Engineering question
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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?
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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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 :-)
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:
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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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No. The reason being
Amarnath S wrote:
match the speed of the wheels
the result of which is that, no matter how hard the plane pushes forward, the conveyor belt pushes the same wheels, with a nonzero mass, back exactly as fast as the wheels are moving forward relative to the conveyor belt, a process which is not limited by anything (say, the wheels will spin at an incredible rate, and they it were a real-world situation, the wheels would explode from centrifugal forces); so that the total speed of the plane relative to the ground is zero. So the plane's thrust is used completely to drive the wheels' reaction force. It cannot be anything different, or else the exactly matching speed in the opposite direction would no longer hold true. So the wings do not catch any wind and the plane stays on the ground. All the force used by the plane for its attempt to take off is put into the wheels spinning. And this conveyor belt, having to make those same crazy speeds in the other direction, would be a mighty impressive piece of work. This is why it is important that you carefully read the question, be precise in what it states and what it does not state, not read carelessly, and not jump to conclusions.
Martijn Smitshoek wrote:
So the plane's thrust is used completely to drive the wheels' reaction force.
Do you happen to have a model (toy) airplane with free running wheels? If you also have an option to mount the the spare tire of you car so it can spin around, it can serve as a model conveyor belt. Now start the tire (/conveyor belt) spinning, with the model plane on top of it, let your hand serve as model engines. You claim that there is no way that your hand can give the plane a thrust forward to throw it into the air. Even though the wheels are free running, in some magical way, they will convey a counter force against your hand making it impossible for the hand (/the engines of the plane) to push the plane forwards at a speed enough to give the plane a lift. Obviously you can push the model plane forwards, even with the tire spinning ahead beneath it. You claim that if your hand is replaced with real engines, providing same thrusting power as your hand did, that thrust will be unable to move the plane ahead the way your hand did. I do not see what makes the principal difference between the thrust from your hand on the plane body and the thrust from the engines on the plane body. You maintain that there is a principal difference, or alternately, that as long as those free running wheels touch the rotating tire conveyor belt, your hand can't possibly move the model plane forward. I'd certainly like to know which one of these two alternatives you go for, along with a good justificaytion.
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The conveyor stuff is just complication. The entire point of the conveyor is to prevent lateral movement. Mythbusters had no budget for a real conveyor and their facsimile wasn't a conveyor and didn't prevent lateral movement. The weight of the plane and the stretch of the material allowed for lateral movement. A dynamometer would have been better. Sure, the wheels don't matter. So take the wheels off and jack the plane onto cinder blocks. Same concept, only without the fake-conveyor nonsense that lets people think you can have lift in no wind with no lateral movement. Jet, plane, whatever... If you prevent it moving forward, it's not going to move upward just because you tilt a control surface. Tie a sea plane off by its rear to a dock. Throttle slightly to get all the slack/stretch of the line out, then push to full. It would remain more or less stationary. "Prop wash" lift is a thing in R/C aircraft where thrust-to-weight blows pretty much all real planes out of the water. You can hover some R/C planes like a helicopter. You might be able to do that with some real sport planes, but I doubt it.
jochance wrote:
If you prevent it moving forward
But what would prevent it from moving forward? The engines, whether jet engines or propellers, thrust the plane in the forwards direction, and there is nothing from stopping it. The free running wheels will not stop it, even if they are spinning at a fairly high speed.
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The ONLY way the airplane can take off is if the speed of air respect of the airplane's speed is equal to the minimum speed the airplane needs to take off when the wind is absolutely calm. That is, because the conveyor makes the plane to be static respect to the ground, the only way the plane will take off is if there is a really hard hurricane that accelerates de wind to the plane's take off speed.
gervacleto wrote:
because the conveyor makes the plane to be static respect to the ground,
How does it do that, with free running wheels? The wheels are the conveyor belt's only contact with the plane, and I cannot see how you can enforce a thrust of the same magnitude (but opposite direction) as the plane engines, through free running wheels.
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The airplane does or does not lift off owing to the upward airflow forces on the wings. If the air does not move relative to the airplane (or vice versa) the plane will stay on the ground. And the speed of the conveyorbelt relative to the air close by will cause some drag, therefore some lift, but it is likely to be way too little, unless you add a quite signifant ventilator to help. That should be pretty obvious, but I miss the joke - if there is one?
