Page:Our big guns.djvu/24

( 18 ) of the increase of the diameter of the circle at the bore; for if not, the percentage of extension would not be equal. But so far from this being the case, it will be found (assuming the density of the metal to remain unaltered that when the gun expands under firing, the actual extension of the imaginary circle at the outside of the gun, will be only one-half of that of the imaginary circle at the bore, and its percentage of extension therefore will only be one-quarter; in other words, the percentage of extension, and therefore the power to resist, decreases, as the square of the distance from the centre of the gun, measured in radii of the bore, increases.

Reverting to the two spiral springs, it is as though you were told to employ the 1-foot and the 2-foot spring together for the lifting of the weight, but with the injunction that whenever you stretched the 1-foot spring a length of "one," you should stretch the 2-foot spring a length of "a half" only—being one-fourth the percentage of extension. If this were done, and the two springs were employed to lift a weight of 1¼ lbs., it would be found that the 1-foot spring was doing one pound of the work while the 2-foot spring was doing only a quarter of a pound.

Let us assume that the designer of the supposed gun of 1-foot bore, with sides 6 inches thick, was of opinion it was not sufficiently strong, and that he determined to add to that strength by doubling the thickness of the walls: these being now 1 foot thick, the outside diameter of the gun would be 3 feet; but from what I have told you, you will be prepared to hear that the value of the imaginary circle of metal at the outside of this three feet diameter, is only one-ninth of the value of the imaginary circle of metal at the bore. The result would be, that while with the gun of 2 feet diameter, the sectional area of the metal, and therefore its weight, would be represented by 2$2$ — 1$2$ = 3, the sectional area, and therefore the weight, of the thickened gun would be represented by 3$2$ — 1$2$ = 8; while it would be found, from the reasons before stated, that this increase of weight from 3 to 8 or 2⅔ times, would only add ⅓rd to the former resisting power; and this is