MinnFlyer

My Feedback: (4)

Exactly, There is no gust of wind. There needn't be.

The question is:

If the conveyor belt moves in the opposite direction of the rolling wheels in order to keep them from moving forward, will the plane take off?

The only thing you need to decide is this:

"What causes the belt to start moving?"

mr_matt

My Feedback: (10)

Here I will try again....

If you ignore the friction in the wheels, then the plane will take off, everytime.

But I see no reason to ignore the friction in the wheels. To me this makes as much sense as ignoring gravity

The rolling friction, especially in the bearings, goes up with speed. Therefore, if you run the conveyor backwards fast enough (hundreds if not thousands of miles an hour) then you get enough force to conteract the thrust of the prop, ALL coming from the rolling resistance of the wheels and tires. There is no other way for a belt (assuming it stays flat and is actually a conveyor) to transmit force to the plane.

So I say again, if you ignore (rolling and bearing) friction at the wheel, the plane can take off, as the conveyor would have no way to apply a counter acting force to the plane. If there is friction, with a velocity dependant component (again, the rolling and bearing friction) AND the conveyor can move fast enough, then the plane cannot take off, as the belt can (by definition in the question) spin up as fast as the wheels. The plane sits motionless, and as the plane attempts to pull itself forward, the conveyor matches that new wheel rotational speed and the plane just sets

And in any real world control system, the plane would need to move a bit (perhaps a tiny bit) forwad before the conveyor could match the speed. I don't see why some people get hung up on this but they do. The conveyor could even move a bit faster than the wheels to move the plane back to the starting point, just like any real world control system could do.

Now if you are fascinated with other real world forces and how they might influence a real experiment, here are some;

1.) The belt would need to be very flat at the point where the plane sits on it. It would need to stay flat at any speed the conveyor might need to run
2.) The boundary layer of air on the belt might become a factor at some high speed, so you would want to keep the wings high enough away from it to minimize the effects
3.) The prop wash could cause enough wing lift to lower the rolling friction, therefore you would want a planform that minimizes lift from the propwash.
4.) You would want bearings that would behave. That is you want their rolling friction coefficient to behave linearly even when they heat up.
5.) The belt control system would need to be very responsive (fast).
6.) The plane would need to be capable of relatively slow acceleration. A more powerful plane just makes the experiment harder to design.

Dukester

My Feedback: (31)

I can't believe I'm even going to step in here with a serious answer.........

For my discussion I am going to specify a plane, my brushless slowstick, which all ready to go weighs in at a whopping 14 ounces and puts out 18 ounces of thrust. The plane is capable of flight at 10 mph. Slow stick has plastic wheels with steel axles unlubricated.

So if we perform the basic free body diagram on the slow stick airframe (prop to axles, leave the wheels out of the FBD). If positive is defined as in the direction the plane is facing, we get the following forces in the horizontal plane with their appropriate signs.
+Fp Prop thrust force
-Fa Wind induced drag
-Fw Wheel resistance
Fnet resultant net force on the plane (unknown)

Such that:
Fnet=Fp-Fa-Fw

Now how do we define these forces. This is a bit touchy as we are going to have do some common sense work supported by direct observation.

Fp is easy as we rigged it up to the fishing scale one time and got 18 ounces directly measured.

Fa is not so easy, nor is Fw. Both involved complex calculations. However, I have another method to use. I have observed that the plane is capable of performing touch and goes without using more than half throttle. Therefore if we use thrustHP program to determine thrust at half throttle setting we get 6 ounces of thrust. So with the plane touching the ground at flying speed only 6 ounces of thrust are required to perform the touch and go. As the resistive forces are a combination of the wind resistance and the wheel axle friction, by extension they can be no more than 6 ounces maximum.

So while we have not determined Fa nor Fw directly, we have set a maximum practical limit for the two forces combined.

We need a couple of other aqcknowledgements:
1. The prop thrust from the plane is capable of overcoming the static friction of the wheel bearing.
2. The rolling friction coefficient does not increase significantly over a wide range of low speeds. As we know the plane will fly at 10 mph, the maximum bearing speed will be 20 mph relative to the wheel (conveyor is matching forward speed remember), we will assume the rolling coefficient of friction stays constant. This is important as the rolling frictional force for hard surfaces (plastic wheel on hard rubber conveyor) is not dependent on angular velocity of the wheel at low speeds and pressures, rather its directly proportional to the weight on the axle and coefficient of rolling friction.
3. Wind speed is zero relative to the ground for our purposes.
4. Slow stick flight speed at half throttle has been determined from timing down a known distance and may vary +-10 to 20%, but for the purposes of the matter at stake, speed variations from 10 to 12mph create no significant variation in the outcome.

