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Senior Member
Registered: 01-06-08
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quote: Originally posted by KSquared:
Erm... How is it not mathematically possible? Solution is airplane's ground speed = 0. Maybe you wanted to say that it isn't mechanically possible? Even then I have to disagree.
It is a trivial matter to compute the acceleration of the belt required to keep the airplane still given a thrust setting. You just use the fact that wheels themselves have some inertia. Now, this will work until either:
1) Belt is mechanically incapable of delivering the acceleration at speed reached.
2) Friction in the wheel finally matches thrust.
3) Belt or wheels catastrophically fail.
First problem can be resolved by adding more power to the belt. Second is not very likely. So if you build it properly, plane will sit still, then the wheels will fail, and whole thing falls apart. Since there can be no takeoff after that, I think it would verify the myth in this particular setting.
You miss one point, the setup that you state is a setup in which the plane would be forced to go so slow that it stays still by matching friction on the wheels, and if applying the same speed on solid ground, the plane wouldn't be able to take off anyway. It's physically impossible to create a treadmill that keeps a plane with its engines going at the power needed to get up to takeoff speed stationary. Soemthing would fail first. A plane using enough engine power that it would be able to take off on solid ground would never be able to be held stationary by a treadmill due to the level of friction being so small on the wheels, thus, it would move forward, and take off. quote:
No, it isn't. Myth in MBs interpretation requires speed to be measured relative to the ground. Airplane's speed indicator doesn't show airplane's speed at all. It shows speed of the air flow under the wing. Neither has anything to do with the throttle setting.
Fine, I'll give you that as I worded it horribly. What I meant and should have said was that there's a setting on the throttle that, when on solid ground, would allow it to take off, and if on that setting, the treadmill set to the same takeoff speed to supposedly counteract it, the plane would still take off. However, if you measure relative to the treadmill, then a plane just sitting on it without applying throttle or just enough so as to stay stationary would be going however fast as the treadmill if it stays still, but either way, you don't measure relative to the treadmill as that defeats the purpose of the myth as the plane isn't even able to take off on solid ground if those circumstances of the plae are replicated. Long story short, my point was you don't measure relative to the treadmill, as that is flawed measurement of speed. |
Senior Member
Registered: 10-28-07
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quote: It is a trivial matter to compute the acceleration of the belt required to keep the airplane still given a thrust setting.
*sigh* And that's been discussed to death and has nothing to do with the question that was being answered. The question quite clearly said that the belt exactly matches the speed of the plane, not that the belt accelerates to match the plane's thrust. For the umpteenth time, this is NOT an alternate interpretation of the question, as worded -- it is a completely different problem altogether. Since it is a DIFFERENT set of conditions, it does NOT "verify the myth" no matter how many times you keep posting to the contrary. The question says quite explicitly that a) the belt and plane move (so the trivial case of v=0 is ruled out), and b) the speeds exactly match. The only thing possibly open for interpretation is what is meant by the plane's "speed". We have about 750 pages (not just this thread, but the previous ones too) explaining why the "wheel speed" condition is invalid, and explaining that even if it wasn't, the plane WOULD STILL TAKE OFF -- it would just break the control system by producing an impossible case. |
Senior Member
Registered: 07-03-07
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quote: Originally posted by Willgamer25:
-rational-, when will you realize that the plane's speed is not measured relative to the belt?
The measurement of speed in the plane is measured by the amount of throttle applied resulting in it's engines powering. The belt is measured in it's speed of rotation. That's how they're measured, and that's how tey pertain in the myth. The situation you speak of is a situation in which the plane is measured when it's sitting still on the treadmill, which is a point at which it's applying an amount of throttle (or none at all) to a point where it wouldn't even take off if it were on solid ground, which does not meet the concept of the myth.
Of you calculate a speed of the plane and use of the throttle that allows the plane to take off, then bring the treadmill's speed up to that same position as the MPH reading in the plane, and the plane were to sit on the treadmill with that throttle fixed to that point, the plane will always take off.
Your problem is how the speed is measured. Simply put, you don't measure the speed realative to the treadmill. It's as simple as that.
Pure gibberish. |
Senior Member
Registered: 07-03-07
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quote: Originally posted by KSquared: quote: Originally posted by -rational-:
When is your stunted brain going to absorb the fact that it is not mathematically possible for the belt to match the speed of the aeroplane if the speed of the aeroplane is measured relative to the belt?
