PROBLEM XXXIII. Given the Inclination of the Plane, the Elevation of the Piece, and the Range; to find the Charge of Powder. RULE. Find the impetus, by Problem XXX., page 663; with which proceed as directed in the last Problem. Example 1. How much powder will throw a 10-inch shell 6760 feet, on an inclined plane which ascends 7:30; the elevation of the mortar being 33:14?? Solution. 33:14 -7:30:25:44 is the elevation of the mortar above the ascending plane. Inclination of the plane = 1690 Log. secant = 542 feet Log. = Log. of charge for 10-inch shell, from Table D = Charge, in pounds = Sum = 5.164 Log. = 2.734375 2.733999 Hence the charge of powder is 5. 164 lbs., or 5 lbs. 2 oz. Example 2. How much powder will throw a 10-inch shell 6760 feet, on an inclined plane which descends 7:30:, the elevation of the mortar being 33:14:? Solution. 33:14: +7:30:40:44: is the elevation of the mortar above the descending plane. Velocity = . To find the Charge of Powder. 442 Log.= Log. of charge for 10-inch shell, from Table D = Hence the charge is 3. 434 lbs., or 3 lbs. 7 oz. nearly. PROBLEM XXXIV. Given the Inclination of the Plane, the Elevation of the Piece, and the Impetus; to find the Time of Flight. RULE. To the logarithmic secant of the inclination of the plane add the logarithmic sine of the elevation above the plane, half the logarithm of the impetus and the constant log. 9.698970; the sum (abating 30 in the index) will be the logarithm of the time of flight, in seconds. Example. In what time will a 10-inch shell strike an object on an inclined plane which ascends 7:30, when discharged with an impetus of 4574 feet, the elevation of the mortar being 33:14; and in what time will it strike another object on a descending plane, with the same impetus and elevation ? Solution. 33:14 -7:30:25:44 is the elevation of the mortar above the ascending plane; and, 33:14: +7:30:40:44 is the elevation of the mortar above the descending plane. To find the Time of Flight on the ascending Plane. To find the Time of Flight on the descending Plane. Time of flight, in seconds. 22.256 Log. = Note.-The constant logarithm made use of in this Rule. is the arithmetical complement of the log. of 2. PROBLEM XXXV. Given the Impetus and the Elevation; to find the Horizontal Range. RULE. To the logarithm of the impetus add the logarithmic sine of twice the angle of elevation, and the constant logarithm 0.301030; the sum (abating 10 in the index) will be the logarithm of the required range on the horizontal plane. Example 1. Let a shell be discharged with an impetus of 1592 feet, at an elevation of 34:49; required its range on the horizontal plane? Let a shell be discharged with an impetus of 1804 feet, at an elevation of 25:; required its range on the horizontal plane? Note. The constant logarithm used in this Rule is the logarithm of 2. PROBLEM XXXVI. Given the Impetus and the Elevation; to find the Time of Flight on the Horizontal Plane. RULE. With the impetus and the elevation compute the horizontal range, by the last Problem; then, with the horizontal range, thus found, and the elevation of the piece, compute the time of flight, by Problem XXVII., page 660. Or, the time of flight may be computed directly, by Problem XXXIV., page 668. Example. In what time will a 13-inch shell strike an object on a horizontal plane, when discharged with an impetus of 6392 feet, the elevation of the mortar being 34:49?? To find the Time of Flight by Problem XXXIV., page 668. Hence the time of flight by both operations is exactly the same, viz., 22. 824, or 23 seconds nearly. PROBLEM XXXVII. Given the Inclination of the Plane; the Range, and the Charge of Powdér; to find the Elevation of the Mortar. As this is one of the most interesting Problems in the art of Gunnery, I shall show the construction of a diagram, which will give the most familiar view of the principles upon which its solution is founded, as thus: arc cd equal to the angle of inclination; then the arc or right angle abc+cd the whole arc a d.-Bisect a d in b; then the angle d Ab (=bA a) is equal to half a right angle plus half the angle of inclination: hence, the arc bc, or the angle b Ac, is half the complement of the angle of inclination. Through the point d, draw the line A B, making it equal to the given range, and it will represent the descending plane. Draw the line BR X parallel to the vertical A Z, making RX equal to AZ, the measure of the impetus :-take this in the compasses; set one foot on the point A, and with the other describe the small arc nn. Take RX (the impetus) + R B, viz., the whole line B X, in-the compasses; set one leg on the point B, and with the other describe the |