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Grumpy
Atl5P

Your puny little tread belt cannot stop any normally powerful plane from moving or taking off. It has no means to do so.

Grumpy cool.gif
Atl5p
QUOTE (egnorant+Feb 23 2007, 02:15 PM)
Correct! Factoring in those items will give you a speed of 55 mph and the plane is still at 50 mph.

Adding the modifier "OVER THE SURFACE OF THE TRAILERBED" (belt..whatever)
Will yield different results.
The original question has no such modifiers.

Bruce

So, lemme get this straight....what you are saying is this:

"A Plane moves through the air......Does the plane move though the air? "
Ans=Yes.

Ummm, ok...go grab a cookie!

Let me ask you a question to clarify....in the late 1960's, man drove a car on the surface of the moon at 5mph wheelspeed.

Question: What was his "SPEED"?

Remember..in order to remain consistant, it goes 'without saying'...you MUST relate his speed to the surface of this planet earth....so, his speedometer says '5mph'....now, what IS his "SPEED"?

And which portion of the earth' surface did you use, in relation to the moon? (closest to the moon, off to the side, far side of the earth?)
Derek1148
Numerous individuals have presented scenarios, which indicate that a plane would be capable of take-off from a conveyor moving in the opposite direction.

The fact that someone can twist and torture logic and state that under his scenario the plane would be incapable of take-off is irrelevant to the question. Even making the assumption that the individual is correct concerning his scenario, it would not prove a negative answer to the question.

Does a plane crash prove flight is impossible?

“The question is: Will the plane take off or not?”

Any plane, not necessarily every plane.

Atl5p
QUOTE (Grumpy+Feb 23 2007, 03:10 PM)
Atl5P

Your puny little tread belt cannot stop any normally powerful plane from moving or taking off. It has no means to do so.

Grumpy cool.gif

Grumpy,

Do you mean to claim that a regular ol'e exercise treadmill can NOT prevent a full sized Cessna 151 from taking off?!?!?!?!

Holee Shee-at! Give this MutherFockingCluckStucker a whole goldamtheriverflowsoverthedam BOX of cookies!!!

I mean, I don't see how the sheer MASS of your BRAIN is able to be SUPPORTED by EARTH'S CRUST, youmustbesosmart!!!

Now....try a nuclear powered treadbelt and your puny little tribute...crank the treadbelt up to 1 billion MPH....let's see if the puny tribute can make any headway..I don't think so.

There is no way that your puny little plane can gain ground on a treadbelt designed to prevent flight....period.

So THERE!
Atl5p
QUOTE (Derek1148+Feb 23 2007, 03:19 PM)
Numerous individuals have presented scenarios, which indicate that a plane would be capable of take-off from a conveyor moving in the opposite direction.

The fact that someone can twist and torture logic and state that under his scenario the plane would be incapable of take-off is irrelevant to the question. Even making the assumption that the individual is correct concerning his scenario, it would not prove a negative answer to the question.

Does a plane crash prove flight is impossible?

“The question is: Will the plane take off or not?”

Any plane, not necessarily every plane.

DikRick, all of the people you have mentioned have barstardised the question into saying the following:
"A plane moves through the air....is the plane moving through the air?"

And I am asking the true intent of the question:
"Can a treadbelt prevent a plane from taking off?"
the answer is yes...many flyboys agree....

yet they just keep saying that they believe the plane's 'speed' means 'speed through the air' ....I personally think that is about as shallow as it gets. You appear to be among that group.

I happen to believe that SINCE the plane is ON the surface of the treabbelt, then THAT is what the plane's speed is based from. If the center lines are rushing by the plane at 100mph, then the plane thinks it's going 100mph.

You guys really roll me! I just imagine you people getting on a treadbelt and reading the instructions: "Walk forward and adjust treadbelt speed to match YOUR speed"....and they you guys keep walking off the front of the treadbelt, scratching your heads, wondering how you broked it. HAHAHAHA!!! So DENSE!
Precursor562
QUOTE
Let’s try again, on a large ship at sea.  The ship is 1 mile long.  It is heading north at 20mph over the bottom.  The wind is heading south at 20mph.  The current is going south at 5mph.
I am walking on the deck at a speed which will take me from stern to bow in 1 hour.
YOU are in a helicopter flying directly above ME as I walk.


Well since your are on a boat moving at 20 mph in order for you to move from the back of the boat to the front (1 mile) in one hour than you must move 1 mph faster than the boat. The boat is already moving you at 20 mph due to you standing on it. So you only need to put the effort to move 1 mph since the ship is doing the effort of 20 mph for you. Is your speed 1 mph or is it 21 mph. Truth is it is 21 mph. You can determine this more clearly by using a map. If you have a map where you can zoom in you can then mark your starting point (not the boat's) on the map. On the same map mark the starting point of the boat (with the same zoom will will have to pick a point on the ship lets say the very tip of the bow). After one hour mark your new location and the boat's new location on the map. From bow tip to bow tip the boat would have traveled 20 miles. You on the other hand started one mile behind the first boat point (since you started at the stern of the mile long boat) and ended at the same point as the second boat point. This is a total distance of 21 miles. One mile more than the boat within the one hour time frame. Hence you physically moved in relation to a fixed point at a speed of 21 mph.

That is what the belt matches its speed to when it comes to the plane. The belt matches it's speed on the distance the plane physically moves through space (again I mean that generically and am not referring to outer space) in a certain amount of time. Speed being distance traveled (physical movement from one point in space to another) divided by the time it took you to do so.

Once again my or anyone else's perception of how fast it "looks" like you are going make absolutely no difference to how fast your are actually moving. All that it means for me in that helicopter is that I am also moving 21 mph and so is the helicopter I am in.

QUOTE (->
QUOTE
Let’s try again, on a large ship at sea.  The ship is 1 mile long.  It is heading north at 20mph over the bottom.  The wind is heading south at 20mph.  The current is going south at 5mph.
I am walking on the deck at a speed which will take me from stern to bow in 1 hour.
YOU are in a helicopter flying directly above ME as I walk.


Well since your are on a boat moving at 20 mph in order for you to move from the back of the boat to the front (1 mile) in one hour than you must move 1 mph faster than the boat. The boat is already moving you at 20 mph due to you standing on it. So you only need to put the effort to move 1 mph since the ship is doing the effort of 20 mph for you. Is your speed 1 mph or is it 21 mph. Truth is it is 21 mph. You can determine this more clearly by using a map. If you have a map where you can zoom in you can then mark your starting point (not the boat's) on the map. On the same map mark the starting point of the boat (with the same zoom will will have to pick a point on the ship lets say the very tip of the bow). After one hour mark your new location and the boat's new location on the map. From bow tip to bow tip the boat would have traveled 20 miles. You on the other hand started one mile behind the first boat point (since you started at the stern of the mile long boat) and ended at the same point as the second boat point. This is a total distance of 21 miles. One mile more than the boat within the one hour time frame. Hence you physically moved in relation to a fixed point at a speed of 21 mph.

That is what the belt matches its speed to when it comes to the plane. The belt matches it's speed on the distance the plane physically moves through space (again I mean that generically and am not referring to outer space) in a certain amount of time. Speed being distance traveled (physical movement from one point in space to another) divided by the time it took you to do so.

Once again my or anyone else's perception of how fast it "looks" like you are going make absolutely no difference to how fast your are actually moving. All that it means for me in that helicopter is that I am also moving 21 mph and so is the helicopter I am in.

Again, due to the tread belt, we MUST use two reference points for speed.


It's not a must but you can if you want to.

The plane sits on the belt and at that very point on the belt is a red circle. At both ends of the belt are towers (tower A and tower B ). The towers are on fixed ground and do not move. The plane accelerates to tower A from 0 up to 100 mph (take off speed). The belt will match the speed of the plane (not its thrust or its wheels) causing the red circle to move toward tower B at an accelerated rate equal to the plane (0 to 100 mph).

Lets say when the plane reaches 50 mph on still ground there is a wheel resistance of X. How big is X? Not big at all. A single prop Cessna could be pulled by an adult of moderate strength. Even then you are not over coming the wheels resistance to turn but rather over coming the resistance to move due to the plane's mass.

On the belt (whose speed would be 50 mph in the opposite direction to match that of the plane) the value of X would NOT become 2X. It would be X+A where A is a value that would be less than 25% the value of X. Since the value of X is small to begin with than the value of A would be next to nothing and so the value of X+A would still be very small and no different whether the plane was on or off the belt.

Could the belt stop the plane? Absolutely BUT the belt would have to be moving much faster than the plane which goes against the OP. The belt IAW the OP has to have a speed equal to the plane. Not greater than (>) or less than (<) but equal to (=). With that said the belt can not stop the plane from moving forward and taking off. Will the engine have to work harder? Yes BUT only a LITTLE bit.

If they wanted to ask if a moving belt could stop a plane from taking off then they would have asked exactly that. Instead they took the time to type that little tin bit of info stating that the speed of the belt would EQUAL the speed of the plane. Which makes the question "Would a belt whose speed equals the speed of the plane in the opposite direction keep the plane from taking off?"

The answer to this is a big fat NO...
Atl5p
QUOTE (Precursor562+Feb 23 2007, 03:36 PM)

Well since your are on a boat moving at 20 mph in order for you to move from the back of the boat to the front (1 mile) in one hour than you must move 1 mph faster than the boat.  The boat is already moving you at 20 mph due to you standing on it.  So you only need to put the effort to move 1 mph since the ship is doing the effort of 20 mph for you.  Is your speed 1 mph or is it 21 mph.  Truth is it is 21 mph.  You can determine this more clearly by using a map.  If you have a map where you can zoom in you can then mark your starting point (not the boat's) on the map.  On the same map mark the starting point of the boat (with the same zoom will will have to pick a point on the ship lets say the very tip of the bow).  After one hour mark your new location and the boat's new location on the map.  From bow tip to bow tip the boat would have traveled 20 miles.  You on the other hand started one mile behind the first boat point (since you started at the stern of the mile long boat) and ended at the same point as the second boat point.  This is a total distance of 21 miles.  One mile more than the boat within the one hour time frame.  Hence you physically moved in relation to a fixed point at a speed of 21 mph.

That is what the belt matches its speed to when it comes to the plane.  The belt matches it's speed on the distance the plane physically moves through space (again I mean that generically and am not referring to outer space) in a certain amount of time.  Speed being distance traveled (physical movement from one point in space to another) divided by the time it took you to do so.

Once again my or anyone else's perception of how fast it "looks" like you are going make absolutely no difference to how fast your are actually moving.  All that it means for me in that helicopter is that I am also moving 21 mph and so is the helicopter I am in.



It's not a must but you can if you want to.

The plane sits on the belt and at that very point on the belt is a red circle.  At both ends of the belt are towers (tower A and tower cool.gif.  The towers are on fixed ground and do not move.  The plane accelerates to tower A from 0 up to 100 mph (take off speed).  The belt will match the speed of the plane (not its thrust or its wheels) causing the red circle to move toward tower B at an accelerated rate equal to the plane (0 to 100 mph). 

Lets say when the plane reaches 50 mph on still ground there is a wheel resistance of X.  How big is X?  Not big at all.  A single prop Cessna could be pulled by an adult of moderate strength.  Even then you are not over coming the wheels resistance to turn but rather over coming the resistance to move due to the plane's mass.

On the belt (whose speed would be 50 mph in the opposite direction to match that of the plane) the value of X would NOT become 2X.  It would be X+A where A is a value that would be less than 25% the value of X.  Since the value of X is small to begin with than the value of A would be next to nothing and so the value of X+A would still be very small and no different whether the plane was on or off the belt.

Could the belt stop the plane?  Absolutely BUT the belt would have to be moving much faster than the plane which goes against the OP.  The belt IAW the OP has to have a speed equal to the plane.  Not greater than (>) or less than (<) but equal to (=).  With that said the belt can not stop the plane from moving forward and taking off.  Will the engine have to work harder? Yes BUT only a LITTLE bit.

Ladies and ..... well, anyways....once again we have it:

"A plane moves through the air......does the plane move through the air"

Correctly answered once again....somebody better get more cookies.

And it's also great that you and ignorant Bruce agree that a treadbelt is able to prevent a plane from flying....now could you please help me explain that to GRUMPY? He says that there is NO WAY that would ever happen.

See, what we have between ATL and Precurser and egrorant is a difference in opinion as to how the plane's speed is determined....that's OK.

But what is NOT ok is when we ask the question 'Can a treadbelt prevent a plane from flying' and some flyboys say "YES" but a few of the lowerslower ones say "NO"....and what is even worse than that is when the enlightened flyboys do not attempt to show the others the light....that is just an abomination....you are not caring for your own kind....how horible!

PS...what is really funny is this...you guys keep saying this:

"If the plane is moving at a speed (IAS) of 50mph, and the treadbelt is spinning at 50mph, the it won't slow down the plane VERY MUCH."

ROTFLOL!!!!

Just what the heckel is THAT supposed to mean? Either the plane is going 50mph IAS or it's NOT. If the plane is really going 50mph, then the treadbelt is NOT slowing the plane down AT ALL from said 50mph....but almost all of the flyboys say somthing to the effect of 'dosn't slow the plane down very much'.

Look....it's either going 50mph IAS or it's NOT...make up your mind.

HAHAHAHAHA!
Derek1148
QUOTE (Atl5p+Feb 23 2007, 08:34 PM)
You guys really roll me! I just imagine you people getting on a treadbelt and reading the instructions: "Walk forward and adjust treadbelt speed to match YOUR speed"....and they you guys keep walking off the front of the treadbelt, scratching your heads, wondering how you broked it. HAHAHAHA!!! So DENSE!

I have previously addressed your intellect, and I am more certain of those appraisals than I am the ability of the plane to take-off.

Due to the above-mentioned deficiency, your opinions only have the ability to annoy but not educate.


Atl5p
QUOTE (Derek1148+Feb 23 2007, 04:17 PM)
I have previously addressed your intellect, and I am more certain of those appraisals than I am the ability of the plane to take-off.

Due to the above-mentioned deficiency, your opinions only have the ability to annoy but not educate.

So Drack....do you think that a treadbelt COULD prevent a plane from taking off?

The thing is, you keep relating the speed of the plane to the calm air....while I relate the speed of the plane to the surface upon which it is traveling.

I'd call that a difference of opinion....not a lack of 'intellect' as you say.

I do realize that a plane moving through the air does have the potential for flight...so I am not deprived in the brains department...I just look at the problem a little differently, and I relate the plane's speed to the surface of the treadbelt....just like at the gym, if you want to run at 10mph, but still watch TV, you get on a treadbelt....hey, you're still running at 10mph....your speed over the treadbelt is 10mph....that is the whole point of a treadbelt ya know.

However, if you feel that it is impossible for a treadbelt to prevent a plane from taking off, THEN we've got an intellect issue on YOUR side.....so....I'd like to see that answer....
adoucette
QUOTE (Atl5p+Feb 23 2007, 02:54 PM)
Yeah, I left that wide open for you on the Var Pitch prop....and you flew through it with flying colors!!! Your variable pitch prop will have NOTHING to do with this situation...the THRUST needed to get moving is in the DIAMETER of the prop....the PITCH can be adjusted for higher TOP END SPEED, that's why they call it 'Pitch speed'.
A variable pitch prop will have nothing to do with the plane's ability to break the deadly grasp of the treadbelt.
Do you not know the difference between thrust and pitch speed?

Or maybe you've discovered a new 'Variable Diameter' prop somewhere that no one knows about? HAHA I am laughing so hard right now, I may choke!


Increasing the pitch for a given RPM increases THRUST (up until the blade slips).

Since you fly RC I checked out an RC Prop calculator for Static thrust, i.e that which gets you moving.

All at 10,000 RPM

11x4 = 49 oz STATIC Trust
11x5 = 61 oz
11x6 = 74 oz
11x7 = 81 oz

Note, only PITCH changed, not diameter or RPM.

Oh, and remember our Aircraft designers motto: There is no DEADLY GRASP of the Treadbelt.

laugh.gif

Arthur
adoucette
QUOTE
I happen to believe that SINCE the plane is ON the surface of the treabbelt, then THAT is what the plane's speed is based from.



laugh.gif


Atl5p's version of the problem:

If a plane on a moving treadbelt only applies just enough power to stay in a fixed position relative to the ground, will it take off?


laugh.gif

Arthur
Grumpy
"A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?"

Yes the plane will take off.

Grumpy cool.gif
Eric England
The treadmill has a firm grip on the plane's tail.

Absent a mechanical failure, it ain't lettin' go.

The only way to get the plane to move off the treadmill, is to reverse the thrust.

Caution: Install rearview mirror before doing so.

Otherwise, I see many here with less than a firm grip.
Grumpy
Dear Cecil:

Please, please, please settle this question. The discussion has been going on for ages, and any time someone mentions the words "airplane" or "conveyor belt" everyone starts right back up. Here's the original problem essentially as it was posed to us: "A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?"

There are some difficulties with the wording of the problem, specifically regarding how we define speed, but the spirit of the situation is clear. The solution is also clear to me (and many others), but a staunch group of unbelievers won't accept it. My conclusion is that the plane does take off. Planes, whether jet or propeller, work by pulling themselves through the air. The rotation of their tires results from this forward movement, and has no bearing on the behavior of a plane during takeoff. I claim the only difference between a regular plane and one on a conveyor belt is that the conveyor belt plane's wheels will spin twice as fast during takeoff. Please, Cecil, show us that it's not only theoretically possible (with frictionless wheels) but it's actually possible too. --Berj A. Doudian, via e-mail

Cecil replies:

Excuse me--did I hear somebody say Monty Hall?

