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Will it take off?

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posted on Feb, 17 2006 @ 06:26 PM

Originally posted by robertfenix
your little pictures with the cars is not complete, you are forgetting the thrid force which is acting on both sides of the hil.

Gravity, is pushing both cars down equally. Both cars having equal mass will not roll down the hill when connected by the chain because the mass+ gravity is equal on both sides.

It has nothing to do with opposing forces. Gravity is pulling both cars down equall (assuming both are equal mass) and the chain and the hill top become a fulcrum. Just like a scale. with equal mass on both sides the scale arms appear to defy gravity by balancing on the fulcrum of the scale.

Take away the hill and the chain is irrelevant because the cars are not opposing forces only gravity is still puching down on each.

So your example is flawed and does not apply here

I'm a little tired of your flaming everyone who attempts to explain this to you, and saying that every example is flawed for not fitting your idea of what is going on.

I'm perfectly aware that my picture of the two cars on a hill involved gravity as a force. It is in fact the ONLY force being applied in the pictures. You may safely consider one force to be that of the propeller, and the other force to be the tread mill. because there is NO mechanical linkage between the two, both vehicles will react to the only forces applied to them.

Now go back to page seven and read my post again.
Did you see it?
here, I'll repeat myself...

but the chain has been removed

I did not remove the hill, I said remove the chain.

At any rate, I know I'm wasting my breath here, as it's increasingly obvious you are deliberately mis reading and misquoteing small portions of posts in order to claim entire posts are flawed and keep this argument going. In other words, you are trolling, and I will instead respond only to those actually putting some thought and effort into understanding this puzzle, rather than those picking fights.

By the way, Kruel, you diagram at the top of page 8 is pretty good. The only thing missing is the fact that someone's winching the rope in. (really fast!)

posted on Feb, 17 2006 @ 07:17 PM

Originally posted by robertfenix
This is just this guys opinion. It is not scientifically proving that the plane would take off.

And, sir/madam, the rest of what you have stated is your opinion and not supported by 'working' fact(s).

As I have attempted to convey, this is nothing new as a physics debate. I have provided an easy explanation previously inside this thread…as you are attempting to ‘reinvent the wheel’ with arguments…. enjoy your search.

Finished.

mg

posted on Feb, 17 2006 @ 08:14 PM

Originally posted by robertfenix

Your pedaling bicycle generates "thrust" at the wheels so your example is flawed.

Originally posted by robertfenix
your "tree" is an outside force indepandant of the forces acting on the plane.

You're so close to understanding it right there.

The air is "independant" too, just like the rope to the tree.

Originally posted by robertfenix
The tree gets to take advantage of a staionary location that then gets to "magically" move into a forward postion.

Forget the "walking tree", just pull on the rope to create the thrust. Since the force is independant, you will not be held in place by the treadmill, and will accelerate forward even if the treadmill "speeds up".

Originally posted by robertfenix
this new "tree" is an additional third force vector.

So is the force of the prop interacting with the air to generate thrust.

Where the thrust is generated does matter.

With a bike, walking, a car, etc. your forward "thrust" is generated by friction/torque against the ground/treadmill.

The treadmill can then effectively cancel out their thrust and hold these objects in place, by moving, because it is required to generate their thrust in the first place.

Suspend any of these objects in the air, by cables, and they are going nowhere because they are dependant on the ground to generate their "thrust".

Now suspend a plane.

It will fly in circles because the only things it is dependant on, to generate it's thrust, are the prop+engine and the air.

The air becomes a "third force vector" in the equation just like the rope tied to the tree.

A conveyer cannot cancel the thrust generated with the prop+air interaction because the thrust is independant of the conveyer, and, is isolated from it by free-spinning wheels .

Since the thrust is independant (and isolated), then there still must be an "equal and opposite reaction" so the plane will accelerate forward and eventually gain lift to achieve flight.

The wheels are spinning freely and will "absorb/convert" the forces of the "treadmill/gravity relationship", by just spinning faster than the planes relative speed (to an independant bystander), to negate their attempt at "pulling/holding" the plane stationary.

