Airplane physics question

Part II : Rear-Wheel Drive Automobile

In a rear-wheel drive car, the friction force between the rear wheels and the ground propels the car forward. The rear tires, through the torque of the motor, push on the road, the road pushes back to accelerate the car forward. There's no slipping of the rear tires, unless the force generated by motor torque exceeds Fmax.

But what about the front tires? What friction force exists between the road and the undriven front tires? The center of the front tires is pushed forward by the accelerating car. If the front tires tires were slightly off the ground, OR if there were slippery grease between the front tires and the road, the front tires wouldn't even rotate ... there would be no force to cause rotation on the fronts, and of course the car would still accelerate forward.

But in a real car, with front tires on the ground, the front tires do indeed rotate (relative to a stationary observer). What force causes this front tire rotation? The friction between the front tires and the road, of course In what direction is this force applied by the road to the front tires? Rearwards, toward the back of the vehicle ... that's the only way to cause the fronts to rotate in a forward-moving direction And, we note that if the road is pushing the front tires back, through friction, to cause the proper rotation, then the front tires must in turn be pushing the road forward

Now, this tire-rotating force acting on the front tires is admittedly small, compared to the force acting on the rear tires causing the vehicle to accelerate forward. But it's interesting to note that the lateral friction force acting on the front's and rear's is also in the opposite direction Like i suggested in Part I, very different lateral friction forces acting on two rolling tires attached to the same vehicle.

Now is also a good time to recognize a classical separation of forces acting on a rigid body ... in this case, the rigid body being our front tires of the rear-wheel drive car. There's the normal or vertical forces : gravity (down), balanced and the vertical force of the road (up). There's also two lateral forces : the lateral force exerted by the accelerating car on the center of the front tires, and the friction force (much smaller) acting in the opposite direction to rotate the tires. In kinematics, it often makes sense to "decompose" these forces into a net lateral force acting on the center of the body (tire), and a rotating force causing the body (tire) to rotate. More later, i think ...

In summary, the road pushes the front tires "rearward" just enough to force rotation without slipping. And the front tires respond, by pushing the road forward. These forces are in opposite direction, but much smaller in magnitude, than the corresponding forces acting between the rear tires & road.

Next up : a typical airplane on a stationary runway

 

Part III: Typical Airplane, Stationary Runway

The wheels of a typical airplane, on a stationary runway, are exactly like the front wheels of a rear-wheel drive car The forward thrust for the plane is provided ... elsewhere. The front wheels are just along for the ride, balancing the vertical force of gravity by a vertical upward force from the runway to the tires.

Laterally, the only force provided by friction is the force necessary to rotate the front tires to prevent slippage at the tire/runway interface. This (relatively small) force must be in rearward direction, to force proper tire rotation as the plane accelerates forward. And, in return, the tires must push the runway slightly forward ... of course, our runway is stationary so it's not moving in response to this slight forward force (the forward force from the tires is balanced elsewhere, in the ground).

Next up, Part IV of V : A plane attempting to take off on a freely moving ... but "passive" ... conveyor belt.

 

Part IV: Plane on a "Passive", Freely-Moving Conveyor Belt

By "passive", i mean a zero (or low) friction conveyor belt. There's no motor trying to move the conveyor under the plane. It's just free to move, however it might be pushed.

If we put a plane on such a conveyor belt, we need to ask how the conveyor might move if the plane attempts to take off. Does it move at all? Does it move forward, or backward?

We can compare to the situation described above, in Part III. If the conveyor belt is "tied down", or temporarily held still for a moment, we have the exact situation described in the previous post. And we saw that the tires exert a (small) forward force on the ground (or conveyor held-still). NOW, we release the conveyor, so that it's free to move. How will it move? Forward, of course, as the tires push it

Will the plane take off? YES it will. the forward force on the conveyor is small, only that need to balance the rotating force exerted on the tires. The thrust of the engines still accelerates the plane forward, relative to stationary observer, and so the wings moving through stationary air provide lift.

BUT ... the freely-moving conveyor does NOT satisfy the constraint, or premise, of the original question (at least, as i read it) Our freely-moving conveyor is moving forward ... there's no slippage at the tire/conveyor interface, but the conveyor is moving forward

To answer the original question, with original constraints (as i understand them), we have to wait until the last and final part, Part V. Thanks for your indulgence so far, guys.

 

Air speed creates the lift.

 

Part V : Answer to the Original Question

First, we have to understand (or interpret) this phrase from the original problem :

"The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation."

