The Straight Dope | The Straight Dope | Chicago Reader

The Straight Dope 

Cecil, always enjoy your column, however you've got this "airplane and conveyor belt business" absolutely wrong. . . . --strafe, via the Straight Dope Message Board

It's all about the interpretation of the question. Unfortunately, Cecil commingled two different interpretations in his column. --zut, via the SDMB

My confidence in Cecil has taken a gigantic hit. . . . Cecil has fallen into the common trap of believing that the velocity of the treadmill in this case is what is important. It's not. What is important is the acceleration of the treadmill. I swear, on pain of retaking physics before I graduate as an ME, that if I accelerate the treadmill at a rate of 2 x (force from engines)/(mass of tires) that the plane goes absolutely nowhere. --treis, via the SDMB

I knew this was going to happen. Everyone else, forgive me. This week's column is for the geeks. Here's the original question, as encountered online: "A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?"

The implicit assumption is that if the conveyor belt's speed backward exactly matches the airplane's "speed" (whatever that means) forward, the plane remains stationary relative to the earth and, more important, to the air. (We assume the winds are calm.) With no wind moving past its wings, the plane generates no lift and can't take off.

But the assumption is false. While the conveyor does exert some modest backward force on the plane, that force is easily overcome by the thrust of the engines pulling the plane ahead. The plane moves forward at roughly its usual speed relative to the ground and air, generates lift, and takes off. Many people have a hard time grasping this (although it can be easily demonstrated in the lab), but eventually they do, smack their foreheads, and move on. We'll call this Basic Realization #1.

Message-board discussions of this question tend to feature a lot of posters who haven't yet arrived at BR #1 talking right past those who have, insisting more and more loudly that the plane won't take off. Then there's a whole other breed of disputants who, whether or not they've cracked the riddle as originally posed, prefer to reframe it by proposing progressively more esoteric assumptions, refinements, analogies, etc. Often they arrive at a separate question entirely: Is there a way to set up the conveyor so that it overcomes the thrust of the engines and the plane remains stationary and doesn't take off?

The answer is yes. Understanding why is Basic Realization #2.

The conveyor doesn't exert much backward force on the plane, but it does exert some. Everyone intuitively understands this. To return to the analogy in my original column, if you're standing on a treadmill wearing rollerblades while holding a rope attached to the wall in front of you, and the treadmill is switched on, your feet will initially be tugged backwards. Partly this is due to friction in the rollerblade wheel bearings, but partly--this is key--it's because the treadmill is accelerating the rollerblade wheels and in the process imparting some angular (rotary) but also some linear (backward) momentum to them. You experience the latter as backward force. Eventually the treadmill reaches a constant speed and the rollerblade wheels cease to accelerate. At this point you can easily haul in the rope and pull yourself forward.

But what if the treadmill continues to accelerate? Different story. In principle it's possible to accelerate the treadmill at a rate that will exactly counteract any forward force you care to apply. (This is a departure from the original question, which said the conveyor belt compensated for the plane's speed, not its force.) The only mathematics needed to demonstrate this is the well-known physics axiom F = ma--that is, force equals mass times acceleration. Given that the conveyor exerts some backward force F on the plane, we simply crank up the acceleration as much as necessary to equal any forward force F generated by its engines. Result: The plane stands still and doesn't take off. Welcome to BR #2.

You may say it's impossible to build a constantly accelerating treadmill, that eventually we run into the limitation imposed by the speed of light, etc. True but irrelevant--BR #2 has an intrinsic elegance that transcends such practical concerns. Why didn't I bring it up in the first place then? You've got to be kidding. It took an entire column to get BR #1 across, and a second one to convey (I hope) BR #2. One fricking thing at a time.

Art accompanying story in printed newspaper (not available in this archive): illustration/Slug Signorino.

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