So this all gets us to here:
Fp = 18 ounces
Fa+Fw = 6 ounces at flying speed of 10 mph airspeed (which is equal and opposite to the conveyor speed or -10 mph, but since rolling friction is not dependent on angular velocity for hard surface interfaces, the rolling friction force is essentially unchanged.

So from our free body diagram we see that:
Fnet = Fp - Fa - Fw
or
Fnet = Fq - (Fa+Fw)
insert values
Fnet = 18 ounces - (6 ounces)
Fnet = 12 ounces

So for the plane moving at an airspeed at which its capable of flying, with the conveyor moving backwards at an equal speed, and the propellor providing maximum thrust, a net force exists of 12 ounces in the positive direction. This net force has no balancing force and will therefore create an acceleration in the plane.

So Yippee you say, why is that important. We know that as plane (and conveyor) speed slows down, the sum of Fa + Fw will decrease. Even adjusting for the differences between static and in-flight (moving plane) thrust, we can see that the maximum thrust output from the prop will always be greater than the sum of Fa+Fw. Therefore, the plane is capable of accelerating against any combination of wind resistance and conveyor motion up to known flying speed at there will always be a net unbalanced force on the plane.

Duke
[standing in uffish thought]

mr_matt

My Feedback: (10)

There are 2 verisons of the question floating around, the one that implies the belt matches the speed on the plane in the reverse direction, and the one where the belt matches the wheels speed.

IMHO, you are correct, when you assume the belt just matches the forward speed of the plane (ignoring the rotational speed of the wheels) then the plane easlity takes off, as the increased friction in the wheels is very low.

The other version of the question has the belt moving back potentially MUCH MUCH faster, hundreds if not thousands of miles per hour surface speed backwards.

Phlip

Right-o! And, as I've already posted many times, the belt that matches the speed of the plane, but in the reverse direction, is how I would interpret it, because to interpret it the other way brings us into fantasy land rather quickly. And when you interpret it this way, it's a no brainer, the plane takes-off with wheels spinning twice the normal rpms but moving forward through the air at the same speed as the plane. However, I think I've just had an epiphany! Look at the question that has everybody all riled up:

"The conveyer belt is designed" is what it says. It says nothing about being flat or rubber or frictionless or anything else. Perhaps the conveyor is designed with massive cleats on it resembling wheel chocks that won't let the plane roll. No wheel speed, no conveyor speed needed, the plane won't take off.

Silly me for assuming we're talking about a flat conveyor with normal physical limitations on speed.

MinnFlyer

My Feedback: (4)

But you have no problem ignoring the friction of the belt's wheels (After all, unless the belt goes around the world, it must be rotating around something)


Here is where you are wrong. The belt does NOT sense that the wheels are about to move, it senses movement. That was not part of the question. Now you are giving this already supernatural belt clairvoyant powers. So basically, you are trying to re-word the question to suit your needs.

You just contrdicted your last sentence, didn't you? So now you admit that that there is a time lag.

Maybe it's because this is the CRUX of the matter which you seem to trivialize so easily

Tsk tsk tsk... The belt CANNOT move the wheels fast enough to move the plane BACKWARD. Where does it say that??? You are again giving the bely greater powers that was stated in the original question - That, in plain and simple English is cheating.

So by your own admission, you see that the belt would have to move faster than the speed necessary to keep it from moving forward so that it may move it BACK to its original position.

No need to bother with the rest of your explaination, your theory is BUSTED

But just for fun, let's take a look:

Obviously

No problem

Stop - - - if you want to customize my airplane, I want to customize your belt. No deal, you don't get this one. I don't even NEED it, but you still don't get to have it. Again it is changing the parameters of the original Question.

I agree

Obviously, but irrelevant

A 40 size trainer with a good ballbearing engine, can you get slower than that? A more powerful plane doesn't make the experiment ANY more difficult, the plane will simply take off sooner.


But in any case, the plane will fly


The only argument you have left is that the wheels would become so hot they would start to melt, but if that happened, the molten plastic would act as a lubricant and the plane would take off sooner

mr_matt

My Feedback: (10)

I really do not know where to begin. You are switching gears so fast I cannot keep up!

I have a new question for you.

