Will it take off?

[Rockets]

QUICK!!! Someone replay the scene from the Blues Brothers where they finally see the LIGHT!!!![]

Not laughing at you Dean, I'm just so HAPPY for you!!! LOL!!

[DizRotus]

Dean (and everyone else who thought for even an instant that the moving belt would prevent take off) don't feel bad. The original problem was so poorly worded as to appear to require the solver to accept impossibility as the premise.

As Myth Busters demonstrated the moving belt does not inhibit takeoff. But the original problem was different. It asked the solver to accept the fact that the conveyor belt was somehow (magic?) able to match the wheel speed of the airplane. As the wheels began to move, so would the belt in the opposite direction and at the same speed. That is impossible, the same as saying X=X+Y.

If it were possible for the belt to do that the plane would not have taken off, it would have remained stationery relative to a fixed point on the ground, but the problem could just as well have said, "A genie is holding the plane's tail, can it take off?" That would have been just as possible as the magical accelerating wheel speed matching conveyor belt.

Myth Busters demonstrated that, irrespective of the speed of a conveyor belt in the opposite direction, a plane will still take off. They did not, and could not, prove or disprove that the magical hypothetical belt that created the impossible premise would have kept the plane from gaining flight.

[oldenough]

As Myth Busters demonstrated the moving belt does not inhibit takeoff. But the original problem was different. It asked the solver to accept the fact that the conveyor belt was somehow (magic?) able to match the wheel speed of the airplane. As the wheels began to move, so would the belt in the opposite direction and at the same speed. That is impossible, the same as saying X=X+1.

Diz, sorry to say but you are wrong, the conveyor could be moving at absolutely ANY speed, it would have absolutely NO consequence in the planes ability to move forward and hence take-off. Back to the corner[]

[Jay481985]

Dean (and everyone else who thought for even an instant that the moving belt would prevent take off) don't feel bad. The original problem was so poorly worded as to appear to require the solver to accept impossibility as the premise.

As Myth Busters demonstrated the moving belt does not inhibit takeoff. But the original problem was different. It asked the solver to accept the fact that the conveyor belt was somehow (magic?) able to match the wheel speed of the airplane. As the wheels began to move, so would the belt in the opposite direction and at the same speed. That is impossible, the same as saying X=X+1.

If it were possible for the belt to do that the plane would not have taken off, it would have remained stationery relative to a fixed point on the ground, but the problem could just as well said, "A genie is holding the plane's tail, can it take off?" That would have been just as possible as the magical accelerating wheel speed matching conveyor belt.

Myth Busters demonstrated that, irrespective of the speed of a conveyor belt in the opposite direction, a plane will still take off. They did not, and could not, prove or disprove that the magical hypothetical belt that created the impossible premise would have kept the plane from gaining flight.

actually mythbusters did more than necessary..... they measured the speed of which the plane needs to take off and pulled the fabric at that speed instead of matching the speed of the wheel with the fabric. So in fact the fabric was going faster than the wheel up to take off in which the wheel was one to one with the fabric. Mythbusters made it even harder for the plane to technically take off since the speed was not matched up but faster than the wheel up to takeoff.

[DizRotus]

Oldenough said,

"Diz, sorry to say but you are wrong, the conveyor could be moving at absolutely ANY speed, it would have absolutely NO consequence in the planes ability to move forward and hence take-off. Back to the corner"

We are not in disagreement. The conveyor can move at any speed in any direction and it alone will not prevent take off. Many pages ago I stated that the plane would take off. Nothing within the laws of physics and mathematics will cause the conveyor belt to prevent take off. Are we clear on that?

Nevertheless, my comments were directed at the faulty word problem and the original fallacious premise of the magical wheel speed matching conveyor. If you accept that premise and ignore the fact that X=X+Y is not possible, then it's not hard to accept that the plane won't take off.

