Can a Plane on a Treadmill Take Off?

"alexy" wrote in message
...
"muff528" wrote:

Uh-Oh!!.....here we go again!

I vote -- 120
Kinda like an escalator. You can still go up on a "down" escalator
if you run. The treadmill would have to be "set" at the 60 mph speed
so that the car's "input" would not affect it.
But that would change the problem. In the problem as stated, you would
have to have a device that determined the cars forward speed (NOT its
wheel speed) and feed that input to the speed regulator on the
treadmill.

OK -- I'll stand corrected on that part of my post. But the car's
speedometer will do just fine. The car's (and the belt's) speed relative to
the ground will be half of the indicated speed if we continue to meet the
"equal speed" condition.

To : Michael Ware

Maybe you should stop replying, as all you give is negitive feedback to
other people. !!!!!

We are in agreement but you took my reply out of context. I was rephrasing
someone else's post who simply claimed that the propulsion system was
irrelevant (and therefore the plane wouldn't fly). In fact, the propulsion
system would be very relevent to this puzzle if it were relative to the
ground as in a wheel driven plane. As you have pointed out, it would then
need to spin the wheels at twice the speed necessary on a staionary runway.

I am beginning to see why some people struggle and even give up getting
their pilot's certificate. I can also begin to see why experienced pilots
crash when something very minor goes wrong in the air. The realities of
physics finally catches up with them.

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Travis
"Kevin" wrote in message
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Travis Marlatte wrote:
The propulsion system is irrelevant as long as it is independant of the
treadmill.

No, it doesn't even have to be "independant of the treadmill." Even
if the wheels of the plane were providing the thrust, all that would
happen is that the wheels would be spinning twice as fast by the time
the plane lifted off.

The key is in the wording of the question. The people here who have
gotten it wrong have misinterpreted the riddle to imply that it means
the aircraft is being held stationary. But that's not true. That's
not what it said. It simply said the belt is moving backwards at the
same speed the plane is moving forward. If the belt were moving
backwards fast enough to keep the plane motionless, then you've just
violated the fundamental rule of the riddle. Vbelt != -(Vplane) in
that case.

Picture it this way:


--- Plane @ 100 mph
Treadmill @ 100 mph ---

Now, what is the TAS of that aircraft? 100 mph. I assure you, it
will fly.

The only braintwister is that one must realize that the WHEELS are
turning at 200 mph, rather than 100 mph.

Kevin.

Tony wrote:
If the car had an airspeed indicator it would, I agree, indicate 60. In
the model I suggested the car is moving to the north at 60, the
treadmill to the south at 60, and the speedometer will indicate 120.

If the car's airspeed indicator said 60 then the speedometer will
indicate 120. But the car would then need to expend the same energy to
accelerate to 60 as it would to accelerate to 120 on a stationary road.

An aircraft would need no additional power to accelerate to 60 on a
treadmill.

I'm not sure what negative feedback you are referring to. I just scanned
back through a few of his posts and they seemed pretty reasonable to me. He
also supports the most popular theory - that the plane will accelerate and
fly in spite of the treadmill spinning the tires twice as fast.

I agree.

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Travis
"The Flying Scotsman" wrote in message
oups.com...
To : Michael Ware

Maybe you should stop replying, as all you give is negitive feedback to
other people. !!!!!

I agree that it would not require much more additional power to overcome the
additional friction drag of wheels spinning at twice the normal speed but it
is not zero.


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Travis
"cjcampbell" wrote in message
oups.com...

Tony wrote:
If the car had an airspeed indicator it would, I agree, indicate 60. In
the model I suggested the car is moving to the north at 60, the
treadmill to the south at 60, and the speedometer will indicate 120.

If the car's airspeed indicator said 60 then the speedometer will
indicate 120. But the car would then need to expend the same energy to
accelerate to 60 as it would to accelerate to 120 on a stationary road.

