Will it take off?, page 4
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reply posted on 15-2-2006 @ 07:07 AM by Popeye
Originally posted by ghost


B.T.W.: Net Speed = F-G (F is the Forward speed of the plane, G is the speed at which the ground is moving backwards)

There the fomula!

Here's a simple test you can do on you own to see the logic I'm explaining:

We all know you need wind to fly a kite (That's a given)!

Now, go get yourself a kite and fallow these steps:

1. Find a room with a treadmill and no indoor draft (Fans, open windows, ECT.)

2. Set up the kite behind you

3. Get on the treadmill and start it.

4. Hold onto the kite.

See if you can get the kite to fly!

Tim



Tim

Really expect more for an experienced poster like you

In your example the forward propulsion is generated by the leg pushing against the treadmill that is moving so the forward motion is negated by the backward motion of the treadmill, thus no movement as you correctly state.

HOWEVER, in the 'trick' question the aircraft propulsion is created by the jet thrust pushing against the air NOT THE TREADMILL, as the air is not moving backward at a speed equal to the thrust (the treadmill is) and the air is NOT connected to the treadmill its forward motion is independant of any motion of the treadmill (I am of course talking a theoretical firctionless system in reality there would be greater friction operating the wheel but not enough to stop taking off unless the wheels blow out).

Remember this is a trick question there would be airflow over the wings as the equal equal and oppoisite force to the jet thrust is operating on a system separate to the treadmill ("the air" not the ground)

Thus the aircraft would travel forward a Xmph throught the air - the tread mill backwards at Xmph the relative speed between the two would be 2Xmph. The airspeed over the wings would be Xmph +/- the current head/tail wind speed.

In reality there would an impact on the take off performance as as there would be an increase in the resistive force caused by the increased friction in the wheels.



PS I was initial caught out by the question

[edit on 15-2-2006 by Popeye]


reply posted on 15-2-2006 @ 02:31 PM by ShatteredSkies
Assuming no head wind of anything of that sort, the plane will never take off, imagine a plane on a treadmill, that's basically what the question is posing.

I asked my teacher(a pilot for over 20 years) and everyone else in the class(who are training to become pilots) said that the plane will never take off, one of the students who is educated enough, said that it is really based on friction, because in the perfect world, there is no friction, no wind, no nothing, so the plane in fact, in the perfect world would take off due to no friction, but because the plane is NOT in a perfect world, friction is one of the biggest factors, and if something defies physics, but is proved by mathetmatics, I believe it doesn't count, physics overpowers math, despite the fact that physics uses math alot, I believe we are using the wrong equations.

Just be because the wheels are moving twice as fast, does not mean that the aircraft will take off faster. I think what people are arguing is that the airplane is on a converyor and the conveyor is moving in the SAME direction of the aircraft(this was posed in my class, and we had a 5 min argument of people yelling back and forth, until we all figured out that half the kids had actually thought the converyor was moving forward, allowing for the aircraft to take off faster and in a shorter distance, of course when this was corrected, everyone came to the same conclusion, the plane will not take off).

Now, like Ghost said, BASIC PHYSICS and Aerodynamics, this is what I'm basing my answer on.

Remember little kiddies, you need lift to get airborne, and you get lift through the motion of air through the air foil, low pressure on top, high pressure on the bottom of the airfoil, and thrust helps with that motion, if the wheels, which are connected to the struts which are connected to the airplane as a whole are not gaining DISTANCE but are gaining a higher VELOCITY, then it's scalar, it's velocity without a direction. So in retrospect, the plane is not gaining DISTANCE which it needs in order for the airmasse around the airfoil to start moving over it. Bernoullis principle tends to help too.

Shattered OUT...


reply posted on 15-2-2006 @ 03:35 PM by Fiverz
I finally understand the question and think I can kinda explain it better.

No assumption was made about the length of the conveyor belt first of all.

Let's take, for example, a plane taking off on ice. There will be minimal friction from skis or whatever, but still it only takes X amount of thrust and Y amount of lift to get it off the ground. The plane will move forward.

THE WHEELS AND CONVEYOR ACT AS THE SKIS ON ICE. Think about it .... the wheels are near-frictionless ... it doesn't matter AT ALL what's moving beneath it (or how fast actually). The plane will still take X amount of thrust and Y amount of lift to get off the ground (there's actually friction, but assume its near the quantity that ice/skis encounter).

People assume that the conveyor is only long enough to hold the plane. In this case, it will not take off .... it will accelerate until it hits the end of the conveyor and then fall off or whatever. But if the conveyor is as long as a typical runway, the plane will take off with no problem. The thrust is independent of the wheels/conveyor ... they could even be air for all it matters (which is why a plane in free fall can still recover ... give it enough forward thrust to recreate lift and you are flying forward again).

Once you realize that the wheels/conveyor are the equivalent of teflon, it's easy to see that the plane will take off.

The trick is in the wording, as someone mentioned above. For the belt to me moving AT ALL the plane has to be moving ... and we've already shown that the wheels and conveyor allow the plane to move freely. So as the plane accelerates the belt does too ... but because of the near-frictionless aspect of the wheels/conveyor, the plane will move forward nearly as normal and take off when the proper lift has been achieved.

EDIT: oops, I see that kilcoo316 already put up a link that explains it the same way I did ... kudos!

[edit on 15-2-2006 by Fiverz]


reply posted on 15-2-2006 @ 06:18 PM by waynos
Don't worry Shattered, I wrote 'simple' at the end of my post as an act of facetiousness after typing a quite convoluted explanation, British humour

I don't know if I can get my viewpoint across to you in simple terms because its not really simple, like you say, however, here goes.

Yes, I agree that the conveyor belt is moving backwards and all that but he question is 'are the wheels trying to push the plane forward?' If the answer was 'yes' then the plane would remain stationary and rooted to the ground because their motion is cancelled out.

BUT the answer to this is 'no' because the propeller is pulling it forward and the wheels are just like the castors on an office chair. Therefore because the wheels are not the driving force it matters not a jot what the belt is doing. The analogy that was posted earlier about it being like a plane on skis taking off from ice sort of works quite well because that illustrates how the prop is pulling the plane along all by itself without any influence from anywhere else, the whole bit about wheels and converyor belts is a red herring, the propeller imparts the speed that makes the plane fly, not the wheels.

Where it gets a bit more complex is that as the plane starts to move forward the reversal of the belt WILL keep it stationary until, that is, the engine could be revved up enough to overcome this, therefore at full throttle the plane will actually move physically forward at an ever increasing speed until take off speed is reached, only the wheels will accelerate twice as fast, but to no avail.

I've probably made the whole thing worse now, oh dear.

edit to remove the evidence of an inability to type and think at the same time.

[edit on 15-2-2006 by waynos]
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