That's why (If we say the take off speed is 160mph):

Originally posted by redmage
Once the speed of the plane (relative to an independant bystander) reaches 160, the treadmill will be going 160 in reverse (the actual wheel speed will be roughly 320 mph) and the plane will still takeoff.

Remember, the question is from a bystander point of view and we are measuring the plane's speed (not the "wheel speed").

If (not that it could) the plane is held "stationary" by the conveyer, it's speed is zero, thus the conveyer cannot be moving.

Or, in a car analogy that does work: You can rev up a car on a dyno till the speedometer says 80 but the car's speed is still zero.

[edit on 2/17/06 by redmage]

posted on Feb, 17 2006 @ 09:27 PM

Originally posted by Travellar
By the way, Kruel, you diagram at the top of page 8 is pretty good. The only thing missing is the fact that someone's winching the rope in. (really fast!)

Nah, the rope wouldn't have to move at all. The only things moving would be the treadmill and the wheels.

What I've been trying to get at, is that in order for a plane/glider to lift, it would have to be moving faster than the air around it.

Some people think that the thrust alone would be strong enough, while I disagree. And I'm positive the glider would NOT lift, since it has no thrust at all. Remember the plane needs to be going a certain air speed, not ground speed, to get off the ground, which is why in general, pilots are encouraged to take off against the wind rather than into it. You'd need to get a higher top speed to lift off the ground while taking off in the same direction as the wind.

posted on Feb, 17 2006 @ 09:44 PM
ah, I see the problem. No, the thrust is not enough to lift the aircraft off by itself. It is however, enough to accelerate it despite the conveyer belt, and as it accelerates, it will gain airspeed and will then be able to generate lift.

posted on Feb, 17 2006 @ 09:46 PM

Originally posted by Kruel
What I've been trying to get at, is that in order for a plane/glider to lift, it would have to be moving faster than the air around it.

You're absolutely right.

Originally posted by Kruel
Some people think that the thrust alone would be strong enough while I disagree.

In your diagram there is no thrust so of course it won't take off.

The conveyer does not create the thrust, it's independant of the thrust in a plane (unlike with walking, bikes, cars, etc.).

Originally posted by Kruel
And I'm positive the glider would NOT lift, since it has no thrust at all.

Exactly, you're back on track.

Originally posted by Kruel
Remember the plane needs to be going a certain air speed, not ground speed, to get off the ground.

Absolutely, and it will be because the thrust is independant of the conveyer, and is actually isolated from it by the free spinning wheels, so the conveyer cannot cancel it out.

Thus if the thrust is not cancelled: for every action, there is an equal and opposite reaction and the plane will accelerate forward to gain the "air speed" needed.

[edit on 2/17/06 by redmage]

posted on Feb, 17 2006 @ 11:27 PM

Originally posted by redmage
Absolutely, and it will be because the thrust is independant of the conveyer, and is actually isolated from it by the free spinning wheels, so the conveyer cannot cancel it out.

Thus if the thrust is not cancelled: for every action, there is an equal and opposite reaction and the plane will accelerate forward to gain the "air speed" needed.

[edit on 2/17/06 by redmage]

In a previous post I referenced the Helica, a unique series of propeller-driven automobiles. Similarly to an airplane, there was no link between the Helicas' wheels and the large, front-mounted prop, which pulled the car down the road.

If the true speed of such a vehicle should be measured as airspeed (the measure of the speed of the air about the car, as generated by the thrust of the prop) then the car should be able to roll/move forward even if the wheels of the car were mounted on a counter-rotating surface. As are the wheels of the plane in the example.

If your analysis is correct, both the plane and the ancient Helica should be able to pull themselves forward off the counter-rotating surface.

It is that condition which must be met in order for our plane to generate lift and take flight, right?

Edited for spelling.

posted on Feb, 18 2006 @ 03:30 AM

If the true speed of such a vehicle should be measured as airspeed (the measure of the speed of the air about the car, as generated by the thrust of the prop) then the car should be able to roll/move forward even if the wheels of the car were mounted on a counter-rotating surface. As are the wheels of the plane in the example.

First, just to clairify, I'm using the term "airspeed" as the rate that the plane moves forward through the air (in relation to surrounding ground, not the conveyer).

