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Will it take off?

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posted on Feb, 19 2006 @ 01:24 AM
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Originally posted by Bhadhidar

Originally posted by redmage
The conveyer cannot hold the plane in place because it's thrust is independant, and isolated by the free spinning wheels.

[edit on 2/18/06 by redmage]


If the thrust is in fact independent, why do the wheels on the car (and a plane while its on the ground) go 'round and 'round?

Let's clarify a few things.

You claim that a conveyor will keep a (normal) car or bicycle in place because they utilize the thrust generated at their wheels to generate forward motion.

Well,...

Actually, they both are merely transfering the kinetic energy, the thrust as you put it, from their respective powerplants to their wheels.

The wheels serve merely to TRANSFER the thrust; they do not generate that thrust.

A conventional automobile's drivetrain is merely an effecient means of transfereing the thrust, the energy produced by the engine, to the pavement; but as the Helica demonstrates, a mechanical drivetrain is not the only way, just an efficient way.

The Helica is still a ground vehicle. To move across the land it must, like every other car, transfer the power of its engine to its wheels, which are in contact with the ground, to move forward. The method employed by the Helica is not as mechanically efficient as a conventional drive train but must accomplish the same feat to propel the vehicle.

Airspeed, Wheelspeed, Groundspeed

Airspeed, as we have already discussed, would only apply once our vehicle became airborn. I'm not sure why wheelspeed has any part in this discussion, except as a misleading indication of groundspeed when our subject vehicle is on the conveyor.

A conveyor is just an elongated roller, so for simplicities sake, let's put the old Helica on a set of rollers.

These rollers have exactly the same diameter as the cars tires, such that one revolution of the car's tire (t) is exactly equal to one revolution of the roller (r) under it: 1 RPM(t)=1RPM(r).

Here's where we may be confusing terms a bit; the measurement of the the rotation of either the tire or the roller under it (or for that matter, the surface of the conveyor belt) is most properly measured in Revolutions Per Minute (RPM's). Groundspeed, the speed of the vehicle as measured by an observer standing next to the vehicle while it's on the roller will be of course 0 MPH (or not, depending on your view of the arguement!).

The speed of the vehicle, as measured on its speedometer, is of course a product of the number of revolutions the tire makes in a given time frame, but assumes that a revolution of the tire covers a certain distance (which would be a product of the tire's circumfrence).

This is why changing the size of your car's tires can throw off your speedometer and odometer readings....But Officer!

As we fire up old Helica the prop begins to turn, generating thrust. The thrust pulls against the mass of the car. This in turn begins to cause the wheels of the car to rotate, because even though there is no direct mechanical linkage between the powerplant and the wheels via a drivetrain, the thrust of the powerplant is being transfered to the wheels due to the fact that the wheels are a part of the same vehicle.

The thrust is neither "independent of" nor "isolated from" the wheels simply because there does not exist a dedicated drive train between them.

The system is not efficient; to move the old hulk, the powerplant must overcome both the vehicle's resting inertia and the frictional coefficient of the contact point between its tires and the ground. A conventional auto drivetrain would put more of the engine's horsepower directly into that point of contact. But even then, if that point of contact is not fixed, allowing the wheels to spin, the car ain't going anywhere.

Now for the problem. As the wheels begin to turn the rollers begin to turn at exactly the same speed as the wheels! RPM for RPM the rollers instantly match the speed of the wheels. Standing next to the assemblage (not too close) we see that to roll forward the car wheel spins counter-clockwise, while the roller spins clockwise ("in the opposite direction" as stated in the excercise).

Expressed another way: clockwise 1RPM(r)= counter-clockwise 1RPM(t)

Since distance would equal the total number of revolutions X the length of the circumference of the wheel (or roller, since the measurements are the same), 1 RPM(t) would equal a given distance (D), while 1RPM(r) could be said to equal the same distance, but in the opposite direction, hence (-D).

D-D=0

The prop is a screaming blur, the speedometer in the car says its going 320 MPH (yeah, right!), but the old buggy hasn't moved.



The problem with this assesment is in the controls, which I did some research on. As the problem states:

This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction) . Can the plane take off?"

