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Atl5p
A plane is sitting on a moveable runway, like a treadbelt. The plane tries to take off in one direction, while the treadbelt moves in the opposite direction. The treadbelt has a control system to measure the plane's grounspeed over the runway, and the treadbelt matches this speed exactly.

Is the plane able to fly? Does it run up and fly?

I say "NO". What do you say, and why?
ImmortalCoil
My answer is the same as before....no, it won't...since it won't gain any airspeed. Someone suggested that it will in the previous thread because it gets a head-start over the belt....but I don't think that's implied by the question.....
it's a hypothetical situation of course, and we assume that the speed of both things will be exactly the same all the time!
Insyght
MODS, can you not prevent this insanity? ... I'll will crush my faith in science!
NoCleverName
Plane takes off, as it does in the "original problem". Engines apply force in direction d, wheels provide rolling resistance in direction -d. Since airplanes have engines designed to overcome all resisting forces likely to be encountered, plane moves forward, as is expected by conservation of momentum (thrust directed backwards, equal and opposite force forwards.

The conveyor belt is merely an unneeded contrivance to trick the gullible; this is a simple force balances problem with engine forces in one direction and resisting forces in another. Planes are designed to overcome resisting forces (wheels, drag, etc.) so off we go. For those who think otherwise, consider that the plane doesn't "know" it has wheels, it just feels resisting forces on its wheel struts that are further transmitted to the body. The fact those resisting forces are generated by "rolling friction" doesn't change the fact they are still simple adverse forces, like having the brakes on or having an uphill runway. The rolling conveyor is just a method of inducing added rolling friction force, it does not impart any rearward motion to the plane as the vector addition of the engine force and the resisting forces balance in favor of the engine.

Again, the key is that engine forces overwhelm adverse force by orders of magnitude in real aircraft. For those who think the conveyor moves the plane backwards, consider a plane confronting an uphill runway. When brakes are released the plane rolls back slightly then engines take charge and plane moves forward. The "thought experiment" of the conveyor somehow implies, equivalently, that the "uphill" runway dynamically get "steeper" as the plane attempts to make progress against it (implication: increasing adverse force). But airplanes, as we understand them, do not meet increasing adverse forces the engines can't overcome (except for aero drag). That is, rolling resistance is negligible.

To pound the point home further, for those who somehow think the plane is "attached" to the conveyor and is moved backwards diabolically just as the plane tries to move forward: the plane's engines do work W = F ds. For the conveyor to move the plane backwards the conveyor has to do the exact reverse. The only "F" the conveyor has to work with is the rolling resistance and shear inertia. Not enough, as demonstrated by planes taking off from non-conveyor runways. For those who think situation is same as runner on a treadmill: not the same. Runner/treadmill is a "closed system" while airplane engines working against air is an "open sysem".

Note to Atl5p: you can attempt to reformulate this problem all you like, but until you constrain the plane to act in non-normal plane-like ways, you're stuck: the plane flys. Forget the dumb conveyor and just do a little vector algebra to see why.
doddsyboy99
QUOTE (NoCleverName+Nov 23 2006, 04:02 AM)
Plane takes off, as it does in the "original problem". Engines apply force in direction d, wheels provide rolling resistance in direction -d. Since airplanes have engines designed to overcome all resisting forces likely to be encountered, plane moves forward, as is expected by conservation of momentum (thrust directed backwards, equal and opposite force forwards.

The conveyor belt is merely an unneeded contrivance to trick the gullible; this is a simple force balances problem with engine forces in one direction and resisting forces in another. Planes are designed to overcome resisting forces (wheels, drag, etc.) so off we go. For those who think otherwise, consider that the plane doesn't "know" it has wheels, it just feels resisting forces on its wheel struts that are further transmitted to the body. The fact those resisting forces are generated by "rolling friction" doesn't change the fact they are still simple adverse forces, like having the brakes on or having an uphill runway. The rolling conveyor is just a method of inducing added rolling friction force, it does not impart any rearward motion to the plane as the vector addition of the engine force and the resisting forces balance in favor of the engine.

Again, the key is that engine forces overwhelm adverse force by orders of magnitude in real aircraft. For those who think the conveyor moves the plane backwards, consider a plane confronting an uphill runway. When brakes are released the plane rolls back slightly then engines take charge and plane moves forward. The "thought experiment" of the conveyor somehow implies, equivalently, that the "uphill" runway dynamically get "steeper" as the plane attempts to make progress against it (implication: increasing adverse force). But airplanes, as we understand them, do not meet increasing adverse forces the engines can't overcome (except for aero drag). That is, rolling resistance is negligible.

To pound the point home further, for those who somehow think the plane is "attached" to the conveyor and is moved backwards diabolically just as the plane tries to move forward: the plane's engines do work W = F ds. For the conveyor to move the plane backwards the conveyor has to do the exact reverse. The only "F" the conveyor has to work with is the rolling resistance and shear inertia. Not enough, as demonstrated by planes taking off from non-conveyor runways. For those who think situation is same as runner on a treadmill: not the same. Runner/treadmill is a "closed system" while airplane engines working against air is an "open sysem".

Note to Atl5p: you can attempt to reformulate this problem all you like, but until you constrain the plane to act in non-normal plane-like ways, you're stuck: the plane flys. Forget the dumb conveyor and just do a little vector algebra to see why.

you dont know what you talking about
NoCleverName
QUOTE (doddsyboy99+Nov 23 2006, 11:40 AM)
you dont know what you talking about

It's not clear from your profile if you've even made it to calculus yet, let alone something like, say, college-level courses. So I wouldn't be so sure about who has a better idea of what they are talking about.

But your counter-argument was crisp and incisive.
rethinker
Hello all
This my first post and I must say that I believe you have no choice but to accept the physics of Mr NoCleverName.

You see one thing you may have overlooked.
All aircraft which may wish to fly west will be taking off with a natural tread belt.
This should show that the air craft no matter how large or small the engine, it will overcome any tread belt under the wheels.

The wheels if let to freely spin, will not hold onto the tread belt.
Thanks for a very good think.

I forget what the speed of the Earth spins at, but it is easy to look up to know.
ImmortalCoil
perhaps ppl are misunderstanding what I wish to say....
I am not opposed to the physics of Noclevername.....yes, the plane will move forward due to propulsion from the air, and yes, it's wheelspeed is not very relevant, but you must understand, unless you strip your aircraft of all weight but the engine/wings/props, it will not be able to take off. It will probably jump forward, bank upwards, then immediately stall and fall to the ground.

anyway, I am busy today, but hopefully I will find time to make a simple 3d simulation with some jumbo-jet.

cheers
NoCleverName
rethinker: unfortunately for your otherwise inspired idea, the real-world earth-air-plane system is all "part" of the same rotating frame and is not the same scenario as envisioned in the problem at hand. The reason why there is no 1000mph wind at the equator (and "none" at the poles) is that the air there, like everything else, has acquired the same linear speed as the ground, with gravity keeping everything (that isn't rigidly attached to the earth) moving in a curve. However, as you may be well aware, this "free" velocity is used to advantage in the space program and is why launch sites are located as close the equator as possible.

Immortalcoil: it's hard to see how you can "accept" the physics but "reject" the consequences (although this can be a common issue with quantum physics ).
Atl5p
Remember, the treadbelt is matching the plane's wheelspeed.

The plane CANNOT have IAS, while the treadbelt is matching wheelspeed. If your version floating around inside your empty head has the plane with any amount of IAS, then you have not done your job correctly...you have not matched the plane's wheelspeed with the treadbelt's speed.

Is this what you are thinking:

Plane has 50mph wheelspeed. Treadbelt spins at 50mph. Plane has 50mph IAS. WRONG!! In order for the plane to have 50mph IAS, and the treadbelt is spinning at 50mph, then the wheels must be spinning at 100mph!! Now, ask yourself: Does 50mph treadbelt speed = 100mph plane wheelspeed?

A ) Yes
B ) No
C ) None of the above
D ) All of the above

(Hint, it's not C or D)
Benny
QUOTE
The plane CANNOT have IAS, while the treadbelt is matching wheelspeed. If your version floating around inside your empty head has the plane with any amount of IAS, then you have not done your job correctly...you have not matched the plane's wheelspeed with the treadbelt's speed.
So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

Wait-a-minute, your version of the question says that the conveyor runway matches the "plane's groundspeed ". Groundspeed, by definition, is always measured against solid ground. Ergo, the plane will fly just like it does in the original question.
Atl5p
QUOTE (Benny+Nov 24 2006, 11:34 AM)
So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

Wait-a-minute, your version of the question says that the conveyor runway matches the "plane's groundspeed ". Groundspeed, by definition, is always measured against solid ground. Ergo, the plane will fly just like it does in the original question.

The question says "grounspeed over the runway" remember the runway...the one that 'moves'...sort of like a band conveyor? The plane's speed is measured as the speed over the surface of the treadbelt's surface...do you not see that? What can I do to help you to read?
NoCleverName
In this "new" definition of how the control system operates the plane still flies, of course, because there are no forces strong enough to oppose its engines.

But interestingly enough, your new "control system" defines a feedback loop that quickly gets the belt moving to "infinite" speed. Since the plane does move, the control system can never satisfy the demand for speed.

Let the plane move forward at 10mph. The control system reading the wheels sees "10mph" and ramps up the conveyor. Now the wheels are moving at 10mph (to earth) against an opposing conveyor runing at 10mph (to earth). So the wheels now read a speed of "20mph" and demand "20mph" from the conveyor ... etc, etc, etc ...

The only stable configuration for this system is engine off, no movement at all. For even the slightest movement ds/dt is going to call for the control system to respond ... it has no reason to respond otherwise. Since it can never match the wheelspeed, which is always at least ds/dt/ ahead of it (otherwise there is no reason for the control system to respond) the system is unstable and runs to infinite speed.

Better open a new thread with a new problem definition. This one's had it.
Benny
QUOTE
So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.
I noticed you completely ignored this part. Not that you can honestly counter something obviously true.

QUOTE (->
 QUOTE So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.
I noticed you completely ignored this part. Not that you can honestly counter something obviously true.

The question says "grounspeed over the runway" remember the runway...the one that 'moves'...sort of like a band conveyor? The plane's speed is measured as the speed over the surface of the treadbelt's surface...do you not see that? What can I do to help you to read?
Groundspeed is groundspeed is speed relative to the solid ground. "groundspeed over the runway" is either gibberish or it means groundspeed. Why not just ask "Can a moving runway stop a plane?" Is it too simple and obvious for you?

So...
Can a moving runway stop a plane?

No. The runway can't spin up quickly enough to stop the plane.
Atl5p
A treadbelt CAN stop a plane...I done it...it works...spin the treadbelt up fast enough, and the plane will not move IAS.

Infinity is a pretty big number....are you sure you want to go there?

Lots of Fly Boys agree...a treadbelt CAN stop a plane...they just like to use IAS to define a planes speed on a treadbelt. What I dont get is when people dont understand how having ANY amount of CRF will result in a fast belt holding the plane to 0 IAS.

It works in a scaled down variation...no reason it wont work with a bigger plane and a bigger treadbelt. No reason at all...

Think of this...put a bicycle on a treadbelt. Go up to 10mph. Now COAST...what happens? YOu slow down on the belt, and begin to move reverse IAS, right?

Now, raise up the back of the treadbelt...so the rear wheel of the bike is above the front. Can you raise the back of the treadbelt just enough so that the rider can "Cosast" indefinatly? WIth the treadbelt coming at you at 10mph, but you are now facing downhill, and coasting...sort of like a never ending hill...got that picture?