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The speed of the wheels is irrelevant - they aren't powered or driven in any way. The speed of the airflow over the wings is the source of lift, along with the wings angle of incidence. Assuming the conveyor is moving at the same speed (but in the opposite direction) as the aircraft would be on a normal runway then the aircraft would actually be stationery and the airflow over the wings would effectively be zero and thus not generating any lift?
Leo56 wrote:
Assuming the conveyor is moving at the same speed (but in the opposite direction) as the aircraft would be on a normal runway then the aircraft would actually be stationery and the airflow over the wings would effectively be zero and thus not generating any lift?
I sure would like to see that airplane sitting there on the runway with the engines running at full power, but the plane is standing completely still because its wheels are spinning around. Nothing else is holding the plane back, and the wheels are free running, but in some magical way they still manage to cancel out the full power of the engines. I'd sure like to see that happen. And also to have a reasonable explanation how it can be possible.
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Why wouldn't the engines provide exactly the same thrust on the plane body, giving it the same forward acceleration as on a non-belted runway?
My understanding was that the engines were not used. If I took that one wrong, sorry. But engines give forward thrust irrespective of the behavior of the wheels (assuming they can turn), so in that case the plane would go forward anyway, the wheels just having to turn faster in response to the belt.
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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:
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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.
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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My understanding was that the engines were not used. If I took that one wrong, sorry. But engines give forward thrust irrespective of the behavior of the wheels (assuming they can turn), so in that case the plane would go forward anyway, the wheels just having to turn faster in response to the belt.
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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?
Maybe I'm understanding something wrong. But to your questions:
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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.
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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)
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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:
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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 whate
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Randor wrote:
Derek Muller was able to show that multiple physics professors at prestigious universities didn't even understand the basic underlying principles.
Along that note, here is one of the worst offenders I've ever seen. Robert Benfer, an anthropology teacher at Missouri, effectively says that once the sun sets, the sky immediately gets dark. Search for the phrase: "Our simulation for that latitude..." I've uploaded an outline of the sky back then at 22° sun depression, highlighting the fox he is talking about. The only way you could say the tail was visible is if you weren't taking twilight into account. If you are going to be studying the astronomy of ancient cultures, understanding twilight should be pretty high on the list of things to comprehend. :laugh: :laugh: :laugh: He even failed to understand when I pointed it out to him. :doh: “And because scientists are first and foremost human beings, they’re loathe to change their theories or their minds because of mere data.” - Glen Hodges
Our Forgotten Astronomy | Object Oriented Programming with C++
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Degree in physics, as if it mattered: No. The lift depends on airflow over the airplane wings. There will be none.
Not quite accurate. The airflow over the wings is only part of the lift needed to fly. Modern aircraft are too heavy to use the Bernoulli Principle to fly. Instead they use the redirected air flow from the belly of the fuselage. Watch an aircraft in flight - the nose is always higher than the tail and the plane is staying aloft from Newton's 3rd Law of motion. The force keeping the plane in the air is the air being deflected down by the slope of the fuselage.
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Martijn Smitshoek wrote:
So the plane's thrust is used completely to drive the wheels' reaction force.
Do you happen to have a model (toy) airplane with free running wheels? If you also have an option to mount the the spare tire of you car so it can spin around, it can serve as a model conveyor belt. Now start the tire (/conveyor belt) spinning, with the model plane on top of it, let your hand serve as model engines. You claim that there is no way that your hand can give the plane a thrust forward to throw it into the air. Even though the wheels are free running, in some magical way, they will convey a counter force against your hand making it impossible for the hand (/the engines of the plane) to push the plane forwards at a speed enough to give the plane a lift. Obviously you can push the model plane forwards, even with the tire spinning ahead beneath it. You claim that if your hand is replaced with real engines, providing same thrusting power as your hand did, that thrust will be unable to move the plane ahead the way your hand did. I do not see what makes the principal difference between the thrust from your hand on the plane body and the thrust from the engines on the plane body. You maintain that there is a principal difference, or alternately, that as long as those free running wheels touch the rotating tire conveyor belt, your hand can't possibly move the model plane forward. I'd certainly like to know which one of these two alternatives you go for, along with a good justificaytion.
