The Treadmill Problem Revisited
Remember the "Treadmill Problem" that I posted about a few weeks ago? Another actual pilot has weighed in on it, and this time an actual physics professor was also involved! What they determined fairly close to my line of thinking, and the physics professor took it past the barrier of my lack of theoretical physics knowledge:
With the airplane and treadmill, there is, at the outset, friction force capable of rotating the tires at the proper speed to keep the plane stationary. However, as the thrust is increased, that force eventually maxes out. (Two separate frictions are at play here, actually, one between the tires and belt, the other between the plane’s axles/bearings and its wheels. The first will max out before the second.)
"And at that point the wheels no longer roll, they slide," says Camp. "Or rather, they roll and slide at the same time. Tire motion is now decoupled from the belt motion. No matter how much you whiz up the treadmill, you won’t add any more rotational velocity to the wheels because friction is already doing everything it is capable of. The plane skids toward takeoff — likely accompanied by much smoke and a powerful rubbery stink."
Of course, we are then left with the realization that it would be impossible for a normal airplane to generate this kind of thrust, not to mention no tires could stand up to it. But regardless, I’m glad this question can finally be put to rest!
Hat tip to BoingBoing for the link.
Related: The Treadmill Problem
October 31st, 2007 at 10:55 pm
A student posed this question to me in my former life as a CFI, and I came up with a slightly different answer.
Since the treadmill is designed to exactly match the tire speed, and we’re starting at a standstill, the treadmill will maintain the tires at 0 while the plane accelerates down the “treadmill runway,” and takes off without the wheels ever spinning.
Just thought I’d chime in.