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Then you did not fully understand this part: designed to exactly match the speed of the wheels This is an extreme requirement and by no means I will claim that it is realistic. However, this is how it was was written down.
The moment that you try to push the model plane, the tire will accelerate with, possibly, hundreds, thousands of Gs, or even more until it satisfies the requirement that the tire's speed equates that of the plane's wheels. If your applied force creates a speed vector v to the wheels, the tire will make a speed of exactly -v to the wheels, which can only result in a standstill.
The result of the equation could be that while you are (maybe gently) pushing the plane, the tire might already be going at Mach 1 the other way, just so that the acceleration (and friction) on the tiny wheel becomes enough to stop that little plane from moving.
And, to get back to your question regarding the difference between hand force and engine thrust, there is no difference, both will not work.
Had the question been that the belt moved as fast as the plane's body, it would have been a different story and the plane would probably be able to take off the conventional way.
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I certainly understood the requirement, and also the lack of realism, but argued based on that premise.
Martijn Smitshoek wrote: he tire will accelerate with, possibly, hundreds, thousands of Gs, or even more until it satisfies the requirement that the tire's speed equates that of the plane's wheels
Why is that? "hundreds, thousands of Gs"?
I do not know the typical acceleration of a plane at take off, but doubt that it exceeds 1G. Whatever the value, it is the acceleration of the wheels during a normal take off. To follow up this, the moving conveyor belt would have to accelerate correspondingly. If you are capable of building a conveyor belt running at (backwards) take off speed under the plane, I am quite sure you can handle that quite moderate acceleration as well.
It also depends on how you read the initial premise: The conveyor belt should match he the speed of the wheels, but at what time? If the belt moves backwards at take off speed when the plane is standing still, and maintains this speed, the relative speed of the wheels to the belt would be twice the take off speed at actual take off. But if the belt's backwards take off speed relative to the wheels should be kept steady, the belt must be braked down when the turbines start working, down to standstill at the moment the plane lift, and is at takeoff speed.
As long as there is physical contact (friction) between the belt / runway and the wheels, you may assume that the wheel thread moves at speed zero relative to the belt / runway - any skidding is (or should be) at the microscopic level).
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trønderen wrote: Why is that? "hundreds, thousands of Gs"?
Because action equals reaction, so that if the plane attempts to accelerate with a typical force to move, say, 40 tonnes, at 100~200 km/h takeoff speed, the reaction force must be generated in its wheels because in these mechanics, the wheels will be free running so the only thing to stop the plane from moving is their rotational inertia. Now, considering that the wheels combined are much lighter than the rest of the plane, they will have
to spin much faster to counteract the engine's thrust and keep the speed equation intact.
trønderen wrote: match he the speed of the wheels, but at what time?
Doesn't say, therefore, at all times.
trønderen wrote: If the belt moves backwards at take off speed when the plane is standing still,
That is based on the way that you put an arbitrary restriction to the rules of the game, so you're trying to answer to a different question than what is being asked. I am not going there.
trønderen wrote: any skidding is (or should be) at the microscopic level).
I agree in that that effect is negligible.
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Martijn Smitshoek wrote: if the plane attempts to accelerate with a typical force to move, say, 40 tonnes, at 100~200 km/h takeoff speed, the reaction force must be generated in its wheels because in these mechanics, the wheels will be free running so the only thing to stop the plane from moving is their rotational inertia.
The engines transfer no energy at all to the wheels to accelerate the plane. The wheels just roll along, with as little friction as possible. The turbines / propellers are what pushes the plane ahead, and work directly on the plane body, with no mechanical coupling to the ground / belt. Action / reaction considerations relate to the plane with its turbines or propellers and the surrounding air, not at all to the wheels / belt.
Sure, if the conveyor belt has managed to accelerate the plane backwards to take off speed, when the turbines are fired up, they must first stop that movement. It takes as much energy to decelerate 40 tons from 100 km/h to zero as it takes to accelerate 40 tons from zero to 100 km/h. But when that is achieved, the turbines continue working, pushing the plane up to (positive) take off speed, relative to the ground and air - but twice the take off speed relative to the belt. That doesn't matter, as the only connection between the belt and the plane is through the free running wheels that just roll along at twice the speed they are used to. Maybe the will be getting hotter than normal, but they will not keep the plane from rising into the air.
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I've been trying to explain this like 3, 4, 5 times and I am not willing to do it again. I have been polite enough and made it perfectly clear that you are frivolously skipping parts of the question just so that you can make up your own story, and I am done with that. I'm sorry.
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Wrong 
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Could be my view is too simplistic but I imagine a plane sort of like a submarine, needing to be buoyant in much lighter fluids (air). For this to happen you need quite a volume of air flowing under the wings as opposed to over it, giving it 'lift'.