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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Maybe I'm understanding something wrong. But to your questions:
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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.
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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)
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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 whate
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 lo
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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.
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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No. The reason being
Amarnath S wrote:
match the speed of the wheels
the result of which is that, no matter how hard the plane pushes forward, the conveyor belt pushes the same wheels, with a nonzero mass, back exactly as fast as the wheels are moving forward relative to the conveyor belt, a process which is not limited by anything (say, the wheels will spin at an incredible rate, and they it were a real-world situation, the wheels would explode from centrifugal forces); so that the total speed of the plane relative to the ground is zero. So the plane's thrust is used completely to drive the wheels' reaction force. It cannot be anything different, or else the exactly matching speed in the opposite direction would no longer hold true. So the wings do not catch any wind and the plane stays on the ground. All the force used by the plane for its attempt to take off is put into the wheels spinning. And this conveyor belt, having to make those same crazy speeds in the other direction, would be a mighty impressive piece of work. This is why it is important that you carefully read the question, be precise in what it states and what it does not state, not read carelessly, and not jump to conclusions.
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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).
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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jochance wrote:
If you prevent it moving forward
But what would prevent it from moving forward? The engines, whether jet engines or propellers, thrust the plane in the forwards direction, and there is nothing from stopping it. The free running wheels will not stop it, even if they are spinning at a fairly high speed.
That's the definition of an actual conveyor belt for purpose of this thought experiment. If the conveyor moved backwards as fast as the plane would otherwise be moving forwards there would be 0 horizontal movement, and so, 0 air flow to generate lift. Prop thrust is what gets the plane moving forward, it has nearly nothing to do with lift. So, simplify it further and take the plane bit out of the equation a moment. Make it a car instead. If X RPM of a propeller will make it top out at 20 mph, use gearing to make that same RPM drive these car wheels at 20 mph. Put that car on conveyor moving the other way at 20 mph. You're telling me that car goes forward? No.
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That's the definition of an actual conveyor belt for purpose of this thought experiment. If the conveyor moved backwards as fast as the plane would otherwise be moving forwards there would be 0 horizontal movement, and so, 0 air flow to generate lift. Prop thrust is what gets the plane moving forward, it has nearly nothing to do with lift. So, simplify it further and take the plane bit out of the equation a moment. Make it a car instead. If X RPM of a propeller will make it top out at 20 mph, use gearing to make that same RPM drive these car wheels at 20 mph. Put that car on conveyor moving the other way at 20 mph. You're telling me that car goes forward? No.
jochance wrote:
If the conveyor moved backwards as fast as the plane would otherwise be moving forwards there would be 0 horizontal movement, and so, 0 air flow to generate lift.
The only force from the conveyor belt on the plane to make it move backwards is the friction in the wheel bearings, which should be rather small. You have the same friction at a normal take of, so the engines are dimensioned to handle it. Compared to the force required to accelerate the plane to take off speed. If the conveyor belt somehow has managed to get the plane up to take off speed in reverse, it must have taken a tremendously long time (and tremendously long conveyor belt) But once done, the turbines have no bigger problem pushing the plane up in speed on top of that conveyor belt. When they have done the same amount of work as is normally required to reach take off speed, the plane is standing still relative to the ground. Any further work done by the turbines will accelerate the plane relative to the ground, in the normal manner. The epsilon force from the conveyor belt, through the friction in the bearings, will not stop it. So getting off the ground will take roughly twice as long (and twice as much fuel), but that doesn't prevent it from happening.
So, simplify it further and take the plane bit out of the equation a moment. Make it a car instead.
That makes a completely different situation. The wheels of a car are not free running, but tightly connected to the car engine. The acceleration of the car is caused by the force of the rotating wheels on the ground / belt. Take the belt away, leaving the car floating in free air, and it can neither speed up nor brake down. Suspend a (motorized) model plane in a string, and start the propellers / turbines: It will pull ahead. So if the original question was talking about a car, then you would be right. But it didn't. That is the essential 'trick question' part: Most of us will think of the way a car accelerates (completely dependent on a solid grip on the road) and overlook that planes are completely different (totally independent of any grip on the runway) with respect to propulsion.
