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Please leave comments if you know other trigger words or know of words on this list that are no longer triggers.
Way back, not long after I got my driver’s license, I was driving around town with a friend in a VW Beetle, manual transmission with a weird auto-fluid clutch arrangement where when you touch the shift stick the clutch disengaged so you could shift.Suddenly, the throttle cable snapped and we had to coast to a stop. We were basically stuck.
I had some string in the glove compartment, so we rigged the throttle with the string through the passenger window and managed to get on our way by having my buddy run the throttle with the string. When we got home I pulled into the garage, which I had to swerve around our van in the driveway to get into the stall. My Dad met us in the garage and asked what was going on. Proud of ourselves for getting the car home, we proceeded to explain what we did. However, we left the engine running and my friend just dropped the string to the garage floor when he got out of the car. Drivers side door open, and for some reason still unexplained today, the car was left in reverse with the parking brake set.
As we were explaining what we did, we were unaware that the car was slowly creeping backwards. It backed over the string, gradually at first but in a logarithmic fashion pulled the string tighter. Suddenly the car rev’d way up and lunged backwards. The drivers side door nearly took out my day who was standing just outside the garage. The only thing that saved him was the frame of the garage door which peeled the door away from the car. The car proceeded to run down the driveway and crashed right into the front of the van behind it.
We stood there in shock. The car went back down to idle and it took us a few minutes to figure out what happened.
Surprisingly, we didn’t get in trouble. Once we figured it all out, my Dad thought it was the funniest thing he’d ever seen and was glad no one was standing in front of the van. Whew!
Ah… Good times…
Very cool image modeling technology. Maybe CSI will actually seem realistic, someday.
A great add-on for FireFox and other browsers. This add-on allows you to select categories that you’d like to surf on the web, then essentially takes you on a tour through those categories. You can also add your own stumbled upon sites for others to stumble to, as well as review sites.
I saw this and immediately thought, “I gotta do that!”.
The site is in Russian, but the images tell the whole story.
The Problem
This site will only concentrate on the classic wording of the Airplane on a Conveyor problem. There are many different wordings of this problem. Most are incorrectly worded by people rephrasing the problem based on their understanding of the solution. If they are a “no-fly”er, they will invariably translate the problem incorrectly.
The Classic Airplane on a Conveyor Problem
A plane is standing on a runway that can move (some sort of band conveyor). The plane moves in one direction, while the conveyor moves in the opposite direction. This conveyor has a control system that tracks the plane speed and tunes the speed of the conveyor to be exactly the same (but in the opposite direction). Can the plane take off?
The TDog Solution
This solution is submitted by tdog514 of the MythBusters discussion board
First the obvious-but-wrong answer. The unwary tend to reason by analogy to a car on a conveyor belt–if the conveyor moves backward at the same rate that the car’s wheels rotate forward, the net result is that the car remains stationary. An aircraft in the same situation, they figure, would stay planted on the ground, since there’d be no air rushing over the wings to give it lift. But of course cars and planes don’t work the same way. A car’s wheels are its means of propulsion–they push the road backwards (relatively speaking), and the car moves forward. In contrast, a plane’s wheels aren’t motorized; their purpose is to reduce friction during takeoff (and add it, by braking, when landing). What gets a plane moving are its propellers or jet turbines, which shove the air backward and thereby impel the plane forward. What the wheels, conveyor belt, etc, are up to is largely irrelevant. Let me repeat: Once the pilot fires up the engines, the plane moves forward at pretty much the usual speed relative to the ground–and more importantly the air–regardless of how fast the conveyor belt is moving backward. This generates lift on the wings, and the plane takes off. All the conveyor belt does is, as you correctly conclude, make the plane’s wheels spin madly.
A thought experiment commonly cited in discussions of this question is to imagine you’re standing on a health-club treadmill in rollerblades while holding a rope attached to the wall in front of you. The treadmill starts; simultaneously you begin to haul in the rope. Although you’ll have to overcome some initial friction tugging you backward, in short order you’ll be able to pull yourself forward easily.