What this phrase can't mean : it can't mean that the point on the airplane tire in contact with the surface underneath exactly matches the speed of the surface underneath. Why? Because, as we've seen, that condition is met by any conveyor speed as long as there is no slippage. The contact point on the tire matches the conveyor speed if the conveyor is held stationary, and it also matches the conveyor speed if the conveyor is freely moving ... in which case, as we've seen, the conveyor surface actually moves forward. So this interpretation of the problem statement is too ambiguous to be meaningful

What this phrase must mean : it must refer to the conveyor speed in relation to the lateral speed of the center of the wheel. Somehow, it must refer to a conveyor that attempts to match the forward speed of the center of the wheel ... by actually moving backwards. But is this even physically possible? We've seen that the conveyor actually moves forward, if it's allowed to freely move underneath an accelerating plane. Clearly, this doesn't match the original problem statement. What's needed to satisfy the original premise, is a motorized conveyor belt.

But what happens if we put a plane on a conveyor belt that motorized so that it's surface moves backwards under a plane? Easiest way to analyze this is in two (2) steps :

1. Plane engines turned off. We put a plane, with engines turned off, on a conveyor belt. At first, the conveyor is stationary. I think we'd all agree that conveyor and plane are standing still at this point Then, we turn on the conveyor so that the surface moves backward. What happens to the plane? Does it stay stationary, while the tires just spin under it? Or does the plane move backward (or even forward)?.

Easiest way to understand what happens is to just imagine an isolated wheel on a conveyor that's suddenly turned on to move, so that the surface accelerates. What happens to the wheel? Through friction, it experiences a lateral force exerted on it's rim. Kinematics is very clear : the lateral force on the rim causes a both a translation of the wheel (just as if that lateral force were acting on the center), as well as a rotation of the wheel (because the off-center force exerts a torque about the center). The easiest way, i think, to understand the need for lateral translation, as well as rotation, is to compare the situation to a newly imagined scenario of a wheel where half the force is applied to the bottom, and half the force is applied to the top ... in this case, the wheel would experience rotation but no translation.

So the answer to this first step becomes clear : If we place a plane, with the engines turned off, on a conveyor and start accelerating the conveyor surface backwards, the tires will rotate forwards, and the plane itself moves backwards ... that lateral backward movement is just the same as if the lateral friction force on the tires were acting on the axle of the wheels, in their center.

Finally, if the conveyor acceleration is constant, so will be the backward acceleration of the plane ... forced backward, under constant acceleration and therefore under constant force, by the friction between the plane's tires and the backward-accelerating conveyor.

2. Turn on the plane engines. We now have a new force, pushing the plane forward. If the force is small, we will reduce the backward acceleration caused by the backward-accelerating conveyor ... but the plane still moves backward. If the engine force is too large, the backward acceleration of the conveyor is overcome, and the plane could actually move forward (relative to stationary ground, and stationary air). But if the balance is just right, the engine force is exactly balanced by the friction force of the motorized, backward-accelerating conveyor ... and the plane stays stationary relative to stationary ground and air. No airspeed over the wings, and the plane can't take off.

As far as i can tell, this is the ONLY unambiguous interpretation of the original problem statement

In summary:

The plane does NOT take off. The forward thrust of the engines is exactly balanced by the friction force of the motorized, backward-accelerating conveyor surface. The airspeed over the wings is zero, because the plane remains stationary relative to the still ground, and still air.

The situation is virtually identical to a plane attempting to take off, with the axle of the wheels tied to the ground by a cable. Except in our case, the force "holding" the wheels is the friction of the motorized, backward-accelerating conveyor ... instead of tension in the cable holding the axle.

Thanks again for the indulgence, guys

 

The wheels in your explanation still have bearings, right?

A pilot will rotate (to take off) bases on airspeed, not wheel speed. Wheel speed is not related to flight.

Your analogy of the cable with the axle and the wheels are two different things. The axle is directly connected to the plane and the movement of one directly moves the other.

A wheel of the plane can spin freely without effect on the airframe itself.

GT

 

No, it can't.

Any single friction force exerted on the surface of the wheel causes two things :

1. Rotation of the wheel, because the force is applied off-center.

2. Translation of the wheel, as if the force acted on it's center of mass.

The only way a wheel can spin freely, with no effect on it's center (or axle), is to "balance" the friction forces ... top & bottom ... so that no net translation force is applied to it's center of mass. That's not our situation.

The motorized, backward-accelerating conveyor exerts a single, "unbalanced" translational force, through friction, on the surface of the wheel, and therefore a net translation force on the center of the wheel ... and therefore on the plane itself. It's this very force that physically moves the plane backward when we turn on the motorized, backward-accelerating conveyor ... before we turn on the plane engines.

 

maybe think about it this way :

Just put a wheel (or log) log on a conveyor belt (no bearings, nothing different about it's center ... uniform material throughout). Turn on the conveyor. Does the wheel (or log) move laterally, or just stay stationary and spin?

