RE: Can It Take Off??  
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RE: Can It Take Off?? - 12/8/2005 1:59:30 AM   
starwoes



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quote:

ORIGINAL: mr_matt

Actually that is not a bad way to more intuitively understand the original question.

Try this:

You are on the skywalk at the airport, on roller skates, facing the wrong direction. You are holding onto a rope that is attached to something rigid, at the end of the skywalk (the attachment point of the rope does not move with the skywalk).

At this point, the skywalk is turned off, and you start pulling yourself forward on the skywalk using the rope. Now the skywalk is turned on. Clearly, it is harder for you to pull against the skywalk.

And now they make the skywalk faster and faster. As you can imagine, at some speed of the skywalk, you either can no longer pull hard enough to move forward (ie no airspeed), or the rope breaks (again, no airspeed, you are now moving backwards!)

In the airplane example, the thrust of the prop functions as the rope. The conveyor belt (or skywalk) is adjusted in speed until the prop can no longer pull forward (like the rope breaking on the skywalk).

Now if the levels of forces are not clear, imagine the skywalk was covered in honey or some sticky substance (increased rolling resistance) and that the rope is like dental floss. Now you can easily see how turning up the speed of the skywalk would at some point cause you to go back and break the dental floss.


excellent example except that at some point,

"And now they make the skywalk faster and faster. As you can imagine, at some speed of the skywalk, you either can no longer pull hard enough to move forward (ie no airspeed), or the rope breaks (again, no airspeed, you are now moving backwards!)"

the rope does not break (the wind does not break as you grab greater and greater bites)....you have the capacity to pull harder and harder because your ability to pull does not depend on conveyor speed but the wind/thrust (rope) and you actually overcome the futile effort of the conveyor working against your rollers..........

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RE: Can It Take Off?? - 12/8/2005 2:03:24 AM   
Phlip



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quote:

In the airplane example, the thrust of the prop functions as the rope. The conveyor belt (or skywalk) is adjusted in speed until the prop can no longer pull forward (like the rope breaking on the skywalk).

BUT THIS IS NOT WHAT THE ORIGINAL CONDITIONS OF THE "BRAIN TEASER" ARE !!!!!

It says nothing about adjusting in speed until the prop can't pull forward, it says the belt moves at the same speed as the wheels, and since they're attached to the plane, the same speed as the plane but in the opposite direction. What plane can't overcome wheels running at twice normal speed? It's not that much!

I'm through.


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RE: Can It Take Off?? - 12/8/2005 2:05:12 AM   
mr_matt



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Status: offline 		The tensile strength of the rope is analagous to the prop thrust. There is most definitley a limit to prop thrust (or strength of a rope) or else our models would pull those starting stands out of the ground.

Imagine the thrust was tiny (like the tensile strength of dental floss or sewing thread). Surely you can see there is a limit to how fast you can run you skywalk on your roller skates before the rope breaks?

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RE: Can It Take Off?? - 12/8/2005 2:16:33 AM   
starwoes



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From: Maple Heights, OH, USA
Status: offline 		i can certainly understand you there regarding prop slippage or limits. but......what if the aircraft hardly had any weight on the conveyor belt? ei, negligent weight such that it could just "scoot" on the belt regardless of direction of rotation of the belt?

got the rest of the week off but............it's been interesting all the same...will just try to observe for a better part of the evening....not sure i can handle the thinking right now

< Message edited by starwoes -- 12/8/2005 2:21:35 AM >

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RE: Can It Take Off?? - 12/8/2005 2:28:22 AM   
mr_matt



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Status: offline 		Yes the lighter the plane is the less effect of the conveyor, as the friction force for has a mass component. So if the plane had no weight, it would just float above the conveyor and take off, no matter how fast the conveyor moved.

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RE: Can It Take Off?? - 12/8/2005 2:46:25 AM   
3dbob37n


 

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Status: offline 		Ah, Matt, but the conveyor has cancelled any forward motion, so how can it take off?
The more the plane tries to move forward, the faster the conveyor belt moves. And it cannot take off until the conveyor slows or stops. The wheels will always maintain enough friction with the ground (read conveyor belt) to cause interference and stop forward movement because the aircraft still has its full weight on the ground.
Yes, yes, I know, my Yak 54 would take off from a dime. We commonly hand launch our Funtana 90's. But we are talking about an ordinary airplane, not an over powered freak.
3dbob

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RE: Can It Take Off?? - 12/8/2005 2:53:56 AM   
mr_matt



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Status: offline 		I am agreeing with you, I was only responding to the question of what would happen if the plane were weightless. Friction (including the rolling kind) has to have a normal force (or weight) in order to have any value. So a weightless plane has no rolling resistance.

For a normal plane there is indeed weight and therefore there is rolling resistance. The conveyor matches the wheel speed, the motor/propellor tries to accelerate the plane, the conveyor goes nuts, running the belt hundreds of miles an hour in reverse and the plane just sit still, does not take off cause it has no airpseed.

