Physics of flight: Plane on a treadmill

[Toro]

not to beat a dead horse. but mythbusters is supposed to test this next week. 1.30.08.

Anybody watch it?

[Guest nightwolf]

Yeah I watched it was pretty interesting. It pretty much touched on most things discussed in this thread. Overall the plane did take off. They used a very large tarp and drug it behind a truck with a smaller type plane on it and, well it took off just as normal. Although, I do agree with JS that the question is kind of ambiguous. With respect to keeping the aircraft in the same point in space on the run way through the conveyer belt obviously the plane would stay grounded. It’s just getting the belt up to the speed to create enough friction on the wheels to negate the forward thrust would be nigh impossible to reach due to such low friction on the wheels. Again though when working with problems sometimes you do look at things in a "perfect" world where anything is possible.

[Guest alfakilo]

car speed is measured by the wheels, not by pitot/static system.

Exactly.

So how fast is the car going?

[RangerMateo]

Exactly.

So how fast is the car going?

According to Einstein, it's all relative...no seriously =) All depends in what you're referencing against. With the car on the dyno you also have to take in account that the purpose of the dyno is to measure power output, so the car is pushing against the roller. The front wheels have to be blocked somehow I would think, I've no experience with those, however, so I can't speak to it.

As far as the friction in the wheels goes, yes, that's one of those things for this experiment you would consider a negligible factor. Simply because we're only concerned with how the relative motion of the treadmill affects the capability of the aircraft to produce thrust that will move it forward relative to the air mass. So you assume massless wheels (ie - no rotational inertia), frictionless bearings, and completely inelastic rolling surfaces (ie - no rolling friction). As evidenced in the mythbusters episode, those are all minor considerations even for the real world.

Here's the show if you missed it. May not be up for real long...

Gotta love the "Copy That" after receiving his TO clearance too...

Edited January 31, 2008 by RangerMateo

[D-ron]

Exactly.

So how fast is the car going?

The car is going zero mph relative to an earth stationary reference frame. Assuming the speedometer is driven by the drive wheels, then the speedometer would say the car is moving at whatever speed depending on RPM and gear ratio. This has absolutely nothing to do with the airplane on the treadmill. I'm not sure why this is so hard for everyone to understand.

[Guest alfakilo]

Here's the show if you missed it. May not be up for real long...

Thanks for that clip. The only problem that I see is that the plane is clearly moving forward across the ground...and therefore through the surrounding air...and in doing so reached flying speed.

If that plane had a speedometer as in a car as well as a conventional airspeed indicator, then the speedometer would have indicated a higher speed (based on wheel rotation) than the airspeed indicator.

Maybe I misunderstood the original question...I thought the plane wasn't supposed to move forward relative to the area around it.

[john]

well i dont know if they actually covered the question really being asked. they said will the plane take off if the treadmill/belt/tarp is moving at the take off speed? but the idea i thought was in question is what would happen if the belt matched the speed with the wheels. if you had a speedo on the wheels and used that to control the speed of the belt. now that would be interesting.

[Guest alfakilo]

well i dont know if they actually covered the question really being asked. they said will the plane take off if the treadmill/belt/tarp is moving at the take off speed? but the idea i thought was in question is what would happen if the belt matched the speed with the wheels. if you had a speedo on the wheels and used that to control the speed of the belt. now that would be interesting.

I agree. I think the issue is really the wording of the question. The initial part of the video seems to indicate that the show is comparing the speed of the belt to the speed of the plane. In their example, we have to decide what they mean by 'plane speed'...wheel speed or airspeed through the air.

In your example, if there was a speedometer on the wheels and it was used by the pilot to match the rearward speed of the belt, then it beats me how the plane is supposed to fly.

[PaddyPilot]

The plane would still fly because the speed of the wheels has no relation to the plane's TAS. The propeller is creating thrust using the AIR, not through the wheels. So no matter what the speed of the wheels, the propeller creates airspeed, which in this case with no wind, is translated into groundspeed. The point with walking on a treadmill that matches your speed doesn't apply here because while a plane with no thrust could be compared to that, a plane with a propeller though, would be like you walking on the treadmill, but then someone else pulls you off with a rope.