On first encounter this question, which has been showing up all over the Net, seems inane because the answer seems so obvious. However, as with the infamous Monty-Hall-three-doors-and-one-prize-problem (see The Straight Dope: "On Let's Make a Deal" you pick Door #1, 02-Nov-1990), the obvious answer is wrong, and you, Berj, are right--the plane takes off normally, with no need to specify frictionless wheels or any other such foolishness. You're also right that the question is often worded badly, leading to confusion, arguments, etc. In short, we've got a topic screaming for the Straight Dope.

First the obvious-but-wrong answer. The unwary tend to reason by analogy to a car on a conveyor belt--if the conveyor moves backward at the same rate that the car's wheels rotate forward, the net result is that the car remains stationary. An aircraft in the same situation, they figure, would stay planted on the ground, since there'd be no air rushing over the wings to give it lift. But of course cars and planes don't work the same way. A car's wheels are its means of propulsion--they push the road backwards (relatively speaking), and the car moves forward. In contrast, a plane's wheels aren't motorized; their purpose is to reduce friction during takeoff (and add it, by braking, when landing). What gets a plane moving are its propellers or jet turbines, which shove the air backward and thereby impel the plane forward. What the wheels, conveyor belt, etc, are up to is largely irrelevant. Let me repeat: Once the pilot fires up the engines, the plane moves forward at pretty much the usual speed relative to the ground--and more importantly the air--regardless of how fast the conveyor belt is moving backward. This generates lift on the wings, and the plane takes off. All the conveyor belt does is, as you correctly conclude, make the plane's wheels spin madly.

A thought experiment commonly cited in discussions of this question is to imagine you're standing on a health-club treadmill in rollerblades while holding a rope attached to the wall in front of you. The treadmill starts; simultaneously you begin to haul in the rope. Although you'll have to overcome some initial friction tugging you backward, in short order you'll be able to pull yourself forward easily.

As you point out, one problem here is the wording of the question. Your version straightforwardly states that the conveyor moves backward at the same rate that the plane moves forward. If the plane's forward speed is 100 miles per hour, the conveyor rolls 100 MPH backward, and the wheels rotate at 200 MPH. Assuming you've got Indy-car-quality tires and wheel bearings, no problem. However, some versions put matters this way: "The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation." This language leads to a paradox: If the plane moves forward at 5 MPH, then its wheels will do likewise, and the treadmill will go 5 MPH backward. But if the treadmill is going 5 MPH backward, then the wheels are really turning 10 MPH forward. But if the wheels are going 10 MPH forward . . . Soon the foolish have persuaded themselves that the treadmill must operate at infinite speed. Nonsense. The question thus stated asks the impossible -- simply put, that A = A + 5 -- and so cannot be framed in this way. Everything clear now? Maybe not. But believe this: The plane takes off.

--CECIL ADAMS

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

Grumpy
cool.gif
Atl5p
QUOTE (adoucette+Feb 23 2007, 04:27 PM)
Increasing the pitch for a given RPM increases THRUST (up until the blade slips).

Since you fly RC I checked out an RC Prop calculator for Static thrust, i.e that which gets you moving.

All at 10,000 RPM

11x4 = 49 oz STATIC Trust
11x5 = 61 oz
11x6 = 74 oz
11x7 = 81 oz

Note, only PITCH changed, not diameter or RPM.

Oh, and remember our Aircraft designers motto: There is no DEADLY GRASP of the Treadbelt.

laugh.gif

Arthur

You also obvioulsy did not use the same MOTOR for each of those figures....because the ONLY way to get the same 10,000 RPM at MAX throttle for differing pitched props is if the motor keeps getting more powerful.

Try it again, except this time use the SAME MOTOR.

Or, better yet, get a performance chart for a SINGLE motor.
Leave the diameter the same each time.
Increase the pitch each time.
Reminder...use the same motor each time!!!! (i really cannot believe I have to say this)
and what happens?

Each time you increase the PITCH, the RPM will DECREASE, and the THRUST will DECREASE!

What you did was keep the RPM at 10,000 for each pitch'ed prop....in order for your agressive 11x7 prop to spin at the EXACT same RPM as the mild 11x4 prop, you would need MUCH more powerful motor to spin that agressive prop, to keep the RPMs the exact same at 10k! You were CHANGING MOTORS! YOU CHEATER!

Need a hint? Go here: http://www.aero-nuts.com/product_info.php?...products_id=286

Check out the 10x6 prop at 10v. Spins at 7200RPM with 24.62oz THRUST.
then, using the EXACT SAME MOTOR (again, can't believe I have to say that)
Look at the 10x8 prop at 10v. Spins at 5910RPM (lower) with onlyl 21.64 ox thrust (Lower)!

Of course your version just kept increasing the size of the motor in addition to the pitch of the prop....

So, in addition to the 'Variable Pitch Prop' solution, you forgot about the 'Variable Engine' setup. WOW, you have a way to change motors while in flight? You'll be a millioniare for sure!

Now, any more comments about my intellect? Didn't think so.

Once again...for a GIVEN MOTOR, increasing the pitch will DECREASE the max RPM and likewise DECREASE the max THRUST.

Prop diameter is the thrust.
Prop pitch is the speed.

Got it?

Dude, you are a RIOT!! I literally fell out of my chair I was laughing at YOU so freaking hard.....'duh, keep increasing the load on the motor, it will stay at 10,000rpm....duhhh....dope, duh....'


ROTFLOL!L!L!L!L!
biggrin.gif rolleyes.gif
yor_on
What if one connected superconducting magnets to the airplanes wheel, f ex a Concorde. The wheels off course would have to be really big and made of pure iron, then one could use the the wheels spinning to build up even more electromagnetic force and then suddenly while f ex singing ' rule Britannica' reverse the magnetic force hmmm. Would the plane take a flying leap or what? What??
Atl5p
QUOTE (Grumpy+Feb 23 2007, 04:58 PM)
Dear Cecil:

Please, please, please settle this question. The discussion has been going on for ages, and any time someone mentions the bsbsbsbsbs

--CECIL ADAMS

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

Grumpy
cool.gif

Grumpy, it is really unneccesary for you to post comments from other threads....If I wanted to comment to cecil's dumb aarse, i would have posted something on his stupid message board...The fact that you know how to 'Cut' and 'Paste' just THRILLS me...you can only imagine!

Now how about this:

Several of your 'FlyBoy' friends are in severe disagreement with you. They are saying that it is obvious to any dumb arse that a treadbelt COULD prevent a plane from taking off....I think they are calling you a dumas I'm not sure.

So, grumpy, what do you have to say to that...I know that you believe it is impossible for a treadbelt to prevent a plane from taking off.....and they are calling you out on this by disagreeing with the Great Grumpy of the Cut and Paste from another thread clan....whattayagoinadoaboutit? huh....HUH???

Also, I would like your opinion of aducate's method of changing motors mid-flight wile at the same time adjusting the variable pitch prop.

Grumpy, would YOU buy a VARIABLE MOTOR'ed plane from someone on a message board? Would you?

Lemme ask you a question, Gurpie....if you used an 11x4 prop on a motor and full power produced 10,000rpm, what do you think would happen if you changed the prop to something more agressive...like say a 11x7 prop? Do you think your same motor and power supply could produce the exact same 10,000rpm? Or do you think the rpm's would be lower or higher?

Is any of this making any sence to you at all?

The faster the treadbelt spins, the more 'backwards force' is applied to the plane....eventually, there WILL BE a speed that will produce enough 'backwards' force to equal the 24.6 oz of thrust coming from the plane.

At that point, aducate will switch the motors to a more powerful set, and will re-set the variable pitch and then take off vertically!!


HAHAHAHAHA!!! How good can this get?!
Atl5p
QUOTE (adoucette+Feb 23 2007, 04:34 PM)


laugh.gif


Atl5p's version of the problem:

If a plane on a moving treadbelt only applies just enough power to stay in a fixed position relative to the ground, will it take off?


laugh.gif

Arthur

The answer is "NO"....

Even better....in this situation, is the treadbelt matching the plane's forward speed as it relates to the surface upon which it is moving?

"YES".

btw---it appears that you are going to leave your 'vairable prop/ variable motor' debackle just blowing out there?
adoucette
QUOTE (Atl5p+Feb 23 2007, 05:08 PM)
You also obvioulsy did not use the same MOTOR for each of those figures....because the ONLY way to get the same 10,000 RPM at MAX throttle for differing pitched props is if the motor keeps getting more powerful.

Try it again, except this time use the SAME MOTOR.

Or, better yet, get a performance chart for a SINGLE motor.
Leave the diameter the same each time.
Increase the pitch each time.
Reminder...use the same motor each time!!!! (i really cannot believe I have to say this)
and what happens?

Each time you increase the PITCH, the RPM will DECREASE, and the THRUST will DECREASE!

What you did was keep the RPM at 10,000 for each pitch'ed prop....in order for your agressive 11x7 prop to spin at the EXACT same RPM as the mild 11x4 prop, you would need MUCH more powerful motor to spin that agressive prop, to keep the RPMs the exact same at 10k! You were CHANGING MOTORS! YOU CHEATER!

Need a hint? Go here: http://www.aero-nuts.com/product_info.php?...products_id=286

Check out the 10x6 prop at 10v. Spins at 7200RPM with 24.62oz THRUST.
then, using the EXACT SAME MOTOR (again, can't believe I have to say that)
Look at the 10x8 prop at 10v. Spins at 5910RPM (lower) with onlyl 21.64 ox thrust (Lower)!

Of course your version just kept increasing the size of the motor in addition to the pitch of the prop....

So, in addition to the 'Variable Pitch Prop' solution, you forgot about the 'Variable Engine' setup. WOW, you have a way to change motors while in flight? You'll be a millioniare for sure!

Now, any more comments about my intellect? Didn't think so.

Once again...for a GIVEN MOTOR, increasing the pitch will DECREASE the max RPM and likewise DECREASE the max THRUST.

Prop diameter is the thrust.
Prop pitch is the speed.

Got it?

Dude, you are a RIOT!! I literally fell out of my chair I was laughing at YOU so freaking hard.....'duh, keep increasing the load on the motor, it will stay at 10,000rpm....duhhh....dope, duh....'


ROTFLOL!L!L!L!L!
biggrin.gif rolleyes.gif

Nope,

YOU got yourself into this sidetrack by THIS:

QUOTE (Atl5p+)
the plane’s thrust will decrease due to the artificial ‘headwind’ it creates by moving through the air.


Which is why I asked you if you had ever heard of Variable Pitch Props.

Then you said

QUOTE (Atl5p+)
the THRUST needed to get moving is in the DIAMETER of the prop..


And I showed you how Thrust can vary by PITCH for the same Diameter and RPM.

You're lost Atl5p.

Everybody and his brother has explained it to you but of course YOU are right and the world is wrong.

laugh.gif

See you in another year or so.

Arthur

Grumpy
Atl5p

What kind of childish display is this???

QUOTE
So, grumpy, what do you have to say to that...I know that you believe it is impossible for a treadbelt to prevent a plane from taking off.....and they are calling you out on this by disagreeing with the Great Grumpy of the Cut and Paste from another thread clan....whattayagoinadoaboutit? huh....HUH???


I will just say that anyone who does not "get it" is simply wrong. The plane flies, and it flies easily, no matter what the treadmill does(within the boundaries of the OP)



QUOTE (->
QUOTE
So, grumpy, what do you have to say to that...I know that you believe it is impossible for a treadbelt to prevent a plane from taking off.....and they are calling you out on this by disagreeing with the Great Grumpy of the Cut and Paste from another thread clan....whattayagoinadoaboutit? huh....HUH???


I will just say that anyone who does not "get it" is simply wrong. The plane flies, and it flies easily, no matter what the treadmill does(within the boundaries of the OP)



Lemme ask you a question, Gurpie....if you used an 11x4 prop on a motor and full power produced 10,000rpm


I don't recall anyone saying anything about full power.

If I have a motor that can turn a 11X8 at 10,000 rpm, it BETTER damn well turn an 11X4 at that same rpm. In fact my Tribute has a controller that turns the motor at a steady RPM, the thrust is varied by adjusting the props blades(I have just as much thrust in a forward direction as I do in reverse. All I do is flip the throttle up switch and put the throttle stick in the middle for 0 thrust.

You can't get a Tribute here anymore(mine's three years old) but this is the same type VPP I have. The Animal

User posted image

User posted image

http://www.hobby-lobby.com/animal.htm

Grumpy cool.gif

PS Some very good videos here.
Precursor562
QUOTE
I do realize that a plane moving through the air does have the potential for flight...so I am not deprived in the brains department...I just look at the problem a little differently, and I relate the plane's speed to the surface of the treadbelt....just like at the gym, if you want to run at 10mph, but still watch TV, you get on a treadbelt....hey, you're still running at 10mph....your speed over the treadbelt is 10mph....that is the whole point of a treadbelt ya know.


With one major problem. A person moves by pushing off the GROUND. A plane moves forward by pushing off the AIR. However that isn't even the biggest mistake you made here. For me to run I must push off the ground. When I do this the foot that is in contact with the ground is moving backward at the speed I wish to go (10 mph to stay consistent with your argument) but since the ground won't move the rest of me does instead. I am now moving forward at 10 mph but if that ground were to move backward by the act of my foot moving it than I wouldn't move and the ground would instead (assuming it has no friction and therefore no resistance to move). On a treadmill the ground is moved for me by the motors of the machine. So now my foot is able to move back without any effort (the machine simulates the frictionless ground mentioned before) and the result is me not moving (by moving I mean traveling from point A to point B ) on the treadmill. The problem here is that the belt is not equal to my speed. The belt is greater than my speed by 10 mph since I am physically not going anywhere and the belt is moving at 10 mph. I would have to decrease the speed of the treadmill to 5 mph while maintaining the same running motion (moving my contact foot rearward at 10 mph) causing me (as a whole) to move forward physically at 5 mph. This would be when the belt speed (5 mph) is equal to my actual speed (5 mph).

QUOTE (->
QUOTE
I do realize that a plane moving through the air does have the potential for flight...so I am not deprived in the brains department...I just look at the problem a little differently, and I relate the plane's speed to the surface of the treadbelt....just like at the gym, if you want to run at 10mph, but still watch TV, you get on a treadbelt....hey, you're still running at 10mph....your speed over the treadbelt is 10mph....that is the whole point of a treadbelt ya know.


With one major problem. A person moves by pushing off the GROUND. A plane moves forward by pushing off the AIR. However that isn't even the biggest mistake you made here. For me to run I must push off the ground. When I do this the foot that is in contact with the ground is moving backward at the speed I wish to go (10 mph to stay consistent with your argument) but since the ground won't move the rest of me does instead. I am now moving forward at 10 mph but if that ground were to move backward by the act of my foot moving it than I wouldn't move and the ground would instead (assuming it has no friction and therefore no resistance to move). On a treadmill the ground is moved for me by the motors of the machine. So now my foot is able to move back without any effort (the machine simulates the frictionless ground mentioned before) and the result is me not moving (by moving I mean traveling from point A to point B ) on the treadmill. The problem here is that the belt is not equal to my speed. The belt is greater than my speed by 10 mph since I am physically not going anywhere and the belt is moving at 10 mph. I would have to decrease the speed of the treadmill to 5 mph while maintaining the same running motion (moving my contact foot rearward at 10 mph) causing me (as a whole) to move forward physically at 5 mph. This would be when the belt speed (5 mph) is equal to my actual speed (5 mph).

There are some difficulties with the wording of the problem, specifically regarding how we define speed


This is what I find difficult to grasp as to how people do not understand what speed is. It's the distance something moves divided by the time to took to move that distance.

If object A starts at point X and is still at point X 5 seconds later than it has a speed of 0 despite what is moving under it, over it and/or around it. If object A starts at point X and end up at point Y 5 meters away 5 seconds later than it has an average speed of 1 meter per second since 5 meters divided by 5 seconds equals 1 m/s.

If you chain the plane in place leaving both the wheels of the plane and the belt it sits on free to move and the belt were to move at 10 mph than the wheels will spin at a speed of 10 mph. What is the speed of the plane? 0 mph. The original question has the belt matching the speed of the PLANE not the wheels or the ground beside or the air above just the plane itself. In order for the plane itself to have speed it must physically travel from point A to point B. Where point A is one volume of space (currently occupied by the plane) and point B is another volume of space (soon to be occupied by the plane). Measure the distance between those two points and divide it by the time it takes for the plane to travel that distance. This will give you the average speed of the plane since the plane must first accelerate then decelerate causing the plane to have a different speed at different points along its travel.

Also note that in the above example I said the wheels will "spin" at 10 mph. They are not actually traveling from point A to point B. So how can it have speed by the way I explained it? Because if your were to take into consideration the thickness of the tire (say 1 inch thick) and were to only look at a 1 cubic inch piece from the whole tire that 1 cubic inch piece or section rather is occupying 1 cubic inch volume of space. Now rotate the wheel one half turn and that same cubic inch section of tire is now occupying a different cubic inch volume of space. Now unlike the plane the section of tire didn't get there in a straight line but rather took a curved path. Despite how it got there you are still taking the distance traveled and dividing by the time it took to get there.

The difference between the plane speed and tire speed is that one is called translation and the other is called rotation. The plane moving from point a to point b would be translational movement and the tire spinning on the spot continuing to occupy the same volume of space as a whole would be rotational movement. The original question refers to the first of the two since the plane isn't doing flips or barrel rolls.
egnorant
QUOTE (Atl5p+Feb 23 2007, 08:17 PM)
QUOTE (egnorant @ Feb 23 2007+ 02:15 PM )



Correct! Factoring in those items will give you a speed of 55 mph and the plane is still at 50 mph.

