With a little consideration of the relative value of the upper and lower scales, the student interested will readily perceive how equations of the third degree may be similarly resolved. The subject is not of sufficient general importance to warrant a detailed examination being made of the several expressions which can be dealt with in the manner suggested; but the author gives the following example as affording some indication of the adaptability of the method to practical calculations. EX.—A hollow copper ball, 7·5 in. in diameter and 2 lb. in weight, floats in water. To what depth will it sink? The water displaced = 27·7 × 2 = 55·4 cub. in. The cubic contents of the immersed segment will be (π)/(3)(3_r_ _x_^2 − _x_^3), _r_ being the radius and _x_ the depth of immersion. Hence (π)/(3)(3_r_ _x_^2 − _x_^3) = 55·4, and 11·25_x_^2 − _x_^3 = 52·9. To solve this equation we place the cursor to 52·9 on A, and move the slide until the reading on D under 1 and that on B under the cursor together amount to 11·25. In this way find 2·45 on D under 1, with 8·8 on B under the cursor _c_, _c_, as a pair of values of which the sum is 11·25. Hence we conclude that _x_ = 2·45 in. is the result sought. With the rule thus set (Fig. 39) the student will note that the slide is displaced to the right by an amount which represents _x_ on D, and therefore _x_^2 on A; while the length on B from 1 to the cursor line represents 11·25 − _x_. Hence the upper scale setting gives _x_^2(11·25 − _x_) = 11·25_x_^2 − _x_^3 = 52·9 as required. [Illustration: FIG. 39.] When in doubt as to the method to be pursued in any given case, the student should work synthetically, building up a simple example of an analogous character to that under consideration, and so deducing the plan to be followed in the reverse process. SCREW-CUTTING GEAR CALCULATIONS. The slide rule has long found a useful application in connection with the gear calculations necessary in screw-cutting, helical gear-cutting, and spiral gear work. SINGLE GEARS.—For simple cases of screw-cutting in the lathe it is only necessary to set the threads per inch to be cut to the threads per inch in the guide screw (or the pitch in inches in each case, if more convenient). Then any pair of coinciding values on the two scales will give possible pairs of wheels. EX.—Find wheels to cut a screw of 1⅝ threads per inch with a guide screw of 2 threads per inch. Setting 1·625 on C to 2 on D, it is seen that 80 (driver) and 65 (driven) are possible wheels. COMPOUND GEARS.—When wheels so found are of inconvenient size, a compound train is used, consisting (usually) of two drivers and two driven wheels, the product of the two former and the product of the two latter being in the same ratio as the simple wheels. Thus with 60 and 40 as drivers, and 65 and 30 as driven, we have, (60 × 40)/(65 × 30) = (2400)/(1950) = (2)/(1·625) as before.