Great, the US Navy could save billions by fitting out every aircraft carrier with a massive conveyor belt! Not!
A parking brake on an airplane is pretty strong. The brakes were locked, solidly; the aircraft skidded through the slush and snow with the brakes locked, and took off. Want to know how I can say this as a fact? Because when we landed on the dry pavement at the other end the brakes were still locked and stayed locked for our very, very short skid down the runway until the tires failed (at about the same time, thankfully) from being dragged down the runway.
I’m quite sure a military jet could do the same thing on a dry runway, both take-off and landing.
Why is this even a question?
Someone (Mythbusters, for example) should set up three evenly spaced conveyor belts in a row and start a plane on the first one. The wheels will be spinning like crazy on the first belt, but the plane will roll off of it. The tires will chirp as they hit the stationary ground, but the plane will continue moving forward as it approaches the next belt, then the tires will be spun fast again, then chirp again on the stationary ground, and repeat until the plane has liftoff. Another plane taking off simultaneously, sans conveyor belts, will travel the same distance and take off exactly the same (but with less annoyance to the pilot).
Do people honestly think a conveyor belt will cause a plane to lift straight up? If anyone believes that, they must also believe planes could LAND on a conveyor belt and do away with all that “unnecessary” rolling around.
Why don’t they ask better questions like “Can a plane with no wheels take off in front of a giant fan?” Yes, it can. But I don’t want to pay to have one built. Nor would I want to land one that way.
Sorry to jump in here… esp on so hot a topic… but goodness so many Engineers and Phycisists and not one comment on the force of friction?
The force of friction comes in 2 flavors… static and kinetic (sometimes called sliding)
The formula is also very simple Ff <= mus N OR Ff = muk N… or the force of friction = coeffiecient of friction (either static or kinetic) * the Normal force
The coefficent of static friciton is always higher than sliding friction (that is the principle ABS works on)…
Just trying to think of a BIG plane… I’ll pick a 747, and assuming it is using perfect ABS while trying to take off (keeping static friction) we can find the maximum force the conveyor can EVER put on the plane…
The 747 has a max take off weight of 870,000 lbs and has 16 ‘main gear’ wheels (we will ignore nose gear to make a ‘worst case’) … we have to assume all 16 wheels share in the N force of the plane… so each one can account for 16th the total weight…
OR N = 870000/16 = 57375 lbs (weight is a measure of the normal force… of sorts)…
So now all we need is the coefficent of static friction for a wheel on this mytical conveyor…
BUT the coefficent of static friction is dependant on material interface… so instead let us see what the coeffiecent of static friction would NEED to be to counter act the force of the thrust…
The Pratt & Whitney JT9D-7R4G2 put out 54,750 lbs of thrust each… and there are 8 of them… for a total of 438,000 lbs… again we need to divide it by 16 and we get a total force of 27,375
so the coefficent of static friction for each tire must be .477 ish… that is EASILY obtainible on lots of surfaces…
SO with the wheels locked in such a way that they do not slide… the tires can easily outpace the force of friction…
That is how you can sit at the end of a runway… run up to full power… and NOT move…
HOWEVER, as soon as you release the wheels… the force equivenlat passed from the tire to the aircraft is controlled by the bearings in the wheels…
The 747’s bearings impart 0.0002% of the force from the wheels to the struts (when freewheeling)… so the coefficent of static friction between the tire and the conveyor needs to be 500,000 times higher… or 238,500 the BEST tires I can find on the stickiest surface… have a coefficent of static friction in the range of 2,000 roughly 2 orders of magnitude too small
The max rotational speed of the bearings for the 747 FAR outpace that of the tires… meaning the tires will explode LONG before the bearings lock up… and you don’t have enough of a coefficent of static friction between the tires and the runway to cancel the forward force (as long as the brakes are not applied AND your tires have not exploded)
Now as long as the runway can not cancel out the force of the jet engines… you WILL have forward momentum… F=ma after all…
Just wanted to mention that, that’s not my quote, but rather my quote of guavajuice737’s quote.
WRT the OP:
Wouldn’t it depend on what kind of plane it was?
If the plane is a jet on a conveyer belt, or a propjob with a single engine-mounted propeller, it is not moving with respect to the air around it, no matter how fast its engine is turning over. That means air is not flowing rapidly over the wings. The conveyer belt is cancelling out any motion imparted to the plane by its engines, relative to the air around it, which I believe is what counts here. So it just sits there, with all the kinetic energy going into the wheels of the plane and the conveyer belt.
In the case of propellers mounted over the wings, the backwash from the props will be forcing air over the wings, even if the plane is not moving forward with respect to the total volume of air around it, and lift will be achieved.
I could be wrong about this, but that’s where my understanding leads me.
What am I missing?
Sigh. This is the whole crux of the question: How, exactly, is the conveyor belt cancelling out the motion? Unless there’s some velocity-dependant friction between the belt and the plane (which to a very good approximation there’s not), or the belt is continually accelerating at a very high rate, the belt cannot stop the plane.
EEMan, friction in all its glory has been covered several times in this thread… Did you read all five pages? It is true that if the wheels are locked, and the coefficient of friction between the tires and the runway is reasonably high, the plane could be prevented from taking off. But this is true both on the conveyor and on a normal runway. If anything, this would make it easier to take off from the conveyor, since it would let you get past the higher static friction.
The fact that the conveyor belt cannot cancel out the motion imparted to the plane by the engines. It can only affect the plane through the wheels, and there isn’t a mechanism to pass that much force through the wheels and axels of a plane.
There’s no connection between the engines and wheels of the plane. Regardless of how fast the belt spins the wheels, the plane will move forward and eventually take off.
