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This problem is being discussed on the AAMT email discussion list. I have a meter long metal ruler. I push the ends together so that they're only 99cm apart, which means the ruler will bow a bit. How tall is that arc?

The problem we haven't been able to solve is 'what is the shape of the arc'?

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Doesn't the answer depend on the nature of the material under consideration? I suspect that the answer needs some 'physical' assumptions about the ruler (e.g., Young's modulus) and hence a purely mathematical answer may not be available. – tards Nov 4 '11 at 5:42
Why wouldn't Euler-Bernoulli apply? – J. M. Nov 4 '11 at 8:45
You can imagine bending a much longer strip into a variety of shapes - a circle, for example, or where the two "ends" cross at right angles. You can then find a chord which is 99% of the length of the arc it cuts off (intermediate value theorem), and scale to match the problem. So a variety of shapes will be possible. What makes the difference, I think (this may be wrong), is the direction of the forces applied at the end of the arc and stuff like gravity (the answer will be a little different in a horizontal v vertical plane). – Mark Bennet Nov 4 '11 at 9:34

Elastic materials assume shapes that are the graphs of polynomials of degree at most 3. Knowing the length of the ruler, the distance between the endpoints, and making an assumtion about symmetry with respect to the middle may give you enough data to get the equation.

Alternatively, if all you want is a bound, you could probably assume some extreme shapes, like the equal sides of an isosceles triangle, and get an answer that way.

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+1 The precise symmetry assumption might make a difference: with reflectional symmetry I would guess you get a U shape as part of a parabola, but with rotational symmetry I can imagine you might get an S shape as part of a cubic. – Henry Nov 4 '11 at 8:07
Why don't you get a circular arc? Assuming that the directions at the endpoints are unconstrained. – TonyK Nov 4 '11 at 8:53
I'd think that the true shape is the solution of a variational problem; e.g., minimize $\int_\gamma \kappa^2(s)\ ds$, and that polynomial graphs are just an approximation to this. – Christian Blatter Nov 4 '11 at 10:01
The method of cubic splines, q.v., is based on the actual physical splines that used to be part of a draughtman's kit. You tied down a thin elastic strip at the points you wanted it to go through, and the physics said that between nodes the 4th derivative would be zero. Hence, polynomials of degree at most 3. – Gerry Myerson Nov 4 '11 at 11:27
@Gerry Myerson: The strip doesn't know what your coordinate system is. Vanishing of the 4th derivative with respect to $x$ is no geometric invariant. – Christian Blatter Nov 4 '11 at 13:08

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