In relation to treadmills, if you had a headwind that matched your airspeed, then your groundspeed would be zero, but you could still takeoff. That's where the treadmill problem would work.

Edited January 31, 2008 by PaddyPilot

[JS]

If the "treadmill" matches the speed of the airplane wheels, the plane will still take off.

If the "treadmill" matches the forward speed of the fuselage, the plane will never take off.

It is all how the problem is defined. I believe it says the treadmill "matches the forward speed of the airplane."

Read my post above for a better explanation.

[D-ron]

If the "treadmill" matches the speed of the airplane wheels, the plane will still take off.

If the "treadmill" matches the forward speed of the fuselage, the plane will never take off.

It is all how the problem is defined. I believe it says the treadmill "matches the forward speed of the airplane."

Read my post above for a better explanation.

The airplane will takeoff regardless of what the treadmill is doing. Their isn't enough friction in the wheel bearings to significantly affect the takeoff performance of the airplane. The treadmill could be moving backwards 1000 times faster than the wheels are moving and the airplane would still takeoff provided the wheels/tires didn't self destruct. You're not understanding that the airplane is not driven by the wheels. Wheels are very low friction devices and there is no way for the treadmill to add a backward force to the airplane through the wheels no matter how fast the treadmill is moving. Watch the episode, suddenly things will become clear. This is one episode of Mythbusters where they actually provided some useful insight. They completely and conclusively busted the myth. The airplane will takeoff no matter how the riddle is worded. period. dot.

[PaddyPilot]

right, the air itself is the treadmill, not anything interacting with the wheels.

[JeepGuyC17]

Awesome! a plane vs treadmill thread! I had never heard of this myth before watching the episode last night.

I think a lot of people get wrapped up arguing that if the plane remains stationary relative to the observer (assuming no wind) then it won't take off, which is indeed true, but miss the point that the plane will NOT remain stationary.

It's a problem of Newton's laws. Assuming frictionless wheels, if the airplane is at rest with the engine off and the treadmill begins to move, there is no outside force acting on the plane and it will remain at rest, regardless of the treadmill's speed. When thrust is then introduced by the engine, this outside force will accelerate the plane no matter how fast the treadmill moves (disregarding things like wheel speed limitations).

Even when you introduce friction in the wheels, the only time the plane would remain stationary would be when just enough thrust was applied to match the relatively small friction of the wheels. (and the treadmill would have to be moving pretty damn fast to produce enough friction to match takeoff thrust).

There are some pretty heated discussion of this stuff out there. I even saw a guy on one board talking about "producing 'x' MPH of thrust", whatever the hell kind of unit of measurement a "mile per hour of thrust" is...

[Riddller]

The airplane will takeoff regardless of what the treadmill is doing.

Definitely.

The problem with using the wheel speed as the controller for the belt speed is that it would instantly become infinite as soon as the plane started moving forward. If a plane takes off at 100 kts, and the belt is moving backward at 100 kts, then the wheels are rotating at 200 kts. So the belt speeds up to 200 kts, but the plane is still moving forward at 100 kts, so the wheel speed is now 300 kts, speeding up the belt. No matter how fast the belt speeds up, the wheels will still be moving faster by whatever speed the plane is moving forward (and it WILL be moving forward).

So the way the problem was worded in the original post, there's no infinite's involved, the belt just moves back at the same speed the plane moves forward, and it takes off.

[RangerMateo]

if you had a speedo on the wheels and used that to control the speed of the belt. now that would be interesting.

By definition the speed of the belt/runway will always match the speed of the wheels, unless they are skidding. The whole principle behind anti-lock brakes is that you want to keep the relative velocity = 0 because Coefficient of Static Friction is >> than Dynamic or Sliding Friction. Explains part of why you have to goose the throttle to get the plane going then back off to idle and keep moving. =)

As far as the other comment about the forward speed of the fuselage matching the backwards speed of the treadmill (Absolute values being equal, I suppose. Both in relation to the ground.), that's exactly what they did on Mythbusters. You just have to remember where your reference is. Truth is that you could do the same thing with the car, what would happen is that you would end up turning the treadmill up to 1/2 the cars original (speedometer) forward speed. So if the car was going 80 mph you would end up with the treadmill at 40 backwards, the car moving 40 forwards in relation to the ground, and the speedo showing 80mph.