Adding the modifier "OVER THE SURFACE OF THE TRAILERBED" (belt..whatever)
Will yield different results.
The original question has no such modifiers.

Bruce 




So, lemme get this straight....what you are saying is this:

"A Plane moves through the air......Does the plane move though the air? "
Ans=Yes.

Ummm, ok...go grab a cookie!

Let me ask you a question to clarify....in the late 1960's, man drove a car on the surface of the moon at 5mph wheelspeed.

Question: What was his "SPEED"?

Remember..in order to remain consistant, it goes 'without saying'...you MUST relate his speed to the surface of this planet earth....so, his speedometer says '5mph'....now, what IS his "SPEED"?

And which portion of the earth' surface did you use, in relation to the moon? (closest to the moon, off to the side, far side of the earth?)


Since you have now shifted the framework of your question to the Moon,
That is our reference.
Rover is moving at 5 mph.

Bruce

P.S. The Rovers were not used until 1971.
Derek1148
Edit
jwax
My first post, and I found this thread hilarious! biggrin.gif
Owen says it best, "This would make airports a whole lot smaller".
No fly.
Precursor562
QUOTE
Let me ask you a question to clarify....in the late 1960's, man drove a car on the surface of the moon at 5mph wheelspeed.

Question: What was his "SPEED"?


Since the ground on the moon has about as much 'give' as the ground on earth (as good as none) then when the tires (in contact with such ground) spun at 5 mph this translated into the car going 5 mph. Once again the 'still' ground of the moon, earth or where ever is not the same as the 'moving' ground of a conveyor belt. One mile, distance wise, is still one mile whether you are on the earth, the moon or in outer space. It's just that in order for you to determine your position before and after in open space you will need a fixed point of reference. Also a car pushing off of the ground to move forward is not the same as a plane pushing off the air to move forward. A car is limited to how fast it can push off the ground, if that ground it moving than the car must push off the ground faster to maintain it's speed through space. A plane pushes off the air to move forward and has wheels that spin freely (as far as these low velocities are concerned) so the plane doesn't need to push off the air any faster since the belt is only affecting the speed of the plane's tires and not the surrounding air (presumed to be still).
Upisoft
QUOTE (Atl5p+Feb 23 2007, 11:17 PM)
Let me ask you a question to clarify....in the late 1960's, man drove a car on the surface of the moon at 5mph wheelspeed.

Question:  What was his "SPEED"?

Since I saw others already answered, I'll not answer, but EXTEND the question.

------------

In the same time a car is moving on Earth with the same speed but in opposite direction.

Question: What is the speed of the car (relative to Earth)? If your answer is 5mph, what is the meaning of the word opposite in that context? If your answer is not 5mph, please explain why not.
RJB
The question is equivalent to asking whether or not the plane would take off while being horizontally anchored to something stationary with a rope. The answer is...it depends. Does the air current being pushed back from the prop cross the wing (or does it simply pass over the fuselage?), AND does it create enough airflow to generate lift? Most planes rely on forward movement to generate the lift across the wings, not the relatively small airflow from the prop, but I'm not ruling that it is impossible.

The short answer, however, is that the plane will not fly.

R.J.
gmilam
QUOTE (RJB+Feb 24 2007, 11:50 PM)
The question is equivalent to asking whether or not the plane would take off while being horizontally anchored to something stationary with a rope. The answer is...it depends. Does the air current being pushed back from the prop cross the wing (or does it simply pass over the fuselage?), AND does it create enough airflow to generate lift? Most planes rely on forward movement to generate the lift across the wings, not the relatively small airflow from the prop, but I'm not ruling that it is impossible.

The short answer, however, is that the plane will not fly.

R.J.

What makes you think the plane's not moving forward?
Johan_K
First of all.. wow! I almost couldn't believe my eyes when I saw this thread.. So many postings.. 283 Thousand views.. this must be some sort of record. I'm Amazed this isn't settled already by someone drawing a diagram with some symbolic forces and what not laugh.gif (Perhaps you did.) Best Thread Ever!

I thought that if everybody else gets a go at this I'm going to pour My useless five cents into the whole discussion as well, if for nothing else just to be able to say "I added to That!" smile.gif I haven't actually bothered reading through all the seven Thousand postings, so these thoughts might be recycled goods... Here we go.

I'm getting the feeling that there are quite the number of different views on the actual setup of the whole experiment. For instance, one camp seems to take the approach where the plane is wholly or partly built with stuff from the same store you get your standard physics experiment equipment like the massless springs and the frictionless planes and what not, whilst others take the more "actual reality" point of view. I come to think of a couple of cases:

"Frictionless wheels"
It seems to me that with the "frictionless wheels" setup, there should be little argument whether the plane takes off or not. With frictionless wheels the conveyor belt won't show to be any resistance against the pulling of the plane forward, so in this case, the plane will take off... I don't think anyone is arguing about this are they? (Assuming that the engine is powerful enough to make the plane take off but I'm thinking that is sort of implied huh?). The plane takes off.

"Wheels with friction and everything made with 'tough enough' materials etc."
If I'm not completely mistaken, the friction force from the wheels will Not depend on the speed with which they're rolling? Pushing a screechy cart slow is as much a pain in the *** as pushing it fast as long as you've gotten it rolling, right? So doesn't it, in this case, come down to - "If the motor is strong enough to overcome the rolling friction, the plane will accelerate and take off"? I guess it becomes a "unstoppable bullet meets indestructible wall" kind of thing, even if assuming that the rolling friction is infinite, the planes motor can push an infinite amount of air backwards, some of that air should stream past the wings and give enough lift to make the plane take off (and boy would it take off then!). The plane takes off.

"Actual reality"
As in the first case, it seems like there should be much doubt about if the plane takes off or not? The rolling friction from the wheels (times the weight of the plane etc. ofc.) Surely much less than the force with which the motor can push/pull the plane forward. (Unless someone forgot to remove the stop-blocks or release the breaks that is). The plane take off.

Thus, I conclude (with a dramatic index-finger thrust upwards), that in all reasonably reasonable (of which the "actual reality scenario"* should be the most important one), and reasonably unreasonable cases, the plane will take off.

Thank you *taking bows*

*Of course you might come saying "but what if it's starting in quick-sand, then surely the resistance to pushing forward increases with speed!", but I hardly think that that was in the spirit of the question to begin with? There are always "what if"s.

Ps. I could of course, and probably are, completely wrong about the whole rolling friction thing laugh.gif
Precursor562
QUOTE
The question is equivalent to asking whether or not the plane would take off while being horizontally anchored to something stationary with a rope.


What are you on huh.gif
and please share cool.gif

English must not be your strong subject. The original question isn't equal to that at all. For starters in the the original question the plane can move. If the plane is anchored then it can not move which means the belt won't move either. So what you'd be left with is a tied down plane on a stationary belt not going anywhere with the engine running.
egnorant
QUOTE (Johan_K+Feb 25 2007, 10:27 AM)

*Of course you might come saying "but what if it's starting in quick-sand, then surely the resistance to pushing forward increases with speed!", but I hardly think that that was in the spirit of the question to begin with? There are always "what if"s.

Ps. I could of course, and probably are, completely wrong about the whole rolling friction thing laugh.gif

The friction question has been used by many folks to prove that the plane will not fly.

It does have a bit of the "unstoppable bullet" aspect to it.
The question does state simply that the plane moves in one direction.
And it does so despite any impediments!

Our favorite is to use 3 planes moving at 30 mph.
First plane is flying east at 30 mph.
Second plane it taxiing with 1/2 brakes on east at 30 mph.
Third plane is towing 2 other planes with full brakes on east at 30 mph.

These planes all fit the description "the plane moves in one direction" and they are
also moving at the same speed.

Should you allow anything to come along and alter the fact that "the plane moves in one direction" then the value of that fact changes also.

One of the current variations that we got (local university guys) has the same setup...BUT the belt moves perpendicular at the same speed as the plane.
I know this would require a wide belt, but deal with it!
The question is to determine what the wheels would be doing in this situation.

Correct answer is that the belt will alter what the wheels do, but not what the plane does. And it flies!

Modern Sophism lives!
Learn or stay stupid.

Bruce

P.S. Welcome to this thread...have fun!










RJB
QUOTE (gmilam+Feb 25 2007, 08:46 AM)
What makes you think the plane's not moving forward?

gmilam, the plane is not moving forward because that's how I read the original proposition - the conveyor belt adjusts its speed to keep the plane stationary. Do you interpret it differently?

R.J.
Guest
QUOTE (Precursor562+Feb 25 2007, 02:34 PM)

What are you on huh.gif
and please share  cool.gif

English must not be your strong subject.  The original question isn't equal to that at all.  For starters in the the original question the plane can move.  If the plane is anchored then it can not move which means the belt won't move either.  So what you'd be left with is a tied down plane on a stationary belt not going anywhere with the engine running.

Precursor, the forward movement of the prop via the wheels is nullified by the backward movement of the belt. They are taken out of the equation, and whether the plane flies is not related to how fast the wheels are spinning (or the friction, wheel speed, etc). It boils down to whether or not the prop generates enough air flow over the wings, and on most planes the answer is no.

If it WAS yes, however, then the plane would be able to lift off of the ground and fly "stationary" while being anchored horizontally.
egnorant
QUOTE (RJB+Feb 25 2007, 06:43 PM)
QUOTE
QUOTE (gmilam @ Feb 25 2007, 08:46 AM)
What makes you think the plane's not moving forward?



gmilam, the plane is not moving forward because that's how I read the original proposition - the conveyor belt adjusts its speed to keep the plane stationary. Do you interpret it differently?

R.J.


Are you saying that the plane is not moving or are you saying it IS moving but not in the direction that is forwards?

Bruce
Precursor562
QUOTE
the conveyor belt adjusts its speed to keep the plane stationary.


That's the problem. The belt doesn't adjust its speed to keep the plane stationary it adjusts its speed to match that of the plane (not wheel speed or thrust). With what your saying you have the belt moving under a plane that isn't moving at all. Then the belt speed would be greater than and not equal to the plane speed and such a thing goes against what is asked in the original question. They specifically mention that the belt has a control that causes the speed to match (equal to) that of the plane's speed.

Grumpy
RJB

QUOTE
the conveyor belt adjusts its speed to keep the plane stationary. Do you interpret it differently?


1) Are the brakes on the wheels on???

2) Is the plane tied to the belt???

3) How does the speed of the belt have an effect on the plane???

Answers

1) No, the wheels are free to turn, producing very little drag.

2) No, the plane is free to move except for a little rolling resistence from the free turning wheels.

So...

3) The belt can only add a little more rolling resistence(drag) to that normally encountered by the plane on takeoff. The plane easily overcomes this additional resistence(remember, the belt is LIMITED to going in the opposite direction at THE SAME SPEED as the plane, not one bit faster).

The plane easily takes off.

Grumpy cool.gif
RJB
Precursor: AHH! I see the confusion. Does the belt "match" the speed of the plane relative to the belt or relative to the ground? I interpreted the question to mean the belt - the belt moves backward just as fast as the plane is moving forward on the belt, so that the plane's speed relative to the ground (and therefore windspeed) is zero. Do you agree that it would not take off in this scenario?

R.J.
Upisoft
QUOTE (RJB+Feb 25 2007, 11:04 PM)
Precursor: AHH!  I see the confusion.  Does the belt "match" the speed of the plane relative to the belt or relative to the ground?  I interpreted the question to mean the belt - the belt moves backward just as fast as the plane is moving forward on the belt, so that the plane's speed relative to the ground (and therefore windspeed) is zero.  Do you agree that it would not take off in this scenario?

R.J.

So, your interpretation is that plane's speed is measured relatively to fixed point on the belt and belt's speed is measured relatively to fixed point on the plane.

The absolute values of these two speeds are always equal, no matter what. So, that device, mentioned in the OP that "tracks" the plane's speed is absolutely useless.

So, if the OP mentioned such a device it is required and that means the speeds have to have the possibility to differ. Thus your "reference frame", if it can be called that, for measuring speeds is against the conditions in the OP.

So, start again with valid reference frame. May I suggest that it has to be independent of the plane and the belt, unless you want one of them to have speed zero all the time.
gmilam
QUOTE (RJB+Feb 25 2007, 02:04 PM)
Precursor: AHH!  I see the confusion.  Does the belt "match" the speed of the plane relative to the belt or relative to the ground?  I interpreted the question to mean the belt - the belt moves backward just as fast as the plane is moving forward on the belt, so that the plane's speed relative to the ground (and therefore windspeed) is zero.  Do you agree that it would not take off in this scenario?

R.J.

I say this isn't possible.

The engines apply thrust to the surrounding air. The plane moves forward, causing the wheels to turn. Conveyor belt moves in the opposite direction, but no matter how it tries it can't do anything to counteract the force of the thrust...

Now the puzzle can take two roads here. Conveyor belt matches the forward motion of the plane (but in the opposite direction) causing the wheels to spin twice as fast as normal. (This is the scenario in the OP.)

Or the conveyor can try to match the speed of the wheels, thereby causing a positive feedback loop. (The movement of the belt will cause the wheels to spin faster, which causes the belt to move faster, causing the wheels to spin faster.) But the thrust will move the plane forward, even if it has to drag the smoking wheels across the evil demon possessed conveyor belt.
MDT
An airplane on the conveyer, it does not have drive wheels like a car. The thrust comes from the prop and the wheel are sort of there for balance, more than anything else. The prop will not only pull the plane forward but it will also move air over and under the wings causing lift.

A good analogy is a wagon on a robe. The wagon is on the conveyer and the wheels are in synch. If I start pulling on the rope, it doesn't matter what the relationship between the wagon and conveyer, it is coming to me since the length of rope defines the separation distance.
Precursor562
QUOTE
1) No, the wheels are free to turn, producing very little drag.

2) No, the plane is free to move except for a little rolling resistance from the free turning wheels.

So...

3) The belt can only add a little more rolling resistance(drag) to that normally encountered by the plane on takeoff. The plane easily overcomes this additional resistance(remember, the belt is LIMITED to going in the opposite direction at THE SAME SPEED as the plane, not one bit faster).

The plane easily takes off.

Grumpy cool.gif


All I have to say about this is....Nicely said. biggrin.gif

QUOTE (->
QUOTE
1) No, the wheels are free to turn, producing very little drag.

2) No, the plane is free to move except for a little rolling resistance from the free turning wheels.

So...

3) The belt can only add a little more rolling resistance(drag) to that normally encountered by the plane on takeoff. The plane easily overcomes this additional resistance(remember, the belt is LIMITED to going in the opposite direction at THE SAME SPEED as the plane, not one bit faster).

The plane easily takes off.

Grumpy cool.gif


All I have to say about this is....Nicely said. biggrin.gif

Does the belt "match" the speed of the plane relative to the belt or relative to the ground?


I say once again that to determine the speed of an object you take the distance it physically travels through space (generically and not outer space) and divide it by the time it took to travel that distance. Now how do you accurately determine how far it traveled? Well the plane is occupying a specific volume of space at the start. After X amount of time the plane is now occupying a new volume of space Y miles away. Divide Y by X and you have the average velocity of the plane. The best way to accurately judge this movement would be to triangulate the position of the plane before and after. Such triangulation is best done with the use of FIXED POINTS. Not only is the belt NOT a fixed point but it is variable whose value is depended on the value of the plane making it one thing you absolutely can not use as a point of reference to accurate determine distance traveled. Since the "ground" is presumed to be a fixed (stationary) point of reference than it can be used as such to determine distance traveled by both the plane and the belt.

QUOTE
Precursor, the forward movement of the prop via the wheels is nullified by the backward movement of the belt.


That is where you are wrong. If the wheels can't stop a plane from reaching take off speed on still ground (say 100 mph) then what makes you think that they will just because they are spinning at 200 mph on the belt? If the wheels offer X amount of resistance at 100 mph they do not offer 2X amount of resistance at 200 mph. They merely create a fraction more. This is such a small amount more that it isn't even worth mentioning. So the plane will reach the speed of 100 mph on the belt. The belt will match this speed accordingly going 100 mph in the other direction. The only contact between the belt and the plane is at the wheels. The wheels are free to turn at any speed providing next to no additional resistance to turn other than the norm that would be felt as if the plane is on still ground. The wheels will therefore spin at 200 mph which is the sum of the plane's speed and the belt's speed.

Plane speed + belt speed = wheel speed
100 mph + 100 mph = 200 mph

100 mph = 100 mph so therefore
Plane speed = Belt speed (in accordance with the original question) and the wheel speed would be 200 mph (the sum of the two).

The real resistance to move would be the plane's own mass. The wheels contribute an insignificant amount in comparison. So there would be an insignificant difference between the plane on the belt compared to off the belt since in either situation the plane is accelerating its own mass (by pushing off the air (thrust)) up to a velocity of 100 mph. The ONLY difference is the wheels spinning speed will go from 0 to 200 mph rather than 0 to 100 mph and the resistance to move placed upon the plane by the spinning of the wheels is soo small that the very little increase (brought on by doubling the spin speed) is not worth including into the whole scheme of things.

The result is a difference between on and off the belt so small that its not worth mentioning and the plane takes off the belt just as it would on still ground.

QUOTE (->
QUOTE
Precursor, the forward movement of the prop via the wheels is nullified by the backward movement of the belt.