Erm... How is it not mathematically possible? Solution is airplane's ground speed = 0. Maybe you wanted to say that it isn't mechanically possible? Even then I have to disagree. It is a trivial matter to compute the acceleration of the belt required to keep the airplane still given a thrust setting. You just use the fact that wheels themselves have some inertia. Now, this will work until either: 1) Belt is mechanically incapable of delivering the acceleration at speed reached. 2) Friction in the wheel finally matches thrust. 3) Belt or wheels catastrophically fail. First problem can be resolved by adding more power to the belt. Second is not very likely. So if you build it properly, plane will sit still, then the wheels will fail, and whole thing falls apart. Since there can be no takeoff after that, I think it would verify the myth in this particular setting. quote: Willgamer25:
The measurement of speed in the plane is measured by the amount of throttle applied resulting in it's engines powering.
No, it isn't. Myth in MBs interpretation requires speed to be measured relative to the ground. Airplane's speed indicator doesn't show airplane's speed at all. It shows speed of the air flow under the wing. Neither has anything to do with the throttle setting.
1) It is mathematically impossible for the belt to match the plane's speed when the plane's speed is measured relative to the belt for the scenario described in the myth because the plane will move in response to the forces provided by the engine. Once the plane is moving the belt cannot match the speed of the plane until the circumstances I described on page 140 or so occur and this does so after a very lengthy and complicated process. 2) I would be very interested in your simple calculation as to what acceleration of the belt would keep the plane stationary. 3) As roofing guy says, to conform to the myth specifications the control system of the belt has to cause the belt to accelerate while it attempts to satisfy the conditions set in the myth and I am not entirely sure that this is possible. |
Senior Member
Registered: 07-03-07
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quote: Originally posted by graham75: quote: Originally posted by -rational-:
When is your stunted brain going to absorb the fact that it is not mathematically possible for the belt to match the speed of the aeroplane if the speed of the aeroplane is measured relative to the belt?
has been proven mathmatically possible depending on how both speeds are measured. the same math applys to ALL objects on a treadmill/conveyor belt. The important FACT is that one speed is being said to have the same magnitude as another and the one doesn't care or know how the other is measured. lets not forget that the Mythbusters frequently modify myths to get the results that desire for the sake of getting the desired results....and the bigger the explosion the better. That point leads to some give and take in the interpretation of the myth given its arguably vague and poor wording. Willgamer25 , you measure the speed of the plane in the manner you are given, whether it be in the original setup of in followup setups being explored. At the very least i;d say you have two options....normal for a plane or normal for an object on a conveyorbelt/treadmill. One leads to a very boring answer while the other doesn't.
No, the mathematics prove that the belt cannot match the speed of the aeroplane if the speed is measured relative to the belt. Logic also proves you are an idiot who is incapable of understanding the simplest of concepts. |
Senior Member
Registered: 07-03-07
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quote: Originally posted by roofingguy: quote: It is a trivial matter to compute the acceleration of the belt required to keep the airplane still given a thrust setting.
*sigh* And that's been discussed to death and has nothing to do with the question that was being answered. The question quite clearly said that the belt exactly matches the speed of the plane, not that the belt accelerates to match the plane's thrust. For the umpteenth time, this is NOT an alternate interpretation of the question, as worded -- it is a completely different problem altogether. Since it is a DIFFERENT set of conditions, it does NOT "verify the myth" no matter how many times you keep posting to the contrary. The question says quite explicitly that a) the belt and plane move (so the trivial case of v=0 is ruled out), and b) the speeds exactly match. The only thing possibly open for interpretation is what is meant by the plane's "speed". We have about 750 pages (not just this thread, but the previous ones too) explaining why the "wheel speed" condition is invalid, and explaining that even if it wasn't, the plane WOULD STILL TAKE OFF -- it would just break the control system by producing an impossible case.
More gibberish. |
Senior Member
Registered: 03-22-07
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quote: Originally posted by -rational-: quote: Originally posted by graham75: quote: Originally posted by -rational-:
When is your stunted brain going to absorb the fact that it is not mathematically possible for the belt to match the speed of the aeroplane if the speed of the aeroplane is measured relative to the belt?
has been proven mathmatically possible depending on how both speeds are measured. the same math applys to ALL objects on a treadmill/conveyor belt. The important FACT is that one speed is being said to have the same magnitude as another and the one doesn't care or know how the other is measured. lets not forget that the Mythbusters frequently modify myths to get the results that desire for the sake of getting the desired results....and the bigger the explosion the better. That point leads to some give and take in the interpretation of the myth given its arguably vague and poor wording. Willgamer25 , you measure the speed of the plane in the manner you are given, whether it be in the original setup of in followup setups being explored. At the very least i;d say you have two options....normal for a plane or normal for an object on a conveyorbelt/treadmill. One leads to a very boring answer while the other doesn't.
