Coworkers and I have a disagreement about the following:
If a plane is positioned on a "magical" runway, that moves backwards
exactly as quickly as the wheels spin forward (ie, net speed of zero),
could the plane ever take
off?

The argument revolves around a given situation in which plane movement
and runway movement are in relation to the ground outside of the
runway, and the air is not moving in relation to the ground.

Hi chrissteele-ga,

First, let me re-state some aspects of your problem, to check that we
are all talking about exactly the same thing. (If I have
misinterpreted your question, please post a Request For Clarification
and I will present a new answer.)

1. The plane's wheels are turning, and are in firmly in
   contact with the runway (i.e. they are not slipping).

2. The wheels are turning as fast as they would normally
   be turning during a regular take-off.

3. The magical runway moves backward as fast as the wheels
   are turning (so that the plane remains stationary relative
   to the ground beyond the runway).

4. The air above the runway does not move backwards with
   the runway (and is therefore also stationary relative
   to the ground beyond the runway).

In this case, the wings of the plane and the air around them are
stationary with respect to each other. Since the lift of the plane is
generated by airflow over the wings, not by the motion of the wheels,
there is no lift and the plane will not take off.

However, if the air above the runway was also magical, and was moving
backwards at the same speed that the runway was moving backwards, then
we would have airflow over the wings and would be able to take off if
the runway speed was sufficient. To an observer beyond the runway, it
would look like the plane was taking off vertically.

However, there is another twist to your question, which is perhaps
phrased a little ambiguously. How and why are the wheels turning? A
plane does not have an engine connected to the wheels; the plane's
engines move the plane through the air, and the wheels turn freely as
the plane rolls along the runway.

So, if we insist that the plane's wheels are turning due to what the
plane is doing (rather than, say, due to friction with a
magically-moving runway) then we are forced to conclude that plane
must be moving forward through the air around it. Then we would have
this situation:

1. The plane's wheels are turning, and are in firmly in
   contact with the runway (i.e. they are not slipping).

2. The wheels are turning as fast as they would normally
   be turning during a regular take-off.

3. The plane is moving forward through the air (because
   we assume that's the only way it can get its wheels
   turning).

4. The magical runway moves backward RELATIVE TO THE PLANE
   as fast as the wheels are turning.

5. The net effect of this is that the runway doesn't need
   to be magical because, relative to the ground beyond
   the runway, it isn't moving at all. It's only moving
   backwards relatively to the forward-moving plane.

In this case, we have an ordinary take-off on an ordinary runway - but
I'm not convinced that the question is trying to refer to this
situation.

I hope this answer helps to settle the disagreement between you and
your coworkers. If anything is not clear, please use the Request for
Clarification facility.

By the way, a close variant of this question has been presented at
"The Straight Dope":
http://www.straightdope.com/columns/060303.html
and discussed at "Digg":
http://digg.com/science/Physics_question:_Can_the_plane_take_off_

Regards,
eiffel-ga


Google Search Strategy:

wing plane take-off
://www.google.com/search?q=wing+plane+take-off

---

Request for Answer Clarification by chrissteele-ga on 24 May 2006 06:59 PDT

Eiffel --
Thanks; I agree with your answer in the first case.  However it is the
second case which I was hoping for, and if I understand correctly, you
are defining the paradox.

That is, the jet causes the wheels to move because of the air
movement, but the runway moves fast enough to cause the wheels not to
move....this seems a paradox which makes this an impossible problem. 
Is it not true that the closer the runway tracks the movement rate,
the quicker the rotation speed would approach infinity?  Of course,
even then, there would be no forward movement, and the force would
need to be absorbed by something.... or am I way off?

---

Clarification of Answer by eiffel-ga on 24 May 2006 07:13 PDT

Well, chrissteele-ga, I don't think it's a paradox so much as an
ambiguous question.

For any precise interpretation of the question, we can give a precise
answer. The answer always depends on just one thing: are the wings of
the plane moving through the air fast enough to provide enough lift
for take-off?

The question states that the magical runway moves backwards exactly as
quickly as the wheels spin forward, but it doesn't state what's
causing the wheels to spin. Perhaps the wheel spin is entirely due to
the movement of the magical runway, with the plane staying at rest
relative to the ground beyond the runway?

We could assume that idling of the engines would create enough thrust
to counteract the friction in the bearings of the wheels, so that the
plane wouldn't actually end up going backwards relative to the ground
beyond the runway.

In this case the plane certainly isn't going to take off.

But suppose the wheels are only allowed to turn due to the engines
causing the plane to move forwards through the air (and relative to
the ground beyond the runway). Then, if the plane is moving at, say,
200 km/hour forwards (relative to the ground beyond), the runway must
be moving at 200 km/hour backwards (relative to the plane). In other
words, the runway is at rest relative to the ground beyond and the
plane can take off as normal.

If you think the question implies any other cause of movement of the
plane's wheels, let me know what the source of the movement is and I
will provide an answer for that situation.

Regards,
eiffel-ga

Thanks Eiffel; we've all been reading your answers and discussing them
amongst ourselves, and I think we're finally reaching a resolution.