As you point out, one problem here is the wording of the question. Your version straightforwardly states that the conveyor moves backward at the same rate that the plane moves forward. If the plane’s forward speed is 100 miles per hour, the conveyor rolls 100 MPH backward, and the wheels rotate at 200 MPH. Assuming you’ve got Indy-car-quality tires and wheel bearings, no problem. However, some versions put matters this way: “The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.” This language leads to a paradox: If the plane moves forward at 5 MPH, then its wheels will do likewise, and the treadmill will go 5 MPH backward. But if the treadmill is going 5 MPH backward, then the wheels are really turning 10 MPH forward. But if the wheels are going 10 MPH forward . . . Soon the foolish have persuaded themselves that the treadmill must operate at infinite speed. Nonsense. The question thus stated asks the impossible — simply put, that A = A + 5 — and so cannot be framed in this way. Everything clear now? Maybe not. But believe this: The plane takes off. -StraightDope Article
rickbrant’s Explanation
This solution is submitted by rickbrant of the MythBusters discussion board
The trouble most people have with this is that they think that the conveyor belt somehow imparts its velocity directly to the airplane.
In other words, they think that if the airplane’s speed without the conveyor belt would be 100 mph, and the conveyor belt is moving “backwards” at 100 mph, that the conveyor belt subtracts 100 mph from the airplane’s true ground speed — leaving it with a net ground speed of zero.
This is not what happens, and that is not the correct way to analyze the problem.
The belt does not impart VELOCITY directly to the airplane because the belt cannot act on the airplane except through the wheels and wheel bearings… which are very very low friction.
Consider that if the tires and wheel bearings had zero friction the belt could not act on the airplane at all. Well, the effect does not suddenly go from “zero” to “imparting the belt’s entire velocity” when the friction goes from zero to very small. Only if the wheel brakes are locked AND the tires don’t skid can the belt impart its entire velocity to the airplane.
Anyway, in simple mechanics (which this is), velocity is not an independent variable. That is, it isn’t what you start with, it’s something you arrive at. Velocity is the RESULT of acceleration applied over time, according to
v = a * t
Acceleration is the result of force applied to a mass, according to
F = m * a
Or, more applicable for this case,
a = F / m
So to find out what is happening to the airplane, we have to determine the forces that act on it.
To do this we draw a force diagram. We draw a box to represent the airplane, and we draw arrows extending from the box to represent the direction and magnitude of the forces.
Acting in the forward direction we have the thrust of the engines. Of course, we have this in a normal takeoff. In our simple 1-dimensional model we can represent this as a positive number, expressed in pounds force or newtons or whatever.
In the rearward direction we have the rolling friction of the wheels, and we have atmospheric drag. We represent these as negative numbers.
The airplane experiences both of these during a normal takeoff, too.
And here is the thing that is difficult for people to “get”:THE FORCE OF ROLLING FRICTION DOES NOT CHANGE WITH WHEEL ROTATION SPEED.
(The force of sliding friction does not change with speed either…)
Therefore the only thing the belt can do, if it matches the speed of the airplane, is to double the rotation speed of the wheels as they turn under the (moving) airplane. This will not result in ANY change in the forces experienced by the airplane… except a bit more heat in the wheel bearings.
If you still don’t get it, work through this:
This solution is submitted by rickbrant of the MythBusters discussion board
1. Consider a cart with freely-turning wheels. (Not necessarily zero friction, but low friction.) Put this on a conveyor belt. Tie a rope from the front of the cart to some fixed point beyond the end of the conveyor belt. Put a force gauge (like a fish scale) in the middle of the rope and start the belt moving at say 80 knots, in the direction away from the front of the cart.
YES, I KNOW: A cart is not an airplane. This is a RELATED example, designed to get you to “see” the actual forces involved, better than you can with the airplane. We’ll get to the airplane soon enough.
2. The cart will (absent an extremely weak rope) be held stationary with respect to (wrt) the true ground, while the belt rolls under it. This is because the rope will go taut and will apply a force to the cart. The force gauge will read some value. This represents the rolling resistance of the wheels of the cart. This is, not coincidently, also the force applied by the belt to the mass of the cart. Since these forces are equal in magnitude but opposite in direction, they add to zero, and the cart doesn’t move.