In order for a wheeled plane sitting on a conveyor belt to move forward relative to the conveyor belt, the plane's wheel speed must exceed the speed of the conveyor belt, at least until lift-off. The original problem asked the solver to accept as fact that the wheel speed and the conveyor speed would ALWAYS BE EQUAL AND IN OPPOSITE DIRECTIONS. The premise was impossibly flawed. Asking the solver to accept the magically offsetting conveyor belt is tantamount to asking the solver to accept the fact that the plane would be absolutely prevented from moving forward relative to the ground, if not by the conveyor belt, then by a magic genie, a brick wall, chains, etc., and then asking if the plane would take off. Without forward motion at a speed sufficient to develop lift greater than the weight of the plane, it cannot take off. Let's not confuse the issue with giant fans blowing air over the plane's wings with sufficient speed and force to generate lift enabling the plane to lift straight up. While that's theoretically possible, it is beyond the parameters of the original poorly worded problem.

If you are forced to accept the impossible premise, then the plane can't move relative to the conveyor belt. It's not the belt and not physics that prevents the plane from moving (Myth Busters demonstrated that) but the words of the problem. When the plane moves forward relative to the conveyor the speeds will not be equal and opposite cancelling each other out as the premise dictates. The only way for the plane's wheel speed to constantly match that of the conveyor, as the premise says IT MUST, is for the plane to not move forward relative to the conveyor. The words keep the pane still and grounded not reality or physics. The words are the problem.

[oldenough]

I stay with my original comment. I think that your argument is flawed. OMHO.

just as an example you use the words "magical wheel speed matching conveyor". Sorry but it dosn't say that anywhere in the problem. What it does say is that it matches the planes speed. However this is still irrelevent, but you have to get it right if your going to pull the argument apart in the way you are.

[DizRotus]

This discussion serves to illustrate the limitations of using words to describe physical phenomena. The actual wording of the original problem might be irrelevant.

Whether expressed in terms of wheel speed or the plane's speed, the premise indicated that the plane's motion would be matched by the motion of the conveyor in the opposite direction and at the same speed. However that might be accomplished, the result would be a plane that did not move relative to the conveyor belt. Any motion of the plane relative to the belt would cause its speed (wheel or otherwise) to exceed that of the conveyor belt.

Take away the prerequisite that the sum of the speed of the plane and the speed of the conveyor always equal ZERO, and the whole problem dissolves. Leave it in and under no circumstances can the plane move forward while the sum of the plane's speed and the speed of the conveyor equal ZERO.

[oldtimer]

Diz that is the whole trick of the question. The motion of the conveyor and the motion of the plane CANNOT cancel each other out, even though the question leads one to believe that it can.

The sum of the speeds can be zero, it does not matter, one speed has nothing to do with the other in terms of counter-force.

[Deang]

The easiest way to get it is like this:

Take a toy car (think hot wheels), and put it on a treadmill so it's facing in the opposite direction. Tie a string to the front of it. Start the treadmill -- and start wrapping the string around your finger. Watch in amazement as the car moves towards you. Now double the speed of the treadmill -- the wheels go around twice as fast, but the car keeps coming to you as you wrap the string around your finger.

[pauln]

The original argument may be flawed in the sense that it not clear. Any mention of a "speed" must be relative and with reference to something. The "speed" of the conveyor is clearly with reference to the ground, but just what does the "speed" of the wheel really mean? Since the wheel is round it goes a different translational displacement "speed" for each constant angle radii measure from the center. Maybe it is meant to be measured from the center point of the hub, making it the same as the speed of the plane with reference to whaterver...

If the assumption of "matching" the conveyor speed to the wheel speed means that the "speed" of the wheel is how fast it would be going with reference to the conveyor if it was rolling down the conveyor at the same speed as the conveyor was moving with respect to the ground, then of course this would mean the translational displacement wheel speed of the center point of the wheel hub (same as the speed of the whole plane) with respect to the conveyor would be the same as the speed of the conveyor but in the opposite direction, so the net speed of the plane to the gound would be zero... which is the source of the X=X+Y problem.

