will it take off?

patrick_g · Nov 26, 2006

Joe, Didn't you appreciate the inertia of the wheels being the "traction" against which the magic conveyor belt worked to hold the plane back? I didn't do the math but spinning the wheels at relativistic speeds is an interesting concept. That puts the conveyor moving at relativistic speeds and the air in contact with it would be accelerated to high speeds, significantly above that required to develop sufficient lift to get the plane out of contact with the conveyor.

Once airborn in this "ground effect" the A/C would begin to accelerate forward. An interesting question is whether or not the A/C would crash when it passed beyond the extent of the conveyor and the high speed airstream generated by it. Inflow air to an air pump is quite difuse and of limited linear extent compared to the outflow. The A/C might even be trapped in ground effect just beyond the end of the conveyor.

Is the action of the conveyor defined for the condition where the A/C lifts off? If the conveyor suddenly stopped when the A/C lifted off the plane would settle back down.

Oh, and it doesn't makek a dimes worth of difference if the plane is a jet, conventional propellor driven (tractor or pusher), turbo prop or propelled by deflating baloons or compressed air tanks.

Patrick

TomKioti · Nov 26, 2006

That plane ain't flying until the wing produces lift. Lift cannot occure unless air pressure exists on the top and bottom of the wing. This only happens with actual motion. Thrust creates the motion. The treadmill prohibits the required motion needed for flight. Myth busted IMHO. I'm sure I've repeated other comments but hey...this is fun.

Cool thread

duck_man · Nov 26, 2006

what color is the plane?

Egon · Nov 26, 2006

It Takes off and it lands!

http://http://youtube.com/watch?v=bGB8CyrQqn4

MossRoad · Nov 26, 2006

joerocker said:
A helicopter is nothing more than a plane that takes off straight up. Would a "magic conveyor belt" stop a helicopter? It's NOT the ground or anything on the ground that affects takeoff.

I KNOW you were kidding...

Not the same at all.

daTeacha · Nov 26, 2006

Does the problem actually state that we are considering a fixed wing craft? It's been so long since I read it and there have been so many interpretations, nuances, implications, improvisations and impossible inventions that I'm not sure anymore.

Anyways, just for fun I presented the thing at the Thanksgiving dinner gathering, which included a couple of engineers and some pretty sharp programmers and other thinking type people. There was no agreement since all the familiar trains of thought emerged. The interesting one was my SIL who at first said it would fly, then later decided it wouldn't if the force acting against forward motion through the wheels and induced by the MCB was sufficient to counteract the force acting to produce forward motion by the engine.

SPYDERLK · Nov 26, 2006

NorthWestBlue, I will address your post point by point:

"Larry,

there is no backward jolt on landing, at least not measurable on an airpeed indicator. There is an almost (Struts, tires, ground all absorb impact) instantaneous arresting of vertical acceleration, but forward velocity remains fairly constant (minus air drag and a very small amount of rolling resistance)." I believe we settled this in a succeeding post; yes, there is a jolt - energy of translational speed is stored rotationaly in the wheels slowing the plane. As you pointed out later, the stored energy is fed back into the airframe as it slows, making it just as hard to stop as if if hadnt slowed as the wheels spun up. - - Good point. Never disagreed before or after you said it. Energy is neither created nor destroyed

"Ok, now somebody help me here. I must be missing something, it's been a long time.

The force due to friction can be determined using various forms of the basic equation: F=muN. Good

where mu (greek letter) is the coefficient of friction. It varies for static or moving (kinetic) and for differing materials. Yes

N is the normal force (in Newtons) which is the mass multiplied by the acceleration due to gravity (9.81m/s^2). F=MA - - yes

Empty weight for U2 in my previous example = 6800kg

Pilot and fuel = 1200kg

Total A/C weight = 8000kg. Good info

mu for rubber is .5 to 1.0. Cool - thats what I thot against road surface

"mu for rolling resistance hard rubber is .01 to .015." Good info,but not germain to the topic of the conveyors action subtracting engine energy from the airframe and storing it in the wheels. This rolling loss would just lessen the wheel acceleration needed to hold the plane still

"mu for steel bearings is .001." Ditto former

"mu for Teflon is .04". Slick stuff. Making skids out of the wheels is, I believe, an abuse of the boundary conditions. The conveyor has always implied wheels. I believe the expected wheels should act AS wheels.

