Can the Plane Take-Off?

[greencactus3]

Originally Posted by KenWittlief

mes in at 200mph for a touch and go landing pass. It flies down on the runway, but maintains its airspeed, so the wheels touch the ground, then it throttles up and flys away.

Ok, so add your moving runway to the picture. The plane comes it at 200mph again, the wheels hit the runway, which is going backwards at 200 mph, so what happens? does the plane instantly stop, ignoring the laws of momentum?!

no, the wheels go BRIRP! and spin up to 400mph, the pilot hits the throttle again and does his 'touch and go', just as before.

The engines on a plane do not react off the runway. If they did, then how do planes in alaska with skis instead of wheels get off the ground?!

whoa i had completely forgotten about those.... i had to actually look at the problem again.... and i found

"A plane is standing on a runway....

so you cannot use that as an example.

[Elgin Clock]

Originally Posted by KenWittlief

The engines on a plane do not react off the runway. If they did, then how do planes in alaska with skis instead of wheels get off the ground?!

The coefficient of friction of the air speed pushing back wards, overcomes the static coefficient of the snow, and lets the ski's move forward, and the snow stays put..

(ignoring sticky snow of course)

For the plane not to move in the snow, then the coefficient of the air speed friction, will have to be equal, or less than the coefficient of the snow/ski's relation.

Also, it has to actually be greater, since there is a funky term (which name escapes me now ) that overcomes the coefficient and let's it move freely..
(it's basically the force necessary to overcome the friction (break the grip between surfaces so to speak) and get it rolling, but it is not equal to a "rolling start" coefficient of frictional speed.

[Madison]

Originally Posted by sciguy125

Ah, I now see the error in my thinking..

Let's say we drop a missile from an airplane. Better yet, from a helicopter that is hovering in place. The missile will still go forward even if it has no contact with the ground. So, the ground has nothing to do with the airplane's ability to take off.

Or, to mix metaphors, strap the missile to Elgin's dyno. The missile isn't going to make the dyno spin.

[greencactus3]

Originally Posted by KenWittlief

ok, now you got me laughing :^)

if the plane cannot move then the conveyor belt cannot move either. When the planes 'speed' is zero the conveyor belt speed is also zero !

exactly... which is why this problem is so confusing.. and fun!

[greencactus3]

Originally Posted by Elgin Clock

Also, it has to actually be greater, since there is a funky term (which name escapes me now ) that overcomes the coefficient and let's it move freely..
(it's basically the force necessary to overcome the friction (break the grip between surfaces so to speak) and get it rolling, but it is not equal to a "rolling start" coefficient of frictional speed.

static and kinetic friction

[Elgin Clock]

Originally Posted by KenWittlief

ok, now you got me laughing :^)

if the plane cannot move then the conveyor belt cannot move either. When the planes 'speed' is zero the conveyor belt speed is also zero !

Define "move".

The wheels may counteract and spin in the opposite axial direction and thus they will "move" but my thoughts are since they cancel each others linear "move"ment out, then technically this is a trick question, cause the plane is not "moving" as most people think of as moving.

When I think of a car "moving" it's displacement is constantly changing. Down the highway. But who's to say that same car on my dyno is not "moving". It's not moving linearly, but the wheels are moving axially.

Originally Posted by greencactus3

static and kinetic friction

Thanks. I should of know that. lol



Oh, and does anyone foresee an episode of mythbusters coming on about this????

[KenWittlief]

Ok, I gotta fess up here and admit that when I took physics 101 in college I had to first unlearn 'roadrunner physics'

our thinking gets so ingrained to the things we experience personally, like driving a car or running on a treadmill, that we have a hard time doing thought experiments on things we are not familiar with (like flying a jet plane).

The question here makes you think the plane will not move at all, because our minds jump to something familiar - like a car on a dyno, or a person running on a treadmill.

Go back to the initial conditions: the plane is at rest, the runway is not moving.

The pilot hits the throttle and the plane has 30,000 pounds of thrust on its airframe. How on earth is the runway going to apply 30,000 pounds of thrust in the opposite direction to STOP the plane from moving? Through 3 wheels with greased roller bearings?

