I don’t think the treadmill shortens the take-off run. The propellor is pulling against the air and the jet is pushing against it. So the airplane moves forward relative to the air until it gets takeoff speed. The wheels and the treadmill are irrelevant.
My assumptions of the OP: 1. The treadmill has it’s own power source. 2. The treadmill is equipped w/ sensors that monitor the forward movement of the wheels on the treadmill and adjust the treadmill speed according to prevent the aircraft from gaining any forward speed through the surrounding air. 3. The maximum speed of the aircraft is limited by it’s engine’s capability, therefore the design of the treadmill is capable of withstanding those limitations. 4. As the engine throttles up the thrust attempts to move the aircraft forward on it’s freely turning wheels. 5. The sensors detect the forward movement of the wheels and increases the speed of the treadmill, thereby preventing forward movement. 6. The propulsion device, whether jet or prop, is creating a pressure difference between the surrounding air in front of the engine and the rear of the engine, however, since the ground (treadmill) is moving backwards at a speed equal to the thrust created the aircraft does not advance through the surrounding air. 7. While there is airflow through the propulsion device, there is no airflow over the wings, no airflow, no lift, no flight. 8. The energy required to move the treadmill will appoximate the energy being expended by the engine, preventing forward movement.
In 5. you are making the assumption that the treadmill and wheels of the airplane are somehow connected. Let’s say that the treadmill is moving 5 times the speed of the airplane and the airplane is sitting still and the wheels are turning. Does the airplane now have to thrust to 5x+<takeoff speed> to gain flight?
The only way the situation you are describing could be possible is if the treadmill asserted enough friction on the wheels that the engine were unable to exert enough force to overcome it. I know you can get this, but the rationale you are currently using is wrong for this experiment.
Here is an easy experiment you can try for yourself.
Get a toy car.
Tape a big balloon to the top of it.
Inflate the balloon and let the car and balloon go on a flat level surface.
The car will move assuming the thrust from the balloon is greater that the weight of the car.
Take the car/balloon contraption to a gym and ask nicely if you can do a physics experiment on their treadmill.
Report back to us with your results.
That’s not actually wrong, but it all depends on your assumptions. If you assume massless wheels riding on frictionless bearings (nothing wrong with that–it’s a thought experiment, after all) then you’re correct. If not, then the friction and mass will have an effect, up to and including keeping the plane stationary (depending on some of your other assumptions).
Thrust and velocity do not scale. Thrust and acceleration scale (F=ma); a misstep A.R Cane appears to be making (note his step 6) Feel free to correct me if I’m wrong.
I assume a standard 747 on a free-spinning treadmill runway that reacts to the movement of the airplane by sliding past the airplane wheels in the opposite direction of travel of the airplane. No massless wheels, and with friction enabled. Yes, there are MANY assumptions being made here by everyoe and not all are necessarily correct given the information in the in original post, but we were given a bit to work with anyway. Given a real-life example - which I believe IS possible to build - I don’t believe the outcome will be any different if the treadmill were powered and “pushing” the wheels as they are also “pushing” the treadmill, or whether it was a freespinning treadmill or whether it was an intelligent magic treadmill. Given that the tires don’t burn or explode and that the bearings don’t seize, the only thing that CAN happen is that the aircraft is pulled backwards by the treadmill as it crashes onto the surface of the treadmill. If the thrust of the airplane is strong enough it will overpower the treadmill and take off if it can also generate enough thrust to carry the treadmill assembly, or it will stand still if it it becomes somehow “glued” to the treadmill surface.
But given that the wheels and tires are working as we expect them to, the plane will take off just as if the treadmill didn’t exist.
The assumption of free-spinning is different than assumptions most other people in the thread were making. But, as you say, the original problem was a bit vague to begin with, so let’s roll with it. With a free-spinning treadmill, there would be a tendency for the treadmill to move in the direction of the 747’s motion (and thus would violate the spirit of the original question–note the treadmill was intended to turn in the opposite direction).
If you’re thinking of a powered treadmill that rotates in the same direction as the plane moves, that’s a reasonable statement (the outcome would be different, but only in degree). However, if you compare a magical treadmill that rotates in the opposite direction, that’s clearly a different case than the freespinning treadmill.
Depends on your assumptions.
[QUOTE=Brutal Tokyo]
These posts are all wrong. See my post 255 for the correct answer.
Why?
My understanding is that the treadmill torques the wheel, which is fine, but that doesn’t translate into force opposing the force of the engine because the engine isn’t applying any torque to the wheel.
I guess I’m asking why the net force is set to zero on page 3.
It’s like this. For the airplane to take off, a certain velocity of airflow over the wing is required. Hence, forward motion. Driving forward, we have the thrust of the engines. Working against the thrust is the rolling friction of the wheels. Theoretically, the plane can be prevented from taking off as long as the conveyor belt can generate enough friction against the wheels, which is mainly a function of the mechanics and composition of the wheels. Any calculation that omits this friction will miss the point of the exercise.
In reality, it’s virtually impossible to imagine any treadmill or wheel that could move this fast without disintegrating, or any energy source sufficient to drive it at the speeds we’re talking about. But if you picture this scenario with wheels that are unusually resistant to turning, it’s quite easy to envision the forces at work, and imagine a treadmill that could prevent the plane from reaching takeoff airspeed.
