Q the direction of the piece so as to hit the object н, will be thus easily found: Take AD = AH, draw DQ perpendicular to An, meeting the semicircle, described on the diameter AP, in q and q: then aq or aq will be the direction of the piece. And hence it appears, that there are two directions AB, AB, which, with the same projectile velocity, give the very same horizontal range AH. And these two directions make equal angles qad, qap with ah and AP, because the arc PQ - the arc aq. 86. Corol. 3. Or, if the range AH, and direction AB, be given; to find the altitude and velocity or impetus. Take AD= AH, and erect the perpendicular DQ, meeting Ab in q; so shall Do be equal to the greatest altitude cv. Also, erect AP perpendicular to a¤, and qp to aq; so shall ar be the height due to the velocity. 87. Corol. 4. When the body is projected with the same velocity, but in different directions: the horizontal ranges AH will be as the sines of double the angles of elevation. Ør, which is the same, as the rectangle of the sine and cosine of elevation. For AD or RQ, which is 1AH, is the sine of the arc ae, which measures double the angle QAD of elevation. And when the direction is the same, but the velocities different; the horizontal ranges are as the square of the velocities, or as the height AP, which is as the square of the velocity; for the sine AD or RQ or AH is as the radius or as the diameter AP. Therefore, when both are different, the ranges are in the compound ratio of the squares of the velocities, and the sines of double the angles of elevation. 88. Corol. 5. The greatest range is when the angle of elevation is 45o, or half a right angle; for the double of 45 is 90, which has the greatest sine. Or the radius os, which is of the range, is the greatest sine. And hence the greatest range, or that at an elevation of 45° is just double the altitude Ar which is due to the velocity, or equal to 4vc. Consequently, in that case, c is the focus of the parabola, and AH its parameter. Also, the ranges are equal, at angles equally above and below 45o. 89. Corol. 6. When the elevation is 15°, the double of which, or 30o, has its sine equal to half the radius ; consequently then its range will be equal to ap, or half the greatest range at the elevation of 45°; that is, the range at 15°, is equal to the impetus or height due to the projectile velocity. 90. Corol. 7. 90. Corol. 7. The greatest altitude cv, being equal to AR, is as the versed sine of double the angle of elevation, and also as AP or the square of the velocity. Or as the square of the sine of elevation, and the square of the velocity; for the square of the sine is as the versed sine of the double angle... 91. Corol. 8. The time of flight of the projectile, which is equal to the time of a body falling freely through GH or 4ev, four times the altitude, is therefore as the square root of the altitude, or as the projectile velocity and sine of the elevation. SCHOLIUM. 92. From the last proposition, and its corollaries, may be deduced the following set of theorems, for finding all the circumstances of projectiles on horizontal planes, having any two of them given. Thus, let s, c, t denote the sine, cosine, and tangent of elevation; s, the sine and versed sine of the double elevation; R the horizontal range; T the time of flight; v the projectile velocity; H the greatest height of the projectile g 16, feet, and a the impetus, or the altitude due to the velocity v. Then, And from any of these, the angle of direction may be found. Also, in these theorems, g may, in many cases, be taken = 16, without the small fraction, which will be near enough for common use. PROPOSITION XXII. 93. 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 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 в; draw EF parallel to AG, and FI parallel to HE; so shall the projectile pass through 1, and the range on the oblique plane will be AI. As is evident by theor. 15 of the Parabola, where it is proved, that 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. 94. Otherwise, without the Horizontal Range. Draw po perp. to AG, and QD perp, to the horizontal plane AF, meeting the inclined plane in K; take AE4AK, draw EF parallel to AG, and Fi parallel to Ap or dq; so shall ar be the range on the oblique plane. For AH 4 AD, therefore EH parallel to Fr, and so on, as above. Otherwise. 95. Draw pq making the angle APq take AG = 4Aq, and draw GI perp. to Aн. to AH, and take AI4Ak. Also кq will greatest height above the plane. the angle GAI; then Or, draw qk perp. be equal to co the For, by cor. 2, prop. 20, AP: ÄG :: AG: 4GI and by sim. triangles, or AP AG :: Aq : GI, AP: AG: 4Aq: 4GI; therefore AG 4Aq; and by sim. triangles, AI-4Ak. cv Also qk, or 4 GI, is to cu by theor. 13 of the Parabola. 96. Corol. 1. If Ao be drawn perp. to the plane ai, and VOL. II. 1.9 Ar be 138 OF MOTION, FORCES, &c. AP be bisected by the perpendicular sro; then with the cen- Ag OF 97. Corol. 2. If there be given the range AI and the velo- 98. Corol. 3. Or, if there be given the range AI, and the 99. Corol. 4. The range on an oblique plane, with a given elevation, is directly proportional to 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 to the square of the cosine of the angle of the plane above or below the horizon. For, put s sin. ▲qar or arq, C c=cos. ▲qAH or sin. PAG, cos. ZIAH or sin. akd or akq or aqp. 100. The range is the greatest when ak is the greatest; that is, when kq touches the circle in the middle points; 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. 101. Corol. 5. The greatest height ev or kq of the projec is as the impetus and square of the sine of direction above the plane directly, and square of the cosine of the plane's inclination reciprocally. For c (sin. Aqr): s (sin. APq) :: AP: Aq, and c (sin. Akq): s (sin. kaq) :: Aq: kq, theref. by comp. c2: s2:: AP: kq. 102. Corol. 6. The time of flight in the curve avi is 28 AP C ✓, where g = 16 feet. And therefore it is as the velocity and sine of direction above the plane directly, and cosine of the plane's inclination reciprocally. For the time of describing the curve, is equal to the time of falling freely through G1 or 4kq or X AP. Therefore, the time being 482 C2 103. From the foregoing corollaries may be collected the following set of theorems relating to projects made on any given inclined planes, either above or below the horizontal plane. In which the letters denote as before, namely, c=cos. of direction above the horizon, cos. of inclination of the plane, sin. of direction above the plane, α the impetus, or alt. due to the velocity v, g= 16 feet. Then, C g gc gc g And from any of these, the angle of direction may be found. PRAC |