The real hangup (Including one of my Aero friends that is refusing to believe Adam and Jamie...lol) seems to be with the forward motion. He keeps saying, well, if there was no forward motion the plane wouldn't lift off. While that is true, it's not really the root question. The root question is will that plane move forward. How does the moving surface beneath the object affect the objects ability to create a net forward force? The argument against the mythbusters experiment is saying that the plane is overcompensating for the treadmill. In essence, with the car analogy, saying that the plane might need 40 mph to take off, so when they turn the treadmill up to 40 mph, the car accelerates it's wheels to 80 mph to still create that net forward force resulting with a 40 mph forward motion. The only way to take care of that is to create a profile that accelerates the treadmill to the cars max speed and have the car go to it's max speed as well. This leaves the car no headroom to create a power/thrust excess...the very argument being made for why the Mythbusters experiment was bad.

Unfortunately, that was already accounted for. Takeoffs are always performed at Max Power (Maybe there are exceptions, but not in this experiment!), so the plane engine had no power excess to 'overtake' additional drag to the level that is purported in this argument. That is why the takeoff speed and distance were so important to the experiment. If you study the free body diagram for this, you can isolate the important parts as being the force pushing the aircraft forward and any forces acting contrary to that. You'll recall from Physics that F=m*a and one of the many ways to express acceleration is (Vf^2 - Vi^2)/(2 * dS/dT). Any deviation in net force will be directly noticeable in the distance used to accelerate (again at Max power for this situation) to a certain speed in a given distance. In our case, it's takeoff speed and takeoff distance both easily measured. Since the forward thrust was the same in both the control and test environments, we can easily isolate the opposing effects by measuring the distance required to attain the same forward velocity. Since the distance was effectively the same, we can calculate that the net force was unchanged...and in fact in a perfect world with massless wheels, perfect bearings and completely inelastic rolling surfaces, it would be exactly the same. You could maybe argue the gravitational waves side of things if you had an incredibly massive and fast treadmill...but I don't think Mythbusters will attack that one ;)

In the real world, we do have rolling friction, bearing friction, and rotational momentum...and given a good enough experiment that would probably become evident in this experiment as a slightly increased takeoff roll. Since the friction increases with speed, we could postulate that there would eventually be a treadmill setting where the entire thrust of the engine was being used just to overcome the friction of the wheel bearings and the rolling friction of the wheels turning on the treadmill. When I used to pull planes around to fuel them, it would take about 20 lbs of force to break the wheels loose on the local training C-150, after that it only took about five pounds, at the most, to keep it rolling at a brisk walk. Figure the engine puts out about 165 lbs of force that means that only about 3% of the thrust is being used to overcome that rolling friction. The rolling friction equation is independent of velocity, but it still includes a coefficient of friction factor that would certainly change (increase) as things got hot and lubricants started breaking down or experiencing turbulent flow in the layer between where the bearing mating surface is. My guess is that the wheels would fly apart or the bearings would literally seize long before there was 165 lbs in backwards force due to friction.

[GrndPndr]

Prof. Bernoulli would roll his eyes.

Most here haven't agreed on the components of the test. I guess I assumed we were all talking about no relative wind over the wing (or any other lifting surface). It sure seems like everyone gets goofed up on wheel friction. Are those folks saying that if the aircraft was tied down it would still fly off?

In the Mythbusters show, it sure looked like the airplane had forward movement (relative to the ground), and therefore ended up with some air moving over the wings.

In the mini-test they did, it looked like there was air pressure (and very little was needed) to get that little thing (sts) airborn.

Hey come on - no relative air over the wing = no fly.

[JeepGuyC17]

Prof. Bernoulli would roll his eyes.