That is where you are wrong. If the wheels can't stop a plane from reaching take off speed on still ground (say 100 mph) then what makes you think that they will just because they are spinning at 200 mph on the belt? If the wheels offer X amount of resistance at 100 mph they do not offer 2X amount of resistance at 200 mph. They merely create a fraction more. This is such a small amount more that it isn't even worth mentioning. So the plane will reach the speed of 100 mph on the belt. The belt will match this speed accordingly going 100 mph in the other direction. The only contact between the belt and the plane is at the wheels. The wheels are free to turn at any speed providing next to no additional resistance to turn other than the norm that would be felt as if the plane is on still ground. The wheels will therefore spin at 200 mph which is the sum of the plane's speed and the belt's speed.

Plane speed + belt speed = wheel speed
100 mph + 100 mph = 200 mph

100 mph = 100 mph so therefore
Plane speed = Belt speed (in accordance with the original question) and the wheel speed would be 200 mph (the sum of the two).

The real resistance to move would be the plane's own mass. The wheels contribute an insignificant amount in comparison. So there would be an insignificant difference between the plane on the belt compared to off the belt since in either situation the plane is accelerating its own mass (by pushing off the air (thrust)) up to a velocity of 100 mph. The ONLY difference is the wheels spinning speed will go from 0 to 200 mph rather than 0 to 100 mph and the resistance to move placed upon the plane by the spinning of the wheels is soo small that the very little increase (brought on by doubling the spin speed) is not worth including into the whole scheme of things.

The result is a difference between on and off the belt so small that its not worth mentioning and the plane takes off the belt just as it would on still ground.

If it WAS yes, however, then the plane would be able to lift off of the ground and fly "stationary" while being anchored horizontally.
blink.gif

For gods sake man..puff, puff, pass.


QUOTE
Or the conveyor can try to match the speed of the wheels, thereby causing a positive feedback loop. (The movement of the belt will cause the wheels to spin faster, which causes the belt to move faster, causing the wheels to spin faster.) But the thrust will move the plane forward, even if it has to drag the smoking wheels across the evil demon possessed conveyor belt.


laugh.gif It's funny because it's true.
Derek1148
QUOTE (MDT+Feb 25 2007, 11:26 PM)
An airplane on the conveyer, it does not have drive wheels like a car. The thrust comes from the prop and the wheel are sort of there for balance, more than anything else. The prop will not only pull the plane forward but it will also move air over and under the wings causing lift.

A good analogy is a wagon on a robe. The wagon is on the conveyer and the wheels are in synch. If I start pulling on the rope, it doesn't matter what the relationship between the wagon and conveyer, it is coming to me since the length of rope defines the separation distance.

Excellent analogy.
egnorant
QUOTE (RJB+Feb 25 2007, 08:04 PM)
Precursor: AHH!  I see the confusion.  Does the belt "match" the speed of the plane relative to the belt or relative to the ground?  I interpreted the question to mean the belt - the belt moves backward just as fast as the plane is moving forward on the belt, so that the plane's speed relative to the ground (and therefore windspeed) is zero.  Do you agree that it would not take off in this scenario?

R.J.

Simply by adding the phrase "in relation to the belt" alters the problem.

Suppose the original question had stated:

A plane is standing on runway that can move (some sort of band conveyor). The plane moves in one direction, while the conveyor does not move.

Your explanation allows for the belts movement to be "in relation to the plane"
This would allow the plane to get up to speed and takeoff with 0 mph wheelspeed because the belt is not moving "in relation to the plane"!

Or is this actually an example of the belt moving at the same speed in the same direction as the plane?

I do find your arguments intelligent and clever.......in relation to the belt!

Bruce
Precursor562
QUOTE
Does the belt "match" the speed of the plane relative to the belt or relative to the ground?  I interpreted the question to mean the belt - the belt moves backward just as fast as the plane is moving forward on the belt,


Ah I see what your saying now. By that however you would end up with a feedback loop. The plane starts out by going 1 mph forward so the belt matches this and goes 1 mph the other way. Since your using the belt to determine speed (wrong to do so but sure lets run with it) then the speed of the plane over the belt would be 2 mph and so the belt will speed up to 2 mph but now the speed of the plane over the belt is 3 mph and so the belt will have to increase in speed again. The speed of the plane will always be 1 mph faster than the belt. This means it would not be equal to the belt and goes against the original question. In all reality you would have a plane going at a constant speed of 1 mph and the speed of the belt forever increasing once again going against the original question.

Also as I just said the speed of the belt is a depended variable and is depended on the value (velocity) of the plane and so can not be used as a point of reference in determining the speed of the plane.
RJBeery
QUOTE (dirak+Jul 19 2005, 09:53 AM)
A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?

OK Precursor, Bruce, et al.

1. ALL movement is relative. There is no such thing as "ground is immobile but belt is a dependant variable". The plane is moving 5 mph in some direction relative to the ground, or belt, or river water, or moon, etc - whatever we specify (or is reasonable to assume). The plane rests on the belt, so I assumed the plane's speed is given relative to the belt. The belt rests on the ground, so I assumed the belt's speed is given relative to the ground.

2. If the "plane speed" meant airspeed (i.e. speed relative to the ground assuming no wind) in the original post, then I agree that it would indeed take off. I believe that the original poster meant wheel speed, however, which would mean speed relative to the belt.

3. Yes, my interpretation would cause an infinite feedback loop of ever-increasing wheel/belt speed (limited by the prop/engine?). Nevertheless, the plane would not take off unless the prop pushed enough air over the wing to allow it to create lift, since it's airspeed would remain zero. Props do not do this, and I would liken it to blowing on a sail to make your boat move.

4. If the plane did indeed take off due to nothing but the air flow from the prop, then it would also be able to sustain lift with a horizontally-tied rope anchor. Yes this sounds ridiculous, and so is the idea that the plane would fly.

5. No, I'm not smoking anything. And no, I won't share. tongue.gif

6. Rather than convince me what the original poster meant, analyze the problem with my interpretation. I've already agreed with you that the plane would fly if "plane speed" meant airspeed.
adoucette
QUOTE
I believe that the original poster meant wheel speed


Except the OP never mentions "Wheel Speed".

It does however say: "THE PLANE MOVES".

How do you interpet that as "The wheels spin"?

Arthur
RJBeery
Adoucette, read #1 above. If you're in a speed boat, making a huge wake, I assume you think you're travelling. What if you're moving upstream and the land isn't moving?

Next, and more importantly, read #6 above.
Derek1148
QUOTE (RJBeery+Feb 26 2007, 03:39 AM)
Adoucette, read #1 above.  If you're in a speed boat, making a huge wake, I assume you think you're travelling.  What if you're moving upstream and the land isn't moving?

Next, and more importantly, read #6 above.

RJBeery,

Even ignoring the fact that the original post limits the speed of the conveyor to that of the speed of the plane (however, in the opposite direction). Once the plane’s engines begin to propel the plane forward, the movement of the conveyor, regardless of the speed, could not preclude the forward motion of the plane.

The movement of the conveyor would not impede the physics that cause a plane to move forward. The wheels of the plane would simply rotate faster or skid along the conveyor's surface. Either way the plane would have forward motion.

If the intent of the author of the original post was to imply that the movement of the conveyor could cause the plane to remain stationary, then the author created a paradox. It is contrary to basic physics.
Grumpy
RJBeery

Since both the plane and the belt move in the OP, neither can be used as a point of reference for the other or nothing can move at all, violating the OP.(without creating a positive feedback loop where the belt accelerates to much faster speeds than the aircraft, also violating the OP)

Why do you ASSUME that aircraft speed means wheelspeed, when no aircraft measures it's speed by the speed of it's wheels??? Aircraft speed is ALWAYS measured by airspeed, though groundspeed is calculated to use in navigation, it has nothing to do with whether the aircraft flies or not.

QUOTE
If you're in a speed boat, making a huge wake, I assume you think you're travelling. What if you're moving upstream and the land isn't moving?


Where is the motive power of the speedboat applied??? If an airplane can fly at 25 mph and is headed into the wind which is blowing at 25 mph, how fast is the ground moving under the plane??? Does the aircraft care that it is not moving over the ground??? If that ground was a tread belt moving at 300 mph, would it make the least bit of difference to the plane???

Grumpy cool.gif
RJBeery
Derek,

I understand your thinking. You think that the wheels are free-spinning, frictionless and therefore irrelevant. Turn the question around though:

A plane sits on a conveyor belt that begins to move backwards. The pilot starts the engine and applies thrust to compensate for his backward movement. As the belt moves faster and faster, the pilot continues to add thrust to compensate. Will this plane ever take off as his ground speed remains zero?

This is NOT a paradox. It is also, I maintain, the original poster's intention.

R.J.
Derek1148
QUOTE (RJBeery+Feb 26 2007, 05:07 AM)
Derek,

I understand your thinking. You think that the wheels are free-spinning, frictionless and therefore irrelevant. Turn the question around though:

A plane sits on a conveyor belt that begins to move backwards. The pilot starts the engine and applies thrust to compensate for his backward movement. As the belt moves faster and faster, the pilot continues to add thrust to compensate. Will this plane ever take off as his ground speed remains zero?

This is NOT a paradox. It is also, I maintain, the original poster's intention.

R.J.

Once the plane’s propulsion system starts the forward motion of the plane, the movement of the conveyor in the opposite direction is irrelevant.

Further, how can movement of the plane be interpreted as a rotation of its wheels? Suppose the plane was on skis, would the revving of its engine be interpreted as speed and movement?

If one accepts your interpretation of the original post, then it is in fact a paradox.

For the third time: Reference: Essays in Science (Albert Einstein); “The movement of a vehicle is considered in reference to the earth’s surface, that of a planet to the totality of the visible fixed stars.”

egnorant
QUOTE (RJBeery+Feb 26 2007, 02:32 AM)
1.  ALL movement is relative.  There is no such thing as "ground is immobile but belt is a dependant variable".  The plane is moving 5 mph in some direction relative to the ground, or belt, or river water, or moon, etc - whatever we specify (or is reasonable to assume).  The plane rests on the belt, so I assumed the plane's speed is given relative to the belt.  The belt rests on the ground, so I assumed the belt's speed is given relative to the ground.

Movement is based on what it is sitting on......incorrect.

A plane is on a trailer that is moving at 50 mph...the plane is also moving at 50 mph.

QUOTE
2.  If the "plane speed" meant airspeed (i.e. speed relative to the ground assuming no wind) in the original post, then I agree that it would indeed take off.  I believe that the original poster meant wheel speed, however, which would mean speed relative to the belt.


This is called reading information into something that is not there.

QUOTE (->
QUOTE
2.  If the "plane speed" meant airspeed (i.e. speed relative to the ground assuming no wind) in the original post, then I agree that it would indeed take off.  I believe that the original poster meant wheel speed, however, which would mean speed relative to the belt.


This is called reading information into something that is not there.

3.  Yes, my interpretation would cause an infinite feedback loop of ever-increasing wheel/belt speed (limited by the prop/engine?).  Nevertheless, the plane would not take off unless the prop pushed enough air over the wing to allow it to create lift, since it's airspeed would remain zero.  Props do not do this, and I would liken it to blowing on a sail to make your boat move.


In order to begin this infinite feedback loop the plane would need to be moving FASTER than the belt at some point.
This violates the original question and is not a valid answer.

QUOTE
4.  If the plane did indeed take off due to nothing but the air flow from the prop, then it would also be able to sustain lift with a horizontally-tied rope anchor.  Yes this sounds ridiculous, and so is the idea that the plane would fly.


Doesn't sound ridiculous compared to the scenarios dreamed up in the past 400 odd pages.
It is incorrect though.

QUOTE (->
QUOTE
4.  If the plane did indeed take off due to nothing but the air flow from the prop, then it would also be able to sustain lift with a horizontally-tied rope anchor.  Yes this sounds ridiculous, and so is the idea that the plane would fly.


Doesn't sound ridiculous compared to the scenarios dreamed up in the past 400 odd pages.
It is incorrect though.

5.  No, I'm not smoking anything.  And no, I won't share.  tongue.gif


Don't care...not relevant

QUOTE
6.  Rather than convince me what the original poster meant, analyze the problem with my interpretation.  I've already agreed with you that the plane would fly if "plane speed" meant airspeed.


Rather than give us your assumption of what the original poster meant,
read back and the "problem with your interpretation" is exposed about every 5 pages...often several times on the same page.
The "problem with your interpretation" has been severely scrutinized and every attempt has been made to try to make it fit the original question.
Result is that your interpretation does not fit the original question unless certain modifiers are added that change the question.

Your interpretation is not new...or innovative...or clever.
It just incorrect.

Bruce
jacobsg
QUOTE (Atl5p+Feb 23 2007, 08:17 PM)
So, lemme get this straight....what you are saying is this:

"A Plane moves through the air......Does the plane move though the air? "
Ans=Yes.

Ummm, ok...go grab a cookie!

So, lemme get this straight....what you are saying is this:

"A Plane [doesn't] move[s] through the air......Does the plane move though the air? "
Ans=[No]

Ummm, ok...go grab a cookie!
gmilam
QUOTE (RJBeery+Feb 25 2007, 09:39 PM)
Adoucette, read #1 above. If you're in a speed boat, making a huge wake, I assume you think you're travelling. What if you're moving upstream and the land isn't moving?

If the land isn't moving then I assume I'm not going anywhere. Size of the wake makes no difference. (Silly me - I associate forward movement with going somewhere.)

Same with my car. My tires may be spinning wildly, but if I'm stuck in the mud - I ain't moving.
Atl5p
QUOTE (adoucette+Feb 23 2007, 05:50 PM)
Nope,

YOU got yourself into this sidetrack by THIS:

QUOTE (Atl5p+)
the plane’s thrust will decrease due to the artificial ‘headwind’ it creates by moving through the air.


Which is why I asked you if you had ever heard of Variable Pitch Props.

Then you said

QUOTE (Atl5p+)
the THRUST needed to get moving is in the DIAMETER of the prop..


And I showed you how Thrust can vary by PITCH for the same Diameter and RPM.

You're lost Atl5p.

Everybody and his brother has explained it to you but of course YOU are right and the world is wrong.

laugh.gif

See you in another year or so.

Arthur

Dude, for a given motor, THRUST is in the diameter of the prop used, and top speed is in the PITCH of the prop.

Are you saying that this is false?

What you did was increase the pitch of a prop, and kept the RPM's the same. The ONLY way you can do this is to use a different motor each time. When you used the BIGGER MOTOR, then the increased THRUST is from the fact that the MOTOR IS BIGGER/More Powerful....

Remember, we're talking about 'Variable Pitch Props' here, which would seem to indicate that the pitch is changed while in-flight, which means the motor CANNOT be changed while IN FLIGHT, meaning that the SAME MOTOR must be used when checking each prop, and if that happens, there is no way you can spin a more agressive prop at the same RPM.

This is it: all things being equal, for a given motor....if you want to increase THRUST, you must get a larger diameter prop....this will also usually entail LESS pitch to compensate for the increased load from the larger diameter prop. This will give you MORE THRUST and LESS TOP SPEED.

Conversly, if you wanted to increase the TOP SPEED, you would use a prop with greater pitch, and usually you have to decrease the diameter to compensate...this will give you GREATER TOP SPEED and LESS THRUST.

Therefore, for a given motor, THRUST is in the DIAMETER of the prop.
SPEED is in the PITCH of the prop.

Now, if you want to disagree with the above, by simply installing bigger and more powerful motors each time, to come up with an increased pitch yeilding increased THRUST.....then you have done so by simply changing out the motor. You have cheated. It is the exact same deal as your entire argument. It is expected.
Have a nice year off!!!
Atl5p
QUOTE (Grumpy+Feb 23 2007, 06:06 PM)
Atl5p

What kind of childish display is this???



I will just say that anyone who does not "get it" is simply wrong. The plane flies, and it flies easily, no matter what the treadmill does(within the boundaries of the OP)





I don't recall anyone saying anything about full power.

If I have a motor that can turn a 11X8 at 10,000 rpm, it BETTER damn well turn an 11X4 at that same rpm. In fact my Tribute has a controller that turns the motor at a steady RPM, the thrust is varied by adjusting the props blades(I have just as much thrust in a forward direction as I do in reverse. All I do is flip the throttle up switch and put the throttle stick in the middle for 0 thrust.

You can't get a Tribute here anymore(mine's three years old) but this is the same type VPP I have. The Animal

User posted image

User posted image

http://www.hobby-lobby.com/animal.htm

Grumpy cool.gif

PS Some very good videos here.

So, you blatently refuse to answer the question...can a treadbelt prevent a plane from taking off...you simply refuse....hey, that is your right....you're a coward. No prob.

About the props...you have clearly avoided this question as well...the question did not have us going BACKWARDS in prop size, we were going FORWARDS.

So, once again....if a motor can spin a 11x4 prop at 10,000rpm with max power...will that same motor be able to spin a 11x8 prop at 10,000 rpm? If you can't understand the correct answer to this, or how it relates to the issue at hand, then you have no real contribution to make to me.
Atl5p
QUOTE (Precursor562+Feb 23 2007, 08:58 PM)

With one major problem. A person moves by pushing off the GROUND. A plane moves forward by pushing off the AIR. However that isn't even the biggest mistake you made here. For me to run I must push off the ground. When I do this the foot that is in contact with the ground is moving backward at the speed I wish to go (10 mph to stay consistent with your argument) but since the ground won't move the rest of me does instead. I am now moving forward at 10 mph but if that ground were to move backward by the act of my foot moving it than I wouldn't move and the ground would instead (assuming it has no friction and therefore no resistance to move). On a treadmill the ground is moved for me by the motors of the machine. So now my foot is able to move back without any effort (the machine simulates the frictionless ground mentioned before) and the result is me not moving (by moving I mean traveling from point A to point B ) on the treadmill. The problem here is that the belt is not equal to my speed. The belt is greater than my speed by 10 mph since I am physically not going anywhere and the belt is moving at 10 mph. I would have to decrease the speed of the treadmill to 5 mph while maintaining the same running motion (moving my contact foot rearward at 10 mph) causing me (as a whole) to move forward physically at 5 mph. This would be when the belt speed (5 mph) is equal to my actual speed (5 mph).