On the other hand, if the assumption of "matching" the conveyor speed to the wheel speed
means that the "speed" of the wheel is how fast it would be going with
reference to the ground if it was rolling down the ground at the
same speed as the conveyor was moving with respect to the ground, then in this case, this would mean the translational displacement wheel speed of
the center point of the wheel hub (same as the speed of the whole
plane) with respect to the ground would be the same as the speed of the
conveyor, so the net speed of the plane to the gound would be the same as that of the conveyor (but in the opposite direction) - but the net speed of the wheel with reference to the converyor itself would be twice the conveyor speed... which violates the constraint of the problem a la X=2X.

The solution is to notice that the verb "match" is misused in this problem. To match the conveyor speed to the wheel speed with reference to the conveyor puts the wheel moving zero with respect to the ground, but to match the conveyor speed to the wheel speed with respect to the ground puts the wheel speed twice the conveyor speed; which violates the constaint to "match" the speeds. The X=X+Y catches this later "logical flaw" and may tend to make one think the wheel and plane itself stays put with respect to the ground (and the air it needs to move through to fly).

In fact, the appearent "logical flaw" is actually the reality - the abuse of the verb "match" is the source of the confusion.

For those that missed the speeds analysis...

The
wheels still seem to be causing some confusion. That's because the role of the
whole thing about the wheels is to provide confusion to make analyzing the
problem challenging and fun, so let's clarify about the wheels and their
speeds...

What the hell, let's analyze all the speeds:

The
fundamental reference is the ground. The speed of the ground under the conveyor
is zero. This is not required for the plane to take off; it could do so during
an Earthquake comprising a single back-wards (or forwards) longitudinal lurch
aligned to the plane's direction of motion that matched the plane's speed (or
didn't match, doesn't matter).

The
speed of the still air (wind speed) is zero with reference to the ground. This
is not required for the plane to take off but it makes all the other speeds less
complicated to understand. The still air speed is equivalent to the still
ground speed.

The
speed of the plane relative to the still air - let's have the plane point West.

The
speed of the conveyor relative to the still air - let's have the top of the
conveyor go East ( the bottom part goes West at the same speed)

Note:
The plane's wheels parts have a variety of speeds depending on the part of the
wheel one is thinking about (because the angular speed of the wheel varies with
radii with reference to it's hub).

The
middle part of the wheels (the hub) has the same speed as the plane with
reference to the still air (with reference to the plane itself the hubs have
zero speed).

The
top part of the wheel is moving West in the direction of the plane's motion
faster than the speed of the plane with reference to the still air. How much
faster varies with how big the radius makes the wheel's circumference, the
bigger the faster, but in all cases where the wheel is of non zero radius the
top is going faster than the plane.

Likewise,
the bottom part of the wheel where it contacts the conveyor is going East with
reference to the plane at slower speed than which the plane is going West with
reference to the still air. The speeds of the top and bottom of the wheel are
equal and opposite in direction with reference to the plane.

Also,
the bottom part of the wheel is dead still with reference to the conveyor -
even though the wheel is turning and rolling on the conveyor. If you will
examine a rolling wheel you can convince yourself that this is true for all
non-skidding rolling wheels (of course all rolling wheels have to skid a tiny
bit - that is how the ground imparts
force to make them turn beyond the angular momentum the wheel already has and
the losses to friction (rolling friction)).

Notice
something interesting and very important here... the contact patch (bottom of
the wheel) is going East with reference to the plane, West with reference to
the still air, and is not moving with reference to the conveyor.