If you slow the belt down, you will roll forwards towards the front of the belt. But if you raised the belt speed, you would roll backwards towards the back of the treadbelt.

OK, now...THAT is what I am talking about...you have come to the equilibrium point. The belt speed, and the downward force from gravity have balanced. Can you see now that varying the treadbelt speed will impart various forces unto the plane?
NoCleverName
I guess I didn't expect you to answer any of the technical questions; it would seem you don't know how. For example:

* You have yet to answer how your magic belt (if we consider it to be "real world") can develop the force needed to oppose the engines (in the case of, say a B-777, that would be some 200,000 lbs of thrust).

* You haven't dealt with the unfortunate requirement that momentum be conserved (all that mass moving at high speed back from the engines). Exactly were is the -mv going?

* You haven't dealt with where the conveyor is going to come up with the work -Fds to oppose the work done by the engines. You're stuck with that because (a) the engines are developing force and (b ) the plane has to move at least ds or the control system would logically never kick in since there was never any movement to measure. (This is where my trick question about the motive power for the conveyor comes in: they need motors at least as powerful as the plane's!)

* You have not explained why your control system, as defined, is not an unstable system.

* etc. ...

In any game, one can find amusement with an equally skilled partner regardless of the skill level. However, it would seem that you do not posess the degree of skills to play this game at the level which would amuse the other participants.

In short, it seems you are semi-literate in math and have no vocabulary in physics. As it is, you're simply unable to defend your argument with anything that resembles an engaging, let alone plausible, position.
rethinker

It is a matter of what creates the most force, the tread belt, or the thrust of the the plane.
If both are equal, yes I agree no forward motion will occur.
Something like equal and opposite.

BTW thanks NoCleverName.
Benny
QUOTE
So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

QUOTE (->
 QUOTE So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

Can a moving runway stop a plane?
No. The runway can't spin up quickly enough to stop the plane.
QUOTE
A treadbelt CAN stop a plane...I done it...it works...spin the treadbelt up fast enough, and the plane will not move IAS.
- If we assume that a Cessna can be stopped with a 400mph conveyor...
- If the plane and conveyor go full throttle at the same time, then the plane will fly because the conveyor is not capable of reaching 400mph.
- If we assume that the conveyor can reach 400mph, then it will take so long to get there that the plane will fly.
- If we assume that the conveyor can reach 400mph quickly enough, then there is not enough traction and either the tires will fail or the plane will fly, but with some extra difficulty.
- If we assume that the traction is not a problem, then the wheels will fail.
- If we assume that the wheels are light and strong enough, then the bearings and tires will fail.
- If we assume that the tires and bearings can handle 400mph, then a 400mph conveyor is no longer fast enough and the plane will fly.
- Repeat.

QUOTE (->
 QUOTE A treadbelt CAN stop a plane...I done it...it works...spin the treadbelt up fast enough, and the plane will not move IAS.
- If we assume that a Cessna can be stopped with a 400mph conveyor...
- If the plane and conveyor go full throttle at the same time, then the plane will fly because the conveyor is not capable of reaching 400mph.
- If we assume that the conveyor can reach 400mph, then it will take so long to get there that the plane will fly.
- If we assume that the conveyor can reach 400mph quickly enough, then there is not enough traction and either the tires will fail or the plane will fly, but with some extra difficulty.
- If we assume that the traction is not a problem, then the wheels will fail.
- If we assume that the wheels are light and strong enough, then the bearings and tires will fail.
- If we assume that the tires and bearings can handle 400mph, then a 400mph conveyor is no longer fast enough and the plane will fly.
- Repeat.

What I dont get is when people dont understand how having ANY amount of CRF will result in a fast belt holding the plane to 0 IAS.
"The force required to overcome the friction is equal to the CRF multiplied by the weight on the tire." What I don't get is when you don't understand your own reference material. (And before you mention your "plane" this equation is useful when the resistance is outmatched by thrust, like a real plane. There are more accurate equations that should be used instead when your vehicle is badly underpowered like your "plane" was/is. For our purposes the extra accuracy is unnecessary and we can stick to the simpler equation.)
Benny
QUOTE (ImmortalCoil+Nov 23 2006, 09:01 AM)
perhaps ppl are misunderstanding what I wish to say....
I am not opposed to the physics of Noclevername.....yes, the plane will move forward due to propulsion from the air, and yes, it's wheelspeed is not very relevant, but you must understand, unless you strip your aircraft of all weight but the engine/wings/props, it will not be able to take off. It will probably jump forward, bank upwards, then immediately stall and fall to the ground.

anyway, I am busy today, but hopefully I will find time to make a simple 3d simulation with some jumbo-jet.

cheers

Can you explain further? I am confused by you stating that the plane can move forward and that wheelspeed is irrelevant, but the plane won't fly anyway and I have no idea what does stripping the aircraft of weight has to do with anything or why it would apparently fly then crash and burn immediately.
tlocity
The lift of an airplane is only the result of the movement of air over the wings and any downward force produced by the engine. Movement over the ground plays no part in the lift of the plane.
egnorant
QUOTE (Atl5p+Nov 22 2006, 07:23 PM)
A plane is sitting on a movable runway, like a treadbelt. The plane tries to take off in one direction, while the treadbelt moves in the opposite direction. The treadbelt has a control system to measure the plane's grounspeed over the runway, and the treadbelt matches this speed exactly.

Is the plane able to fly? Does it run up and fly?

I say "NO". What do you say, and why?

I am having a little trouble figuring the belts speed as measured as "groundspeed over the runway" in the opposite direction.

Another point of clarification is that the plane tries to take off in one direction
This indicates possible success or failure to take off.
Good thing it doesn't say it moves in one direction right?

Groundspeed over the belt......Hmmmm
Can we determine the belts speed as groundspeed under the plane?

Or is this defining the belt AS ground so that the speed of the plane is measured as movement over the ground/belt.
Since the opposite of 10 mph east relative to this ground/belt point is 10 mph west relative to this ground belt point, then the belt is moving 10 mph west relative to itself!!

wait....ummmm...something doesn't look right.
let me study it some more.

Bruce

MikeMonty
QUOTE (Atl5p+Nov 25 2006, 04:00 AM)
A treadbelt CAN stop a plane...I done it...it works...spin the treadbelt up fast enough, and the plane will not move IAS.

Infinity is a pretty big number....are you sure you want to go there?

Lots of Fly Boys agree...a treadbelt CAN stop a plane...they just like to use IAS to define a planes speed on a treadbelt.  What I dont get is when people dont understand how having ANY amount of CRF will result in a fast belt holding the plane to 0 IAS.

It works in a scaled down variation...no reason it wont work with a bigger plane and a bigger treadbelt.  No reason at all...

Think of this...put a bicycle on a treadbelt.  Go up to 10mph.  Now COAST...what happens?  YOu slow down on the belt, and begin to move reverse IAS, right?

Now, raise up the back of the treadbelt...so the rear wheel of the bike is above the front.  Can you raise the back of the treadbelt just enough so that the rider can "Cosast" indefinatly?  WIth the treadbelt coming at you at 10mph, but you are now facing downhill, and coasting...sort of like a never ending hill...got that picture?

If you slow the belt down, you will roll forwards towards the front of the belt.  But if you raised the belt speed, you would roll backwards towards the back of the treadbelt.

OK, now...THAT is what I am talking about...you have come to the equilibrium point.  The belt speed, and the downward force from gravity have balanced.  Can you see now that varying the treadbelt speed will impart various forces unto the plane?

The bicycle analogy was done ages ago - but unless the guy on the bike has a working fan stuck up his arse it's STILL NOT VALID!

Here's a real-life situation that pilots do as an exercise:
The plane will come in to a runway, touch tyres to the runway and take off again before they have to commit to a landing.
What does that tell you?
1. Rolling resistance and even the MASSIVE inertial forces acting on the wheels as they accellerate up to speed are not enough to stop the plane.
2. If the plane touched down at 210 MPH and is capable of lift off at 70 MPH, you can look at this as the final phase of conveyor takeoff with the plane travelling at 70 MPH and the runway travelling at 140 MPH.
(Of course that neglects that the planes airspeed is greater than it should be, but we know it could take off with only 70 MPH across its wings anyway.)

It seems that having lost the battle on the original thread ATLP5 has regrouped and REFORMULATED the ground on what he hopes is more favourable territory.

The question is phrased in even worse and woolier terms than the original and serves only to expose his thought process (or lack thereof) and his commitment to his own flawed opinion.

Looking for fresh fodder ATLP5?

Mike
Atl5p
The treadbelt in this question is like the one at the gym. You get on, and walk 5mph, set the treadbelt at exactly the same speed in the opposite direction...do you get IAS?

Plane has the advantage of static thrust when held at 0 IAS. A plane rolling at 5mph goes onto a treadbelt moving at 5mph. The wheels go to 9mph. The treadbelt increases to 9mph, the IAS goes to 4mph. The wheels go up to 16 mph, the treadbelt follows...eventually the treadbelt is spinning at 50mph, the wheels at 50mph, and plane has 0 IAS. Eventually there will be a speed that holds plane to 0 IAS. It is not infinity.
NoCleverName
You, of course, have been fooled (intentionally) by the problem designer into thinking the plane/conveyor problem is similar to the walker/gym or the boat/river problem.

That was the whole point of the problem.

If it were otherwise, then the original problem would be just a boring repetition of the boat/river and other such problems ... and it would be of absolutely no interest. So the "obvious" answer ("no fly") is clearly false because otherwise the problem would not be interesting.

The real problem is to understand why this problem is different from boat/river or walker/gym. You have not met the challenge. Instead, you keep coming up with reasons why this is actually "a boring problem similar to walker/gym." You have fallen hook, line, and sinker into the trap.

You have spent hundreds of pages caught in this trap; yet others, such as "rethinker", were aware of what was going on after just a couple of posts. Like I said, you aren't playing at the skill level of the other participants.

Atl5p
QUOTE (NoCleverName+Nov 25 2006, 07:33 PM)
You, of course, have been fooled (intentionally) by the problem designer into thinking the plane/conveyor problem is similar to the walker/gym or the boat/river problem.

That was the whole point of the problem.

If it were otherwise, then the original problem would be just a boring repetition of the boat/river and other such problems ... and it would be of absolutely no interest. So the "obvious" answer ("no fly") is clearly false because otherwise the problem would not be interesting.

The real problem is to understand why this problem is different from boat/river or walker/gym. You have not met the challenge. Instead, you keep coming up with reasons why this is actually "a boring problem similar to walker/gym." You have fallen hook, line, and sinker into the trap.

You have spent hundreds of pages caught in this trap; yet others, such as "rethinker", were aware of what was going on after just a couple of posts. Like I said, you aren't playing at the skill level of the other participants.

DId you mean that comment for a different post? See, I have asked my own question...started a new thread...this question sets it up like a walker in a gym.
NoCleverName
The comment is good here, too. You didn't even have the competence to set up the problem is this thread to be different in any meaningful way to the "Original" problem. In fact, you botched it royally with your newly defined unstable "control system" (if you can even use the word "control").

But if you like, you can consider this post as present in the original thread and answer it there.
egnorant
QUOTE (Atl5p+Nov 22 2006, 07:23 PM)
A plane is sitting on a moveable runway, like a treadbelt. The plane tries to take off in one direction, while the treadbelt moves in the opposite direction. The treadbelt has a control system to measure the plane's grounspeed over the runway, and the treadbelt matches this speed exactly.

Is the plane able to fly? Does it run up and fly?

I say "NO". What do you say, and why?

I'm still wondering about "groundspeed over the belt"
I have looked around and all the definitions I can extrapolate from the info given still has the plane speed measured relative to the ground.
Adding "over the belt" indicates a location, but not a reference point.

