The Seven Follies of Science [2nd ed.] / A popular account of the most famous scientific impossibilities and the attempts which have been made to solve them. To which is added a small budget of interesting paradoxes, illusions, and marvels

by John Phin

If the circumference had been given, the diameter might have been found by dividing the circumference into twenty-two parts and setting off seven of them. This would give the diameter. A more accurate method is as follows:

Given a circle, of which it is desired to find the length of the circumference: Inscribe in the given circle a square, and to three times the diameter of the circle add a fifth of the side of the square; the result will differ from the circumference of the circle by less than one-seventeen-thousandth part of it. Another method which gives a result accurate to the one-seventeen-thousandth part is as follows:

Fig. 1.

Let AD, Fig. 1, be the diameter of the circle, C the center, and CB the radius perpendicular to AD. Continue AD and make DE equal to the radius; then draw BE, and in AE, continued, make EF equal to it; if to this line EF, its fifth part FG be added, the whole line AG will be equal to the circumference described with the radius CA, within one-seventeen-thousandth part.

The following construction gives even still closer results: Given the semi-circle ABC, Fig. 2; from the extremities A and C of its diameter raise two perpendiculars, one of them CE, equal to the tangent of 30°, and the other AF, equal to three times the radius. If the line FE be then drawn, it will be equal to the semi-circumference of the circle, within one-hundred-thousandth part nearly. This is an error of one-thousandth of one per cent, an accuracy far greater than any mechanic can attain with the tools now in use.

Fig. 2.

When we have the length of the circumference and the length of the diameter, we can describe a square which shall be equal to the area of the circle. The following is the method:

Draw a line ACB, Fig. 3, equal to half the circumference and half the diameter together. Bisect this line in O, and with O as a center and AO as radius, describe the semi-circle ADB. Erect a perpendicular CD, at C, cutting the arc in D; CD is the side of the required square which can then be constructed in the usual manner. The explanation of this is that CD is a mean proportional between AC and CB.