formed by the tangent AI and AG, is equal to the angle APq, which will therefore stand on the same arc Aq. Cor. 2. If there be given the range and velocity, or the impetus, the direction will hence be easily found, thus: Take Ak = AI, draw kq perpendicular to AH, meeting the circle described with the radius AO in two points q and q; then Aq or Aq will be the direction of the piece. And hence it appears, that there are two directions, which, with the same impetus, give the very same range AI. And these two directions make equal angles with AI and AP, because the arc Pq is equal the arc Aq. They also make angles with a line drawn from A through S, because the arc Sq is equal to the arc Sq. = Cor. 3. Or, if there be given the range AI, and the direction Aq; to find the velocity or impetus. Take Ak AI, and erect kg perpendicular to AH, meeting the line of direction in q; then draw qP making the angle AqP = angle Akq; so shall AP be the impetus, or the altitude due to the projectile velocity. Cor. 4. The range on an oblique plane, with a given elevation, is directly as the rectangle of the cosine of the direction of the piece above the horizon, and the sine of the direction above the oblique plane, and reciprocally as the square of the cosine of the angle of the plane above or below the horizon. For in the triangles APg and Akq we have 33. The range is the greatest when Ak is the greatest; that is, when kq touches the circle in the middle point S; and then the line of direction passes through S, and bisects the angle formed by the oblique plane and the vertex. Also, the ranges are equal at equal angles above and below this direction for the maximum. Cor. 5. The greatest height cv or kq of the projectile above the plane is and Aq kq sin Akq : sin qAI :: cos HAI : sin qAI. : = Cor. 6. The time of flight in the curve AvI = HAI. time of describing the curve is equal to the time of falling freely through GI or sin2 qAI COS2 IAH' ; hence by the laws of gravity, the time of flight ΣΑΡ 2 sin qAISAP. cos HAI 34. SCHOLIUM. From the foregoing corollaries may be collected the following theorems, in which i denotes the inclination of the plane to the horizon, e the HHH angle of elevation above the horizon, and the other letters as in the former 35. The principal properties of a projectile in a non-resisting medium, may be very elegantly deduced, by the method adopted in the following proposition. COS 2 PROP. XIX. 36. A body is projected from "a point A, with a velocity v, in the direction AT, making an angle e with the horizon; it is required to find where it will strike the plane AI, passing through the point of projection, and making an angle i with the horizontal plane AH. Let t=time of flight, or the time in which the body a describes the path AVI. RAI, the range on the oblique plane AI. :: tv = AT space described in t seconds by velocity of projection. g=TI = space described by force of gravity in t seconds. If a = altitude through which a body must fall from rest to acquire the velocity of projection; then v2 = 2ag, and Cor. 1. When the plane AI is horizontal; then i = 0, and we have (4) (5) But x = AK = AT cos TAK = tv cos e, and eliminating t by these two equations we have This is the equation to the curve, and is of great advantage in the solution of equations in reference to projectiles. PRACTICAL RULES IN PROJECTILES. I. The velocity varies nearly as the square root of the charge, when the shot are the same; that is, if V and v are the velocities, and C and c the charges; then II. With equal charges, the velocity varies inversely, as the square root of the weight; that is, if B and b are the weights of two shots; then V √b v ✓B III. When unequal shot are projected with unequal charges, then IV. If the charges are proportional to the weight of the shot; then the velocities will be the same for all shot. For let C = m B and cmb; then we have V. It has been found by experiment, that if the charge be of the weight of the shot; then the velocity is 1600 feet per second nearly. VI. With the same elevation, the range varies as the charge; that is, if R and R, are the ranges, then VII. When the plane is horizontal, and the velocity the same, the range varies as the sine of twice the elevation; that is, if e and e, are the elevations, then R: R, sin 2 e sin 2 e1. : EXAMPLE IN PROJECTILES. Find the velocity and angle of elevation, that the projectile may pass through the two given points I, I'; supposing AK = 300, AK' = 400, KI = 60, and K'I' 40 feet. Also, if in equation (1) we make y = 0, we have the range To determine the greatest height of the projectile above the horizontal plane, we must find the maximum value of y from equation (1); hence by differentiating the equation which is evidently half the whole range; therefore putting this value for x in the equation of the curve we have for y EXAMPLE 1. If a ball of 1lb. acquire a velocity of 1600 feet per second, when fired with 5 ounces of powder; it is required to find with what velocity each of the several kinds of shells will be discharged by the full charges of powder, viz. Ans. The velocities in lbs. 594 584 595 693 693 EXAM. 2.—If a shell be found to range 1000 yards, when discharged at an elevation of 45°; how far will it range when the elevation is 30° 16', the charge of powder being the same? Ans. 2612 feet, or 871 yards. EXAM. 3.—The range of a shell at 45° elevation, being found to be 3750 feet; at what elevation must the piece be set, to strike an object at the distance of 2810 feet, with the same charge of powder? Ans. at 24o 16', or at 65° 44'. EXAM. 4.-With what impetus, velocity, and charge of powder, must a 13 inch shell be fired at an elevation of 32° 12′, to strike an object at the distance of 3250 feet? Ans. impetus 1802, velocity 340, charge 2.95 lb. EXAM. 5.—A shell being found to range 3500 feet, when discharged at an elevation of 25° 12′; how far then will it range at an elevation of 36° 15′, with the same charge of powder? Ans. 4332 feet. EXAM. 6.—If, with a charge of 9 lbs. of powder, a shell range 4000 feet; what charge will suffice to throw it 3000 feet, the elevation being 45o in both cases? Aus. 63 lbs. of powder. EXAM. 7.-What will be the time of flight for any given range, at the elevation of 45°? 32°? Ans. The time in seconds is the square root of the range in feet. EXAM. 8.-In what time will a shell range 3250 feet at an elevation of Ans. 11 seconds, nearly. EXAM. 9. How far will a shot range on a plane which ascends 8° 15′, and another which descends 8° 15'; the impetus being 3000 feet, and the elevation of the piece 32° 30'? Ans. 4244 feet on the ascent, and 6745 feet on the descent. EXAM. 10.—How much powder will throw a 13 inch shell 4244 feet on an inclined plane which ascends 8° 15', the elevation of the mortar being 32o 30'? Ans. 4.92535 lb. or 4 lb. 15 oz. nearly. EXAM. 11.—At what elevation must a 13 inch mortar be pointed, to range 6745 feet, on a plane which descends 8° 15'; the charge being 4¦§ lb. of powder? Ans. 32° 8'. EXAM. 12. In what time will a 13 inch shell strike a plane which rises 8° 30′, when elevated 45o, and discharged with an impetus of 2304 feet? Ans. 14 seconds. 37. Suppose in ricochet firing AK=1600 feet, KI=12 feet, and KH=200 feet; required the elevation and velocity, that the projectile may just clear I and hit H. |