This would have the plane speed as measured as groundspeed,
adding over the belt is restating the fact that the plane is sitting ON th belt and then moving!
We can figure out that the belt is under the plane...over, under...ain't hard!
Location get it??!!

If you mean wheelspeed or speed relative to the belt, don't add the specific reference to the ground.

Bruce

NoCleverName
Why bother to ask him, egnorant. There are only 4 possible control system setups. So let's analyze each.

First, how to measure speed. There are two objects here (plane and conveyor), and there are only two possible frames of reference (external and internal). So let's look at each input to the control system and its output:

1) The input is an external observer's measurement of the plane's speed. The control system adjusts the conveyor such that its speed is also measured by the external observer (i.e., everything's measured against the fixed earth).

- If the observer sees the plane moving by him at, say, 10 mph, the control system starts the conveyor running so that a fixed point on it passes the observer at the same 10 mph. Result: the plane continues forward at 10 mph, it's wheels measure a speed of 20 mph. In otherwords, no opposition to take-off.

- If the plane is, somehow, NOT supposed to move (IAS 0) then there is no movement measured by the observer and by definition the control system will NOT move the conveyor. In otherwords, dead stop, either brakes on or engine off.

2) Again, input is observer's measurement of plane speed, control system responds by moving conveyor so that wheelspeed is observer's speed.

- If observer sees plane moving at 10 mph, control system is directed to move conveyor so that wheelspeed is 10 mph. But the wheelspeed is already 10 mph since the external observer measures the plane moving at that speed. So the conveyor needs no adjustment and remains at zero speed relative to the observer. Again, there is no opposition to takeoff as IAS 10 is present in this case.

- Again, if the plane somehow was supposed to be at IAS 0, that would mean the observer measures zero and signals the control system to maintain zero. Another case of dead stop, engines off.

3) For this and the following case, the plane's speed is measured by it's motion relative to the conveyor; i.e. wheelspeed. In this case, the control system will direct the conveyor to maintain a conveyor speed determined by the conveyor passing the fixed observer.

- The plane reads a wheelspeed of 10 over the conveyor (we aren't going to try to define how this happened). So the conveyor is directed to run at 10 mph relative the observer. If the plane is constained to IAS 0 then after a period of oscillation the control system and the wheels will both be stable at their respective readings of 10 mph.

- If, on the otherhand, the IAS is non-zero (actual motion relative to the observer) the system is unstable since the actual motion must be added to the reverse motion, thus increasing wheelspeed, thus asking for more control motion, etc. The conveyor speed has no upper bound.

4) Again, plane speed measured by wheels, control system responds so that conveyor matches wheelspeed.

- In this state the conveyor never moves because the wheelspeed read is also the target of the conveyor speed as measured at the wheels. For IAS zero, again we have a dead stop system.

So we can see that the only possible control system design that works for the purposes of this problem is case 3: The plane speed is measured by wheelspeed, the conveyor by external speed. Even then, the control system is unstable in every case where the plane is seen to move by the observer.

Further, we did not define how the plane got to any particular wheelspeed. In any "real world" scenerio, the control system would not be able to respond instantly to a plane's motion, so the control system would quickly become unstable.

Like I said more succintly before, this problem setup's a mess.
Atl5p
QUOTE (NoCleverName+Nov 26 2006, 01:10 AM)
Why bother to ask him, egnorant. There are only 4 possible control system setups. So let's analyze each.

First, how to measure speed. There are two objects here (plane and conveyor), and there are only two possible frames of reference (external and internal). So let's look at each input to the control system and its output:

1) The input is an external observer's measurement of the plane's speed. The control system adjusts the conveyor such that its speed is also measured by the external observer (i.e., everything's measured against the fixed earth).

- If the observer sees the plane moving by him at, say, 10 mph, the control system starts the conveyor running so that a fixed point on it passes the observer at the same 10 mph. Result: the plane continues forward at 10 mph, it's wheels measure a speed of 20 mph. In otherwords, no opposition to take-off.

- If the plane is, somehow, NOT supposed to move (IAS 0) then there is no movement measured by the observer and by definition the control system will NOT move the conveyor. In otherwords, dead stop, either brakes on or engine off.

2) Again, input is observer's measurement of plane speed, control system responds by moving conveyor so that wheelspeed is observer's speed.

- If observer sees plane moving at 10 mph, control system is directed to move conveyor so that wheelspeed is 10 mph. But the wheelspeed is already 10 mph since the external observer measures the plane moving at that speed. So the conveyor needs no adjustment and remains at zero speed relative to the observer. Again, there is no opposition to takeoff as IAS 10 is present in this case.

- Again, if the plane somehow was supposed to be at IAS 0, that would mean the observer measures zero and signals the control system to maintain zero. Another case of dead stop, engines off.

3) For this and the following case, the plane's speed is measured by it's motion relative to the conveyor; i.e. wheelspeed. In this case, the control system will direct the conveyor to maintain a conveyor speed determined by the conveyor passing the fixed observer.

- The plane reads a wheelspeed of 10 over the conveyor (we aren't going to try to define how this happened). So the conveyor is directed to run at 10 mph relative the observer. If the plane is constained to IAS 0 then after a period of oscillation the control system and the wheels will both be stable at their respective readings of 10 mph.

- If, on the otherhand, the IAS is non-zero (actual motion relative to the observer) the system is unstable since the actual motion must be added to the reverse motion, thus increasing wheelspeed, thus asking for more control motion, etc. The conveyor speed has no upper bound.

4) Again, plane speed measured by wheels, control system responds so that conveyor matches wheelspeed.

- In this state the conveyor never moves because the wheelspeed read is also the target of the conveyor speed as measured at the wheels. For IAS zero, again we have a dead stop system.

So we can see that the only possible control system design that works for the purposes of this problem is case 3: The plane speed is measured by wheelspeed, the conveyor by external speed. Even then, the control system is unstable in every case where the plane is seen to move by the observer.

Further, we did not define how the plane got to any particular wheelspeed. In any "real world" scenerio, the control system would not be able to respond instantly to a plane's motion, so the control system would quickly become unstable.

Like I said more succintly before, this problem setup's a mess.

No, number 3 is correct. Actually, #4 was good to, except there was no wheelspeed, because you forgot to turn on the plane. Since the plane wasn't turned on, there was no wheelspeed, therefore no treadbelt speed. ANyway, I think you may get it now.

Your problem with #3 is that you think wheelspeed will be more than treadbelt speed. That is solved simply by increasing treadbelt speed. Eventually, there WILL be a speed at which the treadbelt can hold the plane to 0 IAS.

I made it happen, no mistake.
rethinker
If the power induced in the tread belt was directly equal to the power of the plane, and all the power from the plane was directed at the gear or connection to the tread belt,yes it would not fly. However this is a very hard test to build in a real world prototype.

If you can set it up, all power from the plane would be used on the tread belt spin and no power would be let free as thrust.
It is an equal and opposite question.

If you have a video of your test, can you put it on Goggle for us to see.
Was it done with a hobby aircraft or full scale?
MikeMonty
QUOTE (rethinker+Nov 26 2006, 06:50 PM)
If the power induced in the tread belt was directly equal to the power of the plane, and all the power from the plane was directed at the gear or connection to the tread belt,yes it would not fly. However this is a very hard test to build in a real world prototype.

If you can set it up, all power from the plane would be used on the tread belt spin and no power would be let free as thrust.
It is an equal and opposite question.

If you have a video of your test, can you put it on Goggle for us to see.
Was it done with a hobby aircraft or full scale?

I'll be really interested in your reaction to his description of his "simulation" !

Simply put, he proposes a method of controlling the conveyor speed such that for a given amount of thrust he will "tune" the conveyor speed to provide force through rolling resistance to balance the thrust and hold the plane stationary.
Then as thrust is ramped up, the conveyor speed is ramped up too, the purpose being to ensure - by balancing rolling resistance against thrust - that the plane remains stationary and no airspeed is built up thus no flight.

I guess that if he reaches the planes maximum thrust and has succeeded in balancing this against rolling resistance then his conjecture is proven.

Entirely missing the point of the original post on the original thread which limited conveyor speed to equal and opposite to the planes speed (sensible frame of reference implied).

Mike
NoCleverName
QUOTE (MikeMonty+Nov 26 2006, 08:28 PM)
I guess that if he reaches the planes maximum thrust and has succeeded in balancing this against rolling resistance then his conjecture is proven.

This is precisely the point of my previous point about the original problem setup. If the resistance put up by the conveyor was equal to the engine output, this problem would be exactly the same as boat/river where hull drag is a major force. That would make this problem just a tedious repetition of boat/river and of no interest. It is because we understand, in our real-world experience, that rolling resistance is never a major factor we find this problem has some interest.

Only by altering our real-world expectations can this problem be turned into the mundane boat/river style problem alt5p champions. In other words, to ensure the victory of his opinion he changes the rules on-the-fly (so to speak ) to fit his clearly erroneous position. When those rule changes don't result in exactly the outcome he desires, he further modifies them in a quest to twist the problem to fit his answer. Sure, eventually he'll come up with a problem definition where every one will agree the plane won't fly. Oh, happy day!

Until then he remains the petulant child who continues to scream until mommy gives him his candy.

Oh, by the way, Alt, my case 4 doesn't work for you. The conveyor never moves. Case 3 remains unstable with the conveyor reaching unbounded speed. Try to explain that away.
Benny
QUOTE (rethinker+Nov 26 2006, 11:50 AM)
If you have a video of your test, can you put it on Goggle for us to see.
Was it done with a hobby aircraft or full scale?

What Atl claims to have done is take a small, cheap Radio-Shack RC plane that doesn't even have wheels and attached it to Leggo wheels and weighted the contraption down to the point it wouldn't fly and had a max speed of 5mph on the ground, as "measured" by eye. Then he put in on a treadmill and held it at 0mph by turning the treadmill to 5mph.
Benny
QUOTE
So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

QUOTE (->
 QUOTE So you are defining the plane as non-moving, then asking if it can move? That is exactly as stupid as defining the plane as having airspeed, then asking if it has airspeed. Of course, the original question does not ask if the plane just has airspeed, it asks if the plane has enough airspeed, a completely different question.

What I dont get is when people dont understand how having ANY amount of CRF will result in a fast belt holding the plane to 0 IAS.
GayComm
I can't bring myself to read the previous thread on this problem, so apologies if this has already been pointed out before...

You do realize, don't you, that Cecil Adams has already covered this:

http://www.straightdope.com/columns/060203.html

and

http://www.straightdope.com/columns/060303.html
Atl5p
QUOTE (Benny+Nov 26 2006, 04:39 PM)

"The force required to overcome the friction is equal to the CRF multiplied by the weight on the tire." Honestly, what do you not understand about this statement?

The plane moves in relation to the treadbelt's surface....the question then asks if it moves through the air.
Benny
QUOTE (Atl5p+Nov 26 2006, 04:15 PM)
The plane moves in relation to the treadbelt's surface....the question then asks if it moves through the air.

Dodged one and ignored the other. Try again.
egnorant
QUOTE (Atl5p+Nov 26 2006, 11:15 PM)
The plane moves in relation to the treadbelt's surface....the question then asks if it moves through the air.

While the plane may move in relation to the treadbelt..this speed will be different from the "groundspeed over the belt" unless the belt and the ground are not moving in relation to each other.

If I were to suggest a boat moves at X groundspeed over a river this would render the rivers movement as irrelevant.

Earlier someone stated the thrill of a discussion among intellectual equal, I still have found that picking on the retard is a blast!

