I'm no physics expert, but the above scenario resembles a technique used by single engine planes in an effort to 'take off' from a muddy runway (field). Ramp up the power without moving forward until sufficient thrust is achieved. Then, glide along several inches above the muddy field until 'liftoff' speed is reached. Pull Back on the stick, and away you go. Whether a non-Harrier jet can perform this maneuver, I cannot say, but single engine prop planes can, so my answer is yes, the plane can achieve lift-off. - Spydertrader
TL: >Correct, the wheels don't drive the plane. Good, we agree. >But in your scenario the conveyor belt is >moving in the opposite direction at the >same speed. Yes it is. >The equivalent of climbing up a >down escalator. Uhh... I thought we agreed that the wheels weren't driving the plane. If the wheels aren't driven, then they aren't "climbing" anything. When you climb an escalator, you are slowed because the surface you are pushing against is moving away from your desired destination. If your pushing against the air instead of the conveyer/treadmill/escalator, you're set 'cause the air isn't moving against you. >If I'm climbing at the same speed the >escalator is moving downward I'm going >nowhere, I'm stationary. We agree as to the escalator, but you're personally driven through the connection to the ground (legs and feet) as is the car (wheels, tires). The airplane isn't driven from the ground so the comparison isn't valid. >If the plane is stationary there would >be no relative wind, where's the >lift coming from? Yes, if the plane is stationary, there will be not lift -- but the plane will not be stationary, will accelerate down the runway at *very near* normal rate and will take off in *nearly* it's normal rotate distance. Lift will be created just as usual. The wheels however will be rotating at twice the normal speed. JB
You are correct in your ultimate answer, but I don't completely understand your described method so I can't say if your reasoning is correct. JB
Let me reword the question and make the scenario more specific... I am in the cockpit and flying a simple Cessna 172. (single engine, takeoff speed ~75mph, max speed ~150mph). You are in the control tower. In your hand you hold a dial control with which you control the speed of the entire north/south oriented "treadmill runway". This dial in your hand allows you to change the speed of the runway surface from 0 to 151 mph both forward and reverse, smoothly and quickly. I pull out onto the south end of your runway and point my nose north. The winds are calm. I go through my normal checks, run up the engine and release the brakes. When you see me run-up and release the brakes, you decide you don't want me to leave. Can you stop me? Can you stop me from moving down your runway? Can you make me go backwards on your runway. Do you have any control at all over where I go on your runway? The above is the heart of, and the key to, the original question. JB
Bizarre. Its all about relative air speed, the speed of air travelling across the airfoil. The airliner could not take off, due to the higher wing loading, as opposed to a cessna which generates an additional degree of lift vector forces from prop wash, plus vastly lower wing loading. Not much mind you, but it helps a bit. In either scenario, you have only the surface drag of the conveyor belt itself to provide an induced, artificial airpeed-but only very close to the ground, the turbelent effect would likely stall a plane attempting a normal STOL take off, as it may be unable to counter both the turbulence above this zone, and the sudden drop in relative air speed outside of this induced air "flow". I suspect the cessna could take off, if only breifly, using the mentioned ground effect lift-and due to the innefficiency of the belts "artifial airspeed, its likely lift off could only be acheived at wheelspeeds largely in excess of the planes normal take off speed. Remember it was specified, in calm conditions, ruling out additional lift from the wind. That would be my take on it
acronym: >The airliner could not take off, due to the higher >wing loading, as opposed to a cessna which >generates an additional degree of lift vector >forces from prop wash, plus vastly lower wing >loading. Not much mind you, but it helps a bit. Jet, prop, Cessna, Boeing ... all take off as they normally would. After all, other than the rotation speed of the wheels and their associated minimal friction drag, the planes are unaffected by the moving surface underneath. >In either scenario, you have only the surface >drag of the conveyor belt itself to provide an >induced, artificial airpeed-but only very close >to the ground, ... Actually you don't have to rely on any "induced, artificial airspeed" -- just throttle up the plane and mosey on down the runway till you get to normal rotate airspeed. If you were standing some normal distance from this 'treadmill' runway, and watching the take-off, you would not notice *anything* out of the ordinary -- you would never know the runway was moving in the opposite direction of the plane. JB
It's a trick question JB. The wheels and thrust from the jets are disconnected. In this hypothetical situation, the plane would move forward, no matter how fast you ran the conveyor. For instance, if the takeoff speed is say, 150 mph, and you ran the conveyor at 150 mph, then the wheels would have to spin twice as fast, but it would take off. That is, until the wheels fell off from going 300 mph, or the tire exceeded the speed rating. Ha ha...
Hehe, yeah i was just going to edit that-thinking about it, im completely wrong!! !!!!!!!!! Might be true of glider dynamics, but they dont put engines on planes for no reason In fact, the plane would take off faster than normal, with said induced air flow providing a headwind. It wouldnt effect the thrust at all, and thats what counts. That was fun, but i never said i was the sharpest knife in the draw either. Edit-as for the control tower scenario, it may be possible to blow out the tyres, but thats about it.
Even better, consider the infinity situation - conveyor and wheels could spin at an infinite speed. I see the equation like this: INFINITY (belt speed) vs INFINITY (tire speed) + engine thrust. Infinity cancel each other out, leaving you with engine thrust, meaning you'll take off as normal, like Turok describes.
Haroki: >It's a trick question JB. I find it an interesting question, and perhaps counter-intuitive, but personally, I don't find anything "trick" about it. It's a straightforward question with a straightforward answer. >The wheels and thrust from the jets are disconnected. >In this hypothetical situation, the plane would move >forward, no matter how fast you ran the conveyor. For >instance, if the takeoff speed is say, 150 mph, and you >ran the conveyor at 150 mph, then the wheels would >have to spin twice as fast, but it would take off. You have it *exactly* right. >That is, until the wheels fell off from >going 300 mph, or the tire exceeded >the speed rating. Ha ha... YES, the only limitation in the problem IS the speed rating of the tires. Thrust is a non-issue, but it is possible (obviously more of a problem for higher speed planes) to disintegrate the tires before reaching liftoff. JB