For example, if you were to chain the frame of a Helica or plane to an unmovable object and fire up the prop, there would certainly be air moving/blown about it (from the prop) but it's air, and ground, speed (in this case) would still be zero. Like you've stated "An aircraft's propulsion system acts against the mass of the Aircraft" so if you chain the "mass", it can be stopped.

Admittedly, this might be a bad use of the term "airspeed" on my part because if a plane "maxes out at 300mph, and flies into a 300mph head wind, it will "hover" in flight (which in this case would make the "airspeed" zero as well).

I am in no way stating that the prop alone will pull enough air under the wings to attain the needed lift for flight on a "stationary" plane.

The plane will move forward, down the conveyer, to generate the needed lift for flight.

Second, if the Helicas had speedometers, they would most likely be connected to the wheels to measure "wheel speed" because that would be the more relevant measure for the Helica's use.

Third, you lost me at "mounted on".

If you mean that "if you were to place it on/at one end of a runway length counter-rotatable conveyer belt", similar to the brainteaser at hand, then (assuming the wheels and bearings held up) if the conveyer matched the "airspeed" (which would end up being roughly 1/2 the wheelspeed) of the Helica it would accelerate to the other end of the runway/conveyer.

Furthermore, the conveyer belt cannot match the wheelspeed, like in a car, bike, etc., to hold the plane/helica "stationary"
(yup, I said it
)

Because the thrust is independant of the belt, and especially due to the isolation effect of the free spinning wheels, it cannot be cancelled by the belt.

Since the thrust is independant, and isolated by the free spinning wheels, the wheelspeed will always be a combination of the conveyer's speed and the thrust's "pull". 1+1 cannot = 1 so the conveyer cannot match the wheelspeed to hold the vehicle "in place". It (the plane or helica) will move forward (and in the plane's case, will gain airflow over the wings!).

If your analysis is correct, both the plane and the ancient Helica should be able to pull themselves forward off the counter-rotating surface.

Yes, but in the plane's case it would eventually reach the forward speed required, to generate airflow over the wing, and create lift for a takeoff.

It is that condition which must be met in order for our plane to generate lift and take flight, right?

If you're referring to, "be able to pull themselves forward off the counter-rotating surface.", than yes, although the plane shouldn't have to pull itself "off" (the end) of the conveyer.

It should be able to attain the forward acceleration on the conveyer, assuming it is runway length, since they're generally longer than nescessary anyways, and this creates the airflow needed for lift/flight.

[edit on 2/18/06 by redmage]

posted on Feb, 18 2006 @ 03:50 AM
Ok for all of you people that are naysaying and forcing us to continually come up with 3409585 different ways/reasons to explain why the plane does indeed take off, I'm gonna flip the question around.

Please describe to me why the plane WOULD NOT take off. Don't say "the plane goes as fast as the belt" ... give me a description using the laws of physics. Force diagrams are welcome. Then we will take it from there.

posted on Feb, 18 2006 @ 04:15 AM
Ok I'll give you this question.

First I ask you to make two assumptions that are not neccesarily true or could ever be true.

#1 No parts get damaged during this.

#2 The plane doesnt bounce off the conveyor.

Ok now.. If we have a conveyor of reasonable length (1 mile or less) matching groundspeed of a jet aircraft that is about to land going the opposite direction as the plane will it stop it when it comes in to land, without the plane using brakes or any kind of negative thrust?

No it won't the plane will slow down some while its on the conveyor but in the end run off the end. The friction from the wheels to the wheel bearings alone is not enough to the stop the craft.

Or..

We have a conveyor again reasonable runway length that rotates forward at about 400mph and we have a plane sitting on it that will need a ground speed of 375mph to take off. Can this conveyor make the plane airborne without the plane using its own engines or using its brake?

No. There will not be enough friction from the wheels to the wheel bearings to allow the plane to get anywhere near the speed of the conveyor. Therefore it wont acheive the speed needed to take off.

........

If you understand both of those are a NO. Then you must agree that in the original question the plane WILL fly.

....................

If you dont agree with those you can make your own demonstration of the second one. Take a matchbox car, a long strip of paper (dont tape any together the tape will fook it up) put the car on the paper.. pull the paper out from under it quickly. Unless you pull up some the car will sit relatively still. The conveyor does not affect its motion much. The matchbox like the plane has free moving wheels!