Bruce
NoCleverName
QUOTE (egnorant+Nov 27 2006, 12:24 AM)
Earlier someone stated the thrill of a discussion among intellectual equal, I still have found that picking on the retard is a blast!

Hmmm, that could be in challenge in itself ... while still staying within the forum's rules of decorum. Maybe I can get some fun out of this joker yet.

Thanks!

And wouldn't you know it? A private message just arrived from mott.carl! I PM'ed him about the quandry regarding the competing frames of reference, since he, as you well know, is a leading researcher in high energy blather. I invited him to comment on the thread, but he is just too busy right now. In any case, the meat of his reply is:

QUOTE

There is an auto-interactions between the two frames that
control device receive the quantic impression through of a connection with the vibrations,its lenghtwaves and frequencies of ours neurons that can reasone inti
a same frequencies.then is our minds that through of exterior informations,capture
its,and through of its our does generate the space-time,relatives;the quantic measure,already that our mind also are created by quantum,and the quantic mechanical wavefunctions are the variations of space-time,as well as, the shapes
that model the geometric structures

The breakdown of symmetry,that generate the conveyors into anticonveyors,generate a plane that goes "forward" in a direction of time;and
the plane that goes "backward" in a direction.then to each frame has
continuity of space-time that is only one.

Well, there you have it. mott.carl has spoken and has laid the problem to rest. As I'm sure Altl5p knows and respects the work of mott.carl, they being of harmonious mind, I'm sure he will defer to m.c's pronouncement. Unless you care to discuss some of the finer points he raises.
rethinker
QUOTE
Well, there you have it. mott.carl has spoken and has laid the problem to rest.

That sound too Cleaver for me!
Here is the final landing for this plane.

The plane uses thrust against air resistance not ground or tread belt.

The simple over exploited question can be understood by simple physics

Put a spring under compression behind the plane against a stationary anchor, or attach a similar spring on the front of the plane hooked to a stationary anchor.

Now this spring will act in the same way as thrust in the rear or screw in the front by prop theory.

Unless the wheels of the plane have friction equal or greater than the spring, the plane will move forward.
The thrust behind a jet acts the same as placing a large spring compressed and let free.
The spring in the front acts like a screw of a prop in the front.

Let all systems free, and the plane acts as if it is already flying.
Atl5p
Try this at home....put a skateboard on a treadbelt. Attatch the front of the skateboard to a scale of some sort with a rope.

A )Crank up the treabelt to 1mph. Record the 'weight'.

B ) Speed up treadbelt to 10mph. Record the 'weight' again (increased from A)

C )Now, STAND on the skateboard at 1mph. Recort the 'weight' (Increased from A)

D )While standing on the skateboard, speed up to 10mph. Record the 'weight' (increased from C)

Disagree?
NoCleverName
So you have demonstrated that F = uN (force = coefficiant of friction times the normal force). Big deal.

But, just for kicks, did you weigh the skateboad and yourself first to see how the CoF varied under load and speed? It's not always a constant. I would think that because the rubber tread is so deformable the CoF should jump somewhat with the added weight.

Aside from that, you're still laughedly wrong about the plane.
Atl5p
QUOTE (NoCleverName+Nov 27 2006, 03:03 PM)
So you have demonstrated that F = uN (force = coefficiant of friction times the normal force). Big deal.

But, just for kicks, did you weigh the skateboad and yourself first to see how the CoF varied under load and speed? It's not always a constant. I would think that because the rubber tread is so deformable the CoF should jump somewhat with the added weight.

Aside from that, you're still laughedly wrong about the plane.

Oh, I'm sorry....I thought we were talking about CRF (coefficient of rolling friction)
NoCleverName
It's exactly the same concept and has exacty the same formula. It merely differentiates a tire's performance not under braking, where "CoF" is used for braking and cornering. I guess it was too much to hope you had even this tiny nugget of knowledge.
Bloy
QUOTE (Atl5p+Nov 27 2006, 08:06 PM)
Oh, I'm sorry....I thought we were talking about CRF (coefficient of rolling friction)

Touble is....you're not thinking
Benny
QUOTE (Atl5p+Nov 27 2006, 01:06 PM)
Oh, I'm sorry....I thought we were talking about CRF (coefficient of rolling friction)

We are. Coefficient of Rolling Friction is still a Coefficient of Friction. The same formula applies. F = uN. Or, as I have been writing it Resistance = Weight * CRF.
Atl5p
QUOTE (Benny+Nov 27 2006, 05:05 PM)
We are. Coefficient of Rolling Friction is still a Coefficient of Friction. The same formula applies. F = uN. Or, as I have been writing it Resistance = Weight * CRF.

If a vehicles tires use 12hp while moving at 35mph...and the same tires on the same car use 20hp while moving at 55mph...what would cause the increase in HP in step with speed?
NoCleverName
And exactly why is that so surprising? Horsepower is a measure of power equivalent to around 746 watts or 746 joule-sec., where, of course, the joule is the newton-meter.

12 hp is around 9000 watts, 35 mph is around 15.6 m/s, so force is about 576 newtons.
Similarly 35 hp is around 15,000 watts, 55 mph 24.6 m/s, so force is around 609 newtons.

Within the number of decimal points given, 576 is pretty close to 600 so the rolling resistance is about the same at each speed. The only problem with these figures is that gives a CoF in the 0.06 range which is about twice what you'd expect (assuming a 1000 kg car).

But you really don't even have to know any physics to realize that 20 hp is to 12 hp about the same as 55 mph is to 35 mph.

Why wouldn't you think that more power is needed to go faster?
Benny
QUOTE
If a vehicles tires use 12hp while moving at 35mph...and the same tires on the same car use 20hp while moving at 55mph...what would cause the increase in HP in step with speed?
The power output of the engines also increases with speed.

NoCleverName, Atl believes that all of the power dissipated by the wheels is subtracted from the power of the engines. He is missing that the conveyor's motor is also contributing power that is also being dissipated by the wheels. So far I have been unsuccessful in getting the point across.
egnorant
Plane 1 has 3 wheels that require 20 hp to move in one direction at 50 mph.
Plane 2 has 10 wheels and requires 1000 hp to move in one direction at 50 mph.

Both planes are heading north on a runway...which will reach the end of the runway first?

Force required to achieve or negate a certain speed is not relevant.
Plop plane 2 on a belt moving in the opposite direction it may change the power required to maintain 50 mph. All you are doing is changing the power requirements for a certain speed.
Plane 2 may need 2000 hp to move in one direction at 50 mph while the belt is applying an opposing force via the wheels.
Plane is supplying x force to move left, belt is supplying force Y to the right.
x+(-Y)= enough force to move at 50 mph.

Plop plane 1 on a belt moving at 50 mph in the same direction as the plane, then the plane may only require 12 hp to move at 50 mph.
Prop may be supplying x to the left and the belt would supply the Y to the left...both in the same direction. x+(Y)= enough force to move at 50 mph.

Which of these 2 planes will reach the end of the runway first?
(Hint: left is north..the opposite direction of north is south...the opposite direction of left is right)

Plane 1. 20 hp or 12 hp..Plane 2. 1000 hp or 2000 hp.
They all equal 50 mph!

Bruce

NoCleverName
QUOTE (Benny+Nov 28 2006, 05:25 AM)
Atl believes that all of the power dissipated by the wheels is subtracted from the power of the engines.  He is missing that the conveyor's motor is also contributing power that is also being dissipated by the wheels.

Yes, he has a hard time with symmetry, I know. (For example, the force required to pull a rug out from a table is about the same as that required to push the table off a rug).

I have in the past pointed out that if you want the conveyor to do work against the plane, it's going to need engine power commensurate with the plane. But that's just an unimportant detail. The conveyor has no means other than rolling resistance to transmit that power against the plane, so unless the conveyor is set to go much, much, faster than the "speed" (whatever he means by that ) of the plane the conveyor is just a nuisance.

His big problem is that he envisions the conveyor like a frame of reference, which in this problem it is not. It is merely a mechanism deliverying an opposing force (like having the brakes partially on).

Of his many analogies, the runner/treadmill is the most laughable. That situation is completely dissimilar to say, a bike on a treadmill (where you are using motive power to gain against the treadmill). The runner/treadmill is more like the runner moving to continuously maintain balance against a device which is constantly pulling his means of support out from under him. The physiological work in moving the legs to regain balance on the treadmill is about the same as in actual running (which, too, is an act of intentionally losing, then regaining balance).

Like I said earlier, the world can be a subtle place to which, without the right tools, you will be completely oblivious.
Atl5p
All I'm saying is this: If increasing the speed of the treadbelt under the plane will have absolutly zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the treadbelt spins...why is that?

For example. Let's just put a skateboard on the treadbelt. Attatch a rope to the front and then to a scale, scale is nailed to the wall.

Run the treadbelt at 2mph. Then run the treadbelt at 10mph. Obviously the wheels of the skateboard are 'using' more 'Power' at the greater 'Speed' of the treadbelt....that's what you're saying, right?

So, how would that increase in 'Power' be evident in this experiment? Would it make the 'Weight' displayed on the scale 'Increase', as the 'Power' used by the 'Wheels' also 'Increases', as the 'Speed' of the 'Treadbelt' is also 'Increased'?

I was just wanting the opinion of the 'Experts' on this 'Forum'.

I think this would help me 'Understand'.
Atl5p
Well, it's been an hour....about 10 posts on the 'other' plane/treadbelt thread...no answers to this question....also there are 15 views to the Poll, and not one vote.

I do believe I have pinned them down for good this time!! AHHAHAHAHAAHA!!!

Bloy
QUOTE (Atl5p+Nov 28 2006, 04:20 PM)
Well, it's been an hour....about 10 posts on the 'other' plane/treadbelt thread...no answers to this question....also there are 15 views to the Poll, and not one vote.

I do believe I have pinned them down for good this time!! AHHAHAHAHAAHA!!!

Pinned "them"(who is them) down to what?"
Atl5p
QUOTE (Bloy+Nov 28 2006, 11:24 AM)
Pinned "them"(who is them) down to what?"

"Them" is Flyboys....and 'Pinned' means this:

They will NOT answer the poll because they are pusssies, and don't really know WHY they believe what they do, they just FEEL that the plane will fly, because that's what plane's are supposed to do....or whatever the reason.

Pinned means:
I have found a chink in the armor...all flyboys agree it won't fly....logic would dictate that they should all agree as to WHY this is...it's either a TRICK question, or it's PHYSICALLY impossible to happen....this is where FlyBoys will disagree, and as to avoid being seperated and weakened, they choose not to answer the poll...as individuals they are simple pusssy cats...pusssies...they can only function as a group or mob of flyboys...they don't want to risk publicly disagreeing with one another, so they just stick to the party line...nofly nofly nofly......pusssies...pusssies...pusssies...

I mean,,, I really am physically lauging at the cowardness...it is funny! And from a group of people who are "So Right"....HAHAHAHA!!!
Bloy
QUOTE (Atl5p+Nov 28 2006, 04:32 PM)
"Them" is Flyboys....and 'Pinned' means this:

They will NOT answer the poll because they are pusssies, and don't really know WHY they believe what they do, they just FEEL that the plane will fly, because that's what plane's are supposed to do....or whatever the reason.

Pinned means:
I have found a chink in the armor...all flyboys agree it won't fly....logic would dictate that they should all agree as to WHY this is...it's either a TRICK question, or it's PHYSICALLY impossible to happen....this is where FlyBoys will disagree, and as to avoid being seperated and weakened, they choose not to answer the poll...as individuals they are simple pusssy cats...pusssies...they can only function as a group or mob of flyboys...they don't want to risk publicly disagreeing with one another, so they just stick to the party line...nofly nofly nofly......pusssies...pusssies...pusssies...