No, the mathematics prove that the belt cannot match the speed of the aeroplane if the speed is measured relative to the belt. Logic also proves you are an idiot who is incapable of understanding the simplest of concepts.
the mathematics doesn't have to prove the matching speeds.....the is a GIVEN FACT of the setup. That you can't adhere to the given setup is just plain sad, but not at all unexpected. The mathematics has been provided that proves when you measure the speed of the plane as a function of the plane's displacement along the moving belt's surface and the speed of the belt (measured a couple different ways in examples) matches the magnitude of the speed of the plane....the airspeed of the plane is zero. what you seem to say is that an object on a treadmill/conveyor belt that is measuring its speed using the belt surface can never remain stationary from a point outside the treadmill/object interaction is the treadmill/conveyor belt is matching the object's previously described speed. You might want to explain that to all those people on treadmills that are not going flying off the front ends. |
Senior Member
Registered: 10-28-07
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quote: More gibberish.
Not gibberish -rational-. Your accelerated belt scenario has the belt trying to keep up. KSquared has the plane on the belt with the speeds matched, and the belt continuing to accelerate to oppose the thrust. This is not the basis of the question. The belt is supposed to be reacting to the plane, not setting the speed the wheels rotate at. |
Senior Member
Registered: 03-22-07
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the belt may be reacting but at any and every point in time the speeds are the same in magnitude, hence the "match exactly but in opposite direction" condition.
Belt can't be matching the plane if the belt is playing catch-up. matching ≠ catch-up
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Senior Member
Registered: 02-18-08
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quote: Originally posted by roofingguy:
And that's been discussed to death and has nothing to do with the question that was being answered. The question quite clearly said that the belt exactly matches the speed of the plane, not that the belt accelerates to match the plane's thrust.
That's the same freakin' thing. 1) Let belt's speed match plane's speed relative to the belt. 2) Plane's ground speed is trivially zero. 3) Plane's prop generates a force on the plane. 4) Balance of forces for stationary object requires there to be another force that keeps the plane still. 5) The only place that force can come from is the belt. 6) The only way that the belt can apply force on the plane (other than friction) is by accelerating. Ergo, to say that the belt is moving at the speed equal to plane's speed relative to the belt is to say that the belt is accelerating rapidly enough to generate a back force sufficient to cancel thrust. The acceleration for that would be fairly high, but it is trivial to generate. All you have to do is pull/push the surface of the belt with the force equal to the thrust of the plane. That's all. You pull the belt's surface with the force equal to thrust, and the plane isn't going anywhere. Newton's second law. No way around it. Now, most airplane's wheels will fail at just a few times the takeoff speed. So the total power of the motor rotating the belt only needs to have the power these few times higher than power of the plane's engine. That is, again, an entirely reasonable power. Quick summary: Pull surface of the belt with force equal to thrust, and the plane isn't going anywhere until wheels fail. No takeoff, regardless of thrust. quote: rational:
No, the mathematics prove that the belt cannot match the speed of the aeroplane if the speed is measured relative to the belt.
I'm sorry, but what are you smoking? Mathematically, the only requirement for the above is that the plane's ground speed is zero. If you cannot tell the difference between a mathematical proof and something improperly induced from your poor understanding of basic mechanics, try avoid making any "proof" statements whatsoever. |
Senior Member
Registered: 10-28-07
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But why does the belt start moving in the first place???
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Senior Member
Registered: 02-18-08
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quote: Originally posted by roofingguy:
But why does the belt start moving in the first place???
