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[DJ]

Of course a plane couldn't take off without any airspeed. So yeah, if it's staying in one spot and there's no fast headwind, it'll stay put (unless it's a VTOL plane). But that was never the question.

[Urist_McDrowner]

The planes wheels would be spinning irrelevant of it being pulled forward. Imagine a rolling chair with well greased wheels. You would still move in the direction of the conveyor, but less so than if you were just sitting on say, a crate. Unlike in a car, where the wheels are the propulsion, a plane's wheels are mostly irrelevant and are there to reduce friction.

[Facekillz058]

Another though as to an answer to the question.

Spoiler (click to show/hide)

Motion is relative.
You are moving, compared to the conveyer belt, but not compared to the world.

[goblolo]

1-2-3-5! It's the Fibonacci sequence!

Damn, I've lost ability to understant things!
What is the pattern - 1-2-3-4 (natural numbers), or 1-2-3-5 (fibbonacci)? However, fibonnacci is 1-1-2-3-5, so 1-2-3-4 is gonna be a pattern.

Btw, in my first task ( 1-4-10-20...), is it a freaking pattern where you should subtract numbers from each other, then subtract results etc.? It is a "guess what I'm thinking about right now!". What do I want to say: some patterns can overlap (as you've showed with 1-2-3-5), and guessing the answer by the pattern, not by the actual conditions and restrictions, is not the right way to solve (or develop?) tasks.

Yes, there can be tasks to show your knowledge in the world of patterns. In general, it will show that you can think in different ways, find and successfully use patterns (or a "behavior", if we speak about large areas of research, not about number successios, but about experiment results). I think it should be the common task to find patterns used in the code sample when you are going to work as a programmer, or to define result-to-parameters relations of the scientific experiment if you are applying your resume on the sort of a scientific job.
But when I take an IQ test with 10 task of the 1-2-3-5 kind and RRRAGEKILL it with polynomials (writing down the next number to be 0) and get my IQ = 18 (even when people with much higher IQ doesn't know what is Lagrangian polynomial), I am offended a lot!

[goblolo]

I've just created a new question! Less interesting one, but still!

After some sort of a space disaster planet Earth get strong winds. To be precise, all the air mass are spinning around the earth axis. Thus we have light wind blows on the poles and very fast wind on the equator. About 800 km/h.
But the mankind found benefits in it! Assuming that we have enough air flow without even starting an aircraft engine, we can take off even with empty gas tanks! But stop! Will it take off?
And can we use it for plane flights?

[Criptfeind]

I don't see why not. I mean. I've had kites take off without a run up in a strong enough wind. I don't see why it would not work the same way. Good luck with controlling it without any fuel. But it should at least get off the ground.

[WealthyRadish]

Hmm, I actually don't think it will work. Assuming the plane is bolted to the runway and then released facing the wind, the inertia of the plane would resist movement and cause the plane to move slower than the wind, allowing airflow through the wing and raising it, but it would quickly accelerate to the windspeed and lose that airflow. So it might pop up in the air briefly, then roll along the ground at the airspeed, depending on the forces involved (weight and lift from the wings). Could be wrong, though.

[Soadreqm]

It will take off, accelerate until the relative speed of the wind is no longer enough to keep it up and then crash back down. Unless it's tethered to the ground, kite-style. To actually stay up you'd need some kind of anchor to drag behind you, and that sounds like a troublesome engineering problem.

It'd still get used for air travel, of course. If you have a stable 800 km/h wind, you can circumnavigate Earth in about fifty hours in a hot-air balloon. Designing a flying machine to land in conditions like that would be tricky, but if you solved that, flying would be cheap indeed.

[Starver]

Selecting things to reply to...  Some people get it, of course.

Well, if the wheel are firctionless, they won't turn at all. The conveyor belt will break speedlight in a fraction of a second, and collapse into a black hole from all the extra energy. Then suck the plane in.

I think this is as massive a misunderstanding as any other.

