To the versed fine of 20° 48' 8.81361 Half of which added to or subtracted S 58° 26' higher. from 48° 2', gives 237° 38 lower. PROBLEM II. Given tire angles of direction, obliquity of the plane, and amplitude, to find the impeius. As sine ang. dAzx ? 31° 34' = 9.71891 into fine ang. BAD, S 52° 22' = 9.89869 19.61760 Is to the square of the fine of BAZ: } 33° 56' = 9.99756 19.99512 So is AK=1760 3.24557 4 23.24003 To the impetus 4199 3.62303 PROBLEM III. The angles o direction, obliquity of the plane, and impetus being giver, to find the random. EXAMPLE The obliquity of the plane is 6° 4', the angles of the direction 58° 26' 2 {JA}={38 ° 38'} and impetus 4200, to find the distance of the object. As square of the fine DAx 83° 56'=9.99756=19.99512 52° 22' 9.89869 19.01760 So is impetus 4200 3.62325 23.24085 To amplitude 1761 3.24573 PROBLEM IV. The angles of direction, obliquity of the plane, and amplitude being given, to find the amplitude of any given elevation. EXAMPLE. The angle of obliquity KAx is 6° 4', any angle of direction 37° 38', and its amplitude is 7040, any other angle of direction 33° being given, to find the amplitude for that other direction. As the Gine dAz 31° 34'=9.71891 =19.61760 =19.55871 3.84757 To the amplitude required 6147 23.40628 PROBLEM PROBLEM V. The impetus and obliquity being given, to find the greatest rondoin. EXAMPLE Let the impetus be 4200, obliquity of the plane 6° 4', required the greatest random. As tangent 48° 2' 10.04607 3.92428 10.00244 To the greatest random 7596 13.92672 3.88665 If to 45° you add half the angle of obliquity, the sum is the direction that carries farthest up an ascent. If from 45° you subtract half the angle of obliquity, the remainder is the direction which carries farthest on a descent. The greatest distance up an ascent is equal to twice the impetus, wanting the height of the mark above the horizontal plane. And the greatest distance down a descent is equal to twice the impetus, together with the depression of the object below the horizontal line. In actual service, cases on ascents and descents are feldom attended to. COMPUTATION OF Shot. It is customary to pile iron balls and fhells in horizontal rows; the piles are denominated according to the figure of their respective bases. The base is commonly an equilateral triangle, square, or rectangle. Triangular and square piles, when com plete plete, terminate in a Gngle ball, and a rectangular pile in a single row. The two first, when complete, form a pyramid, the last a wedge. To find the number of balls in a triangular pile, RULE. Put n for the number of balls in a side of the base row, then - nx n+1 X n+2, gives the number of balls in the pile. 6 EXAMPLE. I. Required the number of balls in a triangular pile, a side of the base ţire contains 30 balls. 30=n 930 1860 2790 6)29760 4960 balls in the pile. Ex. 2. How many balls are in a triangular pile, the side of the bottom-row being 25? Anf. 2925 Ex. 3. Required the number of balls in a triangular pile, the side of the base-row being 20. Anf. 1540. Ex. 4. How many balls are in a triangular pile, the baserow being 10? Anf. 220. Ex. 5. How many balls are in a triangular pile, whose basetire is 4? Anj. 20 PROBLEM II. To find the number of balls in a square pila RULE. Put n for the number of balls in the side of the square base, then Xn+1 X 2n+1 is the number of balls in the pile. 6 EXAMPLE I. How many balls are in a square pile of 30 balls to the side of the base-row? 30=N 930 930 5580 6)56730 9455 balls in the pile. Ex 2. How many balls are in a pile, the side of th: square base being 15 balls ? Anf. 1240. Ex. 3. How many balls are in a square pile of 13 tires ? Anf. 819. How many balls are in a square pile of s2 tires ? Anj. 650. Ex. 4: |