[edit on 18-2-2006 by Xerrog]

[edit on 18-2-2006 by Xerrog]

posted on Feb, 18 2006 @ 10:21 AM

Originally posted by Fiverz
Please describe to me why the plane WOULD NOT take off.

For the plane to take off it has to push air downwards with an impulse equale to or greater than it's mass. A wing does this by moving through the air, therefor for the plane to take off the wing needs to move through the air.

Lets look at the question again.

The plane moves in one direction, while the conveyer moves in the opposite direction. 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).

But this time lets replace the wheels with skids where the friction between the skid and the conveyer increases linearly with speed, this is a close approximation to reality.

It should be obviouse that with skids the plane will not take off because the thrust from the engin can be cancled out by increasing the speed of the conveyer.

However the question asks about a plane with wheels. Wheels have much lower friction that skids. The question then becomes does the rolling resitance of a wheel increase with the velocity of the ground? If it does then the conveyer can just increase its speed even further and still stop the plane from taking off, if the rolling resitance does not increase with conveyer speed then the plane will take off.

I think that rolling resistance is proportional to velocity. Therefore it is possible for the conveyer to stop the plane.

posted on Feb, 18 2006 @ 10:29 AM
But consider this, the belt can move at the same speed as the plane, it says nothing about cancelling the speed of the plane, in fact, if the belt DID cancel the speed of the plane the belt itself would stop as it is the forward speed of the plane that is matched, not the rotational speed of the wheels, therefore for the belt to match the speed of the plane at all the plane itself MUST be moving.

Likewise you point with the skids, it is never claimed that tjhe belt is counteracting the engine output, only that it is matching the forward speed of the plane, which again therefore must be moving forwards, not in any relative sense, but in a very real sense, or the belt would stop. The very fact that the wheels are free spinning is what allows this phenomenon to occur.

[edit on 18-2-2006 by waynos]

posted on Feb, 18 2006 @ 10:32 AM
Quite

The argument seems to be on the interpretation of the question, ie is the conveyer used to keep the plane in the same possition or is the speed of the conveyer just the ground speed of the plane time -2?

posted on Feb, 18 2006 @ 01:20 PM

Originally posted by Nacnud
The argument seems to be on the interpretation of the question, ie is the conveyer used to keep the plane in the same possition

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

Dependant Thrust

Cars, bikes, etc.

Because their thrust is dependant (generated at/by the wheels), the thrust is the wheelspeed, and it can be cancelled by the conveyer reaching an equilibrium with it.

1 (thrust/wheelspeed are one and the same ) = 1 (conveyer speed) The vehicle will be "held in place".

1=1

Independant Thrust

Plane, helica, etc.

Because their thrust is independant (not generated at/by the wheels), the thrust is not just the wheelspeed, and it cannot be cancelled by the conveyer.

1 (independant thrust) + 1 (conveyer speed) = X wheelspeed (which will always be > 1 so it will not be "held in place" by the conveyer)

Remember: the thrust and conveyer are turning the free-spinning wheels in the same direction so addition (+) is nescessary.

1 + 1 cannot = 1

The conveyer cannot match the wheelspeed (to achieve an equilibrium) because it cannot cancel the thrust.

Since the wheelspeed will always be greater than the conveyer speed, the vehicle will move forward to gain airflow!

Tracking the Speed

Since the conveyer tracks the "plane speed", and we've proven that it can not reach the wheelspeed (due to the independant thrust), we can now effectively substitute speed for thrust in the equations to show us where the "plane speed" is.

Dependant Speed

Because their speed is dependant (generated at/by the wheels), the speed is the wheelspeed, and it can be cancelled by the conveyer reaching an equilibrium with it.

1 (speed/wheelspeed are one and the same ) = 1 (conveyer speed) The vehicle will be "held in place".

1=1

Independant Speed

Because their speed is independant (not generated at/by the wheels), the speed is not just the wheelspeed, and it cannot be cancelled by the conveyer.

