Well that’s the thing. It most definitely can. The original puzzle was meant to trap people who had the perception that aeroplanes were propelled by their wheels, when in fact they are propelled by air.
In other words you can introduce a conveyor belt below the craft, but all it does is increase drag in the form of friction at the wheel/belt interface.
Now imagine placing an aircraft on a conveyor belt, but this time you removed the plane’s wheels. Now it’s just a hunk of metal sitting on the belt. I’m sure we all know nothing is going to take off and fly in this experiment. We assume this because we have prior experience of the difficulty of moving huge objects over rough surfaces, and we intuit that no engine is powerful enough to accelerate an aircraft to takeoff velocity when it’s dragging on tarmac. Well what if we used a hypothetical lubricant that reduces the coefficient of friction to 0? Well that plane would definitely take off! Now what we used wheels instead, since we know such a lubricant cannot exist? Sure, the coefficient of friction is not 0 at the moment, but it is sufficiently low enough for the engines to meet and exceed, this surplus being the net force that produces acceleration.
We should all take note that friction caused by the conveyor belt is the key to this riddle. If we do not take friction into account, the aeroplane would always take off…even if the belt was made of sandpaper and moving at the speed of light. The existence of the belt itself would be inconsequential to the riddle.
If we do however take friction into account, then yes it is possible to drive a belt fast enough such that it’s friction component overcomes the force output of the craft’s engine, such that the plane cannot take off. I believe this is what the original poster was trying to put across. He however grossly overestimated the practicality of such a possibility. Consider that if the presence of wheels of an aircraft decreases its coefficient of kinetic friction to 10% (compared to just the fuselage dragging along on the runway), call it 0.1–this means that the engines have to work out propelled air at an amount of force equal to 10% of the aircraft’s weight just to keep it from sliding backwards with the belt.
This is where all the issues should become clear. The amount of counterproductive friction caused by the belt on the plane is a function of the plane’s weight and the wheel coefficient of friction. Now even if the belt was moving at the speed of light…the friction caused by it against the plane is still the same amount as ever. Its the same amount the engineers factored into power requirements of the engine they placed on the plane. What this means is that though the fast moving belt will cause the plane to move backwards, whether at -10mph or -100,000mph, the engine would still be able to accelerate the plane back to 0mph (stationary with respect to Earth) and then to +400 or whatever the takeoff velocity is. Because the force required to do either is the same in both cases. It just a matter of time and amount of fuel available, etc.
But if we get into that, I’m not coming back to this thread.
I just want to add that the coefficient of kinetic friction is a function of the plane’s weight and so any characteristic of an antagonistic runway moving below it is irrelevant. The plane will always take off.
While I also said that it is possible to manufacture a conveyor belt that COULD in fact prevent an aircraft from taking off, that would however, not be in the spirit of the original question at all since it would require the adding in of various aspects of physics that weren’t required for the answering of the question in the first place.
An airplane faces this belly scenario in flight whenever it is in a headwind. The headwind is essentially acting as the conveyor belt. In a headwind the plane requires more thrust to maintain speed, so would not the plane on the conveyor belt require the same?
No, it it not. One acts on the forward airspeed, the other, on the rotational speed of the wheels. Within practical limits, the rotational speed of the wheels does not affect the forward motion of the plane.
You do know that the wheels aren’t powered, like a car, don’t you?
Is a boat propelling itself forward against the current not the same as the plane propelling itself forward on the conveyor belt? If so then it would require more thrust for the plane to come up to speed just as it would the boat. This being the case, is there enough thrust available to propel the plane to take off speed?
That was my original take as well (I thought people were saying it would indeed take off while stationary, which is absurd of course). It took several rereadings before I grokked the actual point.
You are completely missing the point. A plane propelling itself “against the current” will require more thrust. There has to be some loss due to the conveyor belt moving in the opposite direction of the plane…just as if the conveyor belt was moving in the same direction of the plane it would require less thrust to get up to speed.
example:
[ul]
[li]On pavement a plane requires 100 knots to take off.[/li]
[li]On a conveyor belt moving forward at 50 knots the plane only need produce 50 knots of thrust to take off…and then it would need to increase another 50 knots to maintain flight once airborne.[/li]
[li]So would it be logical to assume that on a conveyor belt moving against the plane at 50 knots the plane would require a thrust of 150 knots to take off?[/li][/ul]
The same distance with respect to the ground? No, because the belt is moving the other direction.
The same distance with respect to a spot on the belt? Yes. If the belt is moving at a constant velocity.
AClockworkMelon said:
This is the root of the issue. The vague statement about the belt moving backwards at the speed of the wheels moving forward makes this an interpretation problem, not a physics problem.
If you interpret that description to mean the belt keeps the plane from moving forward, then the plane cannot take off because it gets no airspeed.
If you interpret that description to just be an attempt to describe how a treadmill responds with a car on it to keep the car in place, then the belt matches engine output, but the wheels rotate faster. Ergo, the plane moves forward, and can take off.
Can you answer your second question now?
Armchair Aviator said:
As Musicat said, a headwind has a different interface to the airplane than a conveyor belt. The headwind is pushing on the airframe. The treadbelt is pushing on the bottom of the wheel. The wheel has the ability to rotate, the airframe does not.
A very tiny amount of rolling friction in the wheel bearings, and another very tiny amount from tire deformation, but that’s all. Nothing compared to propeller thrust.
Nope. The wing lift is based on the airplane’s speed relative to the air, not the ground. A plane will take off faster into a headwind than a tailwind, though.
Thrust is a unit of force, knots are a unit of speed. The question is nonsensical.
That would only be true if the wheel brakes were set. If you had a plane stationary on a treadmill and then accelerated it to 50 MPH the plane’s momentum would keep it relatively motionless and the wheels would spin.
In the original problem the wheels are not braked, they spin freely and impart just a small amount of motion to the plane from bearing friction.
It seems like there are three takes to this question:
A person has no idea of how planes work and just makes silly statements.
A person realizes that the wheels will only impart a negligible amount of movement to the plane.
A person understands the above, but since it is a finite amount of force you can say that there is a speed at which the treadmill’s speed could be an issue (ignoring the physics of building a treadmill that can go several thousands of miles an hour)
But in the end it really comes down to whether you understand the difference between how a car moves forward and how a plane does.
Yes but the plane IS moving at 50 knots on the forward moving conveyor belt before even engaging thrust, thus receiving 50 knots of wind at rest…relative to a bystander it would be moving forward at 50 knots. Anyway this is actually not relevant as the plane in this scenario starts from a stationary position and is not moving backwards.
So the way I see it this entire debate hinges on only two factors:
[ul]
[li]Wheels on a runway are being pulled forward by the planes thrust.[/li][li]Wheels on the conveyor belt are being pulled forward by the planes thrust as well the conveyor belt is slipping underneath them.[/li][/ul]
If this is of no consequence then the plane will take off the same as on a runway.
In a no wind situation a plane needs to reach a speed of 100 kts to create sufficient lift to take off. This would require maximum thrust for the plane in question at full capacity.
In a 50 knot head wind would the same plane still require full thrust to reach 100 kts IAS or 50 kts TAS?
Note that most aircraft are capable of generating more power than the bare minimum necessary to take off.
It will take the same power from the engine(s) to yield the 100-knot airspeed specified as necessary for takeoff.
Compared to the no-wind situation, it will take less time and less total energy to reach that airspeed.
Note: Your phrasing here indicates that you may not understand the correct meaning of IAS (indicated airspeed) and TAS (true airspeed). In particular, TAS is not the same as groundspeed.
