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Question:
Im a hobbyist woodworker and, Im embarrassed to say, a mechanical engineer who has forgotten all the calculus I ever learned. I would really appreciate your help with a geometry/calculus problem.

I make mirror frames using a technique called kerf bending. If your unfamiliar with woodworking, a saw kerf is simply the gap that remains after the table saw blade passes through the board. If the blade is raised to a height slightly lower that the thickness of the wood, then the board can be bent by applying a bending moment until the kerf, or series of kerfs, pinch closed. Using a frame material width of 1-7/8 and by cutting kerfs of 1-5/8(leaving a 1/8 strip of wood around the edge) Ive determined that 20 such kerfs will allow me to bend the material 90 degrees. By cutting a series of 80 kerfs, spaced in a particular way I can make an elongated hexagonal frame out of one board. Simply cut the kerfs, bend the board all the way around until the ends touch, clamp it in place, I then fill the pinched-closed kerfs with tinted epoxy, sand off the strip of wood around the edge (effectively making 80 distict pieces of wood with epoxy wedges in between), and it makes a coo! ! l looking mirror frame. (I hope your still interested at this point.)

Anyway, I'm bored with the hex shape and want to make a elliptical frame using the same technique and wood width (360/80=4.5 degrees of bend per kerf). I found an ellipse in my old calculus book that is esthetically pleasing: x2/25 + y2/9=1. This ellipse is the wrong size of course, but it is the "right" shape. What is the equation (with the length of the arc as a variable) for one quadrant of the ellipse, that will generate the spacings for the kerfs (20 per quadrant) such that the final shape will be close to the desired ellipse. Thanks you very much for your time.

Hi,

You stumbled upon a tough question there; you may never have seen it in calculus. The ellipse you like has the parametric equation

C(t) = (5 cos(t), 3 sin(t)), but this is not in terms of arclength. To find the arclength, you have to integrate sqrt{ 25 sin2(t) + 9 cos2(t) } from 0 to pi/2 if you want the first quadrant, more generally from 0 to theta if you want the arclength from angle 0 to angle theta. This integral cannot be written in terms of the functions we usually deal with, polynomials, rational functions or trig functions. You can try it at the website The integrator: I entered it as Sqrt[25(Sin[x])^2 + 9(Cos[x])^2], and I got the answer 3EllipticE[x,-16/9]. All arclength of ellipses give rise to such contorted elliptic integrals for their arclengths.

Claude
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