Assuming zero friction on the wheels (either on the gear-to-ground interface, e.g. the same as a perfect frictionless ski, or in the bearings of the wheel itself so that the wheel rolls true to the given ground without any resistance) then the plane that gets up to speed in pushing itself forward at 100 knots would only 'force' the conveyor to go backwards at 100 knots.

To go lightspeed backwards, the belt would have to be mirroring (by whatever engineering art we ascribe to both it and the reference aircraft) a plane that is taking off at 'c', itself.


Anyway, I'd already mentioned (many times) the possibility of friction causing problems.  More so than I give it credence.  Last I heard, civil aircraft worked on a safety factor of roughly 1.5 (military aircraft closer to 1.2, for performance benefits, although doubtless the real figures are classified and subsystem-dependant), which means that landing gear should be able to handle half as much again stress as it would encounter at the most stressful time (which is probably landing, the wheel having to suddenly spin up to a maximum rated landing speed) without breaking.  Thus I think that a 'regular' take-off (with unexceptional loading characteristics giving a 'normal' take-off speed in all regards except for wheel rotation which would be doubled) would be handled without problem.

It's quite hard to work out the maths for the wheels, given we are never expected to have any figures, but a friction-suffering wheel that loses a given amount of energy to friction at 100 knots (stationary) ground-speed would probably lose either two times that amount (proportional to speed increase) or four times (square of the doubling) with a 200 knots (retroactively backwards) ground-speed.  A wheel that already accounts for a 25% of efficiency loss under normal conditions doesn't sound very practical to me, never mind 50%, so I doubt we're approaching 100% loss (as it approaches 200kn, asymptotically).  General inertial resistance and air resistance are going to be additional constants, but if there's a nominal 5% (becomes 20%) wheel-friction involved (by the time we reach V2_min, anyway, but of course lower at the lower speeds prior to this) that's 15% extra resistance, and I can't imagine that would take us up to "impossible to reach V2" territory, even adding in those other constants (e.g. that normally 85% of effort is put in to overcome headwind and inertial effects... especially the headwind bit).

Hang on, because you're pushing against the air, the automated treadmill that tracks and matches the speed of the plane won't work?

Bluh!?

Surely because the plane is going forward, the treadmill will match that to cancel it, meaning that despite it pushing against the air, it won't go forward?

  The treadmill isn't moving as fast as it needs to go to drag the plane backwards.  It's moving as fast as it needs to go to match and mirror the forward movement.  As long as this doesn't adversely affect take-off characteristics (see above for why I don't think it doesn't), it'll not slow down the plane significantly.  And if it does slow the plane down then, then the conveyor itself never gets up to full speed, either.  It wouldn't go faster while the plane itself is brought (or kept) to a halt.

I think what the question was aiming for is, if the plane is sitting still but the runway is moving like a conveyer belt would it generate lift?

I'll disagree with this.  That's the misreading of the question that causes the whole fuss.

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?

"plane moves in one direction"
  => V_P, so that x_{plane position}=V_P*t;
"conveyor moves in the opposite direction"
  => V_C (which is negative), so that x_{conveyor position}=V_C*t;
"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)"
  => V_C = -V_P;
  thus x_{conveyor position}=-V_P*t and -x_{plane position}=V_C*t

(For instantaneously attained speed, that is, but you can can also equate "half a.t²" formulae similarly, and the more complex formulae where acceleration is progressively retarded by rolling resistances in some way that needs integration/differentiation...)

For all t that are > 0, x_{plane position} is non-zero (and so is x_{conveyor position}, and thus also x_dif between the two x positions, which leads us to V_dif=2*V_P) so long as the Vs are non-zero.  And the only way that V_C can be non-zero is if V_P is non-zero.  If the plane doesn't move, then its conveyor doesn't, but the plane would never ever be expected to take off from a stand-still, and this only happens wutg the engines not being used (or so low powered that rolling resistance from a stand-still isn't overcome, retrograde runway or not!).  Which isn't part of the question.