Before
figuring out more about all these speeds, lets summarize and give them formal
names. The subscripts will help keep track - they are just like the ones for
speaker perameters:

Vg Ground speed

Vw Wind speed relative to the ground

Vpg Plane speed relative to ground

Vpc
Plane speed relative to the conveyor

Vcg Conveyor speed relative to the ground

Vwtp
Speed of wheel top relative to the plane

Vwtc
Speed of wheel top relative to the conveyor

Vwhp
Speed of wheel hub relative to the plane

Vwhc
Speed of wheel hub relative to the conveyor

Vwbp
Speed of wheel bottom relative to the plane

Vwbc
Speed of wheel bottom relative to the conveyor

So,
what can we know about all these speeds? Let’s do the easy ones first…

First,
let’s assume the ground is not moving so

Vg=0

And
the wind is not blowing so

Vw=0

We
know that the wheel bottom is motionless with reference to the conveyor so

Vwbc=0
(see the calculations below for proof if you don't believe this after just
thinking about it)

And
we know that the speeds of the top and bottom of the wheel are opposite and
equal with reference to the plane so

Vwtp
= (-Vwbp)

And
we know that the wheel hub and plane have the same speed in reference to
anything

Vpg
= Vwhg ...equal with reference to the ground

And

Vpc
= Vwhc ...equal with reference to the conveyor

For
Vwtp and Vwbp, (wheel top and bottom with reference to the plane) we may find their
speeds by dividing the distance traveled around the wheel in one revolution
divided by the time it takes to make one revolution.

We don’t know the radii of the wheels, but we
do know that the circumference of the wheel is 2 pi times the radius; (2pir).
So this is the arc length of one rotation of the wheel.

With
Vwhc (the speed of the wheel hub with reference to the conveyor in feet per
second) we know that the number or revolutions of the wheel will be

Vwhc/2pir

The time for each revolution during that
second at Vwhc therefore is also Vwhc/2pir

Note,
we will use the plane's speed with reference to the conveyor to calculate the
wheel's top and bottom speed with reference to the plane, ground, and the
conveyor.

For
Vwtp (the top of the wheel speed with reference to the plane) we take the
distance traveled around the wheel with one revolution (2pir) and divide it by
the time it takes to do so (1/(Vwhc/2pir))

Therefore
the angular speed of the edge of the wheel with respect to the plane is
(2pir)/(1/(Vwhc/2pir)) which simplifies to just Vwhc, so

Vwtp
= Vwhc

For
Vwbp (the bottom of the wheel) we get the same magnitude of opposite sign
(opposite direction).

Vwbp
= -(Vwhc)

Both
of these speeds are added to the speed of the conveyor to get the wheel’s top and bottom speeds
with reference to the conveyor, Vwtc and Vwbc.

Vwtc
= Vwhc + Vcg

Vwbc
= Vwhc - Vcg ... = 0 I'll show below in an example calculation

Let’s
put in some numbers to see how this works. Let the plane’s speed (same as the
hub’s speed) in relation to the conveyor be 30 feet per second.

Note:
this would be the case when the speed of the plane with reference to the ground
is 15 feet per second and the conveyor is going in reverse at 15 feet per
second.

And
let the wheels be 18 inches in diameter (radii of 9 inches or ¾ foot).

Wheel
circumference is 2pir which is 2*3.14*.75 feet=4.71 feet

The
plane/hub speed is 30feet/second compared to the conveyor.

The
number of revolutions in 30 feet of travel is 30/4.71=6.37 and since the 30
feet of travel took 1 second

The
revolutions per second is 6.37

The
time period for one revolution is 1/6.37 (one second divided by the number of
revolutions during that second) =0.157 seconds

For
Vwtp (the top of the wheel speed with reference to the plane) we take the
distance traveled around the wheel with one revolution (4.71 feet) and divide
it by the time it takes to do so (0.157 seconds) and get 30 feet per second.
Notice that this is with reference to the plane. Since the plan is going 15
feet per second with reference to the ground, the top of the wheel is going 45
feet per second with respect to the ground because you add the speeds.

Likewise,
the same calculation for the bottom of the wheel with respect to the plane will
be –30 feet per second, but since the plane is going 15 feet per second with
respect to the ground you add them and get the bottom of the wheel going –15
feet per second with reference to the ground. Both the bottom of the wheel and
the conveyor are both going –15 feet per second so the bottom of the wheel
really is stationary with reference to the conveyor.

So
we have verified that

Vwbc=0
the contact point at the bottom of the wheel in reference to the conveyor is stationary.