I mean,,, I really am physically lauging at the cowardness...it is funny!  And from a group of people who are "So Right"....HAHAHAHA!!!

AH... now I see you ,Atl5p.
You are simply picking a fight...nothing more or less.

Calling "them" pussies is an attempt to have a reaction.

...reminds me of "back to the future" when I heard BIFF say "chicken".

Sorry, IF that is all you want(a fight) then I digress.

Try re-phrasing your post so that you are more inviting to your poll...while you're at it, try re-phrasing your poll so that IF the OP is re-phrased it still stays within some reasonable bounds.
Atl5p
Once again:

All I'm saying is this: If increasing the speed of the treadbelt under the plane will have absolutly zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the treadbelt spins...why is that?

For example. Let's just put a skateboard on the treadbelt. Attatch a rope to the front and then to a scale, scale is nailed to the wall.

Run the treadbelt at 2mph. Then run the treadbelt at 10mph. Obviously the wheels of the skateboard are 'using' more 'Power' at the greater 'Speed' of the treadbelt....that's what you're saying, right?

So, how would that increase in 'Power' be evident in this experiment? Would it make the 'Weight' displayed on the scale 'Increase', as the 'Power' used by the 'Wheels' also 'Increases', as the 'Speed' of the 'Treadbelt' is also 'Increased'?

I was just wanting the opinion of the 'Experts' on this 'Forum'.

I think this would help me 'Understand'.
NoCleverName
QUOTE (Atl5p+Nov 28 2006, 03:18 PM)

Run the treadbelt at 2mph.  Then run the treadbelt at 10mph.  Obviously the wheels of the skateboard are 'using' more 'Power' at the greater 'Speed' of the treadbelt....that's what you're saying, right?

Power is the measure of work per second and work is the measure of force over displacement.

Say it take about a 1 newton push(about 1/4 pound) on the skateboard to keep it rolling. Over a 1 meter interval (about 3 feet) that would require 1 joule of "work".

If you did that work over a 1 second interval, then you would have expended 1 joule-sec = 1 watt of power. If you did the work in 1/2 second (i.e., you went faster, you covered the 3 feet in less time) that would require 2 watts of power ( about 0.002 hp). However, you still did the same amount of work (1 joule).

A B-777 "weighs" 300,000 kg, has 12 main tires of CRR no more than .03, and has two engines with combined thrust of about 836,000 newtons (188,000 lbs).

Using the F = uN formula, the tires in combination oppose engine thrust with a force of about 88,000 newtons.

If the plane rolls forward 1 meter, the plane expends 836 Kjoules, and the runway expends 88 Kjoules to try to stop it by "moving" in the opposite direction. But if the runway could somehow move fast enough to expend the same 836 Kjoules as the plane, it [/I]could[I] "hold" it in place. 836/88 = 9.5 so the runway needs to "move" 9.5 farther than the plane to prevent it from making any headway. So the "control system" needs to tune the conveyor/runway to move 9.5 times faster than the plane.

There is no way to directly convert jet engine thrust to horsepower unless you take into consideration the velocity you are moving at, so there is no good way to come up with a means of somehow figuring if there was a "terminal velocity" due to more and more horsepower being "blead off" the engines by the wheels: in fact, given Newton's Laws, in an environment where there is nothing but CRR, the plane would not be limited in speed at all! You might say "how can this be?" but it's only because of our "real world experience" do engines come up against limitations: that's because all sorts of other forces (like drag) start to come into play.

So as long as the "push" from one direction exceeds the "push" from another you go in the direction of the greater push with ever increasing velocity: it's the same deal in a car as you feel yourself being pushed back while watching the speedo climb.

Atl5p
NoCleverName...great answer! But what question were you reading, because I don't see the relevence....hmmm...hey, I know you must be smarter than me, hey I'm pretty ignorant with the whole power/speed/thurst deal.

This is just a skateboard on a treadbelt, there is no source of forward propulsion...but if you must, then please try to use a prop plane instead of jet, ok? The piston engine has horsepower rating, and the prop makes thrust. If you can't get horsepower rating from a jet, then why use a jet? Why not use a prop J3 Cub?

You seem to be well informed, this is a simple question, will the scale read a different 'weight' when the treadbelt speed goes up....yes or no...you gave me a 200 word essay and did not answer the question...boy you must be smart! But I DO like the answer you gave...you mathmatically explained the forces at work when a treadbelt actually holds a plane to 0 IAS...I do not have the ability to do that, so I thank you...I may use it in my signature to 'show' the others.

Once again:

All I'm saying is this: If increasing the speed of the treadbelt under the plane will have absolutly zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the treadbelt spins...why is that?

For example. Let's just put a skateboard on the treadbelt. Attatch a rope to the front and then to a scale, scale is nailed to the wall.

Run the treadbelt at 2mph. Then run the treadbelt at 10mph. Obviously the wheels of the skateboard are 'using' more 'Power' at the greater 'Speed' of the treadbelt....that's what you're saying, right?

So, how would that increase in 'Power' be evident in this experiment?

Would it make the 'Weight' displayed on the scale 'Increase', as the 'Power' used by the 'Wheels' also 'Increases', as the 'Speed' of the 'Treadbelt' is also 'Increased'?
[/B]
I was just wanting the opinion of the 'Experts' on this 'Forum'.

I think this would help me 'Understand'.
MikeMonty
QUOTE (Atl5p+Nov 28 2006, 05:02 PM)

So, how would that increase in 'Power' be evident in this experiment?

Would it make the 'Weight' displayed on the scale 'Increase', as the 'Power' used by the 'Wheels' also 'Increases', as the 'Speed' of the 'Treadbelt' is also 'Increased'?
[/B]
I was just wanting the opinion of the 'Experts' on this 'Forum'.

I think this would help me 'Understand'.

You really are totally ignorant!
The scale will not theoretically go up - and here's why:
The force shown by the scale is equal to rolling resistance.(I'd say proportional but it's probably too difficult a word)
this is caused by "squishing" the tyre.
the amound the tyre "squishes" per revolution does not change no matter how fast the wheels turn.
Therefore the force does not change no matter how fast the wheels turn.
What WILL change is the power absorbed by the tyres deformation (sorry, "Squishing")
This is because Work Done = Force times Distance
As Speed increases, Distance per second increases .
Therefore Power increases because Power = Work Done per Second.

Too difficult for you?
NoCleverName
First, you might want to take my comments out of your sig since that's my intellectual property (which I'll gladly license to you )

Second, in the world of theory, that scale on the treadmill is going to read pretty much the same no matter what the treadmill speed. In practice, I doubt that would happen. Too many other factors would creep in.

Same principle as if you rode your bike at two different speeds and touched the brakes the same amount at each speed you would slow at the same rate (ignoring wind resistance).

The force is constant, but the use of that force over distance and time is what measures power: so a "power meter" would go up, but a "force meter" stays constant.

As far as HP ratings jets vs. prop, there is little to gain there, either. Prop engines are rated in HP because propellers are turned by the force of torque. HP is related to torque by the formula HP = torque/time. So it makes sense to describe how well an engine can turn the prop. Unfortunately, that doesn't tell how well the prop is using this power to generate thrust. It could be just wailing away at the air producing no thrust at all!
Bloy
Atl5p says: in his sub-notes of his posts
"------------------------------
The question is:

Will the plane take off or not? Will it be able to run up and take off?
(This is the Original Question or 'OP' for short.
You are a flyboy if: You believe this treadbelt would allow a RUNNER or a CAR to achieve airspeed.
You are a 'NoFly'er if: You believe this treadbelt would hold a runner in place, just like at the gym.)"
------------------------------"

I am a flyboy, but, I do NOT believe the treadbelt would allow a RUNNER or a CAR to achieve airspeed.
IF,IF,IF you are defining airspeed as the ability to take off.
(there is no liftoff airspeed defined for non-flying entities)

I am a flyboy, but, I DO believe the treadbelt would allow a RUNNER or a CAR to achieve airspeed.
IF,IF, IF you are defining airspeed as a simple movement through the air.
(any movement through the air and airspeed can be measured).

Geeesh..he's even defined what a flyboy is. ...according to his own "contrived" definitions.
Benny
QUOTE
All I'm saying is this: If increasing the speed of the treadbelt under the plane will have absolutly zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the treadbelt spins...why is that?
As the tires spin faster they suck power from the conveyor, not from the plane. The conveyor does not get to spin against rolling resistance for free.

Or, from a different reference frame, since both the wheels and engine are fixed to the same airframe, the power of the engine also increases with speed. If you double the wheelspeed you double both the power sucked away by the wheels and the power generated by the engines. (This probably sounds strange, but it is correct.)

QUOTE (->
 QUOTE All I'm saying is this: If increasing the speed of the treadbelt under the plane will have absolutly zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the treadbelt spins...why is that?
As the tires spin faster they suck power from the conveyor, not from the plane. The conveyor does not get to spin against rolling resistance for free.

Or, from a different reference frame, since both the wheels and engine are fixed to the same airframe, the power of the engine also increases with speed. If you double the wheelspeed you double both the power sucked away by the wheels and the power generated by the engines. (This probably sounds strange, but it is correct.)

For example. Let's just put a skateboard on the treadbelt. Attatch a rope to the front and then to a scale, scale is nailed to the wall.

Run the treadbelt at 2mph. Then run the treadbelt at 10mph. Obviously the wheels of the skateboard are 'using' more 'Power' at the greater 'Speed' of the treadbelt....that's what you're saying, right?

So, how would that increase in 'Power' be evident in this experiment? Would it make the 'Weight' displayed on the scale 'Increase', as the 'Power' used by the 'Wheels' also 'Increases', as the 'Speed' of the 'Treadbelt' is also 'Increased'?
The scale would read the same force at 2mph and 10mph. The increase in power would be evident if you attached a power meter to the treadmill. It would show that the treadmill is using 5 times as much electrical power to spin against rolling resistance at 10mph as it does at 2mph.

QUOTE
I was just wanting the opinion of the 'Experts' on this 'Forum'.

I think this would help me 'Understand'.
egnorant
The treadbelt can have an effect on the plane...it can alter its speed and direction.
It can take a plane moving left at 50 mph and slow it to a plane moving left at 5 mph. It can even reduce the speed to 0 mph.
The belt can even cause the plane to move backwards!!!

It cannot do these things with the restraints placed upon it by the original question,
but with a plane with so little power that it cannot overcome the force transferred by the belt AND with no attempt of the belt to match the speed of the plane...it is possible.

QUOTE
Hi my name is Atl5p. Even though I'm wrong, I'll argue with people that have a clue until they finally give up due to my ineptitude. Since noone is left to argue, I must be right. Besides, my reasoning can't be flawed because laws of physics are ludicrous, the problem really means something besides what it says in clear and concise English, and I change my reasoning continuously so as to avoid having any of my wacked out theories identified as the bane of my conclusion.

Please understand that the argument of power and forces have no place in the correct answer to the original question.

His toy experiment shows that his "plane" is incapable of movement due to conditions preventing movement.
I drive my truck at 50 mph with x power used.
I hook up a trailer that requires Y to move at 50 mph.
When the requirement is x+Y to move at 50 mph, I will fail to achieve 50 mph if all I supply is power to move x.

Ignore him when he speaks of forces and power as these are totally outside the bounds of the original question.

Bruce
Atl5p
QUOTE (egnorant+Nov 28 2006, 02:52 PM)
The treadbelt can have an effect on the plane...it can alter its speed and direction.
It can take a plane moving left at 50 mph and slow it to a plane moving left at 5 mph. It can even reduce the speed to 0 mph.
The belt can even cause the plane to move backwards!!!

