the ball is taken 7, which is a little less than that of solid cast iron, as it ought, on account of the air bubble which is found in all cast balls. PROBLEM VI. To find how far a 24lb ball of cast iron will penetrate into a block of sound elm, when fired with a velocity of 1600 feet per second. HERE, because the substance is the same as in the last problem, both of the balls and wood, N = n, and F f; DV2s 5.55 x 16002 × 13 therefore S Dy? = 41 inches nearly, the penetration required. PROBLEM VII. It was found by Mr. Robins (vol. i. p. 273, of his works), that an 18-pounder ball, fired with a velocity of 1200 feet per second, penetrated 34 inches into sound dry oak. It is required then to ascertain the comparative strength or firmness of oak and elm. THE diameter of an 18lb ball is 5'04 inches = D. Then, by the numbers given in this problem for oak, and in prob. 5, for elm, we have From which it would seem, that elm timber resists more than oak, in the ratio of about 8 to 5; which is not probable, as oak is a much firmer and harder wood. But it is to be suspected that the great penetration in Mr. R.'s experiment was owing to the splitting of his timber in some degree. PROBLEM VIII. A 24-pounder ball being fired into a bank of firm earth, with a velocity of 1300 feet per second, penetrated 15 feet. It is required then to ascertain the comparative resistances of elm and earth. COMPARING the numbers here with those in prob. 5, it is £ F 2 x 1500 x 15 x 12 5.55 x 13002 x 13 152 x 24 = = 133 x 0.37 nearly 63 nearly. That is, elm timber resists = about 6 times more than earth. PROBLEM IX. To determine how far a leaden bullet, of of an inch diameter, will penetrate dry elm; supposing it fired with a velocity of 1700 feet per second, and that the lead does not change its figure by the stroke against the wood. HERE D= , N = 11, n = 7. Then, by the numbers and theorem in prob. 5, it is s = = × 11 × 17002 × 13 173 x 13 == DNV'S 2 × 73 × 15002 63869 6600 But as Mr. Robins found this penetration, by experiment, to be only 5 inches; it follows, either that his timber must have resisted about twice as much; or else, which is much more probable, that the defect in his penetration arose from the change of figure in the leaden ball he used, from the blow against the wood. PROBLEM X. A one pound ball, projected with a velocity of 1500 feet per second, having been found to penetrate 13 inches deep into dry elm: It is required to ascertain the time of passing through every single inch of the 13, and the velocity lost at each of them; supposing the resistance of the wood constant or uniform. The velocity v being 1500 feet, or 1500 × 12 = 18000 inches, and velocities and times being as the roots of the spaces, in constant retarding forces, as well as in accelerating 25 1 = part 12 × 1500 9000 692 of a second, the whole time of passing through the 13 inches; therefore as Hence, as the motion lost at the beginning is very small; and consequently the motion communicated to any body, as an inch plank, in passing through it, is very small also; we can conceive how such a plank may be shot through, when standing upright, without oversetting it. PROBLEM XI. The force of attraction, above the earth, being inversely as the square of the distance from the centre; it is proposed to determine the time, velocity, and other circumstances, attending a heavy body falling from any given height; the descent at the earth's surface being 16 feet, or 193 inches, in the first second of time. Put ' cs the radius of the earth, a = ca the dist. fallen from, CP any variable distance, the velocity at P, t = time of falling there, and 8 = 16, half the veloc. or force at s, f = the force at the point P. Then we have the three following equations, viz. JA B x22: 1:ƒ= the force at P, when the force of The fluents of the last equation give v2 = 4gr2 But x when x = a, the velocity v = 0; therefore, by correction, a general expression for the velocity at any point P. When xr, this gives v v = √(4gr × a=*) for the a -) greatest velocity, or the velocity when the body strikes the earth. When a is very great in respect of r, the last velocity becomes (1 − ) × √4gr very nearly, or nearly 4gr only, which is accurately the greatest velocity by falling from an infinite height. And this, when r≈ 3965 miles, is 6.9506 miles per second. Also, the velocity acquired in falling from the the distance of the sun, or 12000 diameters of the earth, is 6.9505 miles per second. And the velocity acquired in falling from the distance of the moon, or 30 diameters, is per second. 6.8927 miles Again, to find the time; since tv = - *, therefore which gives t√ and vers, a √ ax-xx ä 4gr2 :; the correct fluent of × (√ax-xx + arc to diameter a x); or the time of falling to any point p = X (ABBP). And when r =r, this becomes 2r g a AD + DS g SC for the whole time of falling to the surface at s; which is evidently infinite when a or AC is infinite, though the velocity is then only the finite quantity /4gr. When the height above the earth's surface is given = g; becauser is then nearly = a, and AD nearly Ds, the time t for the distance g will be nearly 1 4.gr2 1 X 2DS √ × √ 4gr = 1", as it ought to be. 4.gr If a body, at the distance of the moon at A, fall to the earth's surface at s. Then = 3965 miles, a = 60r, and t = 416806′′ = 4 da. 19h. 46′ 46′′, which is the time of falling from the moon to the earth. When the attracting body is considered as a point c; the whole time of descending to c will be Hence, the times employed by bodies, in falling from quiescence to the centre of attraction, are as the square roots of the cubes of the heights from which they respectively fall. PROBLEM XII. The force of attraction below the earth's surface being directly as the distance from the centre: it is proposed to determine the circumstances of velocity, time, and space fallen by a heavy body from the surface, through a perforation made straight to the centre of the earth: abstracting from the effect of the earth's rotation, and supposing it to be a homogeneous sphere of 3965 miles radius. |