The answer is that the wheel (or log) does two things :

1. Spins, of course, because the force is applied off-center.

2. Moves laterally as well, as if the force were applied to its center-of-mass.

The airplane wheel, with bearings, wants to behave the same way

 

Edit: My following statement applies to the last thing you said too.

The only way that your statement could be true is if the force of system touching the ground (wheels, bearings, friction of the tires on the asphalt etc) equals that of the opposing thrust.

Given "normal" airplane design, no way do I believe that this is true.

GT

 

Plane with ski skids : engine thrust overcomes friction force at ski-skid interface. Plane accelerates forward, relative to still ground and still air. Wings move through air, creating lift.

Remember part IV (if i may) : plane on freely-moving, un-motorized conveyor does take off

My point is that, to unambiguously interpret the original problem statement, we need a motorized, backward-accelerating conveyor ... and this thing applies a rearward friction force designed to exactly balance the forward thrust of the engines. That's what keeps the plane still ... like a cable tied to the wheel's axle, preventing any movement of the plane relative to the still air, and still ground.

 

why not?

The friction force between a road and tires can accelerate a car forward. So, the friction force between a motorized, backward-accelerating conveyor can certainly move a car backwards (with it's engines turned-off, for the moment) with similar acceleration. If one case doesn't exceed Fmax, the other won't either.

Yes, we're talking about a substantial friction force, to achieve the constraint of the original problem statement. But there's no reason to believe it's physically impossible ... we're talking about a very substantial normal force here as well ...

I didn't say that actually building such a conveyor would be easy ...

 

Given your log example, if the plane is on the conveyer and the conveyer begins to move, will the plane stay in the same place? No, I agree, it won't. It will move.

However, let's say that the conveyer produces 100 hp and 100 lb ft of torque. Will it take the exact amount of opposite force to stop the airplane from moving? No, it will take significantly less force to overcome the friction of the wheels and bearings.

GT

 

I agree with you 100%. But, Rob is also correct in that if you set the brakes, so the wheels can not turn, then the conveyor belt would not turn (therefore matching the speed) and, as long as the plane had enough power to overcome the friction of the locked wheels, the plane would move forward and take off. Again, given enough power.

This situation also fulfills all the requirements of the original question.

 

Would an experiment suffice?

Let's say I used a treadmill as the conveyer and an RC airplane. If the airplane takes off with the treadmill on would that put this to rest?

GT

 

To be honest, i had to pull out my old kinematics book to refresh my memory on this Yes, the plane will move ... in the same sense the isolated wheel (or log) would move. Therefore, the motorized backward-accelerating conveyor is definitely exerting a lateral, translation force ... backwards ... on the plane.

Just remember : the friction force acting on the wheel's surface exerts an identical-in-magnitude translational force on the center-of-mass of the wheel. (That's what i needed to refresh for myself). Additionally, there is of course a rotational force on the wheel because it's applied off-center.

I've haven't gone through any numbers, myself ...

 

hard to do, in practice.

The forward thrust of the engines (or prop) is probably approximately constant. To balance such a constant force, you can't use a steady-speed conveyor. With the engines off, the plane moves backward if the conveyor moves backward. But if the conveyor construction is such that the plane moves backward at a constant speed, there is no net backward force exerted by the conveyor (a net backward force implies net backward acceleration of the plane, not constant speed). With no net backward force, you can't balance the forward force of the engines (or prop).

So the conveyor, with plane engines off, needs to not just move the plane backward, but actually, continually, accelerate the plane backward. There's no physical impossibility here, i don't think ... the friction force is capable of doing it. But it's NOT easy to build.

 

It would be virtually impossible for you to make the speed of the treadmill match the wheel speed of the plane. If you could, the treadmill would eventually reach it's maximum speed, likely before the plane has run out of available additional acceleration force.

 

See post 1.

 

Sorry, I mis-wrote but was commenting on his wanting to recreate the situation using a tread mill and RC plane.

 

See Post 2045





(This is getting rediculous)

 

Imagine a plane is sat on the beginning of a massive conveyor belt/travelator type arrangement, as wide and as long as a runway, and intends to take off. The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.
There is no wind.
Can the plane take off?

It doesn't sound like your pilot intends to take off. If I surveyed 1000 pilots and asked them what are the steps to make a plane "take off" I doubt more than 1% would say "apply full brakes to lock the wheels and full thrust to make the wheels skid along the runway." It sounds as if you are imagining a scenario not intended by the question. The scenario proposed is pretty clear in the question being asked. It is basically asking "Does wheelspeed dictate lift?" while at the same time asking "Is relative airspeed necessary for lift?". The answers are no, and yes, respectively. I don't think a Cessna 152 could overcome wheel friction to generate relative airspeed for takeoff, so even in your scenario it would depend on a massively powered airplane with specific loading to lighten the friction on the wheels.

BT

 

Agreed!

BT