It cannot do these things with the restraints placed upon it by the original question,
but with a plane with so little power that it cannot overcome the force transferred by the belt AND with no attempt of the belt to match the speed of the plane...it is possible.

Please understand that the argument of power and forces have no place in the correct answer to the original question.

His toy experiment shows that his "plane" is incapable of movement due to conditions preventing movement.
I drive my truck at 50 mph with x power used.
I hook up a trailer that requires Y to move at 50 mph.
When the requirement is x+Y to move at 50 mph, I will fail to achieve 50 mph if all I supply is power to move x.

Ignore him when he speaks of forces and power as these are totally outside the bounds of the original question.

Bruce

Point of Order:

The OP for THIS thread states that the conveyor matches the plane's wheelspeed.

At any rate, are you saying that in the case of the OP in THIS thread, the plane would NOT FLY? Ehh?
Benny
QUOTE (Atl5p+)
QUOTE (NoCleverName+)
Explination with actual formula:
If the plane rolls forward 1 meter, the plane expends 836 Kjoules, and the runway expends 88 Kjoules to try to stop it by "moving" in the opposite direction. But if the runway could somehow move fast enough to expend the same 836 Kjoules as the plane, it [/I]could[I] "hold" it in place. 836/88 = 9.5 so the runway needs to "move" 9.5 farther than the plane to prevent it from making any headway. So the "control system" needs to tune the conveyor/runway to move 9.5 times faster than the plane.

So, I take this to mean the following:
If a plane can normally go 100mph on static runway, then it would take a 950mph treadbelt to hold the plane at 0 IAS. When this happens, the treadbelt speed will match exactly the plane's wheelspeed.
#1: NoCleverName is wrong.
#2: You have not shown us a runway-sized conveyor capable of 950mph.
#3: You have not shown us airplane wheels capable of 950mph.

In NoCleverName's example, the conveyor would be wasting 836kJ per unit time as wheel heat (not including internal losses) as the plane takes off and flies anyway.
Atl5p
QUOTE (Benny+Nov 28 2006, 03:11 PM)
So, I take this to mean the following:
If a plane can normally go 100mph on static runway, then it would take a 950mph treadbelt to hold the plane at 0 IAS. When this happens, the treadbelt speed will match exactly the plane's wheelspeed.[/QUOTE]#1: NoCleverName is wrong.
#2: You have not shown us a runway-sized conveyor capable of 950mph.
#3: You have not shown us airplane wheels capable of 950mph.

In NoCleverName's example, the conveyor would be wasting 836kJ per unit time as wheel heat (not including internal losses) as the plane takes off and flies anyway.

DAAAAEEEEEEEEMMMMMMMNNNN!

I could feel that burn like a concussion wave!!

NCN....you gonna take that from this kid???
Benny
QUOTE (Atl5p+Nov 28 2006, 01:14 PM)
DAAAAEEEEEEEEMMMMMMMNNNN!

I could feel that burn like a concussion wave!!

Atl, you should be the one feeling that burn. For all your complaints that this is a "real world" problem you don't even blink when someone states that a 950mph conveyor is required to stop the plane.
Atl5p
QUOTE (Benny+Nov 28 2006, 03:22 PM)
Atl, you should be the one feeling that burn. For all your complaints that this is a "real world" problem you don't even blink when someone states that a 950mph conveyor is required to stop the plane.

So, a plane must travel at a speed of 950mph over the surface of a treadbelt, and that treadbelt is moving at a speed of 950mph, and the plane is held to 0 IAS.

Does the plane run up and take off? Can the plane fly?

Ans: No.

Hey, that's all I really need!

I mean, what tire would have trouble spinning at 950mph? It isn't MOVING, so the SPEED RATING has not been violated....did you think the tire would blow up?

We're talking about a treadbelt as big as an airport runway...if I can get past that, why can't I accept this treadbelt can go 950mph? I mean, it IS 100x's as big as a 'normal' treadmill, right? So why not be able to spin 100x's as fast? Nope, dosn't phase me a bit...

Like I said, if one can accept a treadbelt this large, why can't the same person accept this treadbelt spinning at 950mph? Did you think this large treadbelt would only spin at 10mph? Is that where your thought process is breaking down?
NoCleverName
QUOTE (Atl5p+Nov 28 2006, 07:58 PM)
The OP for THIS thread states that the conveyor matches the plane's wheelspeed.

Well, if that's the case then the plane doesn't fly because, by the problem's definition, the plane isn't moving at all (relative to the air/ground/control tower/whatever). That's the only condition that allows beltspeed = wheelspeed (as if the plane were an integral part of the belt itself, like the drive rollers, etc.)

Let me fix up some other issues I probably screwed up (since I haven't touched these problems in over 30 years!). I think we cab get a rough idea of how fast a B-777 can go before coming up against enough rolling resistance (ignoring air resistance) so that it can't go any faster. Note this speed will be far in excess of take-off speed so it's really unrealistic by a long ways, but here goes.

The B-777 has 300,000 kg mass, two really nice GE-90 engines that develop some 836,000 Newtons thrust, and 12 main wheels that (as was figured earlier) have a rolling resistance of 0.03 that results in 88,000 Newtons of resistance.

If we fire-wall our throttles (which was a FAA reg until recently) for takeoff we get all the thrust so by F = ma we have:
(836,000-88,000) = 300,000 * a

or acceleration = 2.5 m/s (about 1/4 g). (Note: that would get us to take-off speed of about 185mph in 33 sec in about 4100 feet of runway.)

Using both this take-off performance data and engine exhaust data (of 200 mph about 100 feet behind the engines) I get a horsepower estimate of some 82,000-90,000 hp (calcs not shown here). I'm going to use 63,000,000 watts, which is about in the middle of these figures.

Using Power = Force * velocity with the opposing power being the 88,000 N provided by the tires, we get a "terminal velocity" of:
63,000,000 = 88,000 * v

or 715 m/s or about 1,600 mph. This exceeds the plane's tire speed limit of 235.

So crank up the conveyor!
Atl5p
QUOTE (NoCleverName+Nov 28 2006, 04:44 PM)
Well, if that's the case then the plane doesn't fly because, by the problem's definition, the plane isn't moving at all (relative to the air/ground/control tower/whatever). That's the only condition that allows beltspeed = wheelspeed (as if the plane were an integral part of the belt itself, like the drive rollers, etc.)

Let me fix up some other issues I probably screwed up (since I haven't touched these problems in over 30 years!). I think we cab get a rough idea of how fast a B-777 can go before coming up against enough rolling resistance (ignoring air resistance) so that it can't go any faster. Note this speed will be far in excess of take-off speed so it's really unrealistic by a long ways, but here goes.

The B-777 has 300,000 kg mass, two really nice GE-90 engines that develop some 836,000 Newtons thrust, and 12 main wheels that (as was figured earlier) have a rolling resistance of 0.03 that results in 88,000 Newtons of resistance.

If we fire-wall our throttles (which was a FAA reg until recently) for takeoff we get all the thrust so by F = ma we have:
(836,000-88,000) = 300,000 * a

or acceleration = 2.5 m/s (about 1/4 g). (Note: that would get us to take-off speed of about 185mph in 33 sec in about 4100 feet of runway.)

Using both this take-off performance data and engine exhaust data (of 200 mph about 100 feet behind the engines) I get a horsepower estimate of some 82,000-90,000 hp (calcs not shown here). I'm going to use 63,000,000 watts, which is about in the middle of these figures.

Using Power = Force * velocity with the opposing power being the 88,000 N provided by the tires, we get a "terminal velocity" of:
63,000,000 = 88,000 * v

or 715 m/s or about 1,600 mph. This exceeds the plane's tire speed limit of 235.

So crank up the conveyor!

Could you show the math for tires total resistance? (88,000 newtons)

(Resistance = CRF x Weight)?

R = 0.03 x 300,000kg?

I'm not really 100% on the whole newtons, joules, watts, HP deal....I am grateful that you are taking the time to explain this.
NoCleverName
Just so there's no mistake, I said that the conveyor had to go 9.5 times faster than the plane to stop it.

Above, I have shown (unrealistically) that the plane have to go 1,600 mph before it got to the point of not being able to overcome tire resistance. By this time, of course, the plane would have long flown away (aside from setting some new passenger plane speed records ... while on the ground! )
Atl5p
QUOTE (NoCleverName+Nov 28 2006, 04:53 PM)
Just so there's no mistake, I said that the conveyor had to go 9.5 times faster than the plane to stop it.

Above, I have shown (unrealistically) that the plane have to go 1,600 mph before it got to the point of not being able to overcome tire resistance. By this time, of course, the plane would have long flown away (aside from setting some new passenger plane speed records ... while on the ground! )

So, if I set your jet down upon at 1600mph conveyor, would the plane achieve positive IAS? I'm now getting confused as to what you are talking about.

It seems that you are saying that the plane will need to move at 1600IAS over a belt that is opposing it at 1600mph, but wouldn't that give a wheelspeed of 3200mph?
NoCleverName
Rolling resistance is a force computed by:

F = uN

where "u" is the coefficient of friction (same as rolling resistance) and "N" is the "normal vector" which is mass * g * sin(theta) where "g" is the gravitational acceleration constant for the earth of 9.8 m/s and, theta is the direction of the force (in case we are on an incline). Luckily theta = 90 degrees in this case (straight down) so sin(theta) = 1 ...

... so since we have u = 0.3 (too high for real world, I think), mass = 300,000 kg

F = (.03) * (300,000) * (9.8) * (1) = 88,000 newtons of force.

Since a newton is about 1/4 lb of force, I think, that gives you about 22,000 lbs of tire resistance.

What F = uN says in words is that "the force of friction is directly proportional to how hard you press down on something that's sliding over something else" and the formula takes into account if you aren't pressing straight down, but at an angle as well. The variable "N" contains "m * g", which is another way of saying "weight", while the vector part of it handles inclination.

Isn't physics neat?
NoCleverName
What I'm saying with the 1,600 mph calc is that the plane's engines can drive the plane to "only" 1,600 mph before the tires bleed away all the engine power, leaving none left to go any faster.

Don't take that calc to the bank, though! It's just a demonstration that the tires can't stop the plane from flying.

Benny
QUOTE (Atl5p+Nov 28 2006, 02:04 PM)
So, a plane must travel at a speed of 950mph over the surface of a treadbelt, and that treadbelt is moving at a speed of 950mph, and the plane is held to 0 IAS.

Does the plane run up and take off?  Can the plane fly?

Ans: No.

Hey, that's all I really need!
A 950mph runway sized conveyor is not possible in the real world, so the real world answer is no, you can't stop the plane.

QUOTE
I mean, what tire would have trouble spinning at 950mph?  It isn't MOVING, so the SPEED RATING has not been violated....did you think the tire would blow up?
Don't be stupid.

QUOTE (->
 QUOTE I mean, what tire would have trouble spinning at 950mph?  It isn't MOVING, so the SPEED RATING has not been violated....did you think the tire would blow up?
Don't be stupid.

We're talking about a treadbelt as big as an airport runway...if I can get past that, why can't I accept this treadbelt can go 950mph?  I mean, it IS 100x's as big as a 'normal' treadmill, right?  So why not be able to spin 100x's as fast?  Nope, dosn't phase me a bit...