I went through all 5 pages (and the 400 post one previous… and a 700+ post one at mythbusters… and serveral others)
In this discussion I had either seen mistakes on what the Normal force would entail… or great descriptions with few numbers (if any)
I thought actually running through what would be required would help… seems it didn’t though 
This problem reminds me of the Monty Hall problem in more ways than just the level of controversy it caused. Both rely on the framing of the problem and playing on people’s assumptions.
In the Monty Hall problem, the answer rests on whether you will ALWAYS be offered a chance to switch, or whether it is on the whim of the host. Why that was not the central issue in the discussions is beyond me.
In this one, a similar diversion occurs. It starts with “put a plane on a treadmill so that the treadmill offsets the forward speed of the engines…”. At that point we should say “STOP!”. Such a setup is impossible, because the tire friction with the ground is irrelevant when compared to the force of the engine. The plane would not stay on the treadmill, it would move forward as if the treadmill were just another runway – whether the ground is moving below it is meaningless. The implied assumption – the one that is causing all the confusion – is false, impossible. The plane would not just sit on the treadmill, it would move forward, same as always. So your moving treadmill would have to be as long as a regular runway.
This question is only controversial because one side is answering the question “Does a plane with zero airspeed take off?” and the other side is answering the question “Does a plan on a treadmill move forward?”. The answer to both is obvious. Cecil has both in his answer, they are just hard to find. I hope he will clarify this soon.
If one were able to construct **a device **to exactly counteract the thrust generated by an airplanes’ engines (either prop or jet) while it attempted to takeoff, would the plane move forward and thus generate enough air speed and lift to achieve flight?
**NO. **
Could a conveyor belt be constructed in such a way to achieve the above?
**NO. **
Let me throw out another scenario which seems much more likely reproducible in the real world than a giant powerful conveyor belt.
What about a wheel inside a wheel? Say you have an inner wheel that’s directly connected to the plane, and around that an outer wheel which is what makes contact with the ground. The inner wheel is free to spin in relation to the outer wheel, with its resistance to the outer wheel approximating the amount of friction as would be encountered by the wheel in relation to the conveyor belt.
If the inner wheel is powered, as in a car, and you floored it, the car would remain almost completely stationary, until after some time the friction would cause it to slowly inch forward. (Like running as fast as you can on ice).
But the jet plane would thrust forward just as it would normally, yes?
To those much more knowledgeable than myself, can this wheel-inside-a-wheel scenario replicate the conditions described in the original question?
My guess is: yes, regarding the question Cecil intended to illustrate, which is all I’m interested in.
I think it could if the amount of friction between the conveyor belt and plane was great enough, like if there were no wheels at all, or maybe even if the wheels were fully braked.
you mean between the plane and the runway… or the wheel and the axle… not the wheel and the runway
BZZZZZZZZT
Try again, here is a hint. The acceleration of your treadmill is 2*(Force from the engines)/(mass of tires).
All this talk about 747’s, supersonic military aircraft, frictionless bearings, infinite speeds, magic conveyors, et.al., is really irrelevant to the original, hypothetical question.
I think that I’ve detected a basic concept that is confusing the issue and I asked this question previously, but no one answered it:
If the aircraft is sitting on a level treadmill, no brakes, no wheel chocks, just the weight (gravity and friction) keeping it in place, and it is facing west. Now the conveyor belt begins to move toward the east, what will happen to the aircraft? Will it move easterly w/ the belt, or will it remain in the same place relative to a given point off the side of the belt?
Whichever assumption you make has a direct effect on your answer to the question.
I think most (all?) would agree that in a real world example, the plane would move easterly with the belt.
But as soon as the engines were turned on and spooled up, even a bit, the plane would move to the west and increasing the speed of the belt wouldn’t have much effect on the situation.
It depends on whether you accelerate the belt enough to overcome the forces tending to keep the plane stationary with respect to the belt. Just how much “enough” is depends on the weight, and gravity, and friction, and tire construction, and so forth.
RaftPeople did a physical experiment on the previous page–post 169.
OK, fair enough. Now since most posters can’t see this as purely hypothetical, let me
try to pose a reasonable scenario.
Suppose we use an ultra light airplane w/ an engine capable of about 40 HP. Assume
that, with a fixed pitch prop, the engine is capable of moving the plane across the
ground at about 75 MPH. Further assume that the takeoff speed is around 40 MPH
w/ minimal prevailing winds.
It’s not outside the realm of possibility that a conveyor could be built to run at 75 MPH.
The belt doesn’t even have to be very long, just enough room for minor movements as
adjustments are made between the plane and the belt.
Now if the unpowered plane moves w/ the conveyor, then it obviously must add
power to maintain a fixed position relative to a given point off the side of the belt. If
the belt increases speed to match the power added to the plane, then there must be
an end point. When the engine, and the propeller, reach their maximum rpm the plane
will be doing about 75 MPH, but the conveyor is moving at 75 MPH in the opposite
direction, so the airplane is still in the same position, relative to the fixed point off to
the side of the belt. Therefore the wings are not moving through the air and cannot
achieve lift and the airplane will not be airborne.
Do the same thing with a car.
Set up a level conveyer belt big enough to hold a car on it with a wall right behind one end of it. Put the car on the belt with it in neutral facing out from the wall. You may need someone at the steering wheel to keep the car aligned. Turn the belt up to 75 mph. The back of the car will, of course, push up against the wall as the wheels spin on the belt.
Now fasten a rope or chain to the front of the car. One average adult healthy man could pull the car off of the belt with just a little more effort than it would take to pull the car across a non-moving level surface.
Do you think that an airplane’s engines acting on the atmosphere are a weaker force than an average adult healthy man?