In making a speed control for an off-the-shelf electric fan, I needed to install a square power receptacle in the phenolic box I am using for the speed dial switches and 7-segment displays. Phenolic is brittle and does not machine well with the woodworking tools that I have, chipping instead of cutting.

A straightforward way to cut the hole would be to drill a hole, disassemble a coping saw and reassemble it with the blade through the hole, and sawing. But the small size of the box would limit the saw strokes to a few centimetres.

A reciprocating "Sabre" saw might do the job but is hard to control.

Twist drill bits seemed to work the best of my tools on the material, and I have a good selection of sizes. How close to a square opening can one create by drilling a small number of round holes? It turns out that I can come fairly close. There are two ways the problem can be posed, the largest opening bounded within the square or the smallest opening just larger than the square. I am interested in the latter; the other solution can be had by scaling.

The idea is to drill holes on the diagonals near the 4 corners such that the corner touches the rim of the hole. One then drills a hole in the center which is enough larger than the square so that its points of contact with the square are points of contact with the 4 smaller holes. If the diameter of the 4 corner holes is reduced, then the diameter of the center hole must be increased in order for it to intersect the smaller holes and the square.

Let L be the length of one side of the square, R be the radius of the center hole, and r be the radius of the corner holes. The center of each corner hole is L/sqrt(2)-r from the center of the square so that the rim lies on the corner point. The furthest that the corner holes exceed the desired square is

r+(L/sqrt(2)-r)/sqrt(2)-L/2 = r-r/sqrt(2) = r*(1-sqrt(1/2))

The furthest that the large center hole exceeds the desired square is R-L/2. Desired is:

R-L/2 = r*(1-sqrt(1/2))

The other constraint is that the rim of the center hole and the rim of the corner hole intersect on the side of the square. The distance from the middle of the side to the intersection of the side with the large hole is x

R^2 = (L/2)^2 + x^2

The distance from the corner to the intersection is r*sqrt(2). So:

L/2 = r*sqrt(2) + x

R, r, and x scale with L. Let L = 1.

R^2 = 1/4 + x^2

1/2 = r*sqrt(2) + x

R-1/2 = r*(1-sqrt(1/2))

Solving this system:

r = 1/(4+sqrt(2))

r = L * 184.69903125906464e-3

R = L * 554.097093777194e-3

The diameters when L = 1.125 are:

d = 415.57282033289544e-3

D = 1.2467184609986863

I made the holes with a 3/8 inch twist drill and a 1.25 inch hole saw. A 7/16 bit would have been closer in size.

The flange on the outlet covers the non-square parts of opening.

Governance by those who do the work.

## Saturday, March 19, 2016

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