Most here haven't agreed on the components of the test. I guess I assumed we were all talking about no relative wind over the wing (or any other lifting surface). It sure seems like everyone gets goofed up on wheel friction. Are those folks saying that if the aircraft was tied down it would still fly off?

In the Mythbusters show, it sure looked like the airplane had forward movement (relative to the ground), and therefore ended up with some air moving over the wings.

In the mini-test they did, it looked like there was air pressure (and very little was needed) to get that little thing (sts) airborn.

Hey come on - no relative air over the wing = no fly.

The real point of the myth is that the "no relative wind over the wing" scenario would not happen. Of course the plane would not fly in such a case, nobody disagrees with that. The point is that unless the plane WAS tied down, it would still move forward through the air and gain lift regardless of the treadmill. The whole point with wheel friction is that there is no other force exerted by the treadmill on the airplane to counteract forward thrust through the air.

Edited January 31, 2008 by JeepGuyC17

[RangerMateo]

The real point of the myth is that the "no relative wind over the wing" scenario would not happen. Of course the plane would not fly in such a case, nobody disagrees with that. The point is that unless the plane WAS tied down, it would still move forward through the air and gain lift regardless of the treadmill. The whole point with wheel friction is that there is no other force exerted by the treadmill on the airplane to counteract forward thrust through the air.

BINGO! What people are really getting caught up on is seeing the treadmill as that tether that holds the plane back. The discussion on wheel friction is as the "Only way it could ever possibly happen" sense. Unrelated to the solution of the original question.

[Rocker]

car speed is measured by the wheels, not by pitot/static system.

It's actually measured through a speedometer cable at the transmission output shaft and is geared for whatever rear axle (or transaxle for front wheel drive) ratio and factory tire size you have to be accurate. If you put your car on a chassis dyno (the kind where the whole car gets strapped down), the speedometer will read whatever it should be reading for that gear/RPM. A motorcycle speedometer cable usually goes into the front wheel hub, so on a chassis dyno, the speedo would read zero. With computer control, everything I just said is probably out the window.

As for all this plane on a conveyor belt hogwash, I'm sticking to my original beliefs whether they are right or wrong, and I will suffer accordingly. The Mythbusters didn't do it right and it won't fly. Remember: you can't change my mind. It's just easier that way.

[sky_king]

This experiment has different outcomes depending on how you set up the situation. With no friction, the airplane will always take off. But then you have to take into account that with out friction, lift won't be produced by the wings, etc... You can't just say, well, there's no friction here but there is friction here.

If the conveyor belt matches the forward speed of the aircraft's fuselage/wing, the airplane will take off with it's wheels spinning two times faster than a normal takeoff. This would maybe blow a tire, maybe not.

If the conveyor belt matches the rotational speed of the aircraft's wheels, as soon as the airplane starts to move forward, the wheels will instantly reach an infinite speed. At infinite speed, the wheels overheat and explode instantly and the airplane falls to the belt without wheels. Now because there are no wheels to control the belt speed, the belt stops and the airplane starts to skid down the belt and eventually takes off. Obviously, to do this you'd have to have a very overpowered airplane to overcome the drag of sliding without wheels.

So regardless of how you set up the question, in real life, the airplane will always take off. Well, that is if you have a big enough engine and the tire schrapnel flying at an infinite speed doesn't destroy your airplane.

[Guest alfakilo]

So regardless of how you set up the question, in real life, the airplane will always take off.

I think it's already been shown that this question is just a matter of what someone thought was asked. If the question was worded this way..."If a plane was on a conveyor belt moving at the plane's takeoff speed and a force, however small, was exerted on it to offset the rearward motion of that belt, would the plane takeoff?"...then the answer, as many have said, is obvious.

Some, like me, read the question that way. Others did not. From our respective viewpoints, everyone was right.

All of this talk about wheel rotation speed and friction was off the point. Had the plane been on skids, the answer would still be the same. Apply enough force and a refrigerator can fly.

[RangerMateo]

Here's the one I really have a problem with...they told us in academics that gross weight has zero effect on your max glide range. Read carefully, not 'a negligible effect', but 'zero effect'. The argument goes that if you refigure L/Dmax and fly that speed for your new weight then you will arrive faster, but in the end you will go the same distance (You're falling faster, but you're also going forward faster so it all equals out...).