Can a conveyor belt system be designed, such that a plane that can other wise take off from the ground cannot take off from the conveyor belt when the belt is running in the opposite direction (from nose to tail)?

rmh

maybe in OZ

MinnFlyer

My Feedback: (4)

I don't think it's physically possible

HighPlains

My Feedback: (1)

Uh, gee, I thought that I made this easy enough for anyone who had not studied physics and calculus to understand. Perhap further amplification is neccessary. In the beginning, the wheels and belt are not moving. If you slice time into very very small segments, the airplane starts to move forward. That is the only way the wheel can start to turn, and the belt to match it's speed. So at this very small increment of time, the aircraft has a forward velocity. We know this, because unless the airplane achieves a forward velocity, then the wheels don't turn and the belt stays in place. Now the belt is unable to remove the forward velocity that the aircraft has achieved in this small increment of time, but it can try to match the speed of the tires. We continue to experience this small increments of time, and in each case the airplane accelerates a bit more and the wheels and belt continue to gain speed. But in each time interval, the airplane accelerates thus gaining speed relative to the free air stream. With enough slices of time, the airplane has accelerated enough to have a velocity required for liftoff.

The truth of the matter is, that the belt can not match the rotational speed of the tire, from that first very very small increment of time.

mr_matt

My Feedback: (10)

When I read the original question, this was the impression I was left with. Not all of these other mental excursions people are taking about how much the plane moves in a microsecond while the belt is accelerating, or which direction the belt is moving (really, unbelievable). The wording of that original question apparently stunk.

I'll bet whomever came up with the original question would agree with me that my version is what was intended.

Dukester

My Feedback: (31)

The wheel speed is kind of a red herring though isn't it. In my example if the plane is moving forward at 10 mph, the axle travels forward at the same speed of 176 in/sec. So then you have to ask how fast is the wheel moving? Well there is angular velocity which for the 3 inch wheels on my plane gives 7 inch circumference for an angular velocity of 25.1 rps. So if the wheel is moving 25.1 rotations per second due to the planes airspeed, then if the conveyor is going backwards at exactly the same speed, then the wheel spins 50.2 rps. So take 50.2 rps where each rotation is 7 inches (circumference of wheel again) and presto you get 351.4 inches/second. Convert that back to mph and surprise you get 20 mph again.

The key here is to remember that if the axle moves forward 1 inch, the wheel's outside surface has to move the exact same amount. If you don't believe this, measure out a line 12 inches long and place the wheel such that the axle is right over the start of the line. Then roll the wheel to the end so that the axle is right above the end of the line. How far did the axle travel? 12 inches right? How much wheel surface touched the paper? 12 inches again.

The only meaningful version of this question involves matching the plane speed.

The other version where it matches wheel speed is an infinite loop type approach. As soon as the wheel starts to roll forward, the conveyor starts also, adding some deltaV to the wheel, which in turn means the conveyor has to speed up again, causing another deltaV, and so on. You simply can't solve a problem where two unknowns are directly defined by each other. For any particular time T after the plane starts rolling, the conveyor speed (Vc) is defined by the wheel speed (Vw) and the wheel speed is defined by the conveyor speed + some constant Vp the plane has at that point. This kind of circular reference is unsolvable. If we put it mathematically:

1. Vc=Vw conveyor speed is equal to wheel speed
2. Vw=Vp+Vc wheel speed is equal to conveyor speed plus speed of plane

So for all cases where Vp><0, there is no solution, lets use Vp = 1 mph to test:
Start with equation 2, Vw = 1+Vc, insert value for Vw into equation 1, so we get Vc = 1 + Vc, subtract Vc from both sides and you get 0=1, sort of a problem there.

And the case where Vp does equal 0 simply leads you to one of the following solutions:
start with equation 1. Vc=Vw , replace Vw with equivalence from equation 2 and you get Vc = Vp + Vc, where Vp=0, so Vc=Vc, hmm doesn't take us far does it. Same thing for going other way.

That's why there is really only one way to interpret the question as the other interpretation is incapable of beign solved.

Duke

Phlip

Dukester,

You are my hero! Although I got straight A's in Math and Physics, it was 25 years ago, and I'm afraid I'm not as sharp as I used to be. You have put it succinctly, and clearly that the "wheel speed" interpretation of the question is mathmetically impossible!

Good job.

Phil

Strat2003

My Feedback: (1)

I somehow get the feeling that the "Can Fly" guys are flyers and the "Can't" guys are lawyers.

starwoes

Nice Job Dukester --
i tried to poke holes in that second solution and can't seem to find any. either i have become simply detached enough from physics or indeed, this is an ironclad solution -- the plane will FLY.

Mr Matt also seems hung up on that friction thingy: Just surfing the web, I came across this interesting statement:

"Rolling bearings are called anti-friction bearing.g. They have high loading capacity and exhibit very low rolling friction torques."

