Will a plane take off on a conveyor belt?

Discussion in 'Physics & Math' started by w00t, Jun 12, 2007.

  1. temur man of no words Registered Senior Member

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    Everytime i enter this thread I become distressed, disturbed, perturbed, and upset.
     
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  3. madanthonywayne Morning in America Staff Member

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    If the plane were a car, you'd be right. But the wheels are not propelling the plane. The plane is propelled by the force of the engines pushing it thru the air. The conveyor belt has no effect on this. The faster the conveyor belt goes, the faster the wheels turn. But this has no (or minimal) effect on the air speed of the plane because its wheels are simply spinning freely.

    Think of a kid pushing a matchbox car on a conveyor belt moving in a direction opposite to that in which he is pushing the car. The speed of the belt will have no effect whatsoever on the speed of the matchbox car because its wheels are spinning freely and imparting no force. The acceleration comes from the kid, not the wheels. The faster the belt goes, the faster the wheels spin. But their spinning is not what is moving the car forward.
     
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  5. Absane Rocket Surgeon Valued Senior Member

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    That's why I'm staying out of this now. I made some silly assumptions at first... now that I got it all sorted, I realize that it's not completely obvious...

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  7. andbna Registered Senior Member

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    Alright, lets get to an agreement here OK?
    First we are going to need to agree on some initial properties. This is what I propose

    The belt cannot apply force directly on the plane, the belt acts on the planes wheel's instead
    The the plane can apply force on its wheels, but not the belt directly
    Thus the the only relationship the belt has to the plane are the wheels

    Furthurmore, can we agree that the terrain is level (for simplicity?)
    And can we agree that the force the belt is applying to the wheels is perpendicular to the wheel surface which is touching the belt?

    from the OP, now we need to agree on what this means:
    Does it mean the perihperal speed of the wheels?
    or
    Does it mean the speed with which the center of the wheels are moving relative to the belt, or the ground or another observer?

    It would seem logical the first option is correct,
    Ok can we agree?

    EDIT: Oh, and that we will use Newtonian physics (rather than Relativity,) for a boat load of added simplicity ok

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    ?

    -Andrew
     
    Last edited: Jun 17, 2007
  8. leopold Valued Senior Member

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    let's also assume the bearings in the planes wheels are frictionless.
     
  9. andbna Registered Senior Member

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    Well, they want a more realistic scenario, because their perfect one fell apart with the fact that all force of the belt on the wheel would resault in torque and therefore not effect the speed of the airplane at all; it wouldnt move if the belt did 100mph and the plane was off, and it would move at 100mph if the plane did 100mph and the belt did 500bazillion mph

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    We could say a minute coefficient of friction (like 0.05 or something)
    However, if they will allow a coefficient of 0, it works for me

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    -Andrew
     
  10. superluminal . Registered Senior Member

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    So this is really easy. You have a plane sitting still on a conveyor belt that makes it's frictionless wheels spin madly. The plane starts to thrust and subsequently takes off.

    Done.
     
  11. andbna Registered Senior Member

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    Close but no cigar, the wheels aren't frictionless, there would be a near-infinite force of friction between the wheel and the belt. The ball-bearings and such would be frictionless (which is between the plane and wheel, or specificaly its axel and the wheel)

    EDIT: I read that as the that conveyer belt is still too... But it could also be read much more logicaly that the belt is moving while only the plan is still, causing the wheels to spin madly. In which case, exactly right, multiple cigars for you

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    and pardon my mistake.
    That was basicaly the point most people are trying to make above... So now that our default 'perfect system' is shown to be working, Time to show how even in a fairly real scenario it would still work... or wouldnt.

    -Andrew
     
    Last edited: Jun 17, 2007
  12. superluminal . Registered Senior Member

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    10,717
    Wheels don't matter at all. A plane thrusts using the atmosphere in which it sits. The plane can be on water, ice, a conveyor belt, dosen't matter. What matters is the relative thrust that the propeller/jet engine produces.
     
  13. andbna Registered Senior Member

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    Yes, however, the wheels do create minute quantities of friction, which they seem to beleive will resault in the plane being bogged down... we both know this wont happen, but they will need super-solid proof first, so thats why im trying to establish some agreement points.