3. Assuming good wheel bearings and tires and good contact with the belt (not bouncing around), this force will be constant, and small. If we have poor bearings, or if the cart is not a wheeled cart but simply a wooden platform, the force will be higher, perhaps much higher. But with good bearings we can make the force very small… and airplane wheels have VERY good bearings.
4. Now, suppose we have a jet or prop or rocket engine on the cart, and we start it up and adjust it to produce an amount of thrust — i.e. force — EQUAL TO what we were reading on the force gauge. The rope will go slack, but the cart will still be stationary w.r.t. the true ground: The force applied by the belt (through the wheel bearings) is matched by the thrust from the engine, so there is no net force applied to the cart, so the cart stays stationary.
Agreed so far?
5. Now here is the part you’re not seeing. Let’s go back to the rope with the force gauge. Increase the belt speed from 20 knots to 30 knots. Or even 40 knots. The reading on the force gauge will not change! Do you know why? The force of rolling friction does not depend on rotational speed! Nor, for that matter, does the force of static friction. It depends on the normal force and the coefficient of friction, right? There is no term there for speed. Nor is there such a term in the formula for the force of rolling friction.
5a. You probably don’t believe that. Please read: Rolling Friction
See the formula at the end? Do you see a term there for rotation speed? I sure don’t.
(The force of sliding friction does not change with speed either. It is simply
force_of_friction = normal_force * Coefficient_of_sliding_friction
– the latter term being a constant for the materials involved.)
5b. You can test it with a good roller skate, a variable-speed benchtop belt sander, and a light-duty fish scale. Just be sure to put enough mass on the skate to keep it from bouncing around. You need good continuous contact… like an airplane’s tires and suspension provide during takeoff. One problem a lot of people have with this is that they think about very lightweight “carts”, like toy cars. Or model airplanes… Of course they bounce around a lot. Real airplanes and their wheel bearings are much, MUCH better.
5c. Yes, real-world bearings do increase their frictional force a little with wheel speed. But if you’re going to bring in wheel bearing failure as a mechanism, I will bring in the fact that no such conveyor belt can possibly be built for anything remotely resembling a full-sized airplane.
So since the conveyor belt in the puzzle is made of impossibly high tensile strength, yet superbly flexible, material; has magic frictioness bearings for all the rollers; and for that matter has zero-diameter idler rollers to maintain a smooth surface, we can also have idealized wheel bearings which will behave exactly as the physics texts say they will — the friction forces WHILE NOT ZERO, will not increase with rotation speed.
5d. Common objection to point 5: What about a boat in a stream? Or for that matter a seaplane? If you’re trying to move upstream, don’t you need more engine power if the stream is moving faster? Yes, you do! But, that’s different. There, you are dealing with fluid flow, the force of friction of which DOES vary with the velocity of the fluid. Wheel bearings aren’t like that.
6. So, we have a FIXED force applied by the belt to the cart, regardless of the belt’s speed (as long as it is nonzero, and as long as the belt applies enough force to the wheels to overcome static friction).
7. But the airplane engine’s force is NOT fixed, is it?
8. Ok, let’s translate this to the airplane. Put the airplane on a normal taxi strip and then runway. While it is stationary, static friction holds it in place until we get enough thrust to overcome this. Rolling friction is LESS THAN static friction, so as soon as this point is reached, the airplane begins to roll forward and continues to do so. Correct?
9. In a normal takeoff, the thrust is increased greatly, so the airplane rolls forward very fast, and the wheels turn much faster than in taxi. BUT THE FORCE OF ROLLING FRICTION REMAINS CONSTANT.
10. Ok, it’s conveyor belt time. Remember that no matter how fast the wheels turn, the force of rolling friction applied to the airplane is constant. So there is a low throttle setting at which the airplane would just be able to hold itself stationary (w.r.t. the true ground) as the belt moved under it. This throttle setting, in fact, corresponds to that used for taxiing.