An aircraft would need no additional power to accelerate to 60 on a
treadmill.

"cjcampbell" wrote:


Tony wrote:
If the car had an airspeed indicator it would, I agree, indicate 60. In
the model I suggested the car is moving to the north at 60, the
treadmill to the south at 60, and the speedometer will indicate 120.

If the car's airspeed indicator said 60 then the speedometer will
indicate 120. But the car would then need to expend the same energy to
accelerate to 60 as it would to accelerate to 120 on a stationary road.
Nope. The same energy as it would take to accelerate to 60 on an
ordinary road, assuming that the mechanical system of the conveyor is
taking care of its motion. The work being done is to accelerate the
same mass to the same velocity in either case. If the car is providing
the energy to move the conveyer (reasonable, if its mass and friction
loads are less than those of the car), how much additional energy it
takes will depend on the conveyer.

An aircraft would need no additional power to accelerate to 60 on a
treadmill.
True. The same laws of physics apply to the car as well.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

What an inane thread! I am amazed at how many people are arguing about
such a silly subject. Get a life people!

Dean

cjcampbell wrote:
Saw this question on "The Straight Dope" and I thought it was amusing.

http://www.straightdope.com/columns/060203.html

The question goes like this:

"An airplane on a runway sits on a conveyer belt that moves in the
opposite direction at exactly the speed that the airplane is moving
forward. Does the airplane take off?" (Assuming the tires hold out, of
course.)

Cecil Adams (world's smartest human being) says that it will take off
normally.

darthpup wrote:
Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??

Wind has nothing to do with it. The airplane will accelerate and move
down the treadmill just as it would a stationary runway. It cannot feel
the treadmill at all. The wheels can, but the wheels spin independently
of the thrust generated by an airplane.

Interesting question. Are you suggesting a wind machine that is capable of
creating a wind that matches the conveyor?

Let's start with no wind. The plane must achieve lift off speed relative to
the stable air mass to take off.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -xVlo (relative to ground)

airspeed = Vlo = Plane motion relative ground + headwind = xVlo + 0Vlo;
solve for x = 1
Wheels spinning = Plane motion - conveyor motion = xVlo - -x Vlo = 2 Vlo

Plane will liftoff at an airspeed that matches its groundspeed of Vlo. The
conveyor is moving backward also at Vlo. The wheels are spinning at twice
that speed.


Now with a headwind. The plane must still achieve lift off speed relative to
the air mass which is now an accelerating headwind.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -x Vlo (relative to ground)
Headwind = Conveyor moving
Headwind = - x Vlo (relative to ground)

airspeed = Vlo = Plane motion relative ground - headwind = x Vlo + x
Vlo; x = 1/2
Wheels spinning = Plane motion - conveyor motion = 1/2 Vlo - -1/2 Vlo =
Vlo

The plane will be able to lift off at 1/2 the ground speed than under normal
circumstances. The wheels are spinning at twice that speed which is just
Vlo. It would be feasible to construct both such a conveyor and the wind
machine.



Now with a tailwind. The plane must still achieve lift off speed relative to
the air mass which is now an accelerating tailwind.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -x Vlo
Tailwind = - Converyor moving
Tailwind = x Vlo (relative to ground)

airspeed = Vlo = Plane motion relative ground - Tailwind = x Vlo - x
Vlo; ******** No solution ********

With this simple formula, the plane will never take off but will continue to
accelerate (until the wheels disintegrate) since the thrust of the engine is
acting on the air mass which is accelerating as a tailwind. Very much the
dog chasing his tail.

Assumptions:
As with most things in life, we need a point of reference. Almost any
will do but I chose the ground upon which the conveyor belt is built and we
are standing watching. It is from this reference that the conveyor belt's
velocity (backwards) is derived.

I arbitrarily chose the positive sign of velocity in the direction of
the plane motion.

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Travis
"darthpup" wrote in message
oups.com...
Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??