This is what I find difficult to grasp as to how people do not understand what speed is. It's the distance something moves divided by the time to took to move that distance.

If object A starts at point X and is still at point X 5 seconds later than it has a speed of 0 despite what is moving under it, over it and/or around it. If object A starts at point X and end up at point Y 5 meters away 5 seconds later than it has an average speed of 1 meter per second since 5 meters divided by 5 seconds equals 1 m/s.

If you chain the plane in place leaving both the wheels of the plane and the belt it sits on free to move and the belt were to move at 10 mph than the wheels will spin at a speed of 10 mph. What is the speed of the plane? 0 mph. The original question has the belt matching the speed of the PLANE not the wheels or the ground beside or the air above just the plane itself. In order for the plane itself to have speed it must physically travel from point A to point B. Where point A is one volume of space (currently occupied by the plane) and point B is another volume of space (soon to be occupied by the plane). Measure the distance between those two points and divide it by the time it takes for the plane to travel that distance. This will give you the average speed of the plane since the plane must first accelerate then decelerate causing the plane to have a different speed at different points along its travel.

Also note that in the above example I said the wheels will "spin" at 10 mph. They are not actually traveling from point A to point B. So how can it have speed by the way I explained it? Because if your were to take into consideration the thickness of the tire (say 1 inch thick) and were to only look at a 1 cubic inch piece from the whole tire that 1 cubic inch piece or section rather is occupying 1 cubic inch volume of space. Now rotate the wheel one half turn and that same cubic inch section of tire is now occupying a different cubic inch volume of space. Now unlike the plane the section of tire didn't get there in a straight line but rather took a curved path. Despite how it got there you are still taking the distance traveled and dividing by the time it took to get there.

The difference between the plane speed and tire speed is that one is called translation and the other is called rotation. The plane moving from point a to point b would be translational movement and the tire spinning on the spot continuing to occupy the same volume of space as a whole would be rotational movement. The original question refers to the first of the two since the plane isn't doing flips or barrel rolls.

Look we don't need a dicertation about how feet push off a treadbelt...

This is all about HOW SPEED IS MEASURED.

The point is this...a runner on a treadbelt is ON THE TREADBELT....he determines his speed via his relationship to the surface of the treadbelt.

Same with anything on a treadbelt....it has SPEED reletive to the treadbelt's surface...regardelss of how that 'thing' is propelled....it does have a 'speed' which can be 'related' to the surface of that belt...and when something is traversing a moving treadbelt, it is customary to related the speed in this way.

Speed is "It's the distance something moves divided by the time to took to move that distance" over the surface of the treadbelt.

Do you not realize that a treadbelt can have a 'Point a' and 'Point B' and that a plane can 'move' between these two points, and that the time it take for said to happen is the SPEED?
Atl5p
QUOTE (egnorant+Feb 23 2007, 11:49 PM)



So, lemme get this straight....what you are saying is this:

"A Plane moves through the air......Does the plane move though the air? "
Ans=Yes.

Ummm, ok...go grab a cookie!

Let me ask you a question to clarify....in the late 1960's, man drove a car on the surface of the moon at 5mph wheelspeed.

Question: What was his "SPEED"?

Remember..in order to remain consistant, it goes 'without saying'...you MUST relate his speed to the surface of this planet earth....so, his speedometer says '5mph'....now, what IS his "SPEED"?

And which portion of the earth' surface did you use, in relation to the moon? (closest to the moon, off to the side, far side of the earth?)[/QUOTE]

Since you have now shifted the framework of your question to the Moon,
That is our reference.
Rover is moving at 5 mph.

Bruce

P.S. The Rovers were not used until 1971.

Why did the fact that the rover was on the moon, all of a sudden require us to use the moon as the 'framework'? Why are you still not the 'outside observer' standing on and relating all things to, planet earth?

After you answer that, tell me...So, why can't we shift the framework of the OP to the treadbelt, and use the 'speed over the treadbelt'?

What is it about an object 'standing on' a certian terra firma that leads us to setting the framework to that certian terra firma?
Atl5p
QUOTE (Precursor562+Feb 24 2007, 09:26 AM)

Since the ground on the moon has about as much 'give' as the ground on earth (as good as none) then when the tires (in contact with such ground) spun at 5 mph this translated into the car going 5 mph. Once again the 'still' ground of the moon, earth or where ever is not the same as the 'moving' ground of a conveyor belt. One mile, distance wise, is still one mile whether you are on the earth, the moon or in outer space. It's just that in order for you to determine your position before and after in open space you will need a fixed point of reference. Also a car pushing off of the ground to move forward is not the same as a plane pushing off the air to move forward. A car is limited to how fast it can push off the ground, if that ground it moving than the car must push off the ground faster to maintain it's speed through space. A plane pushes off the air to move forward and has wheels that spin freely (as far as these low velocities are concerned) so the plane doesn't need to push off the air any faster since the belt is only affecting the speed of the plane's tires and not the surrounding air (presumed to be still).

So, what do you mean about the moon has the same 'give' as the earth? And what does that have to do with where the speed of the rover is based, earth or moon?

QUOTE
One mile, distance wise, is still one mile whether you are on the earth, the moon or in outer space.

If this is so, (which I believe it is), then why isn't one mile distance one mile on the surface of a treadbelt?

Again, what you relate the speed to has NOTHING to do with tires or feet 'pushing' off a certian surface vs rolling over the surface.

ALso, have you figured out what Static Thrust is, and how that relates to the OP, as opposed to your 'feet pushing' theory?

Atl5p
QUOTE (RJB+Feb 25 2007, 12:50 AM)
The question is equivalent to asking whether or not the plane would take off while being horizontally anchored to something stationary with a rope. The answer is...it depends. Does the air current being pushed back from the prop cross the wing (or does it simply pass over the fuselage?), AND does it create enough airflow to generate lift? Most planes rely on forward movement to generate the lift across the wings, not the relatively small airflow from the prop, but I'm not ruling that it is impossible.

The short answer, however, is that the plane will not fly.

R.J.

I like this responce. I'll also give you the 'Fly Boy' answer to this alter-question.

'A plane is tied to the earth with an unbreakable chain...the plane moves one direction, and the chain 'pulls' in the opposite direction....the chain will match the plane's speed exactly opposite. Does the plane take off and fly?'

FlyBoy says 'The question says the plane MOVES, so it MOVES through the air, therefore it FLYs!!!....the unbreakable chain tethering it to the ground is irelevent."
Atl5p
QUOTE (gmilam+Feb 25 2007, 06:25 PM)
I say this isn't possible.

The engines apply thrust to the surrounding air. The plane moves forward, causing the wheels to turn. Conveyor belt moves in the opposite direction, but no matter how it tries it can't do anything to counteract the force of the thrust...

Now the puzzle can take two roads here. Conveyor belt matches the forward motion of the plane (but in the opposite direction) causing the wheels to spin twice as fast as normal. (This is the scenario in the OP.)

Or the conveyor can try to match the speed of the wheels, thereby causing a positive feedback loop. (The movement of the belt will cause the wheels to spin faster, which causes the belt to move faster, causing the wheels to spin faster.) But the thrust will move the plane forward, even if it has to drag the smoking wheels across the evil demon possessed conveyor belt.

There is no such thing as the "positive feedback loop"

There WILL be a speed at which the negetive force from the treadbelt to the wheeled vehicle WILL equal the thrust of the airpropelled vehicle.

Again, the "positive feedback loop" idea is fantasy.
Atl5p
QUOTE (Precursor562+Feb 25 2007, 06:27 PM)

All I have to say about this is....Nicely said. biggrin.gif



I say once again that to determine the speed of an object you take the distance it physically travels through space (generically and not outer space) and divide it by the time it took to travel that distance. Now how do you accurately determine how far it traveled? Well the plane is occupying a specific volume of space at the start. After X amount of time the plane is now occupying a new volume of space Y miles away. Divide Y by X and you have the average velocity of the plane. The best way to accurately judge this movement would be to triangulate the position of the plane before and after. Such triangulation is best done with the use of FIXED POINTS. Not only is the belt NOT a fixed point but it is variable whose value is depended on the value of the plane making it one thing you absolutely can not use as a point of reference to accurate determine distance traveled. Since the "ground" is presumed to be a fixed (stationary) point of reference than it can be used as such to determine distance traveled by both the plane and the belt.



That is where you are wrong. If the wheels can't stop a plane from reaching take off speed on still ground (say 100 mph) then what makes you think that they will just because they are spinning at 200 mph on the belt? If the wheels offer X amount of resistance at 100 mph they do not offer 2X amount of resistance at 200 mph. They merely create a fraction more. This is such a small amount more that it isn't even worth mentioning. So the plane will reach the speed of 100 mph on the belt. The belt will match this speed accordingly going 100 mph in the other direction. The only contact between the belt and the plane is at the wheels. The wheels are free to turn at any speed providing next to no additional resistance to turn other than the norm that would be felt as if the plane is on still ground. The wheels will therefore spin at 200 mph which is the sum of the plane's speed and the belt's speed.

Plane speed + belt speed = wheel speed
100 mph + 100 mph = 200 mph

100 mph = 100 mph so therefore
Plane speed = Belt speed (in accordance with the original question) and the wheel speed would be 200 mph (the sum of the two).

The real resistance to move would be the plane's own mass. The wheels contribute an insignificant amount in comparison. So there would be an insignificant difference between the plane on the belt compared to off the belt since in either situation the plane is accelerating its own mass (by pushing off the air (thrust)) up to a velocity of 100 mph. The ONLY difference is the wheels spinning speed will go from 0 to 200 mph rather than 0 to 100 mph and the resistance to move placed upon the plane by the spinning of the wheels is soo small that the very little increase (brought on by doubling the spin speed) is not worth including into the whole scheme of things.

The result is a difference between on and off the belt so small that its not worth mentioning and the plane takes off the belt just as it would on still ground.

blink.gif

For gods sake man..puff, puff, pass.




laugh.gif It's funny because it's true.

So now we have a BRAND NEW definition of SPEED.

First it was 'Distance traveled between TWO POINTS, divided by time'

NOW, it is 'Distance travled THROUGH SPACE'.

So, if the earth is moving 17,000 mph though orbit, and I am moving 5mph over it's surface, then I am really going 17,005 mph 'Through Space'....right?



adoucette
QUOTE (Atl5p+Feb 26 2007, 02:00 PM)
Dude, for a given motor, THRUST is in the diameter of the prop used, and top speed is in the PITCH of the prop.

Are you saying that this is false?

What you did was increase the pitch of a prop, and kept the RPM's the same. The ONLY way you can do this is to use a different motor each time. When you used the BIGGER MOTOR, then the increased THRUST is from the fact that the MOTOR IS BIGGER/More Powerful....

Remember, we're talking about 'Variable Pitch Props' here, which would seem to indicate that the pitch is changed while in-flight, which means the motor CANNOT be changed while IN FLIGHT, meaning that the SAME MOTOR must be used when checking each prop, and if that happens, there is no way you can spin a more agressive prop at the same RPM.

This is it: all things being equal, for a given motor....if you want to increase THRUST, you must get a larger diameter prop....this will also usually entail LESS pitch to compensate for the increased load from the larger diameter prop. This will give you MORE THRUST and LESS TOP SPEED.

Conversly, if you wanted to increase the TOP SPEED, you would use a prop with greater pitch, and usually you have to decrease the diameter to compensate...this will give you GREATER TOP SPEED and LESS THRUST.

Therefore, for a given motor, THRUST is in the DIAMETER of the prop.
SPEED is in the PITCH of the prop.

Now, if you want to disagree with the above, by simply installing bigger and more powerful motors each time, to come up with an increased pitch yeilding increased THRUST.....then you have done so by simply changing out the motor. You have cheated. It is the exact same deal as your entire argument. It is expected.
Have a nice year off!!!

Nope, when YOU NOW ADD for a given motor, that second statement is mostly correct, except there is a point of diminishing returns that is quickly reached by increasing prop size.

But you continue to IGNORE the ORIGINAL issue:

The ISSUE was YOUR post that said:

QUOTE (Atl5p+)
the plane’s thrust will decrease due to the artificial ‘headwind’ it creates by moving through the air. 


And I pointed out that this issue IS RESOLVED by a Variable Pitch Prop.

The other issue is not germaine to the problem but simply to correct your statement that implied the ONLY thing that determined THRUST was Diameter of the Prop.

Oh, though I'll probably not be posting on this thread, I do enjoy coming here now and again and watching you continue to flounder trying to prop up your untennable position.

Oh, I know, you're right, the rest of the world is wrong.

laugh.gif

Arthur
Atl5p
QUOTE (RJBeery+Feb 25 2007, 09:32 PM)
OK Precursor, Bruce, et al.

1. ALL movement is relative. There is no such thing as "ground is immobile but belt is a dependant variable". The plane is moving 5 mph in some direction relative to the ground, or belt, or river water, or moon, etc - whatever we specify (or is reasonable to assume). The plane rests on the belt, so I assumed the plane's speed is given relative to the belt. The belt rests on the ground, so I assumed the belt's speed is given relative to the ground.

2. If the "plane speed" meant airspeed (i.e. speed relative to the ground assuming no wind) in the original post, then I agree that it would indeed take off. I believe that the original poster meant wheel speed, however, which would mean speed relative to the belt.

3. Yes, my interpretation would cause an infinite feedback loop of ever-increasing wheel/belt speed (limited by the prop/engine?). Nevertheless, the plane would not take off unless the prop pushed enough air over the wing to allow it to create lift, since it's airspeed would remain zero. Props do not do this, and I would liken it to blowing on a sail to make your boat move.

4. If the plane did indeed take off due to nothing but the air flow from the prop, then it would also be able to sustain lift with a horizontally-tied rope anchor. Yes this sounds ridiculous, and so is the idea that the plane would fly.

5. No, I'm not smoking anything. And no, I won't share. tongue.gif

6. Rather than convince me what the original poster meant, analyze the problem with my interpretation. I've already agreed with you that the plane would fly if "plane speed" meant airspeed.

There is no 'Infinite Feedback Loop'.

Increasing the speed of the treadbelt WILL increase the 'backwards force' on the plane....spinning the treadbelt fast enough WILL cause this 'backwards force' to eventually equal the 'Max Static Thrust' of the plane.

WHEN THIS OCCURS, the plane's speed over the treadbelt WILL MATCH EXACTLY the speed of the treadbelt past it's motor.

You are correct....it's just that there is no 'infinite loop'.
Johan_K
QUOTE (Atl5p+)
Increasing the speed of the treadbelt WILL increase the 'backwards force' on the plane....spinning the treadbelt fast enough WILL cause this 'backwards force' to eventually equal the 'Max Static Thrust' of the plane.


Wait, isn't that wrong? The rolling resistance just depends on the weight of the plane and the coefficient of rolling friction..

For instance the "Cessna 172 Q Cutlass" (Link) has a mass of 2550 lbs (seems like a pretty standard 'small' aircraft no? It serves as an example at least.) Apparently the coefficient for rolling friction for rubber tires on concrete (according to my physics textbook (Sears and Zemansky's - University physics with modern physics, 10th edition. Young & Freedman. p137)), is between 0.01 and 0.02. Put the numbers in the cruncher (lets pick the higher friction coefficient just for the hell of it):

Frictional force = (2550/2.2)*9,81*0.02 = 227.41N

But surely the plane can produce enough thrust to overcome a measly 228 Newton? Even I can do that biggrin.gif

Or do you mean that there is a forward force that the plane can "produce" such that it will exactly match the weight*'coefficient for rolling friction' so that it will not move on the treadmill?

I might have misunderstood and/or not read your post carefully enough in which case I do like a politician and take it back
biggrin.gif
Atl5p
QUOTE (RJBeery+Feb 26 2007, 12:07 AM)
Derek,

I understand your thinking. You think that the wheels are free-spinning, frictionless and therefore irrelevant. Turn the question around though:

A plane sits on a conveyor belt that begins to move backwards. The pilot starts the engine and applies thrust to compensate for his backward movement. As the belt moves faster and faster, the pilot continues to add thrust to compensate. Will this plane ever take off as his ground speed remains zero?

This is NOT a paradox. It is also, I maintain, the original poster's intention.

R.J.

You are exactly correct sir....

The 'Fly Boys' would tell you this:

"The treadbelt moves the plane backwards...then you crank the engine so you remain at 0 IAS.....the belt moves faster and faster, but you do NOT have to increase throttle, you will stay at 0 IAS. The treadbelt goes faster still, and you STILL do not have to increase throttle to remain at 0 IAS. The treadbelt is now going 500mph and you STILL don't have to increase the throttle....in fact, you've been going +5mph IAS the entire time. In fact, the propeller is only spinning at 20 rpm, and it gives the plane 5mph IAS while the belt spins at 500mph"

And they would be WRONG. It is BS!
Atl5p
QUOTE (Johan_K+Feb 26 2007, 03:46 PM)
QUOTE (Atl5p+)
Increasing the speed of the treadbelt WILL increase the 'backwards force' on the plane....spinning the treadbelt fast enough WILL cause this 'backwards force' to eventually equal the 'Max Static Thrust' of the plane.