Now you're defining "plane speed" to mean the same thing as it does in a car, i.e. groundspeed. As you stated:


Originally posted by Bhadhidar

The speed of the vehicle, as measured on its speedometer, is of course a product of the number of revolutions the tire makes in a given time frame, but assumes that a revolution of the tire covers a certain distance (which would be a product of the tire's circumfrence).


The problem here is that a plane does not use a speedometer using wheel speed measurements in take off. It uses an airspeed indicator, a measurement of the plane's speed relative to the air around it. If airplanes used auto speedometers, they would never get off the ground in a strong tailwind (assuming they kept the speed steady at takeoff speed according to the guage ... the plane will not have enough lift at the guage's indicated speed). An airspeed indicator is really nothing more than a simple pressure gauge ... when the calclulated value is reached through the guage (compensated for wind speeds), the airplane will achieve lift.

Therefore the conveyor belt must either be connected to:

a) an instrument measuring relative velocity from a stationary postion outside of the plane/conveyor system

-or-

b) from the airspeed indicator

This is what makes the thrust of the aircraft "independent" of the belt. The belt can only move if the airplane is moving. And we know a plane at rest will begin to move when the engines begin creating thrust (at this point there is no "belt" - the ground beneath the plane is staionary). The wheels of the plane will begin to revolve, due to the thrust, and continue at a rate double the conveyor speed (since the conveyor speed is calculated BY the plane's airspeed indicator).

EDIT: spelling, grammar

[edit on 19-2-2006 by Fiverz]




posted on Feb, 19 2006 @ 04:33 AM
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Whether momentum is gererated by thrust or gravity makes little differnece, as long as the surface against which the force is directed is able to counter the vector of the force there will be no movement relative to the ground as the video shows.

www.endlesslope.com...

The endless slope uses gravity to pull the snoboarder "forward" and matches the speed of his descent with an endless conveyor belt.

I'm sure it doesn't take too much imagination to fit the evidence of this video demonstration to our scenario, does it?



posted on Feb, 19 2006 @ 06:19 AM
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The sheer number of complicated pseudo-scientific explanations both for and against is actually hilarious when you consider the actual question. This quote below is a good place to start


I thought about this question some more and it really is in how the question is stated/interpreted.

Assume the question states "... and the wheels are frictionless".


You don't need to even assume that. It is quite specific.

The belt tracks the speed of the plane and matches it, not the relative speed, just the speed.

Therefore if the plane is not moving neither is the belt. The propeller will pull the plane forward if everything is stationary.

As soon as the plane moves the belt tracks this and matches it. Important point number two, the belt matches it, ie goes the same speed in the opposite direction. To read this as 'counteracts' or 'cancels out' is an assumption and one that is wrong because if the plane stops moving SO DOES THE BELT. Think about that, its kind of a trick question in this respect as if the belt continues to get faster and faster it only does so because that is what the plane is doing, and of course, once the plane is going fast enough it will taker off, everything else is red herrings and BS.






[edit on 19-2-2006 by waynos]



posted on Feb, 19 2006 @ 11:57 AM
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Originally posted by Bhadhidar
If the thrust is in fact independent, why do the wheels on the car (and a plane while its on the ground) go 'round and 'round?

Let's clarify a few things.

You claim that a conveyor will keep a (normal) car or bicycle in place because they utilize the thrust generated at their wheels to generate forward motion.

Well,...

Actually, they both are merely transfering the kinetic energy, the thrust as you put it, from their respective powerplants to their wheels.

The wheels serve merely to TRANSFER the thrust; they do not generate that thrust.

A conventional automobile's drivetrain is merely an effecient means of transfereing the thrust, the energy produced by the engine, to the pavement; but as the Helica demonstrates, a mechanical drivetrain is not the only way, just an efficient way.

The Helica is still a ground vehicle. To move across the land it must, like every other car, transfer the power of its engine to its wheels, which are in contact with the ground, to move forward. The method employed by the Helica is not as mechanically efficient as a conventional drive train but must accomplish the same feat to propel the vehicle.