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RE: Can It Take Off?? - 12/8/2005 4:31:56 AM   
Liberator


 

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Status: offline 		Who says we are not talking about "an over powered freak?" Let's assume we are, it's an airplane as well right?

It's a nice try MrMatt, but here is where that skywalk theory fails. The test isn't to see if the plane will simply run down the walk against the turning of the conveyer, but if it will take off. It's like adding the third dimension to a 2 dimensional puzzle. As the prop speed ramps up, more and more air is passing over the flight surface and at some point there wll be enough lift from the prop that a touch of up elevator and the WHEELS BREAK FREE FROM ANY CONFINEMENT!!!! Yay we now have lift off regardless of any wheel friction. Ahh wow that feels good, we are now flying high in the sky instead of being tethered to our earthly shackle.

You're right, Physics do work!


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RE: Can It Take Off?? - 12/8/2005 4:40:28 AM   
3dbob37n


 

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Status: offline 		I'll ask this again. Is the physics professor coming forward with an answer or do we have to agonize over this for eternity???????
My brain is all fuzzy, age I guess. Like the guy who was cursed by a ancient God to push a stone up and hill and let it fall, only to push it up again, for all eternity.
Next thing you know we will be debating about whether wind pushes an airplane or not.

Now where can we borrow a skywalk?

3dbob

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RE: Can It Take Off?? - 12/8/2005 4:48:19 AM   
Liberator


 

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Status: offline 		I think that was hell that was being described as far as pushing the big rock. I am only in Utah, and although some have mistaken this for that other place(usually folks trying to get a Scotch and Soda) The two are not that same.

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RE: Can It Take Off?? - 12/8/2005 5:08:30 AM   
dick Hanson



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My next slide please ---

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RE: Can It Take Off?? - 12/8/2005 5:45:42 AM   
mr_matt



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quote:

ORIGINAL: Liberator

As the prop speed ramps up, more and more air is passing over the flight surface and at some point there wll be enough lift from the prop that a touch of up elevator and the WHEELS BREAK FREE FROM ANY CONFINEMENT!!!!




We are covering old ground here, I already stipulated that if the plane generated any lift in any way OTHER than forward airpeed and resulting wing lift (ie jet packs, helicopters, blown wings/flaps) then yes the plane can take off no problem.


But IF you stipulate that the ONLY way the plane can lift off is via forward airspeed generating lift from the wings, then it will not lift off.

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RE: Can It Take Off?? - 12/8/2005 6:35:28 AM   
BMatthews



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Status: offline 		I've thought about this some more and found I need to change my tune a bit.....

There are two systems at work here joined at the wheel bearings. One is the airplane with the props turning that is going to pull the airplane in the takeoff direction. The other system is the wheels that are free to spin under the influence of the conveyor belt. But the wheel bearings are to all intents frictionless. To my mind the idea here is to play with pure physics and not introduce any real world limitations or assumptions.

The trick here is that the conveyor belt is supposedly set to reflect the wheel's motion and run backwards to try to prevent the wheels from moving forward.

How fast the belt can move depends on how you read the problem.

It can be taken that the conveyor will only reflect the speed of the wheels in which case the conveyor will only accelerate to the same but opposite speed as the aircraft. So the conveyor will only run at the same speed as the plane but in reverse. In that case the wheels only need to turn twice as fast at liftoff as they would normally. So the airplane takes off more or less normally with no real problem. This is one answer

The other way it can be taken is that there is a very, very high limit or no limit to the conveyor speed that is capable of accellerating the wheels up to a very high RPM. It is this assumption that has lead me to alter my previous answer. Let's assume that regardless of speeds nothing will physically fail or blow up despite the fact that the belt can achieve extremely high speeds and the wheels will be spun to extremely high RPMS.

This is where the conveyor is capable of stopping the airplane. To see why consider the following 2 scenarios.

You spin a model wheel on a wire axle. Spin it up good and fast. You added energy to that wheel by flicking it with your finger. The wheel retains that energy and expends it through the friction of the air and contact with the axle as heat until the wheel looses all the spin energy and stops. But if you hold the spinning wheel on the axle just above the floor and suddenly snatch the wire sideways and out of the wheel the spinning wheel falls and hits the floor and immediatly takes off for parts unknown. It does this because the stored rotational kinetic energy in the wheel is converted to linear kinetic by the frictional contact with the floor.

Now for the second part. Do you remember those old gyroscopes so many of us had as kids. The ones with the circular wire frames and the gyro in the middle that we would wind a string around the shaft of and then pull it to spin up the weighted wheel. Remember the resistance and the amount of pull you used on that string? And just as importantly remember how the gyro tried to pull itself out of your other hand that was holding the frame while you pulled on the string with the first? You pulling on the string is like the pull of the conveyor belt on the wheels. The gyro frame trying to pull itself out of your other hand is the force the wheel bearings are going to impart to the airplane to hold it back. The conveyor is going to spin up the wheels of the airplane like that gyro rotor to do this. The resistance the wheels have to being accelerated is what is going to transfer the force from the belt to the airplane axle to hold the plane back just like the gyro rotor being accelerated allowed the pull of the sting to be transferred to the frame in your other hand. But unlike you and the string the conveyor is going to keep on spinning the wheels faster and faster without end as long as that prop keeps pulling in the opposite direction. It has to. The only way the belt can hold back the airplane is through the resistance of the wheels to being accelerated. Either the belt runs out of acceleration or the engines run out of gas.