Wait, isn't that wrong? The rolling resistance just depends on the weight of the plane and the coefficient of rolling friction..

For instance the "Cessna 172 Q Cutlass" (Link) has a mass of 2550 lbs (seems like a pretty standard 'small' aircraft no? It serves as an example at least.) Apparently the coefficient for rolling friction for rubber tires on concrete (according to my physics textbook (Sears and Zemansky's - University physics with modern physics, 10th edition. Young & Freedman. p137)), is between 0.01 and 0.02. Put the numbers in the cruncher (lets pick the higher friction coefficient just for the hell of it):

Frictional force = (2550/2.2)*9,81*0.02 = 227.41N

But surely the plane can produce enough thrust to overcome a measly 228 Newton? Even I can do that biggrin.gif

Or do you mean that there is a forward force that the plane can "produce" such that it will exactly match the weight*'coefficient for rolling friction' so that it will not move on the treadmill?

I might have misunderstood and/or not read your post carefully enough in which case I do like a politician and take it back
biggrin.gif

Could you then explain this for me?

http://auto.howstuffworks.com/tire4.htm

QUOTE
Let's figure out how much force a typical car might use to push its tires down the road. Let's say our car weighs 4,000 pounds (1814.369 kg), and the tires have a CRF of 0.015. The force is equal to 4,000 x 0.015, which equals 60 pounds (27.215 kg). Now let's figure out how much power that is. If you've read the HowStuffWorks article How Force, Torque, Power and Energy Work, you know that power is equal to force times speed. So the amount of power used by the tires depends on how fast the car is going. At 75 mph (120.7 kph), the tires are using 12 horsepower, and at 55 mph (88.513 kph) they use 8.8 horsepower. All of that power is turning into heat. Most of it goes into the tires, but some of it goes into the road (the road actually bends a little when the car drives over it).

From these calculations you can see that the three things that affect how much force it takes to push the tire down the road (and therefore how much heat builds up in the tires) are the weight on the tires, the speed you drive and the CRF (which increases if pressure is decreased).


I DO mean to say "that there is a forward force that the plane can "produce" such that it will exactly match the weight*'coefficient for rolling friction' so that it will not move on the treadmill"
.....AAAAnnnnddd; when the treadbelt is slowed, the plane will have + IAS, and when the treadbelt is made to go faster, said plane will have (-) IAS.
OR...Adding weight to the 0 IAS plane will then cause (-) IAS, and removing weight from said 0 IAS plane will cause + IAS.
FINALLY....Decreasing throttle on plane at 0 IAS will cause plane to have (-) IAS, and Increasing throttle on said 0 IAS plane will cause the plane to have + IAS.

I performed all of these tests with a small model plane and an real life treadbelt....oh, and the plane would propell itself just fine on static ground.

Or, I may have misunderstood your post....were you trying to explain the amount of force necessary to BEGIN rolling those tires? Because I was talking about the increasing force brought on by increasing the speed under those rolling tires....I was not talking about the force necessary to bring those tires from a dead stop to some sort of movement....again, I'm talking about increasing backwards force through increased backwards speed of treadbelt.
Grumpy
Atl5p

QUOTE
So, you blatently refuse to answer the question...can a treadbelt prevent a plane from taking off...you simply refuse....hey, that is your right....you're a coward. No prob.


This is just a flat out lie. I have told you many times that the plane will fly, there is nothing(within the OP) that the treadmill can do about it(other than spin the wheels of the plane faster). The aircraft could care less what the belt does, because the belt has no method of keeping the plane from moving, NO MATTER HOW YOU MEASURE THE SPEED. You even admitted that fact when the wheels were replaced by hover pads. Wheels have A LITTLE BIT of drag compared to the ZERO DRAG of the hover pads, but it is still a very small fraction of the power available in any normal plane.

QUOTE (->
QUOTE
So, you blatently refuse to answer the question...can a treadbelt prevent a plane from taking off...you simply refuse....hey, that is your right....you're a coward. No prob.


This is just a flat out lie. I have told you many times that the plane will fly, there is nothing(within the OP) that the treadmill can do about it(other than spin the wheels of the plane faster). The aircraft could care less what the belt does, because the belt has no method of keeping the plane from moving, NO MATTER HOW YOU MEASURE THE SPEED. You even admitted that fact when the wheels were replaced by hover pads. Wheels have A LITTLE BIT of drag compared to the ZERO DRAG of the hover pads, but it is still a very small fraction of the power available in any normal plane.

About the props...you have clearly avoided this question as well...the question did not have us going BACKWARDS in prop size, we were going FORWARDS.

So, once again....if a motor can spin a 11x4 prop at 10,000rpm with max power...will that same motor be able to spin a 11x8 prop at 10,000 rpm? If you can't understand the correct answer to this, or how it relates to the issue at hand, then you have no real contribution to make to me.


Once again, who said anything about maximum power??? I didn't, neither did adoucette. No need to change the motor, just install one that give the necessary rpm no matter what the pitch.

If my motor can spin a 11X8 prop at 10,000 rpm, why would it not be able to spin an 11X4 prop at the same rpm??? This is why they put variable props on aircraft, so the prop can produce the same thrust as the plane goes faster or, conversely, so the prop can adjust to produce maximum thrust at takeoff speeds without overreving(the tips of the props approaching the speed of sound and becoming very inefficient, not to mention loud). Therefore the engine can turn the prop at it's most efficient speed(10,000 in this case) and the thrust is matched to the speed of the plane by adjusting the pitch.

All turbo props operate this way. They have a ground power setting for startups and taxing, and a flight power setting for takeoff and flying. When set to flight power the turbine spins the prop at a set rpm which stays the same all the time, thrust is determined by prop pitch(they even have a beta range of reverse thrust for braking during landing and ground handling(to back up the plane)).

My VPP prop on my Trbute will stop the plane while nose down and even climb backwards. It maintains the same rpm the whole time(helicopters do too)

So it is obvious you know nothing about aircraft(or tread belts for that matter), you're just a troll. One should not feed a troll(but it sure is fun to poke them through the bars of their cages!!!) and you have certainly wasted enough of my time and the bandwidth you waste on this forum. I would urge all posters to now leave the troll alone in his...utter lack of logical or coherent thought(not to mention, childish displays of NAH,NAH,NAHism). But I will leave you with a little gem of realism.

The plane flies, no matter what your puny, ineffective tread belt can do, and that's a fact.

Grumpy cool.gif
Atl5p
QUOTE (Grumpy+Feb 26 2007, 04:28 PM)
Atl5p



This is just a flat out lie. I have told you many times that the plane will fly, there is nothing(within the OP) that the treadmill can do about it(other than spin the wheels of the plane faster). The aircraft could care less what the belt does, because the belt has no method of keeping the plane from moving, NO MATTER HOW YOU MEASURE THE SPEED. You even admitted that fact when the wheels were replaced by hover pads. Wheels have A LITTLE BIT of drag compared to the ZERO DRAG of the hover pads, but it is still a very small fraction of the power available in any normal plane.



Once again, who said anything about maximum power??? I didn't, neither did adoucette. No need to change the motor, just install one that give the necessary rpm no matter what the pitch.

If my motor can spin a 11X8 prop at 10,000 rpm, why would it not be able to spin an 11X4 prop at the same rpm??? This is why they put variable props on aircraft, so the prop can produce the same thrust as the plane goes faster or, conversely, so the prop can adjust to produce maximum thrust at takeoff speeds without overreving(the tips of the props approaching the speed of sound and becoming very inefficient, not to mention loud). Therefore the engine can turn the prop at it's most efficient speed(10,000 in this case) and the thrust is matched to the speed of the plane by adjusting the pitch.

All turbo props operate this way. They have a ground power setting for startups and taxing, and a flight power setting for takeoff and flying. When set to flight power the turbine spins the prop at a set rpm which stays the same all the time, thrust is determined by prop pitch(they even have a beta range of reverse thrust for braking during landing and ground handling(to back up the plane)).

My VPP prop on my Trbute will stop the plane while nose down and even climb backwards. It maintains the same rpm the whole time(helicopters do too)

So it is obvious you know nothing about aircraft(or tread belts for that matter), you're just a troll. One should not feed a troll(but it sure is fun to poke them through the bars of their cages!!!) and you have certainly wasted enough of my time and the bandwidth you waste on this forum. I would urge all posters to now leave the troll alone in his...utter lack of logical or coherent thought(not to mention, childish displays of NAH,NAH,NAHism). But I will leave you with a little gem of realism.

The plane flies, no matter what your puny, ineffective tread belt can do, and that's a fact.

Grumpy cool.gif

Grumpy....the fact is, other FLy Boys have said that a treadbelt COULD prevent a plane from taking off....you claim it cannot...not ever.

They strongly disagree with you, yet you seem unwilling to confront them....I'm wondering why that is? What would you have to say to someone who said that your idea is completely incorrect? What would you say to someone who told you 'You're wrong grumpy, a treadbelt really COULD prevent a plane from flight"

???

Do you not have any comments for them , just because their end result is still "FLY" for this question? That's rather 'Troll Like' if you ask me....

I mean, if you won't confront a fellow Fly Boy with such a CORE disagreement, then it COULD be said that the rest of YOU are trolling ME!
adoucette
QUOTE (Atl5p+Feb 26 2007, 04:18 PM)
I performed all of these tests with a small model plane and an real life treadbelt....oh, and the plane would propell itself just fine on static ground.

You've made this claim before,

But you've been posting on this topic for a year.

But in all that time you haven't been able to post a video on You Tube in all that time to prove it?

laugh.gif

Arthur

MikeMonty
QUOTE (Atl5p+Feb 26 2007, 07:00 PM)
Therefore, for a given motor, THRUST is in the DIAMETER of the prop.
SPEED is in the PITCH of the prop.

Now, if you want to disagree with the above, by simply installing bigger and more powerful motors each time, to come up with an increased pitch yeilding increased THRUST.....then you have done so by simply changing out the motor.  You have cheated.

I'm sure Adoucette is perfectly capable of defending himself and I swore (again) that I wouldn't get drawn but I'm getting fed up reading the crap you continually peddle.

Some poor guy out there might think you know what you're talking about but you still can't get your head around force and power.

Thrust (FORCE) is affected by the amount of air you can chuck backwards.
This can be affected by BOTH prop diameter and pitch - right up to the point where the prop stalls.

Assuming you have not maxed out your engine or stalled your prop then increasing pitch AND giving it more fuel (or Amps if it's electric) will give you more thrust for a given fixed speed by increasing the amount of POWER to the prop.

I don't intend to - or need to - further defend this statement to you, it's not addressed to you specifically - we just need to redress the balance.

Glad you've found another red herring - thought for a while you'd run out.

Precursor562
QUOTE
First it was 'Distance traveled between TWO POINTS, divided by time'

NOW, it is 'Distance travled THROUGH SPACE'.


Since you choose to ignore what I type in the brackets specifically after the word space I should ignore you but I won't.

As I've said I am using space as a generic term after all space is all around us even if it is filled with things such as air. The plane occupies a specific volume of space (POINT 1) and in 4 hours the plane moved to a different volume of space (the dimensions of the plane do not change and so the volume occupied does not change) 100 miles away (POINT 2). The speed of the plane is an average of 25 mph.

So no I didn't change my definition at all I am merely using different wording but if I'm confusing you than I shall do my best to refrain from such.

As for speed/velocity being relative I think the theory of relativity is the worst thing to have been introduced into the scientific community. Relativity is perception. You are in a car moving at 100 mph but since you can't feel the speed it has the perception of the world outside passing you at 100 mph and you not moving. Don't be fooled, the world is indeed not moving and it is you who are moving in your car.

Don't confuse relative speed with factual speed. Yes I am running at 10 mph on the treadmill but the REALITY is I am not moving on the treadmill and it is the treadmill that is moving beneath my feet. The difference between relativity and actuality.

But if your going to use something as a point of reference DO NOT USE the belt as that reference. The belt is being compared to the plane so don't compare the plane to the belt or you will end up with a loop.
gmilam
QUOTE (Atl5p+Feb 26 2007, 01:54 PM)
There is no such thing as the "positive feedback loop"

There WILL be a speed at which the negetive force from the treadbelt to the wheeled vehicle WILL equal the thrust of the airpropelled vehicle.

Again, the "positive feedback loop" idea is fantasy.

Let's see if you can understand something... The plane is NOT propelled by it's wheels. The wheels are spinning because the plane is moving forward.

When the plane is moving forward and the conveyor tries to match it's "wheel speed" all the conveyor is going to do is cause the wheels to spin faster and then the conveyor has to move faster to try and keep up.

In other words the faster the conveyor moves, the faster it is going to have to move... It is a positive feedback loop.
Atl5p
QUOTE (MikeMonty+Feb 26 2007, 05:23 PM)
I'm sure Adoucette is perfectly capable of defending himself and I swore (again) that I wouldn't get drawn but I'm getting fed up reading the crap you continually peddle.

Some poor guy out there might think you know what you're talking about but you still can't get your head around force and power.

Thrust (FORCE) is affected by the amount of air you can chuck backwards.
This can be affected by BOTH prop diameter and pitch - right up to the point where the prop stalls.

Assuming you have not maxed out your engine or stalled your prop then increasing pitch AND giving it more fuel (or Amps if it's electric) will give you more thrust for a given fixed speed by increasing the amount of POWER to the prop.

I don't intend to - or need to - further defend this statement to you, it's not addressed to you specifically - we just need to redress the balance.

Glad you've found another red herring - thought for a while you'd run out.

I was talking about a 'given' prop, and 'given' power source....

Prop diameter is where you get the 'Thrust'

Prop pitch is where you get the 'Speed'.

aducate was trying to say that the above was not true.

He was trying to say that, using a variable pitch prop, one could increase thrust simply by increasing pitch.....what he FAILED to mention, was that he was actually INCREASING POWER to that prop....and it was this INCREASED POWER from the motor which was creating the EXTRA THRUST.

He was responding to my post where I said that static thrust was the 'most' thrust a prop plane could produce, because when in forward flight it has generated an artificial 'headwind' which will reduce that thrust.
He said that this 'headwind' could be negated by simply increasing the PITCH with a variable pitch prop.

That was untrue...he would only increase his SPEED....NOT his THRUST.

He then tried to rectify himself by claiming that increasing prop pitch alone would result in increased thrust. He was incorrect yet again. His increased thrust was the result of more powerful motor....which was then able to spin a prop with increased pitch at the same RPM.

Again, it's all in how you ask the question.

It's like if I had said 'switching your rear gear from 4.11's to 2.73's will make your car accelerate slower.

And then he responds with a chart that showed a car with 2.73 gear accelerating FASTER than a car with 4.11 gears......only he failed to mention that his car with 2.73 gears has a 611CID big block with a supercharger and nitrous.

So SUUUUUURRRRRE.....A 2.73 geared car will accelerate faster than a 4.11 geared car....when you also replace the engine.
That's what he did....and it's called 'Cheating'.

Look...I'm very sure that there are about 1.7 gazillion ways to make a higher pitched prop show increased thrust....

BUT: For a GIVEN MOTOR and power suppy....remember, the plane is IN FLIGHT....it is the prop DIAMETER that determines max thust and it is PROP PITCH which determines max speed.

Playing with 1 will affect the RPM, which will then affect the OTHER factor.

mggb2001
Imagine this if you will.

A toy car sits on a treadmill and you use your finger to hold it in place. As the speed of the treadmill increases to a very high speed, the force your finger exerts on the car to hold it in place only slightly increases in comparison. The force with which you are pushing the car relative to the car's weight would seem to be perfectly acceptable to apply to the thrust of an engine and the weight of an airplane.

What I am trying to say is: No matter how fast the treadmill is going, you will always be able to overcome it.

The force to bring the car to zero mph relative to the surroundings must converge to a finite number. Tell me if this makes sense.
adoucette
Too bad Atl5p can't remember what he posted:


QUOTE (Atl5p @ Feb 23 2007+ 12:34 PM)

do you know anything about prop diameter, prop pitch, RPM, and STATIC THRUST? Apparently not. As I’ve stated before, a plane has to go pretty slow on static ground to be able to be held at 0 IAS by a 10mph or less treadbelt. When it’s held at 0IAS on the treadbelt, it has all it’s static thrust. But then put it on static ground, and the plane’s thrust will decrease due to the artificial ‘headwind’ it creates by moving through the air. Due to static thrust, the treadbelt speed might need to be higher than you think it should, but when the plane is at 0 IAS, the speed of the plane over the treadbelt, will match EXACTLY the speed of the treadbelt.


So now all you have to do is show you can't compensate for the movement of the plane (up to takeoff speed) by employing a Constant Speed (Variable Pitch) prop.

laugh.gif

Arthur

Johan_K
Hello again smile.gif

QUOTE (Atl5p+Feb 26 2007, 09:18 PM)

Could you then explain this for me?
http://auto.howstuffworks.com/tire4.htm

I shall try it.. with baby steps for my own good laugh.gif Now I'm no physicist (or physician for that matter either), so I'm feeling an incremental approach is sensible here.

I think we all agree on the basic formula

Power=Force*velocity
(With all the add-ons that the forces has to be parallel to the tangent of travel etc. etc.)

If we solve P=F*v for F, we get F=P/v, and indeed, putting the numbers from HowStuffWorks in there we see that in both cases (55 and 75 mph), we get the same result for the force, namely:
F = 12(hp) * 746(W) / 33.53(m/s) = 8.8(hp)hp*746(W) / 24.59(m/s) = 267.2 N.