It detects a slight displacement of the plane and attempts to correct for it. Mathematically, such algorithm will keep speed at precisely zero. Mechanically, there will be a slight error, but so will there be one with any form of measurement. Regardless, if the plane ends up moving at a few cm/s, we can happily ignore that, because that isn't going to make the plane take off. Edit: Don't try to say, "Aha, then the belt doesn't match the speed EXACTLY!" No, it doesn't. Not exactly. No mechanical system will ever match anything exactly. Fortunately, that is never truly required. |
Senior Member
Registered: 03-22-07
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quote: Originally posted by roofingguy:
But why does the belt start moving in the first place???
at any specific time the speed of the belt will match the speed of the plane in magnitude. Notice I never say what the magnitude is, I simply stick to the given info and don't go off on tangents of friction, ect. If the only way to answer the question is with all speeds Zero...then there is your answer. If the speed of the plane as described way above is a non-zero number....the belt matches it and the speed of the plane from outside the system is zero. The 'belt moves' statement is about describing the motion of the belt in relation to the motion of the plane. |
Senior Member
Registered: 03-22-07
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quote: Originally posted by graham75: quote: Originally posted by roofingguy:
But why does the belt start moving in the first place???
at any specific time the speed of the belt will match the speed of the plane in magnitude. That is a given. Notice I never say what the magnitude is, I simply stick to the given info and don't go off on tangents with information not given. The math proves this point. If the only way to answer the question is with all speeds Zero...then there is your answer. If the speed of the plane as described way above is a non-zero number....the belt matches it and the speed of the plane from outside the system is zero. The 'belt moves' statement is about describing the motion of the belt in relation to the motion of the plane.
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Senior Member
Registered: 07-03-07
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quote: Originally posted by roofingguy: quote: More gibberish.
Not gibberish -rational-. Your accelerated belt scenario has the belt trying to keep up. KSquared has the plane on the belt with the speeds matched, and the belt continuing to accelerate to oppose the thrust. This is not the basis of the question. The belt is supposed to be reacting to the plane, not setting the speed the wheels rotate at.
Apologies roofingguy I meant to quote someone else's post. I don't recall you ever speaking gibberish. |
Senior Member
Registered: 07-03-07
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quote: Originally posted by KSquared: quote: Originally posted by roofingguy:
And that's been discussed to death and has nothing to do with the question that was being answered. The question quite clearly said that the belt exactly matches the speed of the plane, not that the belt accelerates to match the plane's thrust.
That's the same freakin' thing. 1) Let belt's speed match plane's speed relative to the belt. 2) Plane's ground speed is trivially zero. 3) Plane's prop generates a force on the plane. 4) Balance of forces for stationary object requires there to be another force that keeps the plane still. 5) The only place that force can come from is the belt. 6) The only way that the belt can apply force on the plane (other than friction) is by accelerating. Ergo, to say that the belt is moving at the speed equal to plane's speed relative to the belt is to say that the belt is accelerating rapidly enough to generate a back force sufficient to cancel thrust. The acceleration for that would be fairly high, but it is trivial to generate. All you have to do is pull/push the surface of the belt with the force equal to the thrust of the plane. That's all. You pull the belt's surface with the force equal to thrust, and the plane isn't going anywhere. Newton's second law. No way around it. Now, most airplane's wheels will fail at just a few times the takeoff speed. So the total power of the motor rotating the belt only needs to have the power these few times higher than power of the plane's engine. That is, again, an entirely reasonable power. Quick summary: Pull surface of the belt with force equal to thrust, and the plane isn't going anywhere until wheels fail. No takeoff, regardless of thrust. quote: rational:
No, the mathematics prove that the belt cannot match the speed of the aeroplane if the speed is measured relative to the belt.
I'm sorry, but what are you smoking? Mathematically, the only requirement for the above is that the plane's ground speed is zero. If you cannot tell the difference between a mathematical proof and something improperly induced from your poor understanding of basic mechanics, try avoid making any "proof" statements whatsoever.
