The results of working at Manchester show that all the visible filth is removed from the Medlock's inky waters, besides which the hardness of the water is reduced to about 6° from a normal condition of about 30°. The effluent is fit for all the varied uses of a dye works, and is stated to be perfectly capable of sustaining fish life. With results such as these the system should have a promising future before it in respect of sewage treatment, as well as the purification and softening of water generally for industrial and manufacturing purposes.--_Iron._ [Illustration: WATER SOFTENING AND PURIFYING APPARATUS.] * * * * * THE TRISECTION OF ANY ANGLE. By FREDERIC R. HONEY, Ph.B., Yale University. The following analysis shows that with the aid of an hyperbola any arc, and therefore any angle, may be trisected. If the reader should not care to follow the analytical work, the construction is described in the last paragraph--referring to Fig. II. Let a b c d (Fig. I.) be the arc subtending a given angle. Draw the chord a d and bisect it at o. Through o draw e f perpendicular to a d. We wish to find the locus of a point c whose distance from a given straight line e f is one-half the distance from a given point d. In order to write the equation of this curve, refer it to the co-ordinate axes a d (axis of X) and e f (axis of Y), intersecting at the origin o. Let g c = x Therefore, from the definition c d = 2x Let o d = D [Hence] h d = D-x Let c h = y [Hence] (2x)² = y² + (D-x)² or 4x² = y² + D²-2Dx + x² [Hence] y²-3x² + D²-2Dx = o [I.] This is the equation of an hyperbola whose center is on the axis of abscisses. In order to determine the position of the center, eliminate the x term, and find the distance from the origin o to a new origin o'. Let E = distance from o to o' [Hence] x = x' + E Substituting this value of x in equation I. y²-3(x' + E)² + D²-2D(x' + E) = o or y²-3x²-6Ex'-3E² + D²-2Dx'-2DE = o [II.] In this equation the x' terms should disappear. [Hence] -6Ex' - 2Dx' = o [Hence] -E = - D/3 That is, the distance from the origin o to the new origin or the center of the hyperbola o' is equal to one-third of the distance from o to d; and the minus sign indicates that the measurement should be laid off to the left of the origin o. Substituting this value of E in equation II., and omitting accents-- We have y² - 3x² + 2Dx - D²/3 + D² - 2Dx + 2D²/3 = o [Hence] y² - 3x² = - 4D²/3 [Illustration: Fig I] [Illustration: Fig II]