So the statement:
QUOTE (http://auto.howstuffworks.com/tire4.htm+)
From these calculations you can see that the three things that affect how much force it takes to push the tire down the road (and therefore how much heat builds up in the tires) are the weight on the tires, the speed you drive and the CRF

Isn't actually correct since the increase in power and velocity cancel each other out in calculating how large the force applied is.

So I think we can both agree on the fact that the actual Force that is needed to hold the plane stationary on the conveyor is constant no matter how fast the conveyor is spinning?
Grumpy
Atl5p

QUOTE
Grumpy....the fact is, other FLy Boys have said that a treadbelt COULD prevent a plane from taking off....you claim it cannot...not ever.


I do not claim it, I know it to be true.

QUOTE (->
QUOTE
Grumpy....the fact is, other FLy Boys have said that a treadbelt COULD prevent a plane from taking off....you claim it cannot...not ever.


I do not claim it, I know it to be true.

They strongly disagree with you, yet you seem unwilling to confront them....I'm wondering why that is? What would you have to say to someone who said that your idea is completely incorrect? What would you say to someone who told you 'You're wrong grumpy, a treadbelt really COULD prevent a plane from flight"


I'll confront any idiot who argues that the treadbelt can stop the Aircraft(within the constraints of the OP, IE limited to the speed of the plane(not it's wheels). Anyone who argues that it can stop the plane is a dumbass who cannot read. It's a simple matter of forces. The prop pulls the plane with 1000 lbs of force(as an example) The wheels produce a force in drag of 10 or 20 lbs normally, maybe twice that as the belt moves backwards at THE SAME SPEED(and not one bit faster), so 40 lbs, leaving 960 lbs to accelerate the aircraft.

The aircraft flies easily off of the conveyor.

QUOTE
Do you not have any comments for them , just because their end result is still "FLY" for this question? That's rather 'Troll Like' if you ask me....

I mean, if you won't confront a fellow Fly Boy with such a CORE disagreement, then it COULD be said that the rest of YOU are trolling ME!


Actually you are the butt of a lot of jokes among the educated segment of those on this forum. Your ignorance is without end("There is no such thing as a positive feedback loop", you can't make this stuff up, you really said such a stupid thing, among others) laugh.gif

As to anyone who does not "get it" after the effort to educate them fails, I would just point out that TS need not BE terminal, but the sufferer must make the effort to learn. Otherwise the abysmal stupidity lasts until your demise.

Grumpy cool.gif
Guest
The plane flies, folks. There is no way the conveyor can counter the thrust of the engine through free-spinning wheels.
Guest
QUOTE (Atl5p+Feb 26 2007, 07:54 PM)
There is no such thing as the "positive feedback loop"

Uh... Yes there is. There are many, many examples. Maybe you need to look up what a positive feedback loop is.
Derek1148
QUOTE (mggb2001+Feb 26 2007, 10:54 PM)
Imagine this if you will.

A toy car sits on a treadmill and you use your finger to hold it in place. As the speed of the treadmill increases to a very high speed, the force your finger exerts on the car to hold it in place only slightly increases in comparison. The force with which you are pushing the car relative to the car's weight would seem to be perfectly acceptable to apply to the thrust of an engine and the weight of an airplane.

What I am trying to say is: No matter how fast the treadmill is going, you will always be able to overcome it.

The force to bring the car to zero mph relative to the surroundings must converge to a finite number. Tell me if this makes sense.

The principles that provide forward motion to a car are different than that of a plane.

The car’s engine supplies power to its wheels. The car is propelled by the power supplied to the wheels. The effect of the conveyor (depending upon the speed of the conveyor in the opposite direction) on a car, with forward rotating wheels, could be to one of the following: to slow, to cause to remain stationary, or to move the car in reverse.

The plane is propelled by engines that push the plane through the air. The wheels are on the plane to reduce friction during this process until take-off is achieved. The effect of the conveyor (moving in the opposite direction) on the plane would be to increase the rotation of the plane’s wheels. Once the plane’s propulsion system starts the forward motion of the plane, the movement of the conveyor in the opposite direction is irrelevant. The plane is moving forward.

The original post states specifically that the plane “moves”.

In order to interpret the scenario as the plane “remaining stationary”, one would have to suspend (or lack) both comprehension of the English language as well as knowledge of basic physics.



Atl5p
QUOTE (mggb2001+Feb 26 2007, 05:54 PM)
Imagine this if you will.

A toy car sits on a treadmill and you use your finger to hold it in place. As the speed of the treadmill increases to a very high speed, the force your finger exerts on the car to hold it in place only slightly increases in comparison. The force with which you are pushing the car relative to the car's weight would seem to be perfectly acceptable to apply to the thrust of an engine and the weight of an airplane.

What I am trying to say is: No matter how fast the treadmill is going, you will always be able to overcome it.

The force to bring the car to zero mph relative to the surroundings must converge to a finite number. Tell me if this makes sense.

What you basically said is this:

"As the speed of the treadmill increases to a ... high(er) speed, the force your finger exerts on the car to hold it in place... [irelevent information] ... increases in comparison.

So YES, there IS a treadbelt speed which will equal the plane's thrust.

Words like 'very fast' and 'slightly increases' are irelevent.

The facts is the essence of what you said....as the treadbelt speed increases, so does the force needed to hold it at 0 IAS. That's all you need to know to understand that a treadbelt CAN prevent a plane from taking off.

Now, what you are left with is the inturpretation of 'what is plane's SPEED'.
When it is related to the surface of the treadbelt (as you seem to like) then the plane no fly.
When it is related to the surrounding air...well, let's look at that for a second..."A plane moves though the air.....Does the plane move though the air...."....see Jane run...does jane run?....the answer to both of those questions is 'Yes' of course....go ask any second grader.
Atl5p
QUOTE (adoucette+Feb 26 2007, 06:05 PM)
Too bad Atl5p can't remember what he posted:




So now all you have to do is show you can't compensate for the movement of the plane (up to takeoff speed) by employing a Constant Speed (Variable Pitch) prop.

laugh.gif

Arthur

OK, I'm not sure exactly what you're talking about, could you please be more specific. 1st, try spelling out exactly what (if any) thing you disagree with the quote you have of mine.
Then you could try explaining what you think would happen, or be more specific as to what you want me to explain.

I see you seem to want me to 'compensate' for the 'movement'(over treadbelt or static ground, I dont' know), by using a variable pitch prop.(?)

The purpose of my post was to show that a plane that can produce a maximum IAS on static ground may go 50mph....that same plane would need a treadbelt spinning at much faster than 50mph in order for the treadbelt to hold the plane at 0 IAS. That is due to the 'static thrust' being higher than thrust with an artificial/self generated headwind a plane has when it is + IAS.

Now you seem to want to put a Variable Pitch Prop (your idea BTW) on the plane....in order to do WHAT?! On Which plane? And WHEN? For what REASON? I don't know.


You SEEM to want to make the plane on static ground go faster than 50mph, by using a variable pitch prop....that's all I can make of it.....WHY you want to do that, and WHAT you're trying to prove, I really wish you'd explain.

Making the plane go faster than 50mph on static ground was not my point.

My point was this: If you have a plane that goes MAX 50mph on static ground, do not expect a 50mph treadbelt to hold it at 0 IAS...it will need to be much faster.

Now you want me to:
QUOTE
"
So now all you have to do is show you can't compensate for the movement of the plane (up to takeoff speed) by employing a Constant Speed (Variable Pitch) prop.
"


And I don't know what the heck you're talking about....please speak english.

Maybe you want to use the variable pitch prop while the plane is on the treadbelt? Not sure why you'd want to do that though...Say you are using turbo-prop setup, the engine computer keeps the prop's RPMs constant, no matter what the pitch is set at? So increasing pitch is basically the same as increasing the power on a 'regular' plane.(?) SO spinning the treadbelt faster will still lead do a speed where the plane's speed over treadbelt = treadbelt speed....plane still at 0 IAS....so I don't understand your point...either way, you're burning more gas to make more power, which is countered by the treadbelt...what's you're point?
Atl5p
QUOTE (Grumpy+Feb 26 2007, 07:46 PM)
Atl5p



I do not claim it, I know it to be true.



I'll confront any idiot who argues that the treadbelt can stop the Aircraft(within the constraints of the OP, IE limited to the speed of the plane(not it's wheels). Anyone who argues that it can stop the plane is a dumbass who cannot read. It's a simple matter of forces. The prop pulls the plane with 1000 lbs of force(as an example) The wheels produce a force in drag of 10 or 20 lbs normally, maybe twice that as the belt moves backwards at THE SAME SPEED(and not one bit faster), so 40 lbs, leaving 960 lbs to accelerate the aircraft.

The aircraft flies easily off of the conveyor.



Actually you are the butt of a lot of jokes among the educated segment of those on this forum. Your ignorance is without end("There is no such thing as a positive feedback loop",  you can't make this stuff up, you really said such a stupid thing, among others) laugh.gif

As to anyone who does not "get it" after the effort to educate them fails, I would just point out that TS need not BE terminal, but the sufferer must make the effort to learn. Otherwise the abysmal stupidity lasts until your demise.

Grumpy cool.gif

Grumpy....OUTSIDE of the OP...."Can a treadbelt prevent a plane from taking off"...lots of FlyBoys say "YES a treadbelt free to spin as fast as it needs to, CAN in fact prevent a plane from taking off".

Yet you say this can NEVER happen...

We're not talking about 'Plane speed = IAS"

We're talking about 'Can a treadbelt prevent takeoff'.

You STILL are avoiding this. They have spoken...what is you're responce to them? Will any of them debate you on that?

Sure, if Plane speed = IAS, then yes it will fly....no one has a problem with that.

But 'Can a treadbelt prevent flight'.....you say NO....lots of others say 'Yes'.....and THAT is the only real disagreement going on around here, but it is systematically ignored....and I'm wondering if the flyboys have the guts to debate this issue against each other? You keep avoiding. This is not to be within the confines of what you believe to be the OP.

And there is no positive feedback loop WITHIN THIS OP, when 'wheelspeed' is used for plane's speed.

I'm sure such a thing exists in the universe....but it does not appy to wheelspeed version. Taken out of context...nice try.
Atl5p
QUOTE (Derek1148+Feb 27 2007, 06:50 AM)
The principles that provide forward motion to a car are different than that of a plane.

The car’s engine supplies power to its wheels. The car is propelled by the power supplied to the wheels. The effect of the conveyor (depending upon the speed of the conveyor in the opposite direction) on a car, with forward rotating wheels, could be to one of the following: to slow, to cause to remain stationary, or to move the car in reverse.

The plane is propelled by engines that push the plane through the air. The wheels are on the plane to reduce friction during this process until take-off is achieved. The effect of the conveyor (moving in the opposite direction) on the plane would be to increase the rotation of the plane’s wheels. Once the plane’s propulsion system starts the forward motion of the plane, the movement of the conveyor in the opposite direction is irrelevant. The plane is moving forward.

The original post states specifically that the plane “moves”.

In order to interpret the scenario as the plane “remaining stationary”, one would have to suspend (or lack) both comprehension of the English language as well as knowledge of basic physics.

So...what you're saying is this:
"A plane moves though the air.....is that there plane moving though the air?"
The answer is obviously 'YES".

I am asking "Can a treadbelt prevent a plane from flying"....the answer is YES.
Derek1148
QUOTE (Atl5p+Feb 27 2007, 04:34 PM)
So...what you're saying is this:
"A plane moves though the air.....is that there plane moving though the air?"
The answer is obviously 'YES".

I am asking "Can a treadbelt prevent a plane from flying"....the answer is YES.

I suppose if the conveyor explodes during the take-off process, you might be right.
Precursor562
QUOTE
Atl5p

QUOTE
Grumpy....the fact is, other Fly Boys have said that a treadbelt COULD prevent a plane from taking off....you claim it cannot...not ever.


I do not claim it, I know it to be true.

QUOTE
They strongly disagree with you, yet you seem unwilling to confront them....I'm wondering why that is? What would you have to say to someone who said that your idea is completely incorrect? What would you say to someone who told you 'You're wrong grumpy, a treadbelt really COULD prevent a plane from flight"


I'll confront any idiot who argues that the treadbelt can stop the Aircraft(within the constraints of the OP, IE limited to the speed of the plane(not it's wheels). Anyone who argues that it can stop the plane is a dumbass who cannot read. It's a simple matter of forces. The prop pulls the plane with 1000 lbs of force(as an example) The wheels produce a force in drag of 10 or 20 lbs normally, maybe twice that as the belt moves backwards at THE SAME SPEED(and not one bit faster), so 40 lbs, leaving 960 lbs to accelerate the aircraft.

The aircraft flies easily off of the conveyor.

QUOTE
Do you not have any comments for them , just because their end result is still "FLY" for this question? That's rather 'Troll Like' if you ask me....

I mean, if you won't confront a fellow Fly Boy with such a CORE disagreement, then it COULD be said that the rest of YOU are trolling ME!


Actually you are the butt of a lot of jokes among the educated segment of those on this forum. Your ignorance is without end("There is no such thing as a positive feedback loop", you can't make this stuff up, you really said such a stupid thing, among others) laugh.gif

As to anyone who does not "get it" after the effort to educate them fails, I would just point out that TS need not BE terminal, but the sufferer must make the effort to learn. Otherwise the abysmal stupidity lasts until your demise.


Well Atl5p I can't talk for the others that said the conveyor COULD stop the plane IF it were to move fast enough under the plane (going against the original question) but I can speak for myself. I merely said the belt could because at high velocities friction will generate enough heat within the wheels (the wheel bearings) causing the grease to breakdown. This will result in a catastrophic failure which would cause the wheels to cease up preventing them to rotate at all. At which point the forces being applied at that moment would most likely rip the wheels clear off the plane and you no longer have free spinning wheels which is what keeps the belt from applying a rearward force directly to the plane. However this is provided that the mechanism which drives the belt (which would be more complex than the bearing in the wheels of the plane) wouldn't fail first. I believe that the belt would cease up first.

So I believe Grumpy is merely excluding such things (as it should be) since such things are outside the limitations of the original question which has the belt moving at an equal speed to that of the plane. Therefore I am a "fly boy" who agrees that theoretically the belt could stop the plane if we were to ignore the limitations of the original question but I do not disagree with what he says. I agree to what he says and I think he explains things rather well. Better than I do for sure.

Edit:

Wow adoucette 60% warning? I thought you trying to turn on my mind's light bulb would have taught you patience with the absent minded tongue.gif and being successful should atleast be a reset back to zero. biggrin.gif
gmilam
QUOTE (Atl5p+Feb 27 2007, 10:34 AM)
I am asking "Can a treadbelt prevent a plane from flying"....the answer is YES.

If your "plane" is an underpowered overweight lego truck with a weak aeroace engine attached to it that is incapable of flight to begin with, then yeah.

If your plane is an F16, no... Unless you found people willing to invest billions in building a super treadbelt from Hades. Then - maybe.

But I suspect your treadbelt would self destruct long before you stopped the plane.
Atl5p
QUOTE (Precursor562+Feb 27 2007, 11:46 AM)

Well Atl5p I can't talk for the others that said the conveyor COULD stop the plane IF it were to move fast enough under the plane (going against the original question) but I can speak for myself. I merely said the belt could because at high velocities friction will generate enough heat within the wheels (the wheel bearings) causing the grease to breakdown. This will result in a catastrophic failure which would cause the wheels to cease up preventing them to rotate at all. At which point the forces being applied at that moment would most likely rip the wheels clear off the plane and you no longer have free spinning wheels which is what keeps the belt from applying a rearward force directly to the plane. However this is provided that the mechanism which drives the belt (which would be more complex than the bearing in the wheels of the plane) wouldn't fail first. I believe that the belt would cease up first.

So I believe Grumpy is merely excluding such things (as it should be) since such things are outside the limitations of the original question which has the belt moving at an equal speed to that of the plane. Therefore I am a "fly boy" who agrees that theoretically the belt could stop the plane if we were to ignore the limitations of the original question but I do not disagree with what he says. I agree to what he says and I think he explains things rather well. Better than I do for sure.

Ummm, I'm not talking about wheels and tires and treadbelts blowing up...I'm talking about an increasing backwards force which increases with treadbelt speed.

It reminds me of a story I heard once. A certian mathemetician was trying to figure out how things 'orbit' around the earth.

To figure this out, he imagined a very tall mountian that went outside our atmosphere...then he put a huge cannon on top and started firing cannon balls at greater and greater force, until one of the balls went so far it 'fell' past the horizon, and continued to 'fall' without hitting the ground. He called that 'Orbit', and helped explain how the moon was in orbit around the earth, and the earth was in orbit around the sun.

And YOU remind me of someone who would say to that man "You can't fire a cannon with that much force, the cannon would explode!....So that means that ORBIT cannot happen...not ever".

Does that sound about right to you?
Atl5p
QUOTE (gmilam+Feb 27 2007, 12:12 PM)
If your "plane" is an underpowered overweight lego truck with a weak aeroace engine attached to it that is incapable of flight to begin with, then yeah.

If your plane is an F16, no... Unless you found people willing to invest billions in building a super treadbelt from Hades. Then - maybe.

But I suspect your treadbelt would self destruct long before you stopped the plane.

Kinda like blowing up cannons, huh?

Let me ask you something: If I took a really big cannon on top of a mountian that reached outside our atmosphere, and I fired a cannon ball horizontally with enough force, do you think the cannon ball could get into 'orbit' and not fall back to earth?

