Math Thought Of The Day: Wobble


This has possibly been making the rounds a bit, but as we are beginning to get into the Carolina fall days, the thought of drinking beer on a porch at some drinkery is quite appealing. The not so appealing aspect of it is when the table you are sitting at decides that only having 75% of its legs in contact with the ground is going to survive the test of drunkenness as the evening progresses. But what if it was possible to solve this problem with math instead of, say, the establishment's menus, coasters, balled up paper, etc...

The first assumption in this problem is that the legs are of equal length, meaning that the problem isn't some defect with the table, but rather that the ground has some defect in that it is not level. In order to get the table to set without wobbling, all you need to do is rotate the table either 1/4 of a turn clockwise or anti-clockwise. At some point during that rotation, all 4 legs will be solid on the ground.

Now, I am not a geologist, and I certainly haven't studied the terrain of every bar table in the world, but it works something like this. When the table is wobbling, 3 out of 4 legs are in contact with the ground. Leg 4 will be some distance above the ground. As you rotate the table (pos 1 -> pos 2, pos 2 -> pos 3, pos 3 -> pos 4, pos 4 -> pos 1), you can conceptually 'force' that legs 1 - 3 maintain their contact with the ground. But since leg 4, which was above ground, is now occupying the space of leg 1, which was even with the ground, the only way for the other three to maintain their contact, leg 4 now has to be below the ground (true, this can't quite happen in real life, but keep playing along).

If one were to measure the distance of leg 4 from the ground, it would start out at some height above the ground at position 1, and as mentioned above, at position 2, it would be at some height below the ground. There is a theory in math called the 'intermediate value theorem' that states that given a continuous function (the measuring from pos 1 -> pos 2), at some point it has to have every value in that interval. Since we go from a positive -> negative height over that quarter turn, at some point, the height has to equal 0, which means at that point, all 4 legs will be in contact with the ground, and no shim will be necessary.

Of course, if for some reason the table is anchored such that rotating it is out of the question you might as well take your beer and go find "all the shawties in the club, let me see you just back it up, drop it down, let me see you just get low, scrub the ground, let me see you just push it up, push it up, let me see you just wobble baby, wobble baby, wobble baby, wobble, yeah"


  • Martin, A. "ON THE STABILITY OF FOUR-LEGGED TABLES." (n.d.): n. pag. Web
  • "Fix a Wobbly Table (with Math)." YouTube

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