Example 2. Required the side of a cubical box that will hold 120 lbs. of gun Note. Since this problem is the converse of the last, the constant logarithm made use of is the logarithm of 30, the established divisor and multiplier for filling rectangular boxes. PROBLEM X. To find how much Powder will fill a Cylinder. RULE. To twice the logarithm of the diameter of the cylinder, add the logarithm of its length and the constant logarithm 8.417937; the sum (abating 10 in the index,) will be the logarithm of the pounds of powder. Example 1. How much powder will a cylinder hold, the diameter of which is 13 inches, and the length 26 inches? Diameter of the cylinder 13 inches; twice its log. = 2.227886 Length of ditto = 26 ditto. How much powder will a cylinder hold, the diameter of which is 4 inches, and the length 12 inches? Diameter of the cylinder = 4 inches; twice its log. = 1.204120 Length of ditto = 12 ditto. Log. = Note. The constant logarithm made use of in this problem is the arithmetical complement of the logarithm of 38. 2, the established divisor for filling cylinders with gunpowder. PROBLEM XI. To find what Length of a Cylinder will be filled with a given Weight of Gunpowder. RULE. To the arithmetical complement of twice the logarithm of the diameter of the cylinder, or caliber of the gun, add the logarithm of the given weight of powder in pounds, and the constant logarithm 1.582063: the sum (abating 10 in the index,) will be the logarithm of the length of the cylinder, in inches. Example 1. What length of a 24-pounder gun, of 5. 66 inches caliber, will be filled with 8 lbs. of gunpowder? Caliber of the gun = 5.66 Ar. comp. of twice its log. = 8.494368 Given weight of powder=8 lbs. Log. =. Constant log. = Length, in inches, = 9.539 Log. 0.903090 Example 2. What length of a 42-pounder gun, of 6. 23 inches caliber, will be filled with 10 lbs. of gunpowder? Note. This problem being the converse of the last, the constant logarithm is the logarithm of 38. 2, the established divisor and multiplier for filling cylinders with gunpowder. PROBLEM XII. To find the Number of Balls in a triangular Pile. RULE. To the logarithm of the number of balls in the bottom row, add the logarithm of that number increased by 1, and also its logarithm increased by 2, and the constant logarithm 9. 221849: the sum (rejecting 10 in the index,) will be the logarithm of the required number of balls. Example 1. Required the number of balls in a triangular pile, each side of its base containing 30 balls? Required the number of balls in a triangular pile, each side of its base Note. The constant logarithm employed in this problem is the arithmetical complement of the logarithm of 6, the established divisor for triangular, square, and rectangular piles of shot. PROBLEM XIII. To find the Number of Balls in a square Pile, RULE. To the logarithm of the number of balls in one side of the bottom row, add the logarithm of that number increased by 1, the logarithm of twice. the same number increased by 1, and the constant logarithm 9. 221849: the sum (abating 10 in the index,) will be the logarithm of the required number of balls. Example 1. Required the number of balls in a square pile, each side of its base containing 30 balls? Balls in one side of the base = 30 Log. 1.477121 Required the number of balls in a square pile, each side of its base To find the Number of Balls in a rectangular Pile. RULE. From three times the number of balls contained in the length of the bottom row, subtract the number of balls, less by 1, contained in the breadth of that row; then, to the logarithm of the remainder, add the logarithm of the number of balls contained in the breadth of the bottom row, the logarithm of that number increased by 1, and the constant logarithm 9.221849: the sum (rejecting 10 in the index,) will be the logarithm of the required number of balls. Example 1. Required the number of balls in a rectangular pile, which contains 46 balls in the base row of its longest side, and 15 balls in that ofi ts shortest side? Balls in length 46×3=138 Required the number of balls in a rectangular pile, which contains 59 balls in the base row of its longest side, and 20 balls in that of its shortest side? To find the Number of Balls in an incomplete triangular Pile. RULE. Find the number of balls in the whole pile, considered as complete, by Problem XII., page 566; and find also, by the same problem, the number of balls answering to the triangular pile, the side of whose base is represented by the number of shot in the side of the top course of the incomplete pile diminished by 1; then, the difference of the two results will be the number of shot remaining in the pile. Example. Required the number of shot in an incomplete triangular pile; each side of its bottom course containing 40 balls, and each side of its top course containing 20 balls? |