4

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'?

J. M. ain't a mathematician
  • 75,051
Rex Boggs
  • 159
  • 3
    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 04 '11 at 05:42
  • 1
    Why wouldn't Euler-Bernoulli apply? – J. M. ain't a mathematician Nov 04 '11 at 08:45
  • 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 04 '11 at 09:34

1 Answers1

4

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.

Gerry Myerson
  • 179,216
  • +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 04 '11 at 08:07
  • Why don't you get a circular arc? Assuming that the directions at the endpoints are unconstrained. – TonyK Nov 04 '11 at 08:53
  • 3
    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 04 '11 at 10:01
  • I agree with Christian; this sounds precisely like the sort of problem that should be done via the calculus of variations. – J. M. ain't a mathematician Nov 04 '11 at 10:28
  • 1
    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 04 '11 at 11:27
  • @Tony, we need to compress the ruler with an external force between the end points. This force has to be in the direction from one endpoint to the other, and must be transmitted through each infinitesimal cross section of the ruler. So at different points along the arc, the angle between transmitted net force and the ruler itself will be different, so you cannot assume all points on the arc are equal. – hmakholm left over Monica Nov 04 '11 at 12:40
  • 3
    @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 04 '11 at 13:08
  • @Christian, you have a point, and I currently have no answer. – Gerry Myerson Nov 04 '11 at 21:58