[Ray Garrison]

Interesting, but I disagree with your conclusion. While at the point of contact the bottom of the wheel is stationary with respect to the conveyor, this is only true for a period of time while they are in contact, which with a nondeforming perfectly round wheel and perfectly flat belt would be tending to zero. Actually, the wheel is scurrying away westword with that contact point defining a cycloid curve.

[pauln]

You are indicating two points.

One is a point fixed to the circumference of the wheel that rotates around the rolling wheel.

The other is the point of contact between the wheel and the conveyor which stays at the bottom of the rolling wheel.

You are identifying a point at the bottom of the wheel (a point fixed to the circumference of the wheel) and the point of contact as two things, because you are saying the wheel turns, and the fixed circumference point at the bottom of the wheel is there only for an instant. That is true.

But then you mix these two defined points when you say the "contact point" defines a cycloid; which is true for the fixed circumference point, but not for the point of contact.

The point of contact remains at the bottom of the wheel - the point of contact with reference to the wheel is not fixed on it's circumference, but rotates around the circumference of the wheel as it turns. But it does not describe a cycliod since both are rotating in opposite directions equally - the point of contact stays at the bottom of the rolling wheel.

It is true that a particular fixed point on the circumference will approach zero time spent at the point of contact at the bottom of the wheel, but here you have a point fixed on the circumference, not the point of contact itself. These are two different things.

Since this is a thought experiment, the imperfections are ignored, but it is true that a real wheel would experience a slight additional non-rolling displacement due to the equivalent of a contact patch slip angle, although instead of a lateral angle it would be in line with the longitudinal displacement of the wheel down the conveyor.

[Jay481985]

This discussion serves to illustrate the limitations of using words to describe physical phenomena. The actual wording of the original problem might be irrelevant.

Whether expressed in terms of wheel speed or the plane's speed, the premise indicated that the plane's motion would be matched by the motion of the conveyor in the opposite direction and at the same speed. However that might be accomplished, the result would be a plane that did not move relative to the conveyor belt. Any motion of the plane relative to the belt would cause its speed (wheel or otherwise) to exceed that of the conveyor belt.

Take away the prerequisite that the sum of the speed of the plane and the speed of the conveyor always equal ZERO, and the whole problem dissolves. Leave it in and under no circumstances can the plane move forward while the sum of the plane's speed and the speed of the conveyor equal ZERO.

hey everybody..... When you try to match the speed of the airplane with the speed of the treadmill, the ball bearings just move faster..... so DiRotus, there is no zeroing of anything the wheel moves faster backwards

[mas]

This discussion serves to illustrate the limitations of using words to describe physical phenomena. The actual wording of the original problem might be irrelevant.

Whether expressed in terms of wheel speed or the plane's speed, the premise indicated that the plane's motion would be matched by the motion of the conveyor in the opposite direction and at the same speed. However that might be accomplished, the result would be a plane that did not move relative to the conveyor belt. Any motion of the plane relative to the belt would cause its speed (wheel or otherwise) to exceed that of the conveyor belt.

Take away the prerequisite that the sum of the speed of the plane and the speed of the conveyor always equal ZERO, and the whole problem dissolves. Leave it in and under no circumstances can the plane move forward while the sum of the plane's speed and the speed of the conveyor equal ZERO.

hey everybody..... When you try to match the speed of the airplane with the speed of the treadmill, the ball bearings just move faster..... so DiRotus, there is no zeroing of anything the wheel moves faster backwards

"Backwards"????[*-)]

Come on guys, I think we are doing a bit too much 'rationalization' and still perhaps missing the point.

I can somewhat understand the problem expressed, but as has been demonstrated by the 50 or so pages that this has persisted as an issue, the plane does not move relative to the conveyor due to the motion of the conveyor! That was stated from the onset. And it is important to finally recognize this.