The wheels on an airplane go "round and round" for the same reason three of the four wheels on a typical grocery cart go round and round, there is another force pushing the cart along. (we won't worrey about the fourth wheel in a grocery cart that always sticks for this argument, airplanes are better maintained.)

You've also made a good point that the wheels in a car do not generate the thrust, they merely transfer the power of the engine. The conveyer belt contraption would be quite effective in preventing this transfer of energy from having any real effect on the position of the automobile.

In an airplane however, the engine does not transfer it's energy to the wheels. It instead transfers it to the surronding air, which is not affected by the conveyerbelt. Wheel rotation in this case is similar to that in the shopping cart, as simply a relatively efficient means to keep from having to drag the cart along the ground.



posted on Feb, 19 2006 @ 03:37 PM
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Originally posted by Bhadhidar

Originally posted by redmage
The conveyer cannot hold the plane in place because it's thrust is independant, and isolated by the free spinning wheels.


If the thrust is in fact independent, why do the wheels on the car (and a plane while its on the ground) go 'round and 'round?


First, the thrust on a car is not independant, and I've never said it was.

Second, why do the wheels go 'round and 'round?

Two different reasons.

In cars, bikes, etc. they "go 'round and 'round" because, as you put it, "they are merely transfering the kinetic energy from their respective powerplants to their wheels."

In a plane they "go 'round and 'round" because gravity holds the plane to the ground untill there is enough airflow for it to gain flight. The wheels are simply free rolling and have no "torque load" that they are transfering. Instead, they are pulled forward by an independant thrust + the effect of gravity on it, instead of being used to directly "transfer the thrust".

Instead of the means of transfer, the wheels only serve to isolate the thrust from the friction that would be caused by gravity holding the belly of the plane to the ground (untill the airflow allows it to overcome that gravity and take flight).

Since gravity is a "constant" in both of my origional "equations" (car and plane) occam's razor allows us to "throw it out.


Originally posted by Bhadhidar
You claim that a conveyor will keep a (normal) car or bicycle in place because they utilize the thrust generated at their wheels to generate forward motion.

Well,...

Actually, they both are merely transfering the kinetic energy, the thrust as you put it, from their respective powerplants to their wheels.

The wheels serve merely to TRANSFER the thrust; they do not generate that thrust.


This is really just a semantics argument.

While you are correct about the powerplant "generating" the energy, the forward movement of a car, or bicycle, is dependant on the wheels, and their interaction with the ground.

I demonstrated the difference between dependant and independant thrust when I used the examples of suspending cars, bikes, and planes by cables.

Cars, bikes, etc, will remain stationary, while suspended, because they are dependant on their transfer of energy to the ground (by the wheels) for them to achieve forward movement.

A suspended plane, or Helica, will still "fly" in circles because it's thrust is independant of any wheel/ground interaction.


Originally posted by Bhadhidar
The Helica is still a ground vehicle. To move across the land it must, like every other car, transfer the power of its engine to its wheels, which are in contact with the ground, to move forward.


Wrong, the Helica doesn't "transfer the power of its engine to its wheels".

As you've also stated "the wheels are free rolling" so there is no dirrect link for it to transfer with.

Instead, the wheels are only there to "absorb" the friction that would be caused by gravity holding the belly of the Helica to the ground.


Originally posted by Bhadhidar
Airspeed, as we have already discussed, would only apply once our vehicle became airborn.


Wrong, airspeed always applies, it's more of a matter of when it is relevant.


Originally posted by Bhadhidar
I'm not sure why wheelspeed has any part in this discussion, except as a misleading indication of groundspeed when our subject vehicle is on the conveyor.


Wheelspeed is important because, for a conveyer to "hold a vehicle in place" it must match the wheelspeed.

The problem, is that when an independant thrust is applied, the conveyer cannot match the wheelspeed, proving that the vehicle will move forward and gain airflow. 1+1 cannot = 1


Originally posted by Bhadhidar
As we fire up old Helica the prop begins to turn, generating thrust. The thrust pulls against the mass of the car.


Right.