Lets say the plane gets up to 10mph like I said before. The runway is only allowed to move backwards at 10mph - to match the forward speed of the plane. There is no way the wheel bearings are going to have 30,000 pounds of friction while spinning at 20mph net speed! the wheels will have a few pounds of friction at the most, and the plane will continue to accelerate

and take off, pretty much like normal.

Picture the tablecloth trick, you yank the tablecloth out from under the dishes, and there is not enough force to pull them off the table.

'yanking the runway' out from under the plane is not going to have enough friction to overcome 30,000 pounds of thrust from the jet engines.

[KenWittlief]

Originally Posted by Elgin Clock

Define "move".

the engines on all types of planes have no connection or effect on the wheels. The engines do not rotate the wheels. Therefore, on an aircraft, the only thing that can 'move' is the aircraft relative to the air

in fact, when you are flying it is impossible to measure your ground speed, unless you know the wind speed and direction. Aircraft do not have speedometers attached to their wheels :^)

[RbtGal1351]

Until I read KenWittlief's post (one above mine) I was pretty sure about this reasoning:

The plane will not take off.

In order for a plane to lift off the ground, the air above the wings needs to be of lower pressure, which happens when it is moving faster, than the air beneath the wings. (The reason that the wings are curved.) In order for the air above to be moving faster than the air underneath, the plane needs to be moving relative to the air. Due to the conveyer belt, the plane DOES NOT MOVE in relation to the air. Thus it does not take off.

But now I'm confused as to whether to plane actually moves... And intuition says the plane will take off...

~Stephanie

EDIT:

Originally Posted by Elgin Clock

Oh, and does anyone foresee an episode of mythbusters coming on about this????

ABSOLUTELY!!

[sanddrag]

You all are driving me insane. The plane will take off just the same as it would on a traditional runway. There will be no episode of mythbusters because this is a stupid myth that requires no experiment whatsoever to disprove. Please, everyone, reread Ken's posts. Other than extremely negligible friction, the runway exerts no force whatsoever on the airplane. Yet the airplane is exerting tremedous force on the air. The airplane will proceed to move forward and take off. You really need to analyze what forces are being applied where and then you'll "get it."

(Don't make me put my RC plane on my treadmill )

Are you guys just being sillyto make this thread go on forever or do you really still not "get it"?

EDIT: I meant the runway exerts virtually no force on the plane in the horizontal direction. In the vertical direction, it exerts a great deal of force until the plane gets some lift.

[Elgin Clock]

Originally Posted by sanddrag

the runway exerts no force whatsoever on the airplane. Yet the airplane is exerting tremedous force on the air.

Hmm.. well.. LIke I said before, I suck at the math and equations behind physics.. but if I recall correctly we have a minimum of 4 forces at work here. (horray for using force digrams in Physics class over and over again )

Fn - Force of Conveyer pushing up at the plane (+Y) (the force comes from the contact of the conveyor to the plane's wheels) and it it is a staple in every force equation which you have apparently forgot about and left out. Also known as Normal Force.
Fg - Force by gravity holding the plane down (-Y) on the conveyer
Ft - For this case, thrust of airplanes jets going in the reverse direction (+X) of:
Fcb - Force being countered by the conveyer belt (via really wicked cool sensors - which I would like to see in the kit next year ) in the equal and opposite direction of the airplane's thrust. (-X)

So.. in this equation.. and using Newton's third law, which states "For every action, there is an equal and opposite reaction."

Fn + -Fg = 0 = no movement in Y direction
Ft + -Fcb = 0 = no movement in X direction

Care to tell me what I am missing and how this hypothetical plane lifts?

The only thing I can think if is since I am not an aerospace engineer, or a hyphothetical physicist (or even a real one), is the combination to add positive force to both Ft and Fn to make the plane raise up which may come from the wings and their relation to the non linear movement of the airplane???

Originally Posted by sanddrag

The airplane will proceed to move forward and take off. You really need to analyze what forces are being applied where and then you'll "get it."
Are you guys just being stupid to make this thread go on forever or do you really still not "get it"?

Is that enough analyzing?

