182. PROBLEM IV. To find the quantity of gunpowder which can be contained in a box. 'Find the content of the box in inches, divide it by 30, and the quotient will be nearly the quantity of powder.' For the specific gravity of shaken gunpowder is 927, and if cthe number of cubic inches of capacity that make 1 lb., then (72) ·927 c' × ·578 ounces = 16 ounces, or '536 c' = 16; hence, 16 = = 30 nearly. Hence, if c = the content in inches of the box, and w= the weight in pounds of the powder, w= c nearly. then When the box is cubical, s being its side in inches, then c=s3, and w=33; and hence s = 30 w. EXERCISES. 1. How much powder will a rectangular box contain, which is 15 inches long, 12 broad, and 61 deep? Ans. 39 lbs. 2. How much powder will a rectangular box contain, which is 15 inches long, 13 broad, and 5 deep. Ans. 32.5. 3. How much powder will a cubical box contain, its side being 8 inches? Ans. 17.06 lbs. 4. Find the side of a cubical box that will contain 15 lbs. of powder. Ans. 7.7 inches. 5. How much powder will a hollow cylinder contain, its internal diameter being 6 inches, and its length 10 inches? Ans. 9.4 lbs. 183. PROBLEM V.-The internal diameter of a shell being given in inches, to find the quantity of gunpowder in pounds that will fill it, and conversely. 'Multiply the cube of the internal diameter by 01745, and the product is nearly the number of pounds. Multiply the number of pounds by 573, and the cuberoot of the product will be the diameter.' For c5236 d3, and wc 01745 ď3; = also, 30 w=57.3 w. EXERCISES. 1. How much powder will a shell contain, whose diameter is 7 inches? Ans. 7.36 lbs. 2. Find the quantity of powder that a bomb-shell can contain, its internal diameter being 13 inches. Ans. 38-34 lbs. 3. A shell contains 1.57 lbs. of powder, what is its internal diameter? Ans. 4.5 inches. 4. A bomb-shell contains 38-34 lbs. of gunpowder, what is its internal diameter ? Ans. 13 inches. PILING OF BALLS AND SHELLS. 184. Balls and shells are usually built into piles of some regular form. The base of a pile is either an equilateral triangle, a square, or a rectangle, and is called accordingly a triangular, a square, or a rectangular pile. The number of balls in a side of each course diminishes by unity upwards; and the pile is said to be complete or incomplete according as it is, or is not, finished. The complete triangular and square piles terminate in a single ball; and the complete rectangular pile in a single row. The adjoining figure represents a square pile; the triangular pile is similar, except that it has a triangular course for its base. 185. In the complete square and triangular piles, which are of a pyramidal form, the number of courses is equal to the number of balls in a side of the lowest course; and in the complete rectangular pile, it is equal to the number of balls in the end of the lowest course. In this last pile, too, the number of balls in the upper row is one more than the difference between the number in a side and an end of the lowest course. 186. PROBLEM I.-To find the number of balls in a triangular pile. 'Find the continued product of the number of balls in a side of the base, that number increased by 1, and the same number increased by 2, and take of the product.' Or, let n the number of balls in a side of the base, then EXAMPLE.-How many balls are contained in a triangular pile consisting of 25 courses? Nn (n+1)(n+2) = × 25 × 26 × 27 The number of balls in the successive courses, reckoning from the top, is evidently 1, 1+2, 1+2+3, 1+2+3+4, &c. or 1, 3, 6, 10, ...; and by the differential method (Algebra 533), the sum of this series is easily proved to be n(n+1)(n+2.) EXERCISES. 1. How many balls are contained in a complete triangular pile of 24 courses? Ans. 2600. 2. What is the number of balls in a triangular pile of 40 balls in one side of the base? Ans. 11480. 187. PROBLEM II.-To find the number of balls in a square pile. Find the continued product of the number of balls in a side of the base, that number increased by 1, and twice that number increased by 1, and take of this product.' N=n(n+1)(2n+1.) Or, EXAMPLE.-How many balls are in a square pile of 30 rows? Nin (n + 1) (2 n + 1) = } × 30 × 31 × 61 = 9455. The number of balls in the courses, reckoning from the top, are 1, 22, 32, 42, ... and (Algebra 533) the sum of this series is n (n + 1) (2 n + 1.) EXERCISES. 1. How many balls are in a square pile of 20 courses? Ans. 2870. 2. How many balls are in a square pile, having 15 balls in a side of the base? Ans. 1240. 188. PROBLEM III.-To find the number of balls in a rectangular pile. 'From three times the number of balls in the length of the base, subtract one less than the number in its breadth; find the continued product of the remainder, the breadth, and the breadth increased by 1, and take of the product." Let = number of balls in the length of the base, b = number of balls in the breadth of the base, then N6(6+1) (37—b +1.) b (b = If p the number of balls in the upper row, pl-b+1,l=p+b−1, and b=l—p+1 and N = : ¿ b (b + 1) (2 7+p.) then EXAMPLE. Find the number of balls in a rectangular pile, the number in the length of the base being 36, and in its breadth 24. N=4b(b+1)(37—b+1)=7×24 × 25 (108—24+1) =100 x 85=8500. D E The rectangular pile ACDEB can be divided into two portions, EGFB and ACDEFG. The former is a square pile, the number of whose courses is b, and the whole number of balls in it (187) is therefore A F B =fb (b+1)(2b+1). Again, the number of balls in the upper row of the portion DGF is lb (for DE= AB-BF), in its second course 2 (b), in its third 3 (l—b), &c., or the whole number of balls in it is = (1-6) (1+2+3+4+ ..... b) = (1 —b) (1 + b) (Algebra 424); hence, for the whole rectangular pile, N=6(6+1) (2b+1+37—36) = zb (b+1)(31—b+1.) b EXERCISES. 1. How many shells are contained in a rectangular pile, the number in the length and breadth of its base being 59 and 20? Ans. 11060. 2. Find the number of balls in a rectangular pile of 20 courses, the number in the length of its base being 24. Ans. 3710. 3. Find the number of balls in a rectangular pile of 8 courses, the number in the uppermost row being 5. Ans. 348. A supplementary pile is one that is required to complete an imperfect pile. 189. PROBLEM IV.-To find the number of balls in an incomplete pile. 'Find the number in the whole pile considered as complete, and the number in the supplementary pile, and their difference will be the number in the incomplete pile." Let N' and N" denote the number in the complete and supplementary piles, and N that in the incomplete pile, then N N'-N". EXAMPLES. 1. Find the number of shot in an incomplete triangular pile, the number in a side of the base and top being 40 and 20. N'n (n+1) (n + 2) = ↓ × 40 x 41 x 42=11480 N"=in (n + 1) (n + 2) = 3 × 19 × 20 × 21 = 1330 Hence, N-N'-N"=10150. 2. How many shells are in an incomplete square pile of 10 courses, the number in a side of the top being 20? N' - : } n ( n + 1 ) ( 2 n + 1) = † × 29 × 30 × 59–8555 N"=jn (n + 1) (2 n + 1) = † x 19 x 20 x 39=2470 N N' = -N"-6085. and 3. Find the number of shot in an incomplete rectangular pile of 11 courses, the number in the length and breadth of its base being 30 and 20. N'b(b+1)(37—b+1) = × 20 × 21 (90-20+1)=4970. |