Having no significant flow of air while basically staying in one place, I think, is not gonna help. Maybe you'd need a vertical vortex?
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JP Reyes wrote: Having no significant flow of air while basically staying in one place,
Are you saying that with the engines under the wings running at full power, those free-running will have the power to hold back the plane with the same force (but in the backwards direction) as the thrust from the engines in the forwards direction, to make the plane stand still?
I guess that would cause so much stress on those wheels that the would break apart. With no wheels on that conveyer belt, the plane would be free to fly away 
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Well if I understood correctly, the conveyer belt is meant to match the speed of the wheels even at full engine thrust. I Don't know if the wheels have a speed threshold with all that weight, one would imagine the rubber does have it's limits (heck I even bet the conveyer belt would buckle long before the jet engines go to full power)
Realistically I can only imagine the most catastrophic take off (I think the wheels would be useless for landing and the huge conveyer belt tarmac, broken and in tatters). Nonetheless I would have to agree with you.
But referring the original (very hypothetical) question:
Quote: If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ...
Can the airplane take off?
Barring the unlikely existence of such a conveyer belt and matching powerful set of wheels, I would still say it doesn't take off. Unless convinced otherwise, it's the volume of the air flowing under the wings that matter, not the volume of air flowing through the turbines.
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JP Reyes wrote: Well if I understood correctly, the conveyer belt is meant to match the speed of the wheels even at full engine thrust.
Sure, but that doesn't null the thrust.
If the conveyor belt is running at takeoff speed before the engines are started, then you fire up the engines and zip down the runway (/conveyor belt), when the plane lifts off the ground the wheels is spinning at twice the takeoff speed (unless the conveyor belt has been slowed down as the plane accelerates, to maintain the 'wheels spinning at takeoff speed).
I am not into construction of air planes, but I wouldn't be surprised if twice the normal takeoff speed is well within the safety margins for the wheels. In any case, it doesn't affect the principal question of whether the plane could take off.
Unless convinced otherwise, it's the volume of the air flowing under the wings that matter, not the volume of air flowing through the turbines. The air flowing through the turbines would give the plane a forward speed that would cause an airflow over and under the wings.
Or are you suggesting that the thrust from the turbines are nulled out because the free running wheels are spinning around? What are the mechanism behind this canceling? Assume that the wheels for some reason started spinning mid-air, would this cancel the thrust from the turbines as well, so the plane crashes? Or does it require the wheels to be in contact with the ground for the thrust to be nulled out?
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Maybe I'm understanding something wrong. But to your questions:
Quote: Or does it require the wheels to be in contact with the ground for the thrust to be nulled out?
If I am to understand correctly, the wheels spin because of the thrust of the turbines. There's no other transmission system attached to them so their speed is directly proportional to the turbines action of moving the plane forward.
The conveyer built is moving backwards at that same speed, but turbine and conveyer belt are playing a tug of war as to which direction the plane should be moving.
Quote: Assume that the wheels for some reason started spinning mid-air, would this cancel the thrust from the turbines as well, so the plane crashes?
No, wheels in the air are negligible, but I do think they do spin anyway at take off speed (unless there's some kind of brake to make them stop vibrating and fold gently into the body)
Quote: Or are you suggesting that the thrust from the turbines are nulled out because the free running wheels are spinning around? What are the mechanism behind this canceling?
No, as in the first answer, the wheels simply provide a means (a medium) for the turbine to move the plane while on the ground. The only thing cancelling the thrust is the conveyer belt itself, using some sort of smart engine that compensates for the force of thrust (for this experiment the plane itself transmits said variable to the tarmac/conveyer belt)
I could be imagining things wrong but in your scenario:
Quote: If the conveyor belt is running at takeoff speed before the engines are started, then you fire up the engines and zip down the runway (/conveyor belt), when the plane lifts off the ground the wheels is spinning at twice the takeoff speed (unless the conveyor belt has been slowed down as the plane accelerates, to maintain the 'wheels spinning at takeoff speed).
The plane already has some thrust in order to taxi itself onto the conveyer belt. This belt is already moving at an incredible speed, backwards, with no weight. The plane would immediately be dragged in the wrong direction the moment the front wheel slips unto the belt (and probably spin and crash).
Say it managed to taxi onto the already rapidly moving belt, without achieving take off speed, the belt would just yank it backwards into whatever is behind the belt (ideally a green field and a ditch).
I think I need more details as to what you're picturing in your head.
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JP Reyes wrote: If I am to understand correctly, the wheels spin because of the thrust of the turbines
They spin due to the friction with the runway / conveyor belt. We may assume that the bearings are reasonably well oiled with moderate friction, and the force thus transferred from the moving runway to the plane is nowhere close to give the plane any significant backwards movement.
