the plane take off?” They did not try to replicate that, because that could just as easily be replicated by keeping the brakes on, the wheel chocks in place, and a chain to the fuselage. As I said in one of the other threads, in this case the question becomes ‘If you make it impossible for the airplane to take off, will it take off?’ ---
**Author:** ![mathius\_dragoon](https://avatars.discourse-cdn.com/v4/letter/m/c67d28/32.png) [@mathius\_dragoon](https://boards.straightdope.com/u/mathius_dragoon)\ **Post date:** [March 5, 2010, 10:42pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/27 "2010-03-05T22:42:01Z")
On a normal runway, the plane’s propeller exerts a backwards force on the air around it, and due to Newton’s third law, the air exerts a forward force on the propeller. This, then creates a forward force on the plane. When this forward force exceeds the opposite force of the static friction of the tires on the ground, the plane will accelerate forward. On a moving conveyor, the wheels still have friction with the surface, albeit kinetic rather then static. The propeller still exerts a force on the air around it, and thus the air exerts a force on the plane. The airplane on the conveyor needs _even less_ force from the propeller to overcome the much smaller kinetic friction (compared to static). ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [March 8, 2010, 5:19pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/28 "2010-03-08T17:19:38Z")
You are incorrect. Rolling friction is static friction. Kinetic friction is sliding between the two surfaces. Rolling is not sliding, it is coming in contact and then going out of contact, but the contact point on the tire stays the same point on the ground for the entire time that part of the tire is in contact with the ground. If not, you do not have a roll, you have a skid. The forces that the treadmill exert are pulls on the wheel hubs through the tires. The friction force does not change. Inertial force of trying to speed up the spin of the wheel and internal friction of the bearings are what resist plane rolling forward. ---
**Author:** ![mathius\_dragoon](https://avatars.discourse-cdn.com/v4/letter/m/c67d28/32.png) [@mathius\_dragoon](https://boards.straightdope.com/u/mathius_dragoon)\ **Post date:** [March 10, 2010, 5:31pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/29 "2010-03-10T17:31:46Z")
> [@Irishman](#): > > You are incorrect. Rolling friction is static friction. > > Kinetic friction is sliding between the two surfaces. Rolling is not sliding, it is coming in contact and then going out of contact, but the contact point on the tire stays the same point on the ground for the entire time that part of the tire is in contact with the ground. If not, you do not have a roll, you have a skid. Doh! I completely missed that. Thanks for pointing that out. I think my main point is still valid though. I was trying to say that the speeds of the plane/ treadmill make no difference because the forces are what are important. I.E. the plane needs acceleration to take off, not a net positive speed (necessarily). ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [March 10, 2010, 10:36pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/30 "2010-03-10T22:36:24Z")
Yes, it has already been mentioned (though perhaps not in this thread) that the belt moving at a constant velocity does not impart a force on the airplane. However, speeding up the belt does transfer force to the airplane. The argument then becomes “What does it take for the speeding up of the belt to overcome the force of the engine pushing on the air which is trying to move the airplane forward?” ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [March 11, 2010, 5:53pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/31 "2010-03-11T17:53:39Z")
> [@Irishman](#): > > Yes, it has already been mentioned (though perhaps not in this thread) that the belt moving at a constant velocity does not impart a force on the airplane. Not so. The belt moving at a constant velocity _does_ apply some force to the airplane, unless the wheel bearings are (impossibly) perfectly frictionless. We can _assume_ they are, if you like, but it is necessary to _say_ that this is an assumption, not a fact. It is unlikely that a treadmill could be produced that would increase this force sufficiently to stop any real airplane from taking off, but that cannot be affirmed with certitude unless an actual quantitative analysis has been performed. ---
**Author:** ![Cheshire\_Human](https://avatars.discourse-cdn.com/v4/letter/c/5f8ce5/32.png) [@Cheshire\_Human](https://boards.straightdope.com/u/Cheshire_Human)\ **Post date:** [March 14, 2010, 6:47am UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/32 "2010-03-14T06:47:55Z")
> [@brewha](#): > > The problem is poorly stated and the conditions are undefined. Depending on how you interpret the question, the plane will or will not take off. > ^^^This is the correct answer. The only correct answer. Also, in fact, any interpretation that has the plane NOT taking off requires materials that are not available in the real world. They violate the laws of chemistry and physics. You can’t have infinite or zero friction, you can’t have infinitely powerful power sources to run your treadmill, you can’t get treadmill material with infinite tensile strength, and it can’t run at infinite speed. If you limit those interpretations to real world materials, then the plane takes off as soon as the treadmill reaches it’s real world limitations, unless the plane’s landing gear breaks first, in which case you get a plane crash. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [March 14, 2010, 12:31pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/33 "2010-03-14T12:31:36Z")
> [@Cheshire\_Human](#): > > Also, in fact, any interpretation that has the plane NOT taking off requires materials that are not available in the real world. …or a “plane” specifically redesigned to increase energy losses in the wheels. Redesigned to _not_ take off, in other words. ---
**Author:** ![Cheshire\_Human](https://avatars.discourse-cdn.com/v4/letter/c/5f8ce5/32.png) [@Cheshire\_Human](https://boards.straightdope.com/u/Cheshire_Human)\ **Post date:** [March 14, 2010, 1:23pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/34 "2010-03-14T13:23:06Z")
> [@zut](#): > > …or a “plane” specifically redesigned to increase energy losses in the wheels. Redesigned to _not_ take off, in other words. True, I hadn’t thought of that one. That would result in a broken plane, because that would be designing the thing to be the first to break. You can either break the plane or the treadmill, unless you assume physically impossible materials. Otherwise, the only alternative is that the plane takes off. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [March 15, 2010, 7:33pm UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/35 "2010-03-15T19:33:25Z")
Yes, you can redesign the airplane to be a lousy airplane, and then prevent the airplane from taking off. There are any number of special cases one can create where the results may be different. Assume an ultralight airplane with overpowered engine and extra large prop - it can get lift just from propwash (i.e. the air the prop pushes flowing over the part of the wings around it). Then you could leave the airplane’s brakes on and still get lift. Same thing for a Harrier (or any plane with VTOL). But if we are trying to examine the general case, not rigged special cases, then you must assume a normal plane with normal equipment - such as a commercial airliner or small, single-engine mass-produced plane. This means the wheels must be designed to work as wheels, not anchors. **John W. Kennedy** said: > [@](#): > > Not so. The belt moving at a constant velocity _does_ apply some force to the airplane, unless the wheel bearings are (impossibly) perfectly frictionless. Hmm, is there some sort of _Guadere’s Law_ for physics? Come in to correct someone’s physics and you’re required to make a physics error yourself? 😉 I was thinking force requires acceleration, and at constant velocity there is no acceleration, so no force. However, I stand corrected, because friction is acting to slow the wheels down, so some force must be countering that slowing force in order for the wheels to continue spinning at the same rate. Ergo, there is a force imparted from the belt moving at a constant velocity. ---
**Author:** ![markci](https://avatars.discourse-cdn.com/v4/letter/m/8491ac/32.png) [@markci](https://boards.straightdope.com/u/markci)\ **Post date:** [March 18, 2010, 4:36am UTC](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224/36 "2010-03-18T04:36:49Z")
> [@nilum](#): > > The original article can be found here: > > The flight of an airplane is based on lift and I don’t think I have to explain that in detail. > > :\ Your error is that you seem to think the plane is stationary. It is not. It is moving forward. The treadmill is just moving backward at the same rate. [Previous page](https://boards.straightdope.com/t/plane-on-conveyor-belt/531224.md?page=1)