This in turn begins to cause the wheels of the car to rotate because even though there is no direct mechanical linkage between the powerplant and the wheels via a drivetrain, the thrust of the powerplant is being transfered to the wheels due to the fact that the wheels are a part of the same vehicle.

Wrong, the thrust is not being "transfered to the wheels".

There is no torque load, they are free-spinning.

There is a big difference between the thrust+mass (and gravity's effect on it) "pulling the wheels along", and the wheels (with a torque load) being used to push the mass forward.


Originally posted by Bhadhidar
The thrust is neither "independent of" nor "isolated from" the wheels simply because there does not exist a dedicated drive train between them.


You're right (sort of), but, the thrust is independant of, and isolated from, the conveyer by the wheels. The wheels have virtually nothing to do with the thrust of a plane, they are free-spinning and only serve to "absorb" the friction that would be caused by gravity holding a plane's belly to the ground.

[edit on 2/19/06 by redmage]



posted on Feb, 19 2006 @ 03:54 PM
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Please read this in full and then tell me if you still think the plane takes off!!

www.aeromuseum.org...

or

cruftbox.com...

God be with you,
Mfourl



posted on Feb, 19 2006 @ 04:08 PM
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Originally posted by mfourl
Please read this in full and then tell me if you still think the plane takes off!!

www.aeromuseum.org...

or

cruftbox.com...

God be with you,
Mfourl



I know it takes off.

The conveyor belt is a distraction, the wheels will rotate quicker, but the aircraft will still move forward and take off.



posted on Feb, 19 2006 @ 04:11 PM
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Originally posted by mfourl
Please read this in full and then tell me if you still think the plane takes off!!


Yes it will.

The author of the second link is not taking into account that the conveyer cannot match the wheelspeed (which is required to "hold it in place").

With an independant thrust, the conveyer cannot cancel it's effects out.

1+1 cannot = 1

So as long as the wheelspeed is greater than the conveyer speed the plane will move forward and gain airflow for lift.



posted on Feb, 19 2006 @ 04:15 PM
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You really think that is the answer to the question at hand.

Personally I don't think your answer is to do with the question. If that was the case we could say that after time, the tyres would heat up and pop, causing a breath moment of elevation, the plane took off!!


God be with you,
Mfourl



posted on Feb, 19 2006 @ 04:22 PM
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Originally posted by mfourl
If that was the case we could say that after time, the tyres would heat up and pop, causing a breath moment of elevation, the plane took off!!



If the conveyer attempted (not that it can) to match the wheelspeed, then yes "the tires would heat up and pop", or the bearings would fail.

Still, he is not realizing, or taking into account, the ramifications that an independant thrust presents. :shk:

[edit on 2/19/06 by redmage]



posted on Feb, 19 2006 @ 04:24 PM
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Originally posted by redmage
There is no torque load, they are free-spinning.

There is a big difference between the thrust+mass (and gravity's effect on it) "pulling the wheels along", and the wheels (with a torque load) being used to push the mass forward.


Originally posted by Bhadhidar
The thrust is neither "independent of" nor "isolated from" the wheels simply because there does not exist a dedicated drive train between them.


You're right (sort of), but, the thrust is independant of, and isolated from, the conveyer by the wheels. The wheels have virtually nothing to do with the thrust of a plane, they are free-spinning and only serve to "absorb" the friction that would be caused by gravity holding a plane's belly to the ground.

[edit on 2/19/06 by redmage]


First, review the video linked here:

www.endlesslope.com...

There is nothing but gravity "pulling the snoboarder down the slope" of the device. The endless slope, by matching the boarder's speed of descent keeps him in relatively the same position (with respect to the gawking crowd) on the moving surface. Otherwise he would either tumble off the bottom or be flung from the top of the machine.

If the boarder were wearing roller-skates his rate of descent would of course be faster (less friction at the contact point with the moving surface), but assuming that the endless slope could compensate for the higher descent rate, the effect would be the same. Right? No need to call the EMT's.

OK. Now let's have our daredevil on roller-skates hold a motor with a spinning prop out in front of himself.