Like I said, if one can accept a treadbelt this large, why can't the same person accept this treadbelt spinning at 950mph?  Did you think this large treadbelt would only spin at 10mph?  Is that where your thought process is breaking down?
I cannot accept a real world 950mph runway sized conveyor. If you want to instead claim that this is not a real world system, then we can model it using standard, Newtonian, high school, textbook physics. If we model the system using standard Newtonian physics rolling resistance is not capable of stopping the plane.

Atl5p, simple question. Is this a real world problem?
Atl5p
QUOTE (NoCleverName+Nov 28 2006, 05:05 PM)
Rolling resistance is a force computed by:

F = uN

where "u" is the coefficient of friction (same as rolling resistance) and "N" is the "normal vector" which is mass * g * sin(theta) where "g" is the gravitational acceleration constant for the earth of 9.8 m/s and, theta is the direction of the force (in case we are on an incline). Luckily theta = 90 degrees in this case (straight down) so sin(theta) = 1 ...

... so since we have u = 0.3 (too high for real world, I think), mass = 300,000 kg

F = (.03) * (300,000) * (9.8) * (1) = 88,000 newtons of force.

Since a newton is about 1/4 lb of force, I think, that gives you about 22,000 lbs of tire resistance.

What F = uN says in words is that "the force of friction is proportional to how hard you press down on something that's sliding over something else" and the formula takes into account if you aren't pressing straight down, but at an angle as well.

Isn't physics neat?

It's not just neat....it's also F=uN!! get it!! Fun? hahaaha!

I would have said:
Plane weighs 300,000kg = 660,000 lbs

CRF = 0.006

Resistance = 660,000 * 0.006 = 3960lbs of resistance.

or, using your CRF of 0.03

Resistance = 660,000 * 0.03 = 19,800 lbs

So, looks good....so, with the 0.03 CRF, the plane is on the treadbelt, a rope is on the front, and it's tied to a digital scale.

The treadbelt starts to move backwards. The scale jumps to 19,800 lbs.
The treadbelt begins to accelerate to 100mph. The scale increases during acceleration (?)
The treadbelt reaches and maintains 100mph. The scale goes back down to 19,800lbs and stays there as long as treadbelt maintains exactly 100mph....

Is that correct?

NoCleverName
As best as I can remember from all the years ago, that's how it works. Kind of like a shopping cart takes roughly the same amount of effort to keep it going no matter how fast you are pushing it (steadily, that is).
Atl5p
QUOTE (Benny+Nov 28 2006, 05:17 PM)
A 950mph runway sized conveyor is not possible in the real world, so the real world answer is no, you can't stop the plane.

Don't be stupid.

I cannot accept a real world 950mph runway sized conveyor.  If you want to instead claim that this is not a real world system, then we can model it using standard, Newtonian, high school, textbook physics.  If we model the system using standard Newtonian physics rolling resistance is not capable of stopping the plane.

Atl5p, simple question.  Is this a real world problem?

SO, what is throwing you off, Benny?
The fact that it's a treadbelt the size of a runway?
OR
The fact that it must be capable of transonic speeds?

BTW...remind me again...where is this whole 'Real world Problem' come into play? Did I say that or something?
Did it have to do with 'Trick Question' vs 'Physics Question'?

Look, I understand that it a treadbelt large enough for an airplane to take off on does not exist...is that what you are asking me?

I mean....if you must have a treadbelt the size of a runway, (2 miles long at Dobbins ARB), in order to think about this question, then why did you even post here in the first place? Obviously one does not exist which is known to me!

I mean, it SEEMS like you cannot accept a 2 mile long treadbelt....so why did you bother?
OR
You CAN accept a 2 mile long treadbelt, but just not one which will spin at 950mph...which leads me to ask 'How can you imagine a 2 mile long runway, but that it cannot go 950mph".

I mean...show me a 2 mile long moveable runway, and I'd believe it would do just about anything!

Have you seen what the F-22A Raptor can do?

Are you one of those people who says "Everything that ever could be invented has already been invented"?
Benny
QUOTE
The treadbelt starts to move backwards. The scale jumps to 19,800 lbs.
The treadbelt begins to accelerate to 100mph. The scale increases during acceleration (?)
The treadbelt reaches and maintains 100mph. The scale goes back down to 19,800lbs and stays there as long as treadbelt maintains exactly 100mph....

Is that correct?
That is correct. (The increase shown on the scale during acceleration is not due to rolling resistance.)
Atl5p
QUOTE (Benny+Nov 28 2006, 05:26 PM)
That is correct. (The increase shown on the scale during acceleration is not due to rolling resistance.)

I'm confuesed again...I though that you said NCN was WRONG...now you agree with him?
NoCleverName
Just to be clear: there are two versions of this problem on the internet:
1. This problem, where the conveyor matches the plane and the plane flys, and
2. The one where you get to make the conveyor go at any speed you want, no matter what the plane is doing. In that version, the conveyor holds the plane back.
Again, this is the first version where the plane flies. The "wheelspeed" version given as an alternate here is really trying to be version 2, which is totally different with a different answer.
Benny
Based on the content and attitude of your posts from the last few weeks I concluded you believed that the original question was intended to be pure real-world with the technological limitations of the real-world for both the belt and plane. If you do not believe that this is intended to be a pure real-world problem then you should have accepted my (and other's) explanations of the Newtonian physics months ago. Actually that can probably be expressed in years by now.

If I was incorrect and you do not believe this is a pure real-world problem, then I apologize for my mistake and ask that you understand and accept that I am explaining the physics involved correctly.
rethinker
QUOTE
All I'm saying is this: If increasing the speed of the tread belt under the plane will have absolutely zero, none, no effect on the plane....but the tires are 'sucking up' more "power" the faster the tread belt spins...why is that?

Ok so now you understand that all you are asking is simply if one force is stronger than another, which force will win?

Should now be clear that the strongest force will always win. as everyone has said.

mighoser
Folks,

Airplanes generate lift when a fluid, in this case air, interactes with a solid body. For this problem we are talking about a pair of wings and fuselage. Since the amount of lift is related to the velocity of the fluid with respect to the solid body, this velocity is our only concern. Let jump to the steady state condition where the airplane isn't moving relative to the "ground" and the "belt" is moving in reverse at constant velocity X relative to the "ground". The plane is full throttle and travels at a velocity X relative to the "belt". So the question is what is the velocity of the fluid relative to the plane. Except at the surface of the belt and areas of the plane which are directly behind the propeller. The latter can be eliminated because airplanes rutinely apply full power and brakes before takeoff. Thus, the lift generated from the propellor propelling the fluid isn't sufficient to levatate the plane. The belt will impart some amount of velocity to the fluid relative to the plane. This will generate some lift in addition to the lift we just disscussed. So will the plane take off. Probably not but a powered glider with wings which are close to the belt might be able to take off. Keep in mind, the glider would loose lift the farther its wing was normally to the belt surface. This is because the velocity of the fluid goes from X to zero the farther it is from the belt.

The solution to this problem isn't about whether the plane flies or not but rather how to organize the problem and apply the laws of physics to answer it.

Thanks,

BTW I'm an Aerospace Engineer who saw this question in the paper and became annoyed.
mighoser
"Except at the surface of the belt and areas of the plane which are directly behind the propeller."

Should say:

Except at the surface of the belt and areas of the plane which are directly behind the propeller the velocity is zero.
Atl5p
This is great! I see that I'm finally getting through to some people...boy, if only Kreegan were here to see me now!! hehe

At least folks are starting to understand that when the plane's speed is defined as 'speed over the runway's surface', and the treadbelt matches this speed, then the plane will never ever be able to fly.

Good to see you kind folks comming around!

Atl5p
QUOTE (mighoser+Nov 28 2006, 07:23 PM)
Folks,

Airplanes generate lift when a fluid, in this case air, interactes with a solid body. For this problem we are talking about a pair of wings and fuselage. Since the amount of lift is related to the velocity of the fluid with respect to the solid body, this velocity is our only concern. Let jump to the steady state condition where the airplane isn't moving relative to the "ground" and the "belt" is moving in reverse at constant velocity X relative to the "ground". The plane is full throttle and travels at a velocity X relative to the "belt". So the question is what is the velocity of the fluid relative to the plane. Except at the surface of the belt and areas of the plane which are directly behind the propeller. The latter can be eliminated because airplanes rutinely apply full power and brakes before takeoff. Thus, the lift generated from the propellor propelling the fluid isn't sufficient to levatate the plane. The belt will impart some amount of velocity to the fluid relative to the plane. This will generate some lift in addition to the lift we just disscussed. So will the plane take off. Probably not but a powered glider with wings which are close to the belt might be able to take off. Keep in mind, the glider would loose lift the farther its wing was normally to the belt surface. This is because the velocity of the fluid goes from X to zero the farther it is from the belt.

The solution to this problem isn't about whether the plane flies or not but rather how to organize the problem and apply the laws of physics to answer it.

Thanks,

BTW I'm an Aerospace Engineer who saw this question in the paper and became annoyed.

Well put sir!

NoCleverName
QUOTE (Atl5p+Nov 29 2006, 01:59 AM)
Good to see you kind folks comming around!

Are we to believe you have deluded yourself into thinking that anyone at all agrees with your position? So far, there are three accepted cases of "no fly":
1. The case where the plane never moves at all due to engine off or tied to the ground while the conveyor belt moves. This is your case, if you haven't noticed.
2. The case where the conveyor belt is allowed to run at any speed it wants to while the plane moves over it at a much slower speed. This is something like a gym treadmill running at a pace far faster than you can keep up with it.
3. The case where the plane's wings don't work and it keeps moving until it can't go any faster. I just threw that one in.

And I thought there was some small hope for you.
Benny
mighoser, this thread is a spin off from Plane on conveyor... Will it ever take off?. What has happened is that Atl is either too stupid or stubborn to learn and understand basic physics and problem solving and has been getting his butt kicked in the other thread for an entire year (I'm serious.) So he just recently tried to rewrite the original problem to fit his mistaken interpretation. Only he screwed up and wrote that the belt matches the plane's groundspeed. Thus, if we state that the belt is moving at Xmph relative to the ground, then the plane must also be moving at Xmph relative to the ground, due to the logical constraints of his problem, and, assuming a windspeed of zero, the plane is generating Xmph worth of lift.

If he had correctly written his question and just asked if a conveyor runway could prevent a plane from taking off his understanding of the situation is still flawed. According to basic physics rolling resistance follows the equation F=uN (Force = coefficient of friction * normal force) and does not increase with speed. Therefore rolling resistance cannot stop the plane and it flies anyway.
egnorant
QUOTE (Atl5p+Nov 28 2006, 07:58 PM)
Point of Order:

The OP for THIS thread states that the conveyor matches the plane's wheelspeed.

At any rate, are you saying that in the case of the OP in THIS thread, the plane would NOT FLY? Ehh?

I am saying that it will fly in this question also.
The belt is not stated as matching the "wheelspeed"!
OP on this thread states that the belt moves at the same speed and in the opposite direction of:
QUOTE
plane's grounspeed over the runway

(grounspeed is really groundspeed isn't it?).

Had you said "speed over the belt" or "planes wheelspeed" or in some way isolated the planes movement from the ground I would have to agree with you.

It even has the same single point that determines the speed of both the plane and the belt......Treadbelt control system!

Plane moves left relative to the Treadbelt control system and the belt moves to the right relative to the treadbelt control system.

Adding the instruction of "plane's groundspeed over the runway" will result in the treadbelt control system seeing the plane as moving Left at 50 mph groundspeed over the belt. Has speed and direction..cool.
Now treadbelt control system plugs this in to move the belt in the opposite direction yields instruction to move RIGHT at 50 mph groundspeed over the belt!!

I have trouble visualizing the belt moving to the right at 50 mph groundspeed over the belt! Please explain!