No it isn't!!!!!! You are having some very basic problems with this scenario and the behaviour of control system of the belt and what it takes as parameters ( the speed of the plane which we are considering to be relative to the belt ) and its output ( power to the motors which would create a specific belt speed ) You are then ASSUMING that the scenario leads to the belt accelerating at a rate which will keep the plane stationary and this is a stable system. This does not follow, for a start you are assuming that it is possible for the control system to cause the belt to accelerate at the right rate to stop the plane and this would be extremely unlikely given that the control system sets the SPEED of the belt, and not its acceleration. If by some amazing coincidence of the system's construction e.g. the control system's sampling rate of the speed of the plane, and the behaviour of the belt motors in responding to directions of the control system, that the belt accelerates at a rate to keep the plane stationary for a moment. What would happen next? The next action of the control system would be to maintain whatever speed the belt will have at this time because it operates on the input parameter of the SPEED of the plane, not whether it is stationary. So instantly the plane becomes stationary relative to the ground the belt will stop accelerating and the plane then no longer be subject to any forces which would oppose the thrust of the engines, so it will start moving again! The plane will NOT be held stationary. The following statement "The acceleration for that would be fairly high, but it is trivial to generate. All you have to do is pull/push the surface of the belt with the force equal to the thrust of the plane. That's all. You pull the belt's surface with the force equal to thrust, and the plane isn't going anywhere. Newton's second law. No way around it." is just top level gibberish. The belt motors do not have to just provide a force that is of the same magnitude as the force of the thrust of the engines! The motors have to provide a force which will accelerate this hugely long belt which will be very massive and therefore require huge forces to be accelerated! The force applied to the aeroplane is the frictional force applied by the belt on the tyre surface and that has nothing to do with Newton's second law. I say that it is mathematically impossible for the speed of the belt to match the speed of the aeroplane relative to the belt because the aeroplane WILL move relative to the ground and this means that mathematically the belt cannot match the speed of the aeroplane because of the equation ( derived in posts above several times ) Vb = Vb + Vp where Vb = velocity of the belt relative to the ground Vp = velocity of the plane relative to the belt which can only be valid for Vp = 0 and Vp is not equal to 0. |
Senior Member
Registered: 07-12-07
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The conditions were met, the plane flew and the myth was busted. Get over it, guys.
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Junior Member
Registered: 02-05-08
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Ksquared wrote: "The acceleration for that would be fairly high, but it is trivial to generate. All you have to do is pull/push the surface of the belt with the force equal to the thrust of the plane. That's all. You pull the belt's surface with the force equal to thrust, and the plane isn't going anywhere. Newton's second law. No way around it." Rational replied: is just top level gibberish. I agree it is wrong but it's a bit harsh to call it gibberish because this is an example of what I was referring to earlier when I said that some people are confusing drag with inertia. Let's take the above example, for instance. Ksquared wrote: ....You pull the belt's surface with the force equal to thrust, and the plane isn't going anywhere.... This would actually be true if the plane had skids as opposed to wheels. But the wheels provide a bearing surface that allow the counter-movement of the belt to be negated so that it does not affect the forward movement of the plane. This can be better illustrated to all if we think of the plane flying at speed a fraction of an inch above the belt in one direction while the belt is moving the opposite direction at the same speed. If the plane had skids instead of wheels and it attempted to touch down, even for a second, the effects would be bad news because the skids would have way to much drag once in contact with the belt and the energy of the belt moving the opposite direction would be instantly transferred to the plane and would counteract the inertia of the plane. Likewise, if the plane had skids and tried to take off, even if the belt was not moving at all, it would have a hard time due to drag. Now, repeat the touchdown experiment but replace the skids with wheels and the results are totally different because the wheels prevent the energy of the belt from being transferred to the plane instantly and the inertia of the plane is not affected. The wheels will simply rotate twice as fast as they normally would but the plane's forward motion will not be affected. If the wheels failed (i.e. bearing failure or lock up) then that would produce the same effect as with skids. Now, if the plane with wheels were sitting on a belt that was already moving an opposite direction that matched take off speed, then the plane as a built up inertia going the opposite direction. Combine that with the reverse airflow over the wings that would negate lift, and the planes engine would have to work very hard, indeed, to start moving the plane forward, much less take off. If the plane has skids, it would never take off because the backwards inertia of the plane would be insurmountable due to the drag of the skids on the surface of the belt. As noted previously, the plane would have a hard time developing forward inertia with skids, even if the belt were not moving. That is why they use wheels!  Roger |
Senior Member
Registered: 07-03-07
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I said above I didn't think you understood what inertia is, and your use of the term in your last post confirms that you haven't the slightest clue what it is.
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Senior Member
Registered: 07-03-07
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quote: Originally posted by graham75: quote: Originally posted by roofingguy:
But why does the belt start moving in the first place???
at any specific time the speed of the belt will match the speed of the plane in magnitude. Notice I never say what the magnitude is, I simply stick to the given info and don't go off on tangents of friction, ect. If the only way to answer the question is with all speeds Zero...then there is your answer. If the speed of the plane as described way above is a non-zero number....the belt matches it and the speed of the plane from outside the system is zero. The 'belt moves' statement is about describing the motion of the belt in relation to the motion of the plane.
Unadulterated first order idiocy, truly stunningly stupid. |
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