"So, we all seem to agree that the wheels will roll at some point after thrust is applied. Let's use mu for rolling of .015

F=muN
F=(.015)*8000kg*9.81m/s(squared)=1177.2 N" OK, but this is only for wheels turning at speed of motion for whatever they are supporting. The loss causing the retarding Force you calculated is caused by the cyclic flexure of the tires. In resisting the plane thrust the conveyor spins the tires very fast and continues increasing the spin rate in order to store the engines energy output. The tires flex MUCH faster -- they "travel" much further than those that validate the assumptions in your equation. However the equation is right in all cases if used scrupulously. We must look at ENERGY of tire rotation, ei Work per revolution, Force x Distance - or 1177Newtons times the circumference. This, in N*Meters is the amount of energy [work] absorbed in each rev of the wheel. The wheel will heat quickly, dissipating many (and increasing) KiloJoules/sec [kilowatts]. This is subtracted from engine energy rate [power]- gone, not stored in the wheels and just lessens the acceleration requirement for the wheel - conveyor interface. Indeed it would finally use ALL engine power and the conveyor cold stop accelerating.

"The standard U-2R with one Pratt & Whitney J75-P-13B turbojet has a sea level thrust of 76kN (17000 lbf)

76000-1177.2=74822.8 N" Im not sre how to deal with energy here since thrust is just force - not energy or power. Yet I know an engine cannot proguce more energt than it consumes oops-see next post

So we have 74.8kN of thrust to oppose this magical contraption. Throw in that F100 engine in my example with 128.9kN of thrust for even more fun.

Air drag is inconsequential because lift will cancel it out as the airplane accelerates through the air and FLYS.

So can anyone see where I went wrong on my thinking.

Oh, if anyone cares I think RobS answered the original question correctly way back on post #5.

SPYDERLK · Nov 26, 2006

Pls see previous post. Hit the wrong button.

76000-1177.2=74822.8 N" Im not sre how to deal with energy here since thrust is just force - not energy or power. Yet I know an engine cannot produce more energy than it consumes nor at a faster rate than it consumes it, and the rate that thetires are absorbing energy is increasing.

So we have 74.8kN of thrust to oppose this magical contraption. Throw in that F100 engine in my example with 128.9kN of thrust for even more fun.
No rockets allowed in this problem. We cant stand that much fun.

Air drag is inconsequential because lift will cancel it out as the airplane accelerates through the air and FLYS.

So can anyone see where I went wrong on my thinking. Well yeah. Significant redefinition of the scenario by converting to skidders and a rocket engine notwithstanding, you introduced real world variables that only make it more difficult for the plane to take off.

Oh, if anyone cares I think RobS answered the original question correctly way back on post #5. Yep. Unless the original was asking us to think a little further on the implication that forward motion was prevented.
Larry

MossRoad · Nov 27, 2006

Egon said:
It Takes off and it lands!

http://http://youtube.com/watch?v=bGB8CyrQqn4

I can't get that link to work.

turnkey4099 · Nov 27, 2006

joerocker said:
MossRoad,

You're KIDDING right? I SURE hope so!

Stand behind a jet when it's sitting at the end of the runway and applies full throttle and you'll end up being blown quite a distance. Maybe killed. I've seen it flip trucks passing too close behind a jet engine.

You were kidding right?

Give the blow drier enough power and it will flail around wildly "trying" to take off, your hand won't fly unless it's powerful enough to lift your entire body. Which is EXACTLY what an airplane/jet engine IS. It's powerful enough to move the plane fast enough to eventually take of.

You were KIDDING right?

A helicopter is nothing more than a plane that takes off straight up. Would a "magic conveyor belt" stop a helicopter? It's NOT the ground or anything on the ground that affects takeoff.

I KNOW you were kidding...