Since the wheels are free to roll on the belt the kinetic energy in the belt is not part of the airplane and wheels system. The belt is just a way of adding opposing energy opposite to the prop generated energy to the airplane and wheels system. This means that the entire energy outputed by the engine and prop has to be matched in the wheels by the belt running against them. One energy must be balanced by the other or the difference will show up as motion.

SO...... if there is a limit to how fast the belt can move then eventually it will be reached and the airplane will be able to accellerate and the plane will takeoff. If you assume that the belt is capable of infinite speed and the wheels can be spun up to an infinite rotational velocity then the aircraft will never move. This is based on the fact that the prop thrust is fixed and the available rotational speed, and thus rotational accelleration, is infinite.

But note that when the engines run out of gas the spinning mass of the wheels now will have the entire fuel tank worth of energy stored in them as rotational spin. Just like that wheel on the wire had your finger's energy stored in it. So the conveyor belt can stop accelerating at that point but it must keep moving at a steady speed. If it was to slow down then the wheels on the plane would act like your model wheel on the wire when it touched the floor after you pulled away the wire and the plane would shoot forward at a rate linked to how fast the belt was slowing down.

I don't want to even THINK about how fast the wheels would be turning to hold back even something like a whimpy little J3 Cub with that little 65hp engine. It would easily be up around a million RPM even after a fairly short time I suspect.

If anyone out there knows the equations for how to convert pull x seconds into some appropriate unit and how to convert that into the rotational acceleration required for the wheels to match that input feel free to post up. Make whatever reasonable assumptions for pull and wheel weight and size you want for a practical example. It may be easier to use model sizes since we are more familiar with them. For example 10 lbs of pull from the prop and three 3 inch wheels (trike gear) that weigh 2 oz each just to make things easier to comprehend. Figure on the fuel lasting for 10 minutes and come up with an RPM figure for the wheels at the end of the engine run.

My brain is fried now.......

< Message edited by BMatthews -- 12/8/2005 6:42:36 AM >


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RE: Can It Take Off?? - 12/8/2005 8:22:20 AM   
multiflyer


 

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Status: offline 		Try this out:

It could be simply asked for the sake of argument: should the rotational inertia of the wheels themselves be included as part of the overall resistance to linear acceleration. Normally not significant, therefore the conveyor belt runway must also be given the ability to continue accelerating infinitely in order to magnify this effect significantly.

A counter to this would be that eventually the speed of the conveyor belt runway will be so great that boundary layer (surface) friction between it and the air above will be great enough to generate enough airspeed for the plane to fly.

Or, what if the plane has sufficient thrust to weight to simply drag the wheels skidding? Then the pilot could simply lock up the brakes and eliminate angular momentum of the wheels as a factor. After all, sliding friction is less than resting friction.

Traditionally, hypothetical questions are posed to accentuate a central point. Therefore ancillary considerations of lesser significance within the setup are traditionally disregarded. When sense of proportion is abandoned, the discussion becomes digressing banter.

So how far will these exaggerations of reasoning propagate? For example, should relativistic effects as tangential velocities of portions of each tire approach the speed of light be debated next?

Conclusion:

On the one hand you could say that such a discussion is beneficial because of the variety of subject matter perused. In which case there should be no limit regardless of how esoteric. But on the other hand success in the real world depends on remaining practical, the vary ability that is being departed from. In that case the second post effectively closes the subject.

Multiflyer

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RE: Can It Take Off?? - 12/8/2005 1:28:16 PM   
Craig-RCU



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Status: offline 		I agree completely with Mutiflyer, but BMathews has presented an interesting problem; what speed will a belt be moving after ten minutes of accelleration maintaining 10lbs of force on an airplane through its 3, 3", 2oz wheels (or something to that effect).

First you'd need to find the rotational inertial force (resistance to rotational accelleration) of a 3", 2oz wheel on an axel under some accelleration conditions. On Earth the accelleration due to gravity is 9.8m/s per second. To measure rotaional inertial force, you could cut a small pie section out of the wheel (lets say a 1/16 pie section), prop up the hub end at the axel axis point and weigh the pie section from the tread end then multiply that weight by 16. That should give you the rotational inertial force of the wheel at an acceleration of 9.8m/s per second. I'm not going to do all that but for convenience sake, I'll just estimate the value to be 1/3 oz. Since there are three wheels, that makes a total of 1oz of inertial force at 9.8m/s per sec. 10lbs = 160oz, so 160x9.8=1568m/s per sec accelleration is needed. 10 minutes =600 second so the speed of the belt would end up at 600x1568 =940,800m/s =2.1million mph. The speed of light is 299,792,458 m/s for reference. The circumference of a 3 inch wheel is about .25meter so the wheels would be spinning at 3,763,200 r