theorems, may in many cases be taken = 16, without the small fraction ; which will be near enough for common use. PROP. XVIII. 30. To determine the range on an oblique plane, having given the impetus or velocity, and the angle of direction. Let AE be the oblique plane, at a given angle, either above or below the horizontal plane AH; AG the direction of the piece, and AP the altitude due to the projectile velocity at A. By the last proposition, find the horizontal range AH to the given velocity and direction; draw HE perpendicular to AH, meeting the oblique plane in E; draw EF parallel to AG, and FI parallel to HE; so P G F k C H E A k K F E shall the projectile pass through I, and the range on the oblique plane will be AI. For if AH, AI, be any two lines terminated at the curve, and IF, HE, parallel to the axis; then is EF parallel to the tangent AG. 31. Otherwise, without the horizontal range. Draw PQ perpendicular to AG, and QD perpendicular to the horizontal plane AF, meeting the inclined plane in K; take AE= 4AK, draw EF parallel to AG, and FI parallel to AP or DQ; so shall AI be the range on the oblique plane. For AH = 4AD, therefore EH is parallel to FI, and so on, as above. OTHERWISE.. 32. Draw Pq making the angle APq= the angle GAI; then take AG= 4Aq, and draw GI perpendicular to AH. Or, draw qk perpendicular to AH, and take AI = 4Ak. Also, kq will be equal to cv, the greatest height above the plane. therefore AG = 4Aq; and, by similar triangles, AI = 4Ak. Corol. 1. If AO be drawn perpendicular to the plane AI, and AP be bisected by the perpendicular STO; then with the centre O describing a circle through A and P, the same will also pass through 9, because the angle GAI, 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 kq 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 A Pq and Akq we have AP: Aq :: sin and Aq : Ak :: sin AqP : sin APq :: sin Akq : sin qAI Akd ... = range on oblique plane. 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 kg of the projectile above the plane is COS HAI V g For the time of describing the curve is equal to the time of falling freely through GI or sin2 qAI COS2 IAH' 2 sin qAI ; hence by the laws of gravity, the time of flight COS HAI AP V g 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 equations. 2v2, cos e sin (e — i) R = g cos2 i. cos2 i cos e cos e sin (e — i) cos2 i v2 sin (2e-i)—sin i = 4a. sin e -i) = 4R. sin (e — i) 2g cos2 i 1): = cos2 i + sin i; whence e may be found. 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. =gt. sin(e-i) = sgt2 = =2 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. space described in t seconds by velocity of projection. =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) (6) (7) Cor. 2. If AK = x and KI = y; then we have 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 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 Cm B and c = mb; 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. Let cb and v1600; then we have 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 e,. : 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. 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 y = 2a sin2 e a sin2 e a sin2 e (7) v2 = sin2 e= sin2 625 e 62.5 feet. ADDITIONAL EXAMPLES FOR EXERCISE. 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 |