    -Andrew
     
    Last edited: Jun 17, 2007
  14. andbna Registered Senior Member

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    316
    A mathematical experiment here, seeing if an F-22 could overcome the force of friction of the wheels and belt, and therby slide its way away.

    Basicaly, the only force the belt exerts on any part of the plane is that of the force of friction between the belt and the wheel. Therefore, perhaps a plane could overcome this at some point and therfore end up sliding allong the belt. Therby, no matter how fast the belt were to spin it could not increase the force acting against the plane any more.

    Now, let's say our plane taking off is an F-22 Raptor
    Well, the weight without anything loaded is 14,379 kg
    add 100Kg for the pilot
    then fuel.. I'm not sure how much would be needed to run a gas guzzler like that, but surely no more than 500KG of fuel for a take off and immediate landing?

    So we say it weighs 15,000KG (rounded up to make a nice number)
    Now the wiki sais the raptor can take off with up to 36,288Kg of wieght, leaving: 21,288 Kg more that it could have. Now if we say
    F=ma then we can figure out based on that, how much force the friction of the wheels would need to put on it to prevent it from taking off:

    given it has 2 engines each providing 155.7 kN or 155700N of thrust, we have a total of 311,400N of force. So if we have that much force, allowing it to take off with 36,288Kg of weight:
    311,400=36,288*a
    and now we need to find how much force is needed to take off with 15,000Kg of weight
    F=15,000*a

    put both in terms of acceleration and sub in:

    311,400/36,288=a
    F/15,000=a
    F/15,000=311,400/36,288
    F=311,400/36,288*15,000
    F=128,720N of thrust needed to lift off for 15,000kg

    311,400N-128,720N=182,680N left to counteract friction

    This is then the amount of force of static friction that would be needed between the wheels and belt to prevent it from taking off.

    So, if the force of friction is: F=un
    and the normal force is equal to Fg
    and the plane weighs 15000kg, then Fg=15,000*9.8=147,000N
    then n=147,000N and thus the coefficient of friction needed to stop the plane would be:
    u=F/n
    u=182,680N/147,000N
    u=1.24

    So the question is, what's our belt made of and how slippery is it? I doubt however it can create such a large coefficient of friction. From a quik search on the internet, rubber on rubber yielded less than 0.9 in all cases I found.
    Furthurmore, after looking at this table
    It appears the only surfaces which were above 1 were two metals in contact (and our belt wont be made out of aluminum, silver or cast iron i should think!)
    Apart from the vauge Solids-Rubber which was stated to be between 1.0 and4.0 ...but what solids? It states coefficients for rubber on concrete and asphalt (neither above 0.8)

    So, in conclusion, It looks like the plane could indeed take of in a real life scenario, simply by sliding along the belt.

    -Andrew
     
  15. leopold Valued Senior Member

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    17,455
    i think i said that about 50,000 times in this thread.

    there is also interaction between the wheel and bearing, and between the bearing and axle.
     
  16. 2inquisitive The Devil is in the details Registered Senior Member

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    andbna,
    What are you talking about?? The tires do not need to 'slide' along the belt. They roll along the belt. Friction is what keeps the tires from sliding, causing them to roll (spin) instead. You seem to be confused about the 'speed of the belt matching the speed of the wheels'. I explained that already in post 106. Assume the top of the belt moves rearward at 120 mph relative to the ground. For the wheels (think axels) to move at the same 'speed', they would have to move forward at 120 mph relative to the ground. These 'speeds' do not cancel out. If a speedometer were connected to the wheels, it would read 240 mph. Some posters are assuming the top of the conveyor belt is moving rearward relative to the wheels for one 'speed', and then assuming the wheel 'speed' is relative to the belt. That tells you nothing about forward motion relative to the ground or the air. Those two 'speeds' will always cancel out regardless of forward motion of the plane. Imagine standing on an escalator that is going down. You can turn around and walk upwards and not have any motion relative to the floors the escalator connects. Your walking speed will exactly match your speed relative to the escalator. You can also run upwards and get back to the top floor. Your running speed will also exactly match your speed relative to the escalator in this case, but you have forward motion relative to the floors. Same with the plane/conveyor belt, you must use both wheel speed and conveyor belt speed relative to the ground or air to have any meaning. In my example, the conveyor belt moves rearward at 120 mph relative to the ground and the wheels move forward at 120 mph relative to the ground. The two speeds exactly match as stated in the opening post, but they do not cancel out as they are relative to the ground, not each other. The 120 mph speed is arbitrary, any speed can be used.
     