Bruce

egnorant
QUOTE (mighoser+Nov 29 2006, 12:23 AM)
Folks,

Airplanes generate lift when a fluid, in this case air, interactes with a solid body. For this problem we are talking about a pair of wings and fuselage.  Since the amount of lift is related to the velocity of the fluid with respect to the solid body, this velocity is our only concern.  Let jump to the steady state condition where the airplane isn't moving relative to the "ground" and the "belt" is moving in reverse at constant velocity X relative to the "ground".  The plane is full throttle and travels at a velocity X relative to the "belt".  So the question is what is the velocity of the fluid relative to the plane.  Except at the surface of the belt and areas of the plane which are directly behind the propeller.  The latter can be eliminated because airplanes rutinely apply full power and brakes before takeoff.  Thus, the lift generated from the propellor propelling the fluid isn't sufficient to levatate the plane.  The belt will impart some amount of velocity to the fluid relative to the plane.  This will generate some lift in addition to the lift we just disscussed.  So will the plane take off.  Probably not but a powered glider with wings which are close to the belt might be able to take off.  Keep in mind, the glider would loose lift the farther its wing was normally to the belt surface.  This is because the velocity of the fluid goes from X to zero the farther it is from the belt.

The solution to this problem isn't about whether the plane flies or not but rather how to organize the problem and apply the laws of physics to answer it.

Thanks,

BTW I'm an Aerospace Engineer who saw this question in the paper and became annoyed.

You just explained why a plane will not fly if it does not move through the air.
You also explained what happens when a belt does move through the air.

We are still working on defining direction and opposite direction.
There is a thought that the movement opposite of 50 mph left, relative to the ground/air is a lack of movement to the right relative to the ground/air.

Enjoy the show!

What paper did you see this in? We could start a worldwide letter-writing campaign.
If we could get the whole world on-line arguing about this plane thing we could stop all the nuclear bomb stuff. I bet Kim Jong whatshisname would be a hoot!

Bruce
rethinker
Sorry NoCleverName, and others who look at this as not possible.

The thing we may be missing is the fact that for the plane to lift off, it needs two things.
1. power to the propeller
2. forward motion to create air flow over wings

What Atl5p has not said clearly is the fact that the two are needed.
He may have it correct after all.

The reverse of the tread belt may indeed act as a brake and thus no forward motion occurs and so no amount of the added flow of air will travel over wings.

A similar experiment may be the ping pong ball in a small funnel. Place a hose on the end of a funnel and blow air while the ball rests in the valley of the funnel. The ball will not fall out of the funnel even while it is inverted.

Fynlcut
QUOTE (rethinker+Nov 29 2006, 11:47 AM)
Sorry NoCleverName, and others who look at this as not possible.

The thing we may be missing is the fact that for the plane to lift off, it needs two things.
1. power to the propeller
2. forward motion to create air flow over wings

What Atl5p has not said clearly is the fact that the two are needed.
He may have it correct after all.

The reverse of the tread belt may indeed act as a brake and thus no forward motion occurs and so no amount of the added flow of air will travel over wings.

A similar experiment may be the ping pong ball in a small funnel. Place a hose on the end of a funnel and blow air while the ball rests in the valley of the funnel. The ball will not fall out of the funnel even while it is inverted.

No one denies the plane needs power, to the prop or by jet, or turbofan, or rocket. etc..etc...

No one denies a plane must have airflow over theunder and around the wings.

I'm not sure how the ping pong ball is relevant, please explain.

QUOTE
The reverse of the tread belt may indeed act as a brake

Ask yourself, CAN the belt provide enough braking? On some planes even the brakes themselves can not hold the plane stationary. Even the traction between the wheel and the ground can not hold some of them.
NoCleverName
QUOTE (Fynlcut+Nov 29 2006, 12:08 PM)
On some planes even the brakes themselves can not hold the plane stationary. Even the traction between the wheel and the ground can not hold some of them.

Quite true; for the B-777 is my examples I believe the engines must be held to no more than about 25% thrust to maintain braking. Full power take-off thrust would probably rip the wheels off. On landing, I don't think you're even allowed to touch the brakes at much more than 100 mph. Which is probably a good argument about how much braking power is in those wheels: why do they need those thrust reversers, speed brakes, and fully extended flaps and slats to slow one of these babys down, anyway? Hell, all they really needed was a moving conveyor runway. I'd run to the patent office!

rethinker: better rethink again. Forget the wings, forget the air (mighoser even brought in the entrained air flow from the moving conveyor ... which would help takeoff), none of that's important. This is a simple motion problem: force one way, force the other: what is the resulting motion from the net force. Unless you, like Limon, believe Newton was wrong about forces. This is one of those problems where "physical intuition" leads you astray; that's why it is a puzzle. Trust in the tools of physics --- let go of experience, use the force!
Atl5p
QUOTE (Benny+Nov 28 2006, 11:20 PM)
mighoser, this thread is a spin off from Plane on conveyor... Will it ever take off?.  What has happened is that Atl is either too stupid or stubborn to learn and understand basic physics and problem solving and has been getting his butt kicked in the other thread for an entire year (I'm serious.)  So he just recently tried to rewrite the original problem to fit his mistaken interpretation.  Only he screwed up and wrote that the belt matches the plane's groundspeed.  Thus, if we state that the belt is moving at Xmph relative to the ground, then the plane must also be moving at Xmph relative to the ground, due to the logical constraints of his problem, and, assuming a windspeed of zero, the plane is generating Xmph worth of lift.

If he had correctly written his question and just asked if a conveyor runway could prevent a plane from taking off his understanding of the situation is still flawed.  According to basic physics rolling resistance follows the equation F=uN (Force = coefficient of friction * normal force) and does not increase with speed.  Therefore rolling resistance cannot stop the plane and it flies anyway.

Uhhh, that's real interesting Benny....because on that 'other thread', I clearly remember you saying the following:

QUOTE
Since this is a "real world" problem (your words) I will concede that a conveyor runway can hold back a plane if you show me a real runway-sized conveyor that is capable of several hundred MPH top speed, is capable of reaching top speed in less than 60 seconds, can handle the weight and thrust of an ordinary Cessna while doing the above, and that the Cessna's wheels can safely handle the speeds and forces involved. (I don't consider running the wheels outside of their safe operating range to be within the intended answer of the question.) Your interpretation of the problem will still be wrong, but I will concede that a plane can be held back.

Benny....the question says the treadbelt can match the plane's speed exactly. It also says the treadbelt IS a giant sized runway. When the question says the plane is on the treadbelt, and can match the speed exactly, why do you question the treadbelt's ability to handle the job? You seem to be saying that the plane would take off simply because there's just no way a treadbelt described in the OP could ever exist.
So, the OP says there's this big powerful treadbelt, and you just can't believe it...so you say the plane will fly. Wow...how insight-full(of BS) that is.

Hey everybody...the plane will fly! This is because the treadbelt doesn't even exist! Gee, none of us ever thought about that!
Atl5p
QUOTE (egnorant+Nov 28 2006, 11:27 PM)
I am saying that it will fly in this question also.
The belt is not stated as matching the "wheelspeed"!
OP on this thread states that the belt moves at the same speed and in the opposite direction of:

(grounspeed is really groundspeed isn't it?).

Had you said "speed over the belt" or "planes wheelspeed" or in some way isolated the planes movement from the ground I would have to agree with you.

It even has the same single point that determines the speed of both the plane and the belt......Treadbelt control system!

Plane moves left relative to the Treadbelt control system and the belt moves to the right relative to the treadbelt control system.

Adding the instruction of "plane's groundspeed over the runway" will result in the treadbelt control system seeing the plane as moving Left at 50 mph groundspeed over the belt. Has speed and direction..cool.
Now treadbelt control system plugs this in to move the belt in the opposite direction yields instruction to move RIGHT at 50 mph groundspeed over the belt!!

I have trouble visualizing the belt moving to the right at 50 mph groundspeed over the belt! Please explain!

Bruce

'Groundspeed over the moveable runway' is that too difficult for you to understand...that does equal wheelspeed...no, the surface is not covered in axle grease either, so the tires DO spin.

If you are having a hard time visualizing this, it's as simple as a runner on a treadbelt. Runner 'runs' at 10mph. Treadbelt digital speed readout says 10mph.

DO YOU GET IT NOW?
egnorant
QUOTE (Atl5p+Nov 29 2006, 02:37 PM)
'Groundspeed over the moveable runway' is that too difficult for you to understand...that does equal wheelspeed...no, the surface is not covered in axle grease either, so the tires DO spin.

If you are having a hard time visualizing this, it's as simple as a runner on a treadbelt. Runner 'runs' at 10mph. Treadbelt digital speed readout says 10mph.

DO YOU GET IT NOW?

Sorry, One little word changes the entire question.
Groundspeed is speed relative to the ground.
Ground speed over the belt is speed relative to the ground that occurs over a belt.

Bruce
NoCleverName
Let me try this new problem which is physically exactly the same as this problem except that I'm hoping to break the "physical intuition grip" that prevents certain people from understanding it. (I'll put in in both threads).

We are in the wide open spaces; perhaps a barren, rocky plain. We see a lone treadmill. And there upon it, mounted on his skateboard, is Alt5p. In front of him, hanging low in the sky, is the Goodyear blimp. A rope is taken from the blimp and handed to Alt5p. Benny is seen at the controls of the treadmill. His job: adjust the speed of the treadmill to match the speed of the blimp. How does he decide the blimp's speed? He sets it to whatever Alt5p tells him it is.

The signal is given, and the blimp "moves"; Alt5p orders the treadmill speed be adjusted to match the "speed" of the blimp.
• You are a "no fly boy" if Alt5p is actually "Clark Kent" and is able to arrest the motion of the blimp. In fact, he's holding the rope in his teeth.
• You are a "fly boy" if Alt5p is towed off the treadmill and dragged unceremoniously over the rocks.
This is the same problem: the blimp is the plane's fuselage and engine, the rope is the struts connecting the plane to its wheels, the wheels are the skateboard, and the conveyor-runway is played by the treadmill.

Now when you look at it you see a different physical reality, don't you?
Atl5p
QUOTE (NoCleverName+Nov 29 2006, 01:48 PM)
Let me try this new problem which is physically exactly the same as this problem except that I'm hoping to break the "physical intuition grip" that prevents certain people from understanding it. (I'll put in in both threads).

We are in the wide open spaces; perhaps a barren, rocky plain. We see a lone treadmill. And there upon it, mounted on his skateboard, is Alt5p. In front of him, hanging low in the sky, is the Goodyear blimp. A rope is taken from the blimp and handed to Alt5p. Benny is seen at the controls of the treadmill. His job: adjust the speed of the treadmill to match the speed of the blimp. How does he decide the blimp's speed? He sets it to whatever Alt5p tells him it is.

The signal is given, and the blimp "moves"; Alt5p orders the treadmill speed be adjusted to match the "speed" of the blimp.
• You are a "no fly boy" if Alt5p is actually "Clark Kent" and is able to arrest the motion of the blimp. In fact, he's holding the rope in his teeth.
• You are a "fly boy" if Alt5p is towed off the treadmill and dragged unceremoniously over the rocks.
This is the same problem: the blimp is the plane's fuselage and engine, the rope is the struts connecting the plane to its wheels, the wheels are the skateboard, and the conveyor-runway is played by the treadmill.

Now when you look at it you see a different physical reality, don't you?

How much mass does the blimp have? Is that mass resting upon the treadbelt? The full mass of the blimp...does it's mass affect the CRF of the skateboard's wheels?

If not, then it is nothing like the OP at all....but it is a
QUOTE
different physical reality