True. Which is essentially the same as if the plane were bolted to the ground, ie: a rocket on a concrete and steel test bed.
But if you go the airport today, put the plane on a treadmill that applies some “normal and expected” amount of friction to the wheels, the plane will still take off from the treadmill without anyone noticing any difference. This is what the original post was trying to get at, but the wording of it has given us all kinds of fun generated a few more healthy brain cells.
Ah, I see the connection… it’s the slipping.
Here’s how it goes:
As it turns out, this is called “rolling without slipping” (as I read in University Physics, Vol 1, 10th edition, p302). If this condition is always true, then we have:
V[sub]t[/sub] = R*w = V[sub]cm[/sub]
(V[sub]t[/sub] being the speed of the treadmill, R being the radius of the wheel, w being the angular speed of the wheels - RPMs, essentially - and Vcm being the velocity of the wheel at it’s center of mass)
Because V[sub]cm[/sub] is function of the engine’s output, the speed of the treadmill and the output of the engine define each other. So, define the engine’s force at any particular time, and the speed of the treadmill instantly follows.
I don’t like it, but I think that’s it.
The net force is set to zero becuase we don’t want any motion. That means we want no acceleration and thus no unbalanced forces.
The important value here is not the torque the treadmill applies to the wheels rather the linear force from friction. As you can see the final answer for the acceleration of the treadmill does not depend on the radius of the wheels. That means the torque can be anywhere from very small (tiny wheels) to very large (very big wheels). Torque and angular acceleration just happens to be in this instance the way to find the acceleration of the treadmill. The treadmill in this case doesn’t apply a torque rather it applies a linear force that due to the make up of the plane results in a torque.
Basically the way to do the analysis is to draw a big box around the plane and see what forces apply on the box. In this case we have the force from the engines and the force from friction. If we don’t want that box to move those two forces must be equal and opposite. Once we have that we need to find out (1) if its possible for that to occur and (2) under what conditions will that occur.
Well it depends on what type of plane and what sort of treadmill you can design. You probably aren’t going to affect a F-22 taking off but with a good deal of effort I think we could keep a Cessna on the ground.
No, the acceleration of the treadmill as a function of the force from the engines is defined. The velocity of the treadmill or the wheels isn’t particularly relevant.
True, that’s a little more correct. But my problem in the first place was finding the link between the treadmill and the engine that made it all come together - which was the lack of slipping.
I am not trying to nitpick but its a lot more correct. One of the biggest problems for a lot of people in learning physics is seperating the velocity/speed of an object from forces and acceleration. You could spin the wheels up to 100,000 RPM and then start the engines without changing anything in the governing equation.
I’m not trying to nitpick either - I didn’t have the best math education, so I’m really wondering if I’m incorrect with this - but aren’t we both looking at different sides of the same coin? Assuming the governing equation (from the question) is:
V = Wr
Can’t I get to your
A = alpha*r
by differentiation wrt time?
Yes and no, you can get that do that and get that equation but you can’t solve the problem by using velocity. Its important to be clear on this though becuase a lot of people are asking the wrong question here. They are asking how fast the treadmill has to go which is the wrong question to be asking. The right question is what acceleration does the treadmill need.
Yes, that’s true, I agree. I also realized that in the original question, there seemed to imply some kind of false correlation between the thrust of the engine and the speed of the conveyor belt. For example, taking 150mph as the minimum takeoff speed “If the engines are thrusting forward 150mph, but the conveyor belt is going backward 150mph, will the plane take off?” Of course there’s no such thing as “thrusting forward 150mph” but we can imagine “thrust normally needed to reach 150mph on the ground in absence of wind.” And the answer in this scenario, as in most scenarios, is that the plane will still take off at essentially the normal distance, albeit with the wheels passively spinning much faster than normal.
If you imagine, on the other hand, that the belt is matching the linear speed of the wheel, then what you have is a magical positive feedback loop that will accelerate the wheel/treadmill system to infinite speed nearly immediately, making all aerodynamic concerns irrelevant.
Maybe it has been addressed before, but there is still one thing I still don’t understand:
If the distance the plane travels from of one revolution of the wheel is canceled by the distance the treadmill travels in the opposite direction, how can the plane ever move forward, regardless of any external forces, if no slippage of the wheel occurs?
It seems to me that the force from the jet engine would be canceled by the force of friction from the wheels against the treadmill at any speed.
Yes you can, but you also need to use some other equations, and one of those other equations has to relate motion to the force coming from the jet engines:
F = ma
That equation relates the force to acceleration. Your equations above are correct, yes, but there’s no particular scaling between V and A, or between alpha and omega: the wheel can be be accelerating at any rate (positive or negative) while having any velocity (positive or negative).
You are correct, but what a lot of discussion has revolved around is not this particular question, per se, but rather the next step in the chain: How does the treadmill move in order to cancel the force? Constant velocity? Constant acceleration? Instant acceleration? And, will the coupling (friction or otherwise) between the plane and the treadmill transmit a “large enough” force to the body of the airplane to resist the engine thrust?