You might find this to be interesting:

http://www.answers.google.com/answers/threadview?id=428718

Thanks pinkfreud; this question is slightly different than the linked
question.  In the linked question, the conveyor matches the land speed
of the plane, but in this question, it matches the spinning rate of
the wheels, which is an important distinction.

Hi Chris,

I've thought about this since I first ran into the question in the
thread that Pink provided a link to.  The arguments are largely due to
the ambiguity of the question.  Your first paragraph seemed to clarify
the ambiguities of the linked question, but then I read your second
paragraph, and it seems inconsistent with what you said in the first
paragraph.

There are three ways to look at this.

1)  The speed of the plane, as measured by the spinning of the wheels,
is exactly matched by the speed of the runway in the opposite
direction.

In this case, if there is no slippage of the wheels, your speed
relative to the ground is exactly zero and you will not be able to
take off.  If the plane has  sufficient thrust to cause the wheels not
only to spin, but also to slide forward on the runway, the plane will
move forward relative to the ground by the amount of slippage.  The
runway will not match and reverse the slippage because it is not the
result of spinning wheels.  Will you be able to skid forward on the
runway fast enough to take off?  It seems doubtful.

2)  The speed of the plane, as measured by its speed relative to the
stationary ground, is exactly matched by the speed of the runway in
the opposite direction.

In this case, if the plane is moving forward relative to the
stationary ground at, say, 100 mph, the runway will move backwards at
100 mph relative to the stationary ground.  That means the plane is
moving forward at 200 mph relative to the backwards-moving runway. 
Whatever speed the plane is moving relative to the runway, the runway
will take half of that speed away and the other half will result in
forward speed relative to the ground.  In this case the plane should
be able to generate suffient forward speed relative to the ground to
get the lift it needs to take off.

3)  The speed of the plane, as measured by the spinning of the wheels
and slippage of the wheels, is exactly matched by the speed of the
runway in the opposite direction.

In this case it is impossible to achieve any speed relative to the
ground.  You will not be able to take off.

I hope this clarifies things.

Ansel

Hello Chrissteele,

I'd say that jstrong (in the other question) has it right. The plane
takes off because the air drag and wheel friction is far less than the
engine thrust. The wheels would spin far faster than on a normal take
off, but the drag would still be far less than engine thrust (until
the wheels burst - but I even have an answer for that).

Let's take it to the limiting case (a Helicopter or Harrier). The
force from the engine lifts the vehicle straight (or almost straight)
up and the forces from the runway are NIL. The "magic" runway does not
have any effect in that case.

At the other extreme, I can make the argument that if the pilot used
full thrust w/ brakes engaged (like a takeoff at SNA - part of "noise
abatement"), the "magic" runway would have to move with the plane and
the plane would appear to rise vertically in the frame of reference of
the runway (obviously not in the frame of reference of the air /
ground outside the runway).

  --Maniac

The plane will not take off. 


When do I receive my $20?

As ansel001 says, if the wheels don't slip, the speed of the plane is
zero, and it won't take off.  Eiffel is mistaken that the plane must
move forward through the air if the plane thrust is the cause of the
spin of the wheels.  The thrust provides a forward force, and the
magic runway provides an equal and opposite backward force, keeping
the plane at rest.  The magic runway can do this even if the wheels
are frictionless, as long as they have mass.  The backward force from
the runway accelerates the spin of the wheels.

Bottom line: there is no ambiguity, beyond whether or not the wheels
can slip.  If they don't slip, then the answer to the question, as
asked, is unequivocally NO.

The mass of the plane is much greater than that of the wheels.  The
plane doesn't start to move until the engines acting against the air
overcome the inertia of the mass of the plane  - the friction of the
wheel bearings being relatively negligable.  The runway does not begin
to move until the plane does, at which time the engines have put the
great mass of the plane in motion.  Immediately  - but slowly - the
runway moves backwards, but only doubles the speed at which the wheels
turn, which has no significant effect on the engines' accelerating the
plane forward.
At a takeoff ground speed of 200 mph, the wheels are spinning as
though the plane were moving at 400mph, and the plane takes off.

Wrong wrong wrong.  This problem explicity states that the speed the
wheels are turning forward is equal to the speed that the runway is
moving backward relative to the ground.  It is therefore obvious that
as long as the wheels don't slip on the runway, the forward motion of
the plane through the air is zero: If the wheels are rolling forward
at V, then the plane is moving forward over the runway at V.  Also,
the runway is moving backward at V.  Myoarin, if the plane is moving
over the runway at V, and the runway is moving backward relative to
the ground at V, how fast is the plane moving relative to the ground? 
What is V - V?  Your statement that the plane moves forward half as
fast as the wheels is only true if the speed of the runway is equal to
the speed of the plane relative to the ground. That is not the case
here.

The point is that the treadmill CANNOT exert equal force on the body
of the plane through the wheels. That's why the plane can move
relitive to the air around it. The thrust of the plane is greater than
the amount of drag caused by the wheels at ANY speed up to bearing
falure.