# rolling wheel problem

To achieve this: http://en.wikipedia.org/wiki/Square_wheel, what should $L$ be?

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If the road the square wheel is rolling on is made up of catenary pieces, then the center traces a straight line. On the other hand, how would a square wheel roll on a single catenary? –  Ｊ. Ｍ. Sep 7 '11 at 16:41
@J.M.: it rolls along the bottom of a single catenary. i.e. It starts from when the up-right corner of the wheel touches the catenary, rolls along it, and ends when the up-left corner touches the catenary. –  ninja Sep 7 '11 at 16:44
In any event, have you seen this, this, and this? –  Ｊ. Ｍ. Sep 7 '11 at 16:49
Now the question is less clear. $L$ is supposedly what? –  Ｊ. Ｍ. Sep 7 '11 at 17:48
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## 1 Answer

The height of the catenary of a square of side length 1 is $(1/\sqrt2)-(1/2)$ since the diagonal of a square is $\sqrt2$ times as long as the side length and the centre of the square must remain at a constant height. The length of the catenary is 1 because the square rolls without slipping.

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Probably more precise to say "catenary arc" instead of "catenary", as the latter is in fact unbounded. –  Ｊ. Ｍ. Sep 8 '11 at 11:47
@J.M.: If you're going to be pedantic, it's an "inverted catenary arc". –  TonyK Sep 8 '11 at 11:52
@Tony: In the first version of the question (see edit history), the OP's visualization had the hump directed downward (he said $y=\cosh\,x$ and not $y=-\cosh\,x$), so... :) –  Ｊ. Ｍ. Sep 8 '11 at 11:54
@J.M.: You can't get out of it like that! ninja's catenary was inverted at least seventeen hours before your comment. –  TonyK Sep 8 '11 at 12:20
@Tony: I know; you win. :) –  Ｊ. Ｍ. Sep 8 '11 at 12:28
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