Which is not to say that V_P1 with the resistance isn't < V_P0 without the resistance (for a given engine power), but V_P1 << V_P0 such that it's not possible to reach V2_min with all engines going full blast?  No, probably not for any decent plane...  (I worry more about the effective runway length, which necessarily has to be at least as long as a regular one in all its conveyor belt glory, and the the pilot who has to be more careful keeping orientated to the moving runway's centre-line.)

[Itnetlolor]

If I remember this correctly, the engine is independent from the wheels, therefore the power comes solely from the props/jets; with that in mind, the treadmill also works as a compressed runway, matching the speed of the aircraft as it speeds up; overall, reducing the amount of takeoff (since the engine only moves faster, and any friction from the treadmill and the wheels drags it back more; but only limited to the maximum friction relative to the weight of the aircraft itself (heavier craft yielding more friction, naturally)), and maybe even landing required for the plane on the strip.

That's at least as far as my understanding goes; and the Mythbusters did do a pretty good job at tackling this myth, even illustrating the explanation above. Overall, I think the takeoff efficiency gets up to a 25% boost (or 1/4 of the normal amount of takeoff distance required to get off the ground). Add some wind in favor of the aircraft, and up that to potentially 33%-50% (Wild guessing here), depending on velocity of the wind and the total power behind the engine. If anything, as long as there's plenty of airflow going past the wings, then it will go up (I think when the Mythbusters took it on, they made sure to subtract the wind variable; so with the full-scale, they made sure there was minimal to no wind; but even so, the wind actually shortened the takeoff further). The treadmill can only keep the plane still for so long, but then the engine and wind will take over not long after (with enough lift to render the craft weightless, and no longer hindered by the treadmill; thus essentially no more friction to hold the plane down, and impossible to pull back by the moving floor.).

[Starver]

I'm going to come over all sarcastic, I think, but I'm wondering if you aren't coming up with the right (FCVO "right") answer for totally erroneous reasons.  Let's break it down/

If I remember this correctly, the engine is independent from the wheels, therefore the power comes solely from the props/jets;

This is how planes work.  Your memory does you great honour, sir.

with that in mind, the treadmill also works as a compressed runway, matching the speed of the aircraft as it speeds up;

I think you're imagining the runway as moving in the same direction as the plane.  Such that less (moving) belt is being used because it's "keeping up" with the plane.  (c.f. the comment about "I don't know why they developed catapults" comment from that web-site, that I quoted a while ago...  Sounds like you have the wrong end of the wrong stick, here...)  Of course the amount of conveyor belt infrastructure (i.e. the fixed (but, of course rotating) rollers that guide the belt one way or another) is going to be at least as long as the standard static airstrip it is replacing.

(As an airstrip is normally not going to be exactly as long as needed for a standard plane to take off from static ground. I'm going to say that a similar length of conveyor-top ('replenished' at the fixed front end as it moves backwards, of course) is probably going to be long enough even to deal with a slightly-more frictional take-off situation with an extra percentage point or two of friction through the wheels, so it would still be a safe take-off situation, albeit with slightly lower margins than before.)

overall, reducing the amount of takeoff (since the engine only moves faster, and any friction from the treadmill and the wheels drags it back more;

...except here you're ascribing to it some sort of air-moving ability?  To provide an assisting headwind?  Personally, I would assume unless there was mention of beltside-mounted baffles, or somesuch, that (except for a very close-to-ground(/belt) zone), that air (insofar as that which the airscrews/jets works against, and that the wings and other bits of fuselage move through) is not moving at all, w.r.t. stationary ground.

To that end, I still subscribe to it being a normal(ish) ground-length.  But maybe a little longer (for the given friction factors) but no shorter at all.

but only limited to the maximum friction relative to the weight of the aircraft itself (heavier craft yielding more friction, naturally)), and maybe even landing required for the plane on the strip.

I'm lost on this one.  I'm assuming a same-weight aircraft under both standard and belt-borne take-offs.  And landings aren't discussed.

That's at least as far as my understanding goes;

..and yet I don't understand your understanding, I think.  Please correct me where I've gotten it wrong.

and the Mythbusters did do a pretty good job at tackling this myth, even illustrating the explanation above.

