can the plane fly?

[hemijustin]

*** Brain Teaser ***



Imagine an airplane is on the beginning of a massive conveyor belt, as wide and as long as a runway, and intends to take off. The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.

There is no wind.

~Edit. ~ Assume 100% traction

Can the plane take off?

Plane cannot vertical take off!!!!!!!!!!



Sorry. . had to do it here.

 

[SRadke]

Is the plane moving forward then? And the conveyer belt moves in the opposite direction at the same speed?



It'll warm up the wheel bearings but I see no reason why she wouldn't lift off.

 

[RustyJC]

If the bearings are assumed to be frictionless and the brakes aren't applied, then yes, the plane will take off and fly.



If the bearings are frictionless, the conveyor belt cannot apply any fore/aft force that would restrain the plane's acceleration. The only thing the wheels are doing is counteracting gravity by holding the body of the plane off the ground (vertical force only). Wheel speed is irrelevant. The thrust of the engines is working against the atmosphere, and the restraining forces working against acceleration are the plane's mass (inertia) and aerodynamics - this is no different than a normal takeoff.



Think vector algebra.



Rusty

 

[hemijustin]

Sorry but your both wrong. . Reread the question. The belt is matching the speed of the wheels not the plane itself.



Justin

 

[MMeier]

Yes it will, the treadmill has nothing to do with this equation !!

 

[RustyJC]

The conveyor belt is absolutely irrelevant. I explained why. It's a red herring!!



Rusty

 

[hemijustin]

here is some math to think about...

A= speed of plane. since the wheel and belt are the same but opposite B=-c

B= speed of wheel.

C=speed of belt.

if the speed of the airplane times 10 = the rpm of the wheel then. 10A=A now lets add the belt in. the speed of the airplane times 10 = the rpm of the belt. so that is matches the wheel.

so now you add B and the opposite of B (C).

(10A)-(10A)= A the speed of the plane right?



remember it is said the wheel and belt are the opposite of each other at all times... and traction is 100%

 

[RustyJC]

and trac(k)tion is 100%

Which means absolutely nothing unless the brakes are applied (in which case the plane's speed will always be the conveyor's speed) or the bearing drag in the wheel is enormous. With frictionless bearings in the wheels, the conveyor belt cannot apply any fore/aft forces to the airplane. Let that soak in awhile. When it does, you'll understand why the conveyor belt is totally irrelevant.



Rusty

 

[hemijustin]

Which means absolutely nothing unless the brakes are applied (in which case the plane's speed will always be the conveyor's speed) or the bearing drag in the wheel is enormous. With frictionless bearings in the wheels, the conveyor belt cannot apply any fore/aft forces to the airplane. Let that soak in awhile. When it does, you'll understand why the conveyor belt is totally irrelevant.



Rusty

your still not thinking Illogically. the belt is NOT matching the speed of the PLANE. but the speed of the Wheels. the Belt and Wheel relationship must remain the same.

 

[fkovalski]

Forward motion is required to create negative pressure above the wing providing the lift for take off. The conveyor belt moving at opposite direction to the wheels will not allow the wheels to move forward as a whole although they will rotate ( no forward motion. ) Hence, no lift, no take off. The conveyor belt is totally relevant. Friction is needed for the wheels to rotate. the conveyor belt is in essence creating a friction less surface.

 

[hemijustin]

Forward motion is required to create negative pressure above the wing providing the lift for take off. The conveyor belt moving at opposite direction to the wheels will not allow the wheels to move forward as a whole although they will rotate ( no forward motion. ) Hence, no lift, no take off. The conveyor belt is totally relevant. Friction is needed for the wheels to rotate. the conveyor belt is in essence creating a friction less surface.

Sorry but you are not even in the ballpark...

 

[RustyJC]

The wheels are not powered. Nor are they braked. Therefore, they are not transmitting any fore/aft forces to the plane.



Given that, the thrust of the engine acting against the atmosphere will accelerate the plane. The takeoff velocity is referenced as airspeed, not ground speed, so wheel speed and/or the speed of the conveyor belt passing under the plane is meaningless.



Rusty

 

[fkovalski]

*** at any given time, moving in the opposite direction of rotation.

These negate each other and the plane will take off normally. Key word: OPPOSITE direction

 

[hemijustin]

I hate to spoil the fun so early. How many pages should i let this go on until i let out the real answer ????



justin

 

[RustyJC]

And what validates the "real answer"?



Rusty

 

[fkovalski]

I hate to spoil the fun so early. How many pages should i let this go on until i let out the real answer ????



justin

Don't do it! Let the riddle be answered!

 

[hemijustin]

And what validates the "real answer"?



Rusty

ME... the real answer is going to **** you off... lol

 

[RustyJC]

ME

With all respect, just because you say it won't fly isn't sufficient for me in and of itself. Present the engineering calculations.



Rusty

 

[fkovalski]

Oh, This is one of those non- sense riddles that must be read litterally. Your original statement says that it "intends to take off. " Until the pilot (assuming there is one since you do not mention) INTENDS TO ACTUALLY fly the plane, it will just sit there.





EDIT

Ironic: I just heard that there was a shooting on a plane in Miami just now! (not a joke)

 

[hemijustin]

i took it off a post on DTR almost 40 pages in like 2 weeks lol. only a couple of people have the right awnser for the right reason... .

Justin