1 (independant speed) + 1 (conveyer speed) = X wheelspeed (which will always be > 1 so it will not be "held in place" by the conveyer)

1 + 1 cannot = 1

The "1 (independant speed)" is our "plane speed" that the conveyer tracks (and matches) so the wheelspeed would be, for all practical purposes, double.

[edit on 2/18/06 by redmage]

posted on Feb, 18 2006 @ 01:56 PM
If you put the plane on skids.. No it will not take off.

However this is referring to a normal plane. Normal planes have wheels. A plane with wheels WILL take off.

posted on Feb, 18 2006 @ 08:24 PM

Originally posted by Nacnud
Quite

The argument seems to be on the interpretation of the question, ie is the conveyer used to keep the plane in the same possition or is the speed of the conveyer just the ground speed of the plane time -2?

The original question as I read it was the conveyer matches the aircraft's speed in the oppisite direction.

Not, as some people choose to read it, "Conveyer moves fast enough in the opposite direction to generate enough friction at the wheels to prevent the plane from achieving forward momentum."

The wheels may not be completely frictionless, however, they are by design very low in resistance to rolling forwards. That was the design requirement that caused people to start putting wheels on airplanes in the first place. Obviously, the airplane has enough power in it's engine to overcome the rolling resistance of it's wheels, otherwise it would never be able to accelerate and achieve flight in the first place.

posted on Feb, 18 2006 @ 08:48 PM
To re-visit this thread (groan),consider the original postulate;

"A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. 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 consider;

"This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly 100 times as fast (but in the opposite direction). Can the plane take off?"

This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly 100 times as fast in the same direction. Can the plane take off?"

It will fly in each case, if the wheels can stand the rotations.
See? The ground is irrelevant.

posted on Feb, 18 2006 @ 09:27 PM
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".

1) If it means 'the wheels, bearings, and struts/support are frictionless (but there is friction between the tires and belt)', then the plane will take off. Any friction that would be between the tires and belt DOES NOT GET TRANSFERRED to the plane. The tires "cause" the wheels to spin, and since the bearings and/or supports have no friction, they are allowed to rotate freely (hence no 'connection' to the belt).

2) If it means 'the tires are frictionless (but not the wheels, bearings, and struts/support)' then the plane will take off. This is identical to the 'skids on ice' case. There is no part of the entire group we call a "plane" that would feel the forces of the belt. Think of a car downhill on ice ... even with the brakes on and wheels stationary, it will still move (the force is simply gravity in that case, not thrusters). This is the case that I assumed when stating the plane would take off in all my previous posts.

3) The case where there's "no friction at all in any of the wheel assembly (tires, wheels, bearings, and supports)" is trivial. As a combination of the two previous cases, the plane will take off.

However, if nothing at all is stated about the wheels, you have a different beast.

4) To takeoff*, a plane with "wheels of normal friction" would take the normal level of thrust + an additional amount to overcome the increase in friction caused by the conveyor belt. But since the belt is only moving as fast as the plane at any given point, an outside stationary observer will not see any difference in the take off. The added force of friction on the plane would create the need for a greater force of thrust (to overcome that friction), but the SPEEDS still match to the observer. The same LENGTH of the belt is needed to take off as in the other cases. The wheels will be spinning at twice their normal speeds (because there normally IS friction on a static runway, there is no need to add/subtract any other forces here ... it's just as if the static runway was moving by twice as fast as normal at any given point).

*There will also be additional stress and heat generated, so the wheel assembly would have to be able to withstand that. I am positive that wheel assemblies on modern aircrafts would be rated to exceed all standard takeoffs/landings (emergency landings often have speeds that are much greater than ever found on a takeoff).

- - - - -

Remember, this is all assuming that the plane will have a velocity relative to a stationary viewer outside the system.

It really does get into a question of semantics in regards to how the question is stated. But in all cases (assuming the * conditions are met), the plane will move forward down the runway (in spite of the belt), the wings will create lift, and it will take off.

posted on Feb, 18 2006 @ 10:20 PM

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.

posted on Feb, 18 2006 @ 10:45 PM
Lets just end this right now, we will put a channel wing on the belt that way you do not have to worry about it needing to much forward speed to take off.

[edit on 18-2-2006 by ULTIMA1]

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