So does that mean that a C-182 with no fuel and a scrawny kid piloting it will glide as far as one with my old CFI (Who weighed close to 280!) and a 55 gallon drum of Uranium in the back (About 8700 pounds)? It just seems to me that if you strapped a boxcar (and don't count the new wind friction on the boxcar, just it's weight) onto a DA-20, that plane isn't going to get a 11:1 glide ratio, no matter how fast you fly it.

The only place I see an argument for this is that really in a glide you're converting your PE to KE and when you solve those two equations mass simply drops out (V=Sqrt(2gh) if I remember right), but there's the problem of which direction you're converting it into and how efficient you are at that. All falling objects have a terminal velocity and the 'lost' energy is accounted for as heating to the air through friction, etc. Wings help us to redirect that downward vector into a vector that is moving forward and down (at say, an 11:1 ratio...). I know the equations don't change, so I still think that the overall glide ratio would have to change. If I remember right, the heavier you are the further L/Dmax shifts to the right and the faster you have to fly, which means a steeper angle down....eventually that angle would have to approach and pass 45 degrees, and your glide ratio would flip around (1 foot forward for 2 feet down for instance now).

Thoughts?

[D-ron]

Here's the one I really have a problem with...they told us in academics that gross weight has zero effect on your max glide range. Read carefully, not 'a negligible effect', but 'zero effect'. The argument goes that if you refigure L/Dmax and fly that speed for your new weight then you will arrive faster, but in the end you will go the same distance (You're falling faster, but you're also going forward faster so it all equals out...).

So does that mean that a C-182 with no fuel and a scrawny kid piloting it will glide as far as one with my old CFI (Who weighed close to 280!) and a 55 gallon drum of Uranium in the back (About 8700 pounds)? It just seems to me that if you strapped a boxcar (and don't count the new wind friction on the boxcar, just it's weight) onto a DA-20, that plane isn't going to get a 11:1 glide ratio, no matter how fast you fly it.

The only place I see an argument for this is that really in a glide you're converting your PE to KE and when you solve those two equations mass simply drops out (V=Sqrt(2gh) if I remember right), but there's the problem of which direction you're converting it into and how efficient you are at that. All falling objects have a terminal velocity and the 'lost' energy is accounted for as heating to the air through friction, etc. Wings help us to redirect that downward vector into a vector that is moving forward and down (at say, an 11:1 ratio...). I know the equations don't change, so I still think that the overall glide ratio would have to change. If I remember right, the heavier you are the further L/Dmax shifts to the right and the faster you have to fly, which means a steeper angle down....eventually that angle would have to approach and pass 45 degrees, and your glide ratio would flip around (1 foot forward for 2 feet down for instance now).

Thoughts?

Its pretty close as long as the weight doesn't change too drastically. If, say, you add aboxcar on a DA-20, the DA-20 will have to fly VERY fast to be at the best glide AOA. At that speed, compressibility comes in to play. Also, there are Reynold's number effects that they are ignoring. However, if you are talking about any realistic load then the assumptions they are making are probably good assumptions because the best glide speed won't change *that* much. Just don't over think it.

One more thing that I forgot to add... I think in a headwind, the heavier guy will win in glide. Look up "speeds to fly" for gliders, and you'll learn some interesting stuff. Unfortunately, I've forgotten most of it by now, haha.

[Skitzo]

How is this situation any different than running with a kite on a day when there is no wind vs running with a kite on a treadmill with no wind present?

[JeepGuyC17]

How is this situation any different than running with a kite on a day when there is no wind vs running with a kite on a treadmill with no wind present?

Because when you run, your forward motion is caused by interaction between your legs and the ground, while for an airplane (or airboat or hovercraft or rocket car or any other such vehicle) this is not the case. Their forward motion is caused by creating a force of air which propels them, not by pushing against the surface they are sitting on like a runner, car, boat, etc.

Edited February 2, 2008 by JeepGuyC17