The added thrust (which, as in the previous examples, is not 'connected' to the wheels of the roller-skates; they are 'free-wheeling' and have no torque-load as you put it) would naturally increase his rate of descent drastically.

However, would you not agree that, if the endless slope were able to compensate for this increase in descent rate as well, that our boarder would still stay on the slope?



posted on Feb, 19 2006 @ 04:27 PM
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Yes I understand what you are saying but I dont think this was the intention of the question, the wording is bad so it leaves two options depending on how look at it. yes the plane does take off because it will fall of the end of the conveyour but I dont think that was the answer the question was seeking.



posted on Feb, 19 2006 @ 04:53 PM
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Originally posted by Bhadhidar
There is nothing but gravity "pulling the snoboarder down the slope" of the device.


Almost, he is also using the bar (as an independant "thrust") to compensate/adjust.

He could pull himself right off it if he applied enough independant force to.


Originally posted by Bhadhidar
If the boarder were wearing roller-skates his rate of descent would of course be faster (less friction at the contact point with the moving surface), but assuming that the endless slope could compensate for the higher descent rate, the effect would be the same. Right?


The example is somewhat flawed because gravity alone, although independant, is a relatively "weak" force in comparison.

Still, the skater would acceelerate down the hill (barring bearing failure which would be a real problem with such a "weak" independant thrust) because the conveyer cannot "cancel" gravity to "compensate".

Also, "assuming that the endless slope could compensate" is where the logic fails, because that's where you throw the consequences/ramifications of an independant thrust/force "out the window". :shk:

When an independant force is applied, the conveyer cannot match the wheelspeed (the equalibrium required to "hold something in place") but can only add to it.


Originally posted by Bhadhidar
OK. Now let's have our daredevil on roller-skates hold a motor with a spinning prop out in front of himself.

The added thrust (which, as in the previous examples, is not 'connected' to the wheels of the roller-skates; they are 'free-wheeling' and have no torque-load as you put it) would naturally increase his rate of descent drastically.


Yup, now there are 2 independant "thrusts" that the inclined conveyer cannot cancel.


Originally posted by Bhadhidar
However, would you not agree that, if the endless slope were able to compensate for this increase in descent rate as well, that our boarder would still stay on the slope?


I would not agree, the conveyer cannot cancel the 2 independant thrusts, it can only add to the wheelspeed, not match it.

Again, "assuming that the endless slope could compensate" is where the logic fails, because that's where you throw the consequences/ramifications of an independant thrust/force "out the window".

When an independant force is applied, the conveyer cannot match the wheelspeed (the equalibrium required to "hold something in place") but can only add to it.

[edit on 2/19/06 by redmage]



posted on Feb, 20 2006 @ 02:08 AM
link   

Originally posted by mfourl
Please read this in full and then tell me if you still think the plane takes off!!

www.aeromuseum.org...

or

cruftbox.com...

God be with you,
Mfourl



Originally posted by mfourl
First, review the video linked here:

www.endlesslope.com...

There is nothing but gravity "pulling the snoboarder down the slope" of the device. The endless slope, by matching the boarder's speed of descent keeps him in relatively the same position (with respect to the gawking crowd) on the moving surface. Otherwise he would either tumble off the bottom or be flung from the top of the machine.

If the boarder were wearing roller-skates his rate of descent would of course be faster (less friction at the contact point with the moving surface), but assuming that the endless slope could compensate for the higher descent rate, the effect would be the same. Right? No need to call the EMT's.

OK. Now let's have our daredevil on roller-skates hold a motor with a spinning prop out in front of himself.

The added thrust (which, as in the previous examples, is not 'connected' to the wheels of the roller-skates; they are 'free-wheeling' and have no torque-load as you put it) would naturally increase his rate of descent drastically.

However, would you not agree that, if the endless slope were able to compensate for this increase in descent rate as well, that our boarder would still stay on the slope?


