To achieve this: http://en.wikipedia.org/wiki/Square_wheel, what should $L$ be?
rolling wheel problem
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0If 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? – 2011-09-07
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0@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. – 2011-09-07
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2In any event, have you seen [this](http://www.maa.org/pubs/sampMMA.pdf), [this](http://dx.doi.org/10.1119/1.17401), and [this](http://dx.doi.org/10.1119/1.17675)? – 2011-09-07
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2Now the question is less clear. $L$ is supposedly what? – 2011-09-07
2 Answers
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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0Probably more precise to say "catenary arc" instead of "catenary", as the latter is in fact unbounded. – 2011-09-08
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0@J.M.: If you're going to be pedantic, it's an "inverted catenary arc". – 2011-09-08
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0@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... :) – 2011-09-08
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0@J.M.: You can't get out of it like that! ninja's catenary was inverted at least seventeen hours before your comment. – 2011-09-08
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0@Tony: I know; you win. :) – 2011-09-08
The square wheel is a special case of an old general problem known since James Gregory (1668) who defined a transformation between plane curves one in polar coordinates "the wheel" (rho,theta) and one orthonormal frame "the ground" (y,x). Gregory's Transformation (GT) defines a generalized ground : y=rho and x= Integral[(rho).d(theta)]. Inverse transformation GT-1 defines a generalized wheel : rho=y and theta=Integral[dx/y] if y<>0. GT gives the ground if we know the wheel (rho, theta) and GT-1 gives the wheel if we know the ground (y, x). In each case there is only one integration. Cesaro in NAM 1888 has given many properties of these associated curves which have same arc length. When the polar curve rolls on the ground (with initial conditions) the pole O runs along the x-axis (the trivial example is the circle wheel (pole O at the center) [rho=R, any theta] the ground is [y=R, any x] A theorem of Steiner-Habich about pedal of a curve plays an important role in the theory. For details about "Gregory's transformation" and geometry of plane curves see here: http://christophe.masurel.free.fr/#s9 (all papers on this site are open access) and also in "Nouvelles Annales de Mathematiques" http://www.numdam.org/numdam-bin/feuilleter?j=nam (NAM in french language) after 1886 (Cesaro, d'Ocagne,etc. de la Goupilliere and others) and also Habich, Cesaro in "Mathesis (google books).
C. Masurel