I am taking the geometry approach. We know from intuition that more than three legs on a chair will make it unstable if any of the legs have a different length than the others. So by "wobble" I mean the possibility that at least one of the legs will be in the air when one or more legs are made shorter/longer than others. Also, the "surface" must be perfectly flat.

A three legged chair is unaffected by any amount of change we make to its legs. So to prove this I started out connecting lines between each legs (diagonals). So far I haven't made any progress.

For a triangle there are no diagonals. Is it enough to show that all the legs must be in the same plane for the chair to be stable?

  • $\begingroup$ A badly designed three-legged chair can fall over. $\endgroup$ – Henry Apr 22 '15 at 21:06
  • $\begingroup$ @Henry Nevertheless, it wouldn't wobble! $\endgroup$ – user41235 Apr 22 '15 at 21:07
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    $\begingroup$ There is always a plane that passes through 3 points. Not so for more than 3 points. $\endgroup$ – Simon S Apr 22 '15 at 21:08
  • $\begingroup$ You can have a chair with more than 3 unequal legs that isn't wobbly, if the surfaces of the feet touch a common plane. $\endgroup$ – John Bentin Apr 22 '15 at 21:10
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    $\begingroup$ In fact, even a four-legged chair can always be rotated to be stable on any smooth floor. The proof involves the intermediate value theorem. $\endgroup$ – vadim123 Apr 22 '15 at 21:36

If “wobble” means “one leg in the air”, then “doesn't wobble” means “all legs touch the floor”. For a mathematician, it should be enough that all legs lie in the same plane, since you can rotate the chair in such a way that this plane coincides with the plane of the floor.

From an egineering point of view, you might have additional constraints like the center of gravity should be above the convex hull of the points touching the floor and so on, but that has little to do with the number of legs, and only affects whether your chair is likely to fall over instead of wobble.


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