Your 'consistant' answer should be:
"No, the cannon would blow up before you could fire a cannon ball with that much force".
Precursor562
QUOTE
Ummm, I'm not talking about wheels and tires and treadbelts blowing up...I'm talking about an increasing backwards force which increases with treadbelt speed.


No instead you are talking about the belt moving at velocities that not only go against the original question by going faster than the plane instead of equal to but are impossible to achieve using todays mechanics.

Oh and that backwards force applied by the belt is applied to the plane's wheels and since those wheels are free to spin that is where the force stops. It doesn't get transfered to the plane and therefore it doesn't affect how the thrust moves the plane.

QUOTE (->
QUOTE
Ummm, I'm not talking about wheels and tires and treadbelts blowing up...I'm talking about an increasing backwards force which increases with treadbelt speed.


No instead you are talking about the belt moving at velocities that not only go against the original question by going faster than the plane instead of equal to but are impossible to achieve using todays mechanics.

Oh and that backwards force applied by the belt is applied to the plane's wheels and since those wheels are free to spin that is where the force stops. It doesn't get transfered to the plane and therefore it doesn't affect how the thrust moves the plane.

To figure this out, he imagined a very tall mountian that went outside our atmosphere...then he put a huge cannon on top and started firing cannon balls at greater and greater force, until one of the balls went so far it 'fell' past the horizon, and continued to 'fall' without hitting the ground. He called that 'Orbit', and helped explain how the moon was in orbit around the earth, and the earth was in orbit around the sun.


Another irrelevant scenario just as your relative speed theory was.
Atl5p
QUOTE (Precursor562+Feb 27 2007, 12:51 PM)

No instead you are talking about the belt moving at velocities that not only go against the original question by going faster than the plane instead of equal to but are impossible to achieve using todays mechanics.

Oh and that backwards force applied by the belt is applied to the plane's wheels and since those wheels are free to spin that is where the force stops. It doesn't get transfered to the plane and therefore it doesn't affect how the thrust moves the plane.



Another irrelevant scenario just as your relative speed theory was.

First of all, the way the plane's 'Speed' is determined is up for debate....I do not deny that if IAS is used as plane's speed, then obviously it will fly.

What I am talking about is this: what IF the OP meant to use the 'speed over the surface of the treadbelt' to mean plane's speed. (keep in mind, the OP is NOT specific, either way...that is fact....your 'version' is your OPINION, not FACT...get that right JACK)

When this 'wheelspeed' version is used, we have to ask ourselves 'does the backwards force increase with velocity of treadbelt.' I say the answer to this is 'Yes'. Several others on this board have said 'Yes'. And even those who study such things and write papers about wheelchair wheels...they say 'Yes' as well.

And you are correct, that if say an F16 were the plane, the speed needed would probably blow the tires up, and there is not yet a treadbelt which will spin this fast. But a tire blowout does not have ANYTHING to do with the fact that the force increases with velocity.

The following statement is the ONLY thing I am debating in this regards:
QUOTE
Oh and that backwards force applied by the belt is applied to the plane's wheels and since those wheels are free to spin that is where the force stops.

I think you are 100% INCORRECT on this.

And the 'orbit' story is VERY relevent. You are trying to disprove my theory by saying that the tires and treadbelt will explode...because of this, my theory is incorrect.

That is the EXACT same as saying 'The cannon would blow up before you could fire a cannon ball that far to put it into orbit...therefore ORBIT is IMPOSSIBLE"...

that is EXACTLY the same thing you are saying when you explode the tires....you explode the cannon....it's the same thing, I can't see how you can't see the relevence.
Fynlcut
QUOTE (Atl5p+Feb 27 2007, 06:09 PM)
When this 'wheelspeed' version is used, we have to ask ourselves 'does the backwards force increase with velocity of treadbelt.' I say the answer to this is 'Yes'. Several others on this board have said 'Yes'. And even those who study such things and write papers about wheelchair wheels...they say 'Yes' as well.

With your "wheelspeed theory" you don't need any resistance at all to stop the plane. In your question the plane doesn't move. Why try to go any farther with it?
Precursor562
QUOTE
First of all, the way the plane's 'Speed' is determined is up for debate


There is only one way to determine the speed of an object. Take its physical movement from point to point and divide the distance by the time it took to travel that distance. It's not something that is open to debate. Under normal circumstances we use what is available to us to determine such movement. I stand in a room, I then stand somewhere else in that room. In order for this to happen I will have to physically move from one spot to the other. We can see this change in location by comparing my position to the rest of the room. It's all we have available but even if you couldn't see the room and could only see me then that doesn't mean I'm not moving just because you don't have anything to compare my motion to.

So in this case we have to compare using what is available to us and no it is not the belt. The speed of the belt matches that of the plane and so you can't use the belt to determine the speed of the plane. You can't use the depended variable to find the value of the independent variable. You find the value of the independent variable first then find the value of the dependent variable whose value is depended on the value of the independent variable. So we use the plane itself (its starting point) or we add a fixed point of reference (ground, tower etc.). This is done just to 'see' the movement.


QUOTE (->
QUOTE
First of all, the way the plane's 'Speed' is determined is up for debate


There is only one way to determine the speed of an object. Take its physical movement from point to point and divide the distance by the time it took to travel that distance. It's not something that is open to debate. Under normal circumstances we use what is available to us to determine such movement. I stand in a room, I then stand somewhere else in that room. In order for this to happen I will have to physically move from one spot to the other. We can see this change in location by comparing my position to the rest of the room. It's all we have available but even if you couldn't see the room and could only see me then that doesn't mean I'm not moving just because you don't have anything to compare my motion to.

So in this case we have to compare using what is available to us and no it is not the belt. The speed of the belt matches that of the plane and so you can't use the belt to determine the speed of the plane. You can't use the depended variable to find the value of the independent variable. You find the value of the independent variable first then find the value of the dependent variable whose value is depended on the value of the independent variable. So we use the plane itself (its starting point) or we add a fixed point of reference (ground, tower etc.). This is done just to 'see' the movement.


what IF the OP meant to use the 'speed over the surface of the treadbelt' to mean plane's speed.


Then he would have added that but since he didn't you must take the question for what is actually said.

He asks if the belt can keep the plane from taking off if the speed of the belt matches the speed of the plane period. Not if the belt matches the speed of the plane whose speed is determined by how fast its wheels are spinning or anything along those lines.

QUOTE
does the backwards force increase with velocity of treadbelt.


This backwards force that does increase with velocity of the belt is applied to the plane's wheels only since the wheels are the only thing in contact with the belt. Since these wheels are free to spin the plane doesn't feel this backwards force. The force which can be seen as energy is sent from the belt to the wheels. From there it goes into making the wheels spin faster instead of countering the thrust energy of the plane.

QUOTE (->
QUOTE
does the backwards force increase with velocity of treadbelt.


This backwards force that does increase with velocity of the belt is applied to the plane's wheels only since the wheels are the only thing in contact with the belt. Since these wheels are free to spin the plane doesn't feel this backwards force. The force which can be seen as energy is sent from the belt to the wheels. From there it goes into making the wheels spin faster instead of countering the thrust energy of the plane.

I think you are 100% INCORRECT on this.


Please explain. If your talking about the little force that does transfer to the plane through the wheel's bearings (as a result of friction) than such a force exists when the plane is taking off from still ground. The increase from still ground to moving ground equal to the speed of the plane is so small it's not worth mentioning.

plane speed + belt speed = wheel speed
100 mph + 0 mph = 100 mph (take off speed) and so the resistance to move applied to the plane through the wheel's bearings are given the value of X

100 mph + 100 mph = 200 mph

Plane speed is still 100 mph and so take off speed is reached also the belt speed is equal to the plane speed (100 mph) in the opposite direction and the result is a wheel speed of 200 mph instead of 100 mph. This DOES NOT give you a value of 2X. It would give you a value of X+A where A is such a small number (smaller than X) and X is a small number to start the end result is a situation virtually unchanged whether the plane is on or off the belt.

Just like to note that the equation I use isn't accurate. The actual equation is
Plane speed - Belt speed = Wheel speed

Plane speed is X and belt speed equals X but in the opposite direction so you get -X.

X - -X = X + X = 2X
100 - (-100) = 100 + 100 = 200

QUOTE
With your "wheelspeed theory" you don't need any resistance at all to stop the plane. In your question the plane doesn't move. Why try to go any farther with it?


Good point.

Let's just throw in a question that needs answering. Why do planes have brakes?
Atl5p
QUOTE (Fynlcut+Feb 27 2007, 02:26 PM)
With your "wheelspeed theory" you don't need any resistance at all to stop the plane. In your question the plane doesn't move. Why try to go any farther with it?

With 'wheelspeed' version, the plane IS moving in relation to the treadbelt...what is your point? Do you have a point?
Grumpy
Atl5p

QUOTE
First of all, the way the plane's 'Speed' is determined is up for debate....I do not deny that if IAS is used as plane's speed, then obviously it will fly.


The ONLY way an arcraft's speed is determined is IAS. It has no speedometer on the wheels. Nor does it matter how many times the wheels rotate. Or even if it has wheels at all(skids, skis, floats, hoverpads, or like the Wright Brothers, pieces of polished steel riding on a steel rail + a little grease).

Now, what force on Earth would cause the situation to be different??? The belt can do whatever it likes(within the bounds of the OP), the plane will take off anyway. What magically changes just because you measure the movement of the aircraft differently???

If the speed of the aircraft is measured by a reference point on the belt that leads INSTANTLY to an infinite positive feedback loop as soon as the plane moves an inch. This violates the conditions of the OP and is therefore ridiculous.

Whereas, if both the plane's speed and belt's speed are measured by a fixed point on the ground, the plane flies easily off the belt. So...

QUOTE (->
QUOTE
First of all, the way the plane's 'Speed' is determined is up for debate....I do not deny that if IAS is used as plane's speed, then obviously it will fly.


The ONLY way an arcraft's speed is determined is IAS. It has no speedometer on the wheels. Nor does it matter how many times the wheels rotate. Or even if it has wheels at all(skids, skis, floats, hoverpads, or like the Wright Brothers, pieces of polished steel riding on a steel rail + a little grease).

Now, what force on Earth would cause the situation to be different??? The belt can do whatever it likes(within the bounds of the OP), the plane will take off anyway. What magically changes just because you measure the movement of the aircraft differently???

If the speed of the aircraft is measured by a reference point on the belt that leads INSTANTLY to an infinite positive feedback loop as soon as the plane moves an inch. This violates the conditions of the OP and is therefore ridiculous.

Whereas, if both the plane's speed and belt's speed are measured by a fixed point on the ground, the plane flies easily off the belt. So...

A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?


The plane flies.

Our work here is done gentlemen. Alt5p has admitted that under the conditions of the OP the plane will fly.

Grumpy cool.gif

Oh, by the way. If the belt goes into an infinite positive feedback loop it immediately will disintegrate, if the aircraft is not damaged, it is then free to take off normally.

Precursor562
QUOTE
The cannon would blow up before you could fire a cannon ball that far to put it into orbit...therefore ORBIT is IMPOSSIBLE"


First off just because the cannon would blow up does not disprove orbit it just makes putting something into orbit by that means not possible using todays technology. Just like you couldn't get the belt moving fast enough using todays technology. If you could get it going fast enough the speed of the belt would have to be FASTER than the speed of the plane. In the original question the speed of the belt EQUALS the speed of the plane.
Atl5p
QUOTE (Grumpy+Feb 27 2007, 02:36 PM)
Atl5p



The ONLY way an arcraft's speed is determined is IAS. It has no speedometer on the wheels. Nor does it matter how many times the wheels rotate. Or even if it has wheels at all(skids, skis, floats, hoverpads, or like the Wright Brothers, pieces of polished steel riding on a steel rail + a little grease).

Now, what force on Earth would cause the situation to be different??? The belt can do whatever it likes(within the bounds of the OP), the plane will take off anyway. What magically changes just because you measure the movement of the aircraft differently???

If the speed of the aircraft is measured by a reference point on the belt that leads INSTANTLY to an infinite positive feedback loop as soon as the plane moves an inch. This violates the conditions of the OP and is therefore ridiculous.

Whereas, if both the plane's speed and belt's speed are measured by a fixed point on the ground, the plane flies easily off the belt. So...



The plane flies.

Our work here is done gentlemen. Alt5p has admitted that under the conditions of the OP the plane will fly.

Grumpy cool.gif

Oh, by the way. If the belt goes into an infinite positive feedback loop it immediately will disintegrate, if the aircraft is not damaged, it is then free to take off normally.

an infinate positive feedback loop will NOT occur when wheelspeed is used.

You are saying that the spinning belt will have NO EFFECT on the plane AT ALL....and that is INCORRECT.

I was at an airshow recently where the announcer said that the F18 was going to fly past us at 600mph. That can only mean 'speed over the ground'...NOT airspeed...if he meant 'airspeed' then he would have said 'knots'.
He was relating the speed of the plane to US standing on the ground...NOT related to the air around us.

So, you might want to let the NAVY know that they are incorrect.
mggb2001
QUOTE (Derek1148+Feb 27 2007, 11:50 AM)
The principles that provide forward motion to a car are different than that of a plane.

The car’s engine supplies power to its wheels. The car is propelled by the power supplied to the wheels. The effect of the conveyor (depending upon the speed of the conveyor in the opposite direction) on a car, with forward rotating wheels, could be to one of the following: to slow, to cause to remain stationary, or to move the car in reverse.

The plane is propelled by engines that push the plane through the air. The wheels are on the plane to reduce friction during this process until take-off is achieved. The effect of the conveyor (moving in the opposite direction) on the plane would be to increase the rotation of the plane’s wheels. Once the plane’s propulsion system starts the forward motion of the plane, the movement of the conveyor in the opposite direction is irrelevant. The plane is moving forward.

The original post states specifically that the plane “moves”.

In order to interpret the scenario as the plane “remaining stationary”, one would have to suspend (or lack) both comprehension of the English language as well as knowledge of basic physics.

I'm sorry if I worded this in a confusing way, but what I was trying to say was that the plane WOULD take off because the force pushing the toy car forward is your finger. Similar to the engine of a plane, this force has nothing to do with the tire speed. The only forces stopping the plane from moving are; air resistance, friction between the tires and the treadmill, and friction between the axle and wheel bearings. All of which are negligible in comparison to the enormous force provided by the engine of the plane.
Atl5p
QUOTE (Precursor562+Feb 27 2007, 02:34 PM)

There is only one way to determine the speed of an object. Take its physical movement from point to point and divide the distance by the time it took to travel that distance. It's not something that is open to debate. Under normal circumstances we use what is available to us to determine such movement. I stand in a room, I then stand somewhere else in that room. In order for this to happen I will have to physically move from one spot to the other. We can see this change in location by comparing my position to the rest of the room. It's all we have available but even if you couldn't see the room and could only see me then that doesn't mean I'm not moving just because you don't have anything to compare my motion to.

So in this case we have to compare using what is available to us and no it is not the belt. The speed of the belt matches that of the plane and so you can't use the belt to determine the speed of the plane. You can't use the depended variable to find the value of the independent variable. You find the value of the independent variable first then find the value of the dependent variable whose value is depended on the value of the independent variable. So we use the plane itself (its starting point) or we add a fixed point of reference (ground, tower etc.). This is done just to 'see' the movement.




Then he would have added that but since he didn't you must take the question for what is actually said.

He asks if the belt can keep the plane from taking off if the speed of the belt matches the speed of the plane period. Not if the belt matches the speed of the plane whose speed is determined by how fast its wheels are spinning or anything along those lines.



This backwards force that does increase with velocity of the belt is applied to the plane's wheels only since the wheels are the only thing in contact with the belt. Since these wheels are free to spin the plane doesn't feel this backwards force. The force which can be seen as energy is sent from the belt to the wheels. From there it goes into making the wheels spin faster instead of countering the thrust energy of the plane.



Please explain. If your talking about the little force that does transfer to the plane through the wheel's bearings (as a result of friction) than such a force exists when the plane is taking off from still ground. The increase from still ground to moving ground equal to the speed of the plane is so small it's not worth mentioning.

plane speed + belt speed = wheel speed
100 mph + 0 mph = 100 mph (take off speed) and so the resistance to move applied to the plane through the wheel's bearings are given the value of X

100 mph + 100 mph = 200 mph

Plane speed is still 100 mph and so take off speed is reached also the belt speed is equal to the plane speed (100 mph) in the opposite direction and the result is a wheel speed of 200 mph instead of 100 mph. This DOES NOT give you a value of 2X. It would give you a value of X+A where A is such a small number (smaller than X) and X is a small number to start the end result is a situation virtually unchanged whether the plane is on or off the belt.

Just like to note that the equation I use isn't accurate. The actual equation is
Plane speed - Belt speed = Wheel speed

Plane speed is X and belt speed equals X but in the opposite direction so you get -X.

X - -X = X + X = 2X
100 - (-100) = 100 + 100 = 200



Good point.

Let's just throw in a question that needs answering. Why do planes have brakes?

1) There are many reference points by which you can determine speed…you can determine speed via a stationary object (like a tower), or a moving object (like reading IAS while flying in a 20mph wind)

2) When I say ‘what if the question meant wheelspeed’, I mean to say ‘what do YOU think would happen if wheelspeed was used’….sorry about the confusion…care to answer?

3) You are incorrect in saying that the treadbelt does not inflict a backwards force onto the plane.

4) FYI, all those calculations at the bottom of the post are describing ‘IAS’ version….I am well aware of this. If a plane goes 100mph IAS, and the belt goes 100mph iAS, then the wheels go 200mph….that is NOT NEWS.
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