To transmit friction through an anti-friction device to hold the plane seems kinda on the "extreme" side...

starwoes

so what will the answer be if you are both a lawyer and a flier??? A can cannot fly?

bkdavy

My Feedback: (1)

The solution is both simple and elegant. To design a conveyor that exactly matches the wheel speed at all times, simply put a wheel block in front of the wheels. Wheel speed and conveyor speed will be 0 for all conditions. Now how much lift can be generated is a function of air speed across the wings. The only source of air speed is the prop wash. If you don't limit the motor to practical power limits, there is a prop speed at which there will be sufficient lift generated to lift the plane off the ground.

There is a real world example of this - its why we hold our planes while running at full throttle.

Brad

David Cutler

This is the essence of the error in the logic of those who say it can't fly.

If you put a block on the wheels, then of course, the belt does indeed take part in the equation as to whether it flies or not as it has an input into the airspeed, which is the only important speed that should be considered.

However, it doesn't have that block and no matter how elegant the mathematics or physics surrounding the discussion of the wheels and belt, it doesn't take part in the base issue.

The belt and wheels simply do not affect the airspeed; one way or the other as there is no mechanism to let it (unless this new block or a sand surface is added, of course).

No connection therefore no effect. It's that simple!!

-David C.

air mail rcu

Are you saying the wheels are not connected to the plane?

David Cutler

Erm no.

I am saying the wheels have no effect on the acceleration of the plane, as they are free wheeling.

They work on the plane in two of the three available axes and all three rotational degrees of freedom. Sideways (to control the direction the plane goes when it's on the ground) and vertically, (to stop it dropping on its belly), pitch roll and yaw while on the ground. That just leaves the forwards motion, which is what we are discussing.

Of course, when the aircraft gets airborn the control surfaces take over and control five of the 6 degrees of freedom. The sixth, again the forward direct, is controlled by the prop and, of course, drag forces. The wheels are only there to take over the function of the control surfaces when there is no air passing over them. It only takes over the ones that are controlled by the control surfaces, so whatever happens in that direction (e.g. our fictitious belt) have no effect, as the drag is so low that it can be ignored.

Just like any other analysis of any control system, you have to look at what controls all of the 6 degrees of freedom to see what's going on.

-David C.

air mail rcu

Simple question: According to the original question the wheels can not rotate faster then the treadmill. If that is true. Then how can the wheels starting at the back of the treadmill advance to the front in order for the plane to gain airspeed over the wing? Without going fast then the treadmill.

David Cutler

Well, they won't unless there is continuous slippage (skidding).

If, by some miracle, (which hasn't so far been explained, so it's all speculation) the belt does have an effect on the plane's forward motion, through a freewheeling mechanism, then obviously the plane's speed will be controlled by the belt.

If it isn't then it won't.

Simple really.

[sm=bananahead.gif]

-David C.

CHassan

My Feedback: (13)

If the plane accelerates at 2mph

The conveyor matches the speed of the wheels (can't but according to the question)

The engine will constantly produce the same amount of thrust. (since the airspeed neverchanges this is what happens right?)

So then the wheels will be in a constant state of acceleration yes/no?

In 30 second the wheels will be going 216,000mph (or whatever the equvilant "wheel speed" of a plane moving at that speed would be)

David Cutler

I'm sorry, I didn't understand any of that, mainly because it started with the statement:-

Which is a false premise as acceleration isn't measured in mph (that's a velocity or speed, not an acceleration).

Would you care to rewrite it?

-David C.

PowerPlay

My Feedback: (11)

I think that the "wheel speed" is one of the factors that trips people up here. How fast the wheels are turning (RPM's) and how fast they are moving from point A to point B are two separate factors. If a plane is flying at 100 mph but the pilot has his foot on the brakes, what is the wheel speed? The wheels are moving 100 mph but they have 0 rpms.

The other trick part of the question is that the conveyor belt could match the speed of the wheels in the opposite direction. People assume that the conveyor would have to move the opposite direction that the plane is moving, but the interesting thing is that the belt would then be turning in the same direction that the wheel are spinning and you can't match the speed of the wheels (RPMS) in the opposite direction if you are turning them in the same direction. can you?

And of course the bottom line is that the conveyor belt determines the speed (RPMs) of the wheels not the aircraft, The belt can't "match" the speed that the wheels are spinning because it controls how fast they will spin to begin with.

The belt could match the ground speed of the plane in the opposite direction, but that would mean that the plane is moving through the air as well, and once it got moving fast enough, it would take off. Wheeeeee!