Did I miss something? If I did, I guess I am stupid or rather just uninformed.

Care to inform me?


Oh, and please refain from using words such as stupid on these forums. We're all here to learn and help each other, not to call each other names.. Thank you!

[RogerR]

ok, looking at this from a relative velocity standpoint, if we analyze just the plane and treadmill in relation to the runway, ken is right; the plane can't be sitting still and moving at an equal but opposite velocity. physics prevents this. i know you've said that some physics rules need to be bent to solve this problem, but i refuse to accept that; if you get to bend the rules that manage relative velocity, then i'm going to bend the rules that manage gravity, and float the plane off the ground.

now, if we take the linear velocity of the wheel at a point 180 degrees from it's contact patch (instead of the velocity of the stationary plane), and the velocity of the treadmill, both relative to the ground, then we can satisfy both the 'equal but opposite treadmill' and the 'stationary plane' requirements, assuming that the wheels generate enough of a drag force (the pilot left the parking brake on?).

[dlavery]

Ken is right. The airplane will take off. The only thing that is different than a "normal" take off is that the wheels will be freewheeling at twice the lift-off speed.

The problem is that you have to remember that the wheels are free spinning, and DO NOT provide motive force (other than the minute amount of drag due to bearing friction, also accounted for by Ken). The acceleration forces for the airplane are due to the thrust of the engines against the surrounding atmosphere - NOT by the wheels transmitting force to the ground.

Originally Posted by Elgin Clock

Fn - Force of Conveyer pushing up at the plane (+Y) (the force comes from the contact of the conveyor to the plane's wheels) and it it is a staple in every force equation which you have apparently forgot about and left out. Also known as Normal Force.
Fg - Force by gravity holding the plane down (-Y) on the conveyer
Ft - For this case, thrust of airplanes jets going in the reverse direction (+X) of:
Fcb - Force being countered by the conveyer belt (via really wicked cool sensors - which I would like to see in the kit next year ) in the equal and opposite direction of the airplane's thrust. (-X)

So.. in this equation.. and using Newton's third law, which states "For every action, there is an equal and opposite reaction."

Fn + -Fg = 0 = no movement in Y direction
Ft + -Fcb = 0 = no movement in X direction

In Elgin's analysis, the assumption is made that Ft is equal to Fcb. This is where the mistake creeps in. The problem statement does NOT state that the counter-rotating conveyor belt is applying a force to the airplane that is equal and opposite to the engine thrust. It simply states that the conveyor runs at a velocity inverse to the (ground) speed of the plane. The actual force transmitted to the airplane is limited to the relatively minute amount of drag due to friction in the wheel bearings (under the presumption that the pilot is not standing on the brakes). This drag force is a very tiny fraction of the forward thrust force applied by the airplane engines. In other words, Fcb is defined as the wheel bearing drag, and Ft >Fcb. And since Ft >Fcb, then Ft +-Fcb>0 => movement in X direction.

If you still think that the plane will not take off, then think through this: imagine exactly the same stating conditions. The airplane engines ramp up to full thrust. If you are assuming that the airplane (relative to an off-conveyor observing position) is "standing still" the explain the following: the pilot quickly retracts the landing gear, so the airplane is no longer in contact with the conveyor belt, the engines are still at full thrust, and the airplane is at zero velocity relative to the ground. What happens next? When you try to reconcile this self-conflicting condition, you will realize that in the original problem, the airplane takes off.

-dave

[EddieMcD]

I just have one question about the treadmill: is the runway moving itself (to match the plane's forward speed) via motors, or is it the plane's force against the runway causing it to move?

[sanddrag]

Originally Posted by EddieMcD

I just have one question about the treadmill: is the runway moving itself (to match the plane's forward speed) via motors, or is it the plane's force against the runway causing it to move?

The plane would not be able to move the runway in the opposite direction unless the wheels were driven (like an automobile). So, I'm assuming the runway has some really freaking huge motor driving it in a continuous loop like a treadmill.

Now, for anyone else who doesn't get why the plane does take off, draw your free body diagrams You are mingling your separate forces into two different systems that have very little to do with each other. Isolate the system, and then determine the forces.