Even on a plain, no conveyor belt, runway, when the turbines push the plane into speed, the wheels will turn, but not because the turbines are exercising any force on the wheels, but, again, because of the friction with the runway.
The "tug of war" is an extremely uneven one, with the turbines having a firm grip on the plane body, the runway ideally has none: If the wheel bearings had no friction at all, the wheels would spin like crazy, but the inertia of the plane would keep it in place. We do not have a perfect oil giving no frictions in the bearings, but the force would be magnitudes below that of the turbines.
No, as in the first answer, the wheels simply provide a means (a medium) for the turbine to move the plane while on the ground. Are you similarly saying that for a sea plane, "the water simply provides a means (a medium) for the turbine to move the plane while on the water"? Neither the sea nor the wheels contribute at all to the acceleration of the plane (rather to the contrary, in both cases, although we should not overestimate this effect). I cannot see how either serves as any "means for the turbines" to get the plane into the air.
This belt is already moving at an incredible speed, backwards, with no weight. The plane would immediately be dragged in the wrong direction the moment the front wheel slips unto the belt (and probably spin and crash). If the wheels got stuck in their bearings so that they did not rotate, then the belt might be able to accelerate the plane backwards. I guess it would take the belt some time to get the plane into (backwards) take off speed, though; a plane has quite some inertia.
Assuming a long enough belt: After a while, the friction in the bearings - whether tiny, thanks to high quality oil, or almost infinite, because the wheels got stuck - the plane body may have been accelerated into backwards take off speed. (Especially with well working bearings, this would take an incredibly long conveyor belt!) But then the wheels would no longer be spinning.
We have another situation: When the plane is moving backwards at take off speed, will then the turbines have enough power to both break this movement and furthermore kick the plane up to forward take off speed? I think it is quite obvious if they can take the plane from zero to take off, then they can also take it from minus take off to zero - that amounts to the same total energy. If there is still fuel left (and there ought to be, after a normal take off procedure), the turbines can continue push from the now achieved zero speed and upwards. There is no reason why they should not be able to get the plane body from zero to (forwards) take off speed. The free-running wheels will spin at twice the normal rate when the plane lifts off the ground, which may cause slightly more friction in the bearings, but that will be negligible compared to the thrust of the turbines.
If, the wheels are stuck in their bearings, not rotating, but for that reason the belt is able to accelerate the plane up to backwards take off speed in significant shorter distance, the situation is somewhat different: Then the turbines, in addition to breaking down / accelerating the plane, will have to work against the friction of the locked wheels skidding on the conveyor belt. Most likely, the turbines would have the reserve power to handle this. But after the trip, the plane would most likely be without working landing wheels, and it would have to do a belly landing.
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If memory Serves Mythbusters did that and tried it with a very light plane and a really long tarp, the pilot was amazed. If I recall the wheels are not powered and only the engine which forces wind over and under the wing matters. Non-trival question! Full marks have a look on Amazon+ or You Tube for the episode. 
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Unlawful means against the law and illegal is a sick bird.
If you can't laugh at yourself - ask me and I will do it for you.
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but I don't know any French, is there any quick way to pick some basic math terms to understand this book?
diligent hands rule....
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"All the best mathematics books are in German." -- A professor my father spoke about
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Many great mathematicians were French, so maybe it's a case of teaching when you can't do. 
And in a fair turnabout, the French should then write the philosophy texts.
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it seems like the only way at this time.
diligent hands rule....
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From my experience with Google Translate, you aren’t going to get anywhere near a good understanding of the material. If the subject is elementary math (trig, calculus, linear algebra), Google Translate might fair better but then, you have many good books in English.
Maybe you should have asked the question before buying the books.
Mircea
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this is the book I got from Amazon.
diligent hands rule....
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During my younger days in India, (I still live in India), we used to get Soviet Mathematics and Physics books published by Mir Publishers. Deep mathematical content, and very educative.
There were some among us who liked American/British books as compared to Soviet books, but I found Soviet books also quite nice. Learnt a lot from them. Some of them are now available online for download at archive.com - Mir Titles : Free Texts : Free Download, Borrow and Streaming : Internet Archive[^]
Examples are Piskunov - Differential and Integral Calculus, and Demidovich - Problems in Mathematical Analysis, beautiful book with a great set of problems.
modified 9-Apr-22 23:27pm.
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thanks for this great link on Mir books!
diligent hands rule....
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I am not that old, but I am just old enough to have been gifted a few books from Soviet publishers.
Oddly enough, a couple were English books (as in, they taught English) but I probably lost them now and a few were from the Folk Tales from the Soviet Union series, which I still have.
Thank you for the link, I will save it and check it out 
Cheers,
Vikram.
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It is better you view the site on a laptop, rather than mobile. Because on a laptop browser, it refreshes itself with newer books at the bottom of the page, and then it refreshes itself further and further. There is a large set of books here.
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