PROBLEM XXV. Given the Range for one charge; to find the Range for another Charge. RULE. As the logarithm of the first charge, is to the logarithm of its corresponding range; so is the logarithm of the other charge, to the logarithm of its corresponding range; the elevation being the same in both cases, Example 1. If a shell be projected 5334 feet by a charge of 12 lbs. of powder, at what distance will it strike an object when discharged with 6 lbs. of powder; the elevation being the same in both cases? If a shell be projected 4000 feet by a charge of 9 lbs. of powder, at what distance will it strike an object when discharged with 64 lbs. of powder; the elevation being the same in both cases? Given the Range and the Elevation; to find the Impetus.. RULE. As the logarithmic sine of twice the angle of elevation, is to the logarithm of half its corresponding range; so is radius, or the logarithmic sine of 90%, to the impetus. Example 1. With what impetus must a shell be discharged at an elevation of 34:49, to strike an object at the distance of 2986 feet? With what impetus must a shell be discharged at an elevation of 25%, to strike an object at the distance of 2760 feet? Log.=. 3.202096 Given the Elevation and the Range; to find the Time of the Flight. RULE. As radius, is to the logarithmic tangent of the elevation; so is the logarithm of the range, in feet, to a logarithmic number; which, being divided by 2, will give the logarithm of 4 times the number of seconds taken up in the flight. Example 1. In what time will a shell range 11986 feet, at an elevation of 34:49: ? Example 2. In what time will a shell range 3250 feet, at an elevation of 32 degrees? Note. From this it is manifest that when the elevation of the piece is 45%, half the logarithm of the range will be the logarithm of 4 times the number of seconds taken up in the flight. PROBLEM XXVIII. Given the Range and the Elevation: to find the greatest Altitude of the As radius, is to the logarithmic tangent of the elevation; so is the logarithm of one-fourth of the range, to the logarithm of the required altitude. Example 1. If a shell range 11986 feet, when projected at an elevation of 34:49'; required the greatest altitude which it acquires during its flight? If a shell range 4760 feet, when projected at an elevation of 45;. required the greatest altitude which it acquires during its flight? Note. From this it is manifest, that when the elevation of the mortar is 45 degrees, one-fourth of the range will be equal to the greatest altitude at which the shell can arrive. PROBLEM XXIX. Given the Inclination of the Plane, the Elevation of the Piece, and the Impetus; to find the Range. RULE. To twice the logarithmic secant of the inclination of the plane, add the logarithmic sine of the elevation of the piece above the plane, the logarithmic co-sine of the elevation of the piece above the horizon, and the logarithm of 4 times the impetus: the sum of these four logarithms (rejecting 40 in the index,) will be the logarithm of the required range. Example. How far will a shell range on a plane which ascends 10°15', and also on another plane which descends 10:15; the impetus being 2000 feet in both cases, and the elevation of the mortar 31:45:? Solution. 31:45-10:15-21:30, the elevation of the piece above the ascending plane; and, 31:45 +10:15:42: 0, the elevation of the piece above the descending plane. To find the Range on the ascending Plane:— Twice the log. secant=20.013974 Inclination of the plane = 10:15 Given the Inclination of the Plane, the Elevation of the Piece, and the Range; to find the Impetus. RULE. To twice the logarithmic co-sine of the inclination of the plane, add the logarithmic co-secant of the elevation of the piece above the plane, the logarithmic secant of the elevation of the piece above the horizon, and the logarithm of the one-fourth of the range: the sum of these four logarithms (abating 40 in the index,) will be the logarithm of the impetus. Example. With what impetus must a shell be discharged to strike an object at the distance of 2575 feet, on an inclined plane which ascends 10:15', and, also, another object at the distance of 4701 feet, on an inclined plane which descends 10:15; the elevation of the piece being 31:45 in both cases? Solution. 31:45-10:15-21:30, is the elevation of the piece above the ascending plane; and, 31:45 +10:15:42: 0, is the elevation of the piece above the descending plane. |