The problem you still fail to see is that the question is poorly worded. If the belt's sensors are connected to some speedometer that measures WHEEL SPEED then no, the plane will not take off. But in real life planes use AIRSPEED INDICATORS to determine if they are going fast enough to achieve lift. This is an air pressure guage that is compensated for wind. If an object is not moving through space, it will not register a velocity via the airspeed indicator. This is the real-world situation. If the belt's sensors are connected to the airspeed indicator, the plane will take off as normal. Please re-read my previous post for more detail if you wish.



posted on Feb, 20 2006 @ 02:21 AM
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Originally posted by Fiverz
The problem you still fail to see is that the question is poorly worded. If the belt's sensors are connected to some speedometer that measures WHEEL SPEED then no, the plane will not take off.


I will agree, the wording is tricky, but it doesn't change the fact that with an independent thrust (or force, like gravity on an incline) the belt cannot "match" the wheelspeed. It can only add to it. So the entire idea of the belt matching the wheelspeed of the plane is wrong.


Originally posted by Fiverz
But in real life planes use AIRSPEED INDICATORS to determine if they are going fast enough to achieve lift. This is an air pressure guage that is compensated for wind. If an object is not moving through space, it will not register a velocity via the airspeed indicator. This is the real-world situation. If the belt's sensors are connected to the airspeed indicator, the plane will take off as normal.


Absolutely


[edit on 2/20/06 by redmage]



posted on Feb, 20 2006 @ 09:50 AM
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The endless slope isn't the greatest example, as it uses friction to compensate, something which an aircraft's tires are specificly designed to minimise. Also, the endless slope is using it's conveyer to cancel the snowboarder's speed, not match it.

As for matching tire speed, that is paradoxal if the plane is moving at all. If the plane rolls forwards at 1 MPH, the wheels will spin at 1 MPH, and the belt will counter at 1 MPH, which causes the wheels to spin at 2 MPH, forcing the belt to accelerate to 2MPH, which causes the wheels to spin at 3 MPH... The only way to match wheelspeed is to tie the aircraft in place so it cannot move, then run the conveyer at any speed you like. This is why it's so important to note the original question does not say the belt matches the wheel's speed, but the plane's speed.



posted on Feb, 20 2006 @ 11:29 AM
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Interpretation: I assumed by the question that the conveyor would move fast enough to hold the plane in place.

But what yall are saying is that the conveyor would NOT keep the plane in place and it would move forward regardless. The wheels would just spin faster.

Makes sense if the wheels are frictionless enough and the conveyor doesn't have enough power to compensate.



posted on Feb, 20 2006 @ 05:20 PM
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Originally posted by Kruel
Interpretation: I assumed by the question that the conveyor would move fast enough to hold the plane in place.

But what yall are saying is that the conveyor would NOT keep the plane in place and it would move forward regardless. The wheels would just spin faster.

Makes sense if the wheels are frictionless enough and the conveyor doesn't have enough power to compensate.


That's just what I meant by the bad wording. There are usually no speedometers on a plane that would measure wheel speed in real life, so it's easy to assume the question means the belt matches the airspeed indicator's reading of speed. If it matched wheel speed, it's easy to see the plane wouldn't move at al in spacel, would not generate any lift, and will not take off.



posted on Feb, 20 2006 @ 07:38 PM
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But its not down to interpretation surely, the question states 'matches the speed of the plane'. Nothing else, not wheels, not relative ground speed, not anything. Just the speed of the plane - which must therefore be moving itself at the same speed as the belt . Have I said that before?


Because of this, if the planes actual forward speed is zero then the belt stops, which would allow the plane to move and so the belt goes and so on. In this case it can be seen that the notion that the belt can cancel out the forward speed of the plane cannot work, therefore the only logical conclusion is that the plane continues to move forwards and eventually takes off.

[edit on 20-2-2006 by waynos]



posted on Feb, 20 2006 @ 08:56 PM
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Originally posted by waynos
But its not down to interpretation surely, the question states 'matches the speed of the plane'. Nothing else, not wheels, not relative ground speed, not anything.
[edit on 20-2-2006 by waynos]


"Matches the speed of the plane" is quite vague. This could be the speed of the plane in relation to the ground, the belt, or the air. But which? It doesn't specify.

Oh well, at least now I know that you guys aren't thinking a plane can take off with an air speed of 0.



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