Brain teaser
#41
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Changing my answer. God-damn geniuses!!!
As sam and Joe pointed out (others too) The fact that the plane has wheels allows the belt to be an insignificant factor in this equation.
So, the thrusters push the plane foward relative to the ground/air. The wheels spin backwards on the belt but that's irrevelant. The only thing that changes is that the wheels are spinning backwards (at an increased rate) as the plane accelerates down the runway. I'm going with pilot/engineer on this one.
Plane takes off
Quote FTW.
As sam and Joe pointed out (others too) The fact that the plane has wheels allows the belt to be an insignificant factor in this equation.
So, the thrusters push the plane foward relative to the ground/air. The wheels spin backwards on the belt but that's irrevelant. The only thing that changes is that the wheels are spinning backwards (at an increased rate) as the plane accelerates down the runway. I'm going with pilot/engineer on this one.
Plane takes off
Quote FTW.
Last edited by UofACATS; 12-08-2006 at 09:22 AM.
#48
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you have to move relative to the air, not ground.
The wheels allow the ground to move "backwards" as the plane stays still. (ignoring friction since we're in make-believe mode) The speed the wheels are spinning is an insignificant factor.
The wheels allow the ground to move "backwards" as the plane stays still. (ignoring friction since we're in make-believe mode) The speed the wheels are spinning is an insignificant factor.
#54
If you take friction of wheel bearings into account, The engines just have to work a little harder before the plane starts moving.
<-- 1st year engineering student, but this is stuff I learned last year in high school physics.
#55
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It does not matter what speed or direction the wheels are travelling, thus the moving runway is a non-factor.
#56
F= M*A
A = F/M
The force in this situation is jet thrust (Ft) - frictional force of the tires to the treadmill (Ff) - wind resistance (Fr).
making the equation:
A = (Ft-Ff-Fr)/M
Now. Ff = uN (mu * normal force) and airplane tires have a fairly low mu if I recall (hard compound). However, a very large mass so there is a signifigant Ff.
The normal force is equal to Mass (M) * gravity(g) (or weight)
Equation:
A = (Ft-(mMg)-Fr)/M
For now we will ignore the wind resistance because some of you seem to think the plane is sitting still (which it's not), and therefor has no wind resistance.
A = (Ft - (mMg))/M
A = Ft/M - mMg/M
A = Ft/M - mg
Notice that A (the acceleration of the aircraft) is NOT dependant on the velocity of the ground. It is only dependant on thrust, mass, the tire friction coefficient, and gravity.
If you take into account wind resistance it is dependant on velocity, however this affects the craft the same regardless of the treadmill.
#60
Yeah, I guess I got caught up in the idea of this magical conveyor belt that would be able to move fast enough to overcome what little amount of friction there is from the plane's wheels against the conveyor belt's surface.
But the conveyor belt would have to be going thousands of miles an hour in the opposite direction...it's pretty much impossible...but I thought that was a stipulation in the original questions...magical conveyor belt.
I stand by the fact that if you have no airflow across the wings, you have no lift.
But the conveyor belt would have to be going thousands of miles an hour in the opposite direction...it's pretty much impossible...but I thought that was a stipulation in the original questions...magical conveyor belt.
I stand by the fact that if you have no airflow across the wings, you have no lift.