11. But if we push the throttles forward from that point, the airplane WILL accelerate (w.r.t. the true ground) and WILL achieve a normal takeoff. The only thing the belt can do is make the airplane’s wheels turn faster — but that doesn’t matter; because this does not increase the FORCE applied to the airplane by the belt.
Once more, with feeling: the force of rolling friction doesn’t change with wheel rotation speed. Since the airplane can obviously overcome wheel rolling friction during a normal takeoff, and since the belt can’t make this frictional force any higher, the airplane can equally well overcome this friction despite anything the conveyor belt does. Match the airplane’s speed, move faster, slower, whatever. Doesn’t matter.
If you still don’t see it, perhaps you could explain at exactly what point in the above you disagree, and why.
nurflugelfrund’s Perspective
This solution is submitted by nurflugelfrund of the MythBusters discussion board
For an airplane on the ground there’s one propelling force and three retarding forces.
(A) rolling resistance is equal to about 4% of the propulsive force available in a general aviation plane regardless of wheel speed
(B) Inertia which varies according to F=MA
(C) Aerodynamic drag which varies as the square of the speed
So let’s sum our vectors. Let’s see… The rolling resistance is -4%. Inertia is… well the plane is not accelerating yet so inertia is zero. Um… well… aero drag must be huge since it’s exponential! Well no, the air drag isn’t equal to the thrust until AFTER the plane gets up to takeoff speed and the pilot rotates to the climb AoA. Before rotation drag doesn’t even equal 1/2 of the available thrust and of course the drag would be zero if the plane was stationary.
So the retarding forces are:
A=4%
B is null
C is null
Assuming the thrust is 100% of normal it looks to me like the forces are unbalanced and the plane will accelerate. Just for fun let’s square the wheel friction. Now A=16%, still not enough
Since the increased speed of the conveyor increases the distance that the wheels have to roll for a given movement of the plane the amount of work done in spinning them is doubled. But that work is accomplished by the same amount of force applied over a longer time. Acceleration still occurs but at a slightly slower rate.
tdog’s Ice Analogy
This solution is submitted by tdog514 of the MythBusters discussion board
If you put a plane on a perfectly smooth sheet of ice, so that there is no friction between the wheels and the ice, would the plane take off?
The answer is yes, because the turbines/propeller push off/pull through the air. The effect on the wheel, in the frictionless environment, is that they don’t move at all. The wheels “skid” down the ice.
slashdot’s Poetic Explanation
This solution is submitted by slashdot of the MythBusters discussion board
Why do the no-fly’s fail to see,
what is explained so perfectly?
That airplanes on a treadmill too,
do what planes are designed to do.
The jets and props provide the thrust,
the wheels just spin, because they must.
And while the treadmill’s speed is added
to the wheel’s rotation, granted;
the friction falls upon the bearing
which makes the treadmill a red herring.
-slashdot
davjosmes’ Explanation
This solution is submitted by davjosmes of the MythBusters discussion board
Planes aren’t cars. A car would remain stationary in this scenario because it applies force to the roadway through it’s wheels to move forward. Planes don’t work that way. The treadmill can’t counter the force of the engine thrust because the treadmill can’t apply a counter-force through free spinning wheels.
-davjosmes
greatnt249’s Analogy
This analogy is submitted by greatnt249 of the MythBusters discussion board
Remember back in school when you used to have math tests, and the teachers would put a bunch of word problems on the tests? Remember how sometimes, the teacher would put information in the problem that seemed to have something to do with solving the problem, but in reality, it didn’t? That’s the role the conveyor belt plays in this myth.
-gretant249
davjosmes’ Counter Question
This item is submitted by davjosmes of the MythBusters discussion board
A car is standing in a wind tunnel. The car moves in one direction, while the wind tunnel moves air in the opposite direction. This wind tunnel has a control system that tracks the car’s speed and tunes the speed of the wind to be exactly the same (but in the opposite direction). Can the car drive forward?
Think about it…
-davjosmes
Interesting Points
Analysis
This analysis is used by permission from Precursor562 of the PhysOrg Forum
Ok time to punch some numbers. I will go with a Cessna T-37 and the associated specs.
