In a short overload trial the pump delivered about two million gallons of water per day at a head of 1000 feet (see Engineering, vol. ii. p. 86 (1903). The centrifugal pump in the near future will probably play a very much more important part amongst water-raising appliances than it has done in the past. Until quite recently it was universally considered that it was only applicable to the lifting of large quantities of water through low lifts, but now such pumps are employed for lifts up to 1000 feet, with an efficiency as high as that of many reciprocating pumps. The wear and tear is very small, and there are no sudden variations in the flow which produce such troublesome effects in the mains of plunger pumps. APPENDIX THE reader should get into the habit of checking the units in any expression he may arrive at, for the sake of preventing errors and for getting a better idea of the quantity he is dealing with. A mere number or a constant may be struck out of an expression at once, as it in no way affects the units: thus, the length of the circumference of a circle is 2′′r (p. 22); or— The length = 2′′ (a constant) × r (expressed in length units) = a certain number of length units Similarly, the area of a circle = r2 (p. 28); or The area = (a constant) × r (in length units) × r (in length units) = a certain number of (length units)2 Similarly, the volume of a sphere The volume = 4π 3 = 3; or (a constant) × r (in length units) × r (in length units) × r (in length units) = a certain number of (length units)3 The same relation holds in a more complex case, c.g. the slice of a sphere (p. 44). The volume = "[3R(Y,2 – Y,2) − Y ̧3 + Y‚3] π 3 (a constant) {3 \a constant) × R (in_length_units) [Y, in length units Y in length units ] - Y2 (length units) + Y3 (in length units } 3 = a constant {a constant [(length_units 3 = a constant x (length units)3 = a certain number of (length units)"} length One more example may serve to make this question of units quite clear. The weight of a flywheel rim is given by the following expression (p. 176) : : [N.B.-The (length units) cancel out, and the constant is omitted, as it does not affect the units.] The weight = acceleration x mass × space2 × time2 = mass x acceleration of gravity (see p. 9) Thus showing that the form or the dimensions of our flywheel equation is correct. Operations of Differentiating and Integrating.-Not one engineer in a thousand ever requires to make use of any mathematics beyond an elementary knowledge of the calculus, but any one who wishes to get beyond the Kindergarten stage of the principles that underlie engineering work must have such an elementary knowledge. The labour spent in acquiring sufficient knowledge of the calculus to enable him to solve practically all the problems he is likely to come across, can be acquired in a fraction of the time that he would have to spend in dodging it by some roundabout process. There is no excuse for ignorance of the subject now that we have such excellent and simple books on the calculus for engineers as Perry's, Barker's, Smith's, Miller's, and others. For the benefit of those who know nothing whatever of the subject, the following treatment may be of some assistance; but it must be distinctly understood that it is only given here for the sake of helping beginners to get some idea of what such processes as differentiation and integration mean, not that they may stop when they can mechanically perform them, but rather to encourage them to read up the subject. Suppose we have a square the length of whose sides is x units, then the area of the square is a = x2. Now let the side be increased by a small amount A, and in consequence let the area be increased by a corresponding amount ▲a; then we have FIG. 614. a + sa = (x + ▲x)2 = x2 + 2x^x + (▲.x-)2 Subtracting our original value of a, we have-▲α = 2x▲x + (A.x)2 Thus by increasing the side of the square by an amount Aư, we have increased the area by the two narrow strips of length x and of width ▲r, and by the small black square at the corner (Ar)2. varies, i.e. For many problems we want to know how the area increases or with any given variation, i.e. decreases increase or in the length of the side; or, exdecrease pressed in symbols, we want to know the value of obtain from the expression above thus- which we can Δ. Δα Ax Now, if we make ar smaller and smaller, the fraction will become more and more nearly equal to 2x, and ultimately (which we shall term the limit) it will be 2x exactly. We then substitute da and dx for ▲a and ▲r, and write it thus or, expressed in words, the area of the square varies 2x times as fast as the length of the side. This is actually and absolutely true, not a mere approximation, as we shall show later on by some examples. The fraction is termed the differential coefficient of a with da dx regard to r. Then, having given a = r2, we are said to differentiate Subtracting our original value of V, we have ▲V = 3x2▲x + 3x(Ax)2 + (▲x)33 The increase is shown in the figure, viz. three flat portions of |