Can't see this (national limitations, on the Youtube link, cited), but as I understand it they dealt with it how I see the problem, and succeeded,  ICBW on the first count, though it does appear they succeeded at least.

(This interposting quotes thing is getting sillier than I might imagine, let's try it freestyle...)

And... no, no boost.  And I'm assuming no wind (more or less than over a stationary runway, at least).  Airflow over the wings is proportional to absolute speed (forward), which is significantly unaffected by the equal (but opposite) absolute speed (backward) of the surface the plane is rolling over during take-off.

The treadmill cannot keep the plane still at all, and you're falling into the trap some of the nay-sayers seem to have done.  It (the conveyor) only travels backwards when the plane goes forwards.  If the plane isn't travelling forwards then the treadmill isn't travelling backwards, thus even if it were a car in the conveyor it wouldn't stop the car from moving.  It would go as fast backwards (based on absolute stationariness) as the car is going forwards (based on absolute stationariness, not based upon relative speed to the conveyor) and so at all times the attempted speed of the car would result in half this speed by 'absolute' measurement, the difference being the other half of the matched speed being that of the retrograde conveyor movement.  Here, V_{car wheels} is equated to V_{car absolute}-V_{conveyor absolute} (which is negative, thus it's just the sum of the two |V|s, thus the car wheels are going twice as fast as either the absolute car speed or the (equal) conveyor retrograde speed).  But in a plane then you have |V_{plane engines}| equalling |V_conveyor|, only oppositely signed, with the plane's landing gear passing over the belt at V_{plane engines}-V_{conveyor absolute}, thus V_{plane wheels} == V_{plane absolute}-V_{conveyor absolute} == V_{plane absolute}+V_{conveyor retrograde} which is just another V_wheels=2*(|either V_absolute|).  IYSWIM.

Friction to hold the plane down?  No, sorry, I'm lost.  Friction to make the plane slower, I would understand.  And consider either absent (in a simplified example) or sufficiently inconsequential (in a true-to-life example) to mean there's still no practical problem with take-off (maybe uses 10% more runway or something, plucking an arbitrary figure out of the air which is probably an exaggeration anyway).


...so, I'm probably looking very rude, but I wanted to make sure that there were no misunderstandings left.


To the continued nay-sayers, the possibility that the problem is "the conveyor moves backwards as fast as is needed to counteract (via exponential friction on the plane wheels) all forward force by the plane given by the engines" is a totally different problem, I'll allow and would (assuming this does not also drag enough air across the wings at near-ground level to effect lift).  But that isn't even the problem as quoted on page 1 of this thread, and I'm afraid you've got no chance of convincing me that it is.  You're welcome to your answer to your problem, but I'm (as you might be able to detect) quite agitated that anyone would think that this other scenario (or any part of this scenario) is a part of the originally posited question.

Rant over?  Maybe.  (It wasn't supposed to be a rant, but may end up looking like one, I'll allow.)  Having gotten it (whatever 'it' is) out of my system I shall now do my best to stay quiet except for any direct challenge.  But the mathematical adrenaline is still coursing through my veins, so I may not be accountable for any immediate reactions I might make.

(Still, I regret any perceived abrasiveness.  Not my intention, but a possible interpretation nonetheless.  [note=to self]Shut up about it![/note])

[10ebbor10]

As for the conveyor belt making the plane take of on a shorter distance, the only thing it does is giving the engine more time to warm up. Same thing can be estabilished by just locking the brakes till the engine is at full power/overcomes the friction. In fact, it might even lengthen the distance, since the conveyor belt slows the arceleration down untill the critical point where the lift starts to negate the friction. If the slowing effect of the conveyor belt is strong enough to stop arceleration completely (which is extremely unlikely, considering a plane applies force through it's engines, and not on it's wheels) it won't be able to take of. However, at this point you are applying a ridicilous amoint of friction to the wheels, and they'll probably catch on fire.

((Ie, mostly just restating things))