Cessna T-37 (aka Tweety Bird, Tweet)
Type- Trainer
Empty weight (meaning just the plane) = 1, 755 kg
Take off speed (IAS) = ~26 m/s
No. of Engines = 2
Thrust per engine = 935 lbs ( 1870 lbs total)
Wing Area = 17 m^2 per wing
Equation for finding lift;
L = C_l * .5 * p * V^2 * A
L = lift force
C_l = lift coefficient
p = air density (1.2kg/m^3 at 20degC and at sea level)
V = IAS
A = wing surface area (17m^2 for the T-37)
But how do you find the lift coefficient?
By the equation C_l = L / (.5pV^2 * A)
We know that at 26m/s we get the necessary lift force to lift a 1, 755kg plane.
1, 755kg on earth applies a force on the ground equal to 3, 869 pounds due to the pull of Earth’s gravity. This means a 1, 755kg plane when sitting on the ground applies a force of 17, 210 N to the ground. So the lifting force must at least equal 17, 210 N at 26 m/s for it to be take off speed.
C_l = 17, 210/(.5(1.2)(26^2) * 17
C_l = 17, 210/(.6)(676) * 17
C_l = 17, 210/6895.2
C_l = 2.5 rounded off.
So now we can calculate the lift force at different velocities.
However we now need to know the drag applied to the plane brought on by friction within the bearings in the wheels. Lets keep things simple and say that all three wheels are the same size and shape and each supports 1/3 of the plains total mass.
Now the wheels will have roller bearings. The friction coefficient of steel on steel for cylinders is very small. Considering that bearings roll instead of slide this is understandable since they are bound to slide a bit. The coefficient of friction does not increase with speed. So the resistance force to move cause by friction between two solid objects is only affected by the pressure they exert (which would decrease as the plane increases in speed) on each other and the material they are made of. However lubrication is added to put a film between the contact points so that if sliding does occur you do not encounter this friction coefficient. Instead it is drag of the individual cylinders as they move through the grease.
So the equation for this is…
F_d = -1/2 pV^2 A C_d
F_d = force of drag
p = density of the fluid or gas (aeroshell grease with a density of 860kg/m^3)
V = velocity of the object moving (bearing cylinders move at half the speed of the wheel so in the case of a plane on solid ground 26/2 but since the belt is moving in the opposite direction at the same speed the wheels will spin at 52m/s when the plane takes off with an IAS of 26m/s giving the bearing cylinders a speed of 26m/s)
A = area in contact with the fluid or gas (leading side only)
C_d = Drag coefficient
So where do we find C_d? It a value given depending on size of the object, surface smoothness, viscosity of the fluid the object is passing through. I am uncertain what the value would be for such a bearing so I’ll give it a value of 0.01. This would be an estimate in comparison to other values I have seen for other things.
F_d = -1/2(860)26^2 * 0.000471 * 0.01
F_d = -430 * 676 * 0.000471 * 0.01
F_d = -1.36N per cylinder per bearing
The total (assuming were using a 16 cylinder bearing) would be -65.71N this is equal to -15 pounds rounded up. Where the – means a force applied in a direction opposite to travel.
So the difference between a plane taking off from a belt moving in the opposite direction at the same speed as the plane is moving forward (in accordance with the OP) and a plane taking off from solid ground is 15 lbs. The plane will have to produce 15 lbs more thrust at take off speed.
If the max thrust is 1870 so how fast will the belt have to move in order to counter this thrust
Well we already know the thrust which converted to Newtons is 8, 318N and so the same force must be applied by the bearings in the opposite direction to travel. What we are finding is the velocity of a single cylinder so 8318/3=2773; 2773/16=173.31
-173.31 = -1/2(860)V^2 * 0.000471 * 0.01
-173.31 = -430 * V^2 * 0.00000471
-173.31 = V^2 * -0.0020
-173.31/-0.0020 = V^2
86655 = V^2
V = 294 m/s
26m/s is equal to 58 mph
294m/s is equal to 657 mph
The speed of sound is 760 mph. Now odds are I went too high with the drag coefficient of a small steel cylinder (10mm in diameter and 30mm long with a highly polished surface) passing through the grease.
657 mph does not equal 58 mph.
