Devlin's Angle

How to stabilize a wobbly table

This problem - as a math problem - has been around since the 1960s, when a British mathematician called Roger Fenn first formulated it, presumably while in a restaurant sitting at a wobbly table.

In 1973, the famous math columnist Martin Gardner wrote about the problem in his Scientific American column, presenting a short, clever, intuitive argument to show how rotation will always stop the wobble. Here is that argument.

Two caveats. This only works for a table with equal legs, where the wobble is caused by an uneven floor. If the table has uneven legs, you probably need the folded napkin. Also, the floor can be bumpy but has to be free of steps - if the table is on a staircase, as you often find in outdoor cafes in Tuscan hill towns like San Gimignano, the math doesn't work. But then, if you are sitting on a restaurant terrace in Tuscany, who cares?

However uneven the floor, a table will always rest on at least three legs, even if one leg is in the air. Suppose the four corners are labeled A, B, C, D going clockwise round the table, and that leg A is in the air. If the floor were made of, say, sand, and you were to push down on legs A and B, leaving C and D fixed, then you could bring A into contact with the floor, but leg B would now extend into the sand. Okay so far?

Here comes the clever part. Since all four legs are equal, instead of pushing down on one side of the table, you could rotate the table clockwise through 90 degrees, keeping legs B, C and D flat on the ground, so that it ends up in the same position as when you pushed it down, except it would now be leg A that is pushed into the sand and legs B, C, and D are all resting on the floor. Since leg A begins in the air and ends up beneath the surface, while legs B, C, and D remain flat on the floor, at some point in the rotation leg A must have first come into contact with the ground. When it does, you have eliminated the wobble.

This argument seems convincing, but making it mathematically precise turned out to be fairly hard. (Not that the problem attracted a great deal of attention - I doubt it was even considered as a possible Millennium Problem.) In fact, it took over 30 years to figure it out. The solution, presented in the paper Mathematical Table Turning Revisited, by Bill Baritompa, Rainer Loewen, Burkard Polster, and Marty Ross, is available online at

http://arxiv.org/abs/math/0511490

As most regular readers of this column will have guessed immediately, the result follows from the Intermediate Value Theorem. But getting it to work proved much harder than some other equally cute, real-world applications of the IVT, such as the fact that at any moment in time, there is always at least one location on the earth's surface where the temperature is exactly the same as at the location diametrically opposite on the other side of the globe.

As the authors of the 2005 solution paper observe, "for arbitrary continuous ground functions, it appears just about impossible to turn [the] intuitive argument into a rigorous proof. In particular, it seems very difficult to suitably model the rotating action, so that the vertical distance of the hovering vertices depends continuously upon the rotation angle, and such that we can always be sure to finish in the end position."

The new proof works provided the ground never tilts more than 35 degrees. (If it did, your wine glass would probably fall over and the pasta would slide off your plate, so in practice this is not much of a limitation.)

Is the theorem any use? Or is it one of those cases where the result might be unimportant but the math used to solve it has other, important applications? I have to say that, other than the importance of the IVT itself, I can't see any application other than fixing wobbly tables. Though I guess it does demonstrate that mathematicians do know their tables.