  17. Challenger78 Valued Senior Member

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    This may sound like a stupid question but would the movement of the conveyor belt and the heat from the engines, wouldn't that create the wind ?... If that were cancelled out then no, it would not take off.
     
  18. Vern Registered Senior Member

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    This is what is left when you restrict controversial posts as we have done in this forum

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  19. andbna Registered Senior Member

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    That is between the wheel and the plane, not the belt, so It is irrelevant in my scenario.
    This friction would exist for anything with wheels moving, and in any good wheel system, it's negligable, as it will not increase with wheelspeed.

    2inquisitive :
    No they dont need to slide, but my post was about the possibility of it sliding as :
    There is a maximum speed the belt can move before the tires will simply slide, its like a tablecloth and plates, if you move the cloth slowly, the plates come off with it, but if you rip it out, they stay relativly untouched. why? Because the coefficient of static friction is overcome by the force of the table cloth moving under the plates. And that was what my experiment demonstrated, that the jet can overcome and take off.

    However 2inquisitive: that is an excelent way of putting it, but I think the OP wanted the speedometre's speed to determine how fast to move the belt, which case, the convayer is going to make itself speed up

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    -Andrew
     
  20. leopold Valued Senior Member

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    the friction, although almost negligible, must still be considered.

    2inquisitive,
    at a wheel speed of 240 MPH the plane would already be airborn.
     
  21. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    Why would the friction have to be considered in this gedankin?
    Wheel speed alone has nothing to do with whether or not the plane lifts off from a moving conveyor belt. You are using the wrong frames of reference as I stated earlier. An observer at rest on the conveyor belt sees the wheel move past at 240 mph. An observer on the wheel/plane sees the belt move past at 240 mph. Neither of those frames of reference measures airspeed or ground speed. Airspeed/groundspeed are all that matters. Since the opening post stated there was no wind, airspeed and groundspeed would be the same in this exercise. You must deduct speed the conveyor belt moves relative to the ground to arrive at the air/ground speed of the plane.
     
  22. Janus58 Valued Senior Member

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    1,787
    As has already been pointed out, it is the propeller that causes thrust for the plane, so in an ideal situation. (no wheel bearing friction, etc.) the plane takes off with no problem.

    If you try to introduce real world physics, then the propblem becomes a tad more complicated.

    Assume that you have a speedometer connected to the wheels and this is what the conveyor belt uses to adjust its own speed (You can't use existing plane instruments because its speed indicator measures air speed).

    Now the engine starts and the plane starts to pull forward. The wheels begin to turn and sends a signal to the conveyor, which begins to travel in the opposite direction. But with the plane moving forward and the conveyor moving backward, the wheels have to spin even faster (if the plane is traveling forward at 1kt, and the belt backwards at 1kt, the wheels have to turn at 2kts)
    But if the wheels turn faster, so does the conveyor belt, which means the wheels turn faster.... Its an ever increasing repeating loop. No matter how slowly the plane starts to move forward, the conveyor belt will accelerate up to infinite speed.

    Here's where the real world physics rears it head.
    A real conveyor belt system will have a maximum speed. either it will just not go any faster or it will fail.
    A real conveyor belt will have a fixed rate at which it can accelerate (Change its speed)
    In the real world, there will be a finite response time between a change in wheel speed and a change in conveyor belt speed.

    In the real world,there is a finite speed at which the planes wheels can spin before they overheat and fail (either the bearing will fail or you will throw a tire.

    All these factors determine whether the plane will take off. If the combination of top speed of the conveyor belt, its acceleration rate and response time allow the plane to reach take-off airspeed before the wheels fail, the plane will take off. If the wheels fail first, the plane will not. (if it is a catastrophic failure, the plane will likely nose over and crash.)

    So, in the real world, you can't answer the question without knowing the above factors.
     
  23. nietzschefan Thread Killer Valued Senior Member

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    Well said Janus, I would agree with all you just said(better than I could).
     

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