This point keeps being brought up as if it is some indication of a flawed description of the problem and it is not. The fact is, the statement should have been sufficient for most to move beyond that subsystem and to look at the larger system whereupon the force is applied by the plane (via motor) to the air, thus providing motion relative to the air and subsequent lift.

The conveyor and the wheels have NOTHING to do with the problem - no more than what color jacket the pilot was wearing, his heart rate, date of birth, or astrological sign; and ANY persistance in focusing on the wheels or conveyor is the real fundamental problem. And that was exactly the purpose of mentioning it in the first place. Its a red herring. Without it, the problem becomes a simple question of how sufficient lift can be generated!

The irony is that, far from recognizing that the wheels and conveyor are superfluous, we are STILL obsessed with the wheels and conveyor.

[]

I am more interested in hearing about this mystical force that keeps additional forces applied by the body of the airplane upon the environment (other than through the wheels to the conveyor) from increasing its velocity relative to the air!

So, if an aircraft carrier is traveling at, say, 1000 mph, it is impossible for a plane to take off from its deck? Not to mention an underlying planet traveling through space at ~795 mph....

Or for that matter, as our galaxy speeds towards Andomeda at 80 mps, or that our solar system's movement about the Milky Way at 12 mps system - and that this is just the sideways drift superimposed on our main forward speed around the Milky Way, which is a brisk 144 miles per second, or that the expansion rate of the universe is around 14 miles per second for each million light-years of distance....

And to think, if we just sit on our butt, our velocity relative to the earth is zero. Whew! You would think we would all be a bit dizzy!

But instead, we're going nowhere....Fast! None of it matters...Just like the airplane's velocity due to rates of wheel spin relative to the speed of the conveyor results in no net motion...

The plane will fly.

[CECAA850]

Mas, you have PM.

[Jay481985]

umm backwards as the plane moves relative to forwards..... remember mas..... the car wheel moves backwards for a car to move forwards..... newton's third law of motions?

[mas]

umm backwards as the plane moves relative to forwards..... remember mas..... the car wheel moves backwards for a car to move forwards..... newton's third law of motions?

.

Nope.

The wheel experiences no net forward motion due to the conveyor, per the given state.

The wheel spins in a forward direction relative to the airplane The conveyor's surface at the point of contact moves towards the rear of the airplane, thus the conveyor moves backwards.

Its all relative...

And still the preoccupation with the red herring conveyor which has nothing to do with whether the plane can fly.[]

You guys would be an ideal audience for a magician! []

(and TV 'wrassling' - sure its real!) []

[Ray Garrison]

umm backwards as the plane moves relative to forwards..... remember mas..... the car wheel moves backwards for a car to move forwards..... newton's third law of motions?

.

Nope.

The wheel experiences no net forward motion due to the conveyor, per the given state.

The wheel spins in a forward direction relative to the airplane The conveyor's surface at the point of contact moves towards the rear of the airplane, thus the conveyor moves backwards.

Its all relative...

And still the preoccupation with the red herring conveyor which has nothing to do with whether the plane can fly.

You guys would be an ideal audience for a magician!

(and TV 'wrassling' - sure its real!)

Just a thought on the TV wrestling reference... think about this. A 250 pound guy climbs 6 feet up a corner post, jumps 15 feet through the air, and lands on top of another 250 pound guy without killing either of them. That, to me, seems like it would take quite a bit of athletic ability. The outcomes are scripted and the acting is usually a bit over the top, but the athleticism is real.

[mas]

TV wrassling (I hesistate to call it wrestling, as I did that in HS and college) is most definately physical theater.

I just wish it was done well enough for someone to be able to suspend their disbelief sufficiently to actually get into it.

But then i remember the days when they refused to acknowledge that it was not 'real' but choreographed.

And even Andy Kaufman exploited that - and many STILL believe that his bouts and injuries incurred with Jerry Lawler were real! (And even the extended duel that followed was all part of the act , folks!) It was all publicity! How many still hate Jerry Lawler for 'hurting' poor Andy!? LOL!

[oldtimer]

The Three Stooges were a little more subtle with physical theater, even lol! more believable.