PROJECTILE$.
Prob. I. The impetus of a ball, and the horizontal distance of an object aimed at, with its perpendicular height or depres sion, if thrown on ascents or descents, being given, to determine the direction of that ball.
From the point of projection A (Plate VI. Miscell. fig. 8, 9, 10, 11) draw A m_ re presenting the horizontal distance, and B m the perpendicular height of the ob ject aimed at: bisect A m in H, and A H on H and f erect H T, f F perpen dicular to the horizon, and bisecting A B the oblique distance or inclined plane in D, and A D in F. On A raise the impetus A M at right angles with the horizon, and bisect it perpendicularly in c, with the line G G. Let the line A C be normal to the plane of projection A B, and cutting G G in C; from C as centre, with the ra dius C A, describe the circle A GM, cut ting, if possible, the line F S in S, a, points equally distant from G; lines drawn from A through S, a, will be the tangents or di rections required.
Continue A S, A a to T, t; bisect D T, A t, in V, v; and draw lines from M to S, s; then the angle A SF = angle MA S = angle A M a= angle a A F; and for the same reason angle A a F = angle MAs angle A M S = angle A F; where fore the triangles M A S, S A F, a A F are similar, and A M: A a:: A 3:8 F= t v; consequently AT is a tangent of the curve passing through the points A, v, and B; because t v — v D, A D is an ordinate to the diameter T 11, and where produced must meet the curve to B.
In horizontal cases (fig. 10.) v is the highest point of the curve, because the diameter T v H is perpendicular to the horizon.
When the mark can be hit with two di rections (the triangles S AM, s A F being similar) the angle which the lowest direct. tion makes with the plane of projection is equal to that which the highest makes with the perpendicular A M, or angle s A F= angle S A M. And the angle S A a, comprehended between the lines of direction, is equal to the angle S C G, and is measured by the arch S G.
When the points S, a coincide with G, or when the directions A S, A a become A G (fig. 11.) A B will be the greatest distance that can be reached with the same impetus on that plane; because S F coin ciding with G g, the tangent of the circle at G, will cut off A g, a fourth part of the greatest amplitude on the plane A B. The
rectangular triangles m A B, c A C are si milar, because the angle of obliquity mAB=cAC; wherefore mA:mB:: one-half impetus : c C, and m A : A B Ac: AC.
Horizontal Projections (ibid. fig. 10, 11.) When the impetus is greater than half the amplitude, there are two directions, T A H and t A H, for that amplitude; when equal to it, only one; and when less, none at all; and conversely. For in the first case the line F S cuts the circle in two points S, a, its the second case it only touches it, and in the last it meets not with it at all; and conversely. When there is but one direction for the amplitude A rn, the angle of elevation is 450; and when the angle of elevation is of 45°, Am is the greatest amplitude for that impetus, and equal to twice the impetus. The impetus remaining the same, the amplitudes are in proportion to one another as the sines of double the angles of elevation, and con versely. For drawing a N (fig. 10.) pa rallel and equal to A F, a fourth part of the amplitude, and supposing lines drawn from a to the points C and M, the angle AC a. 2 A M a= 2 s A F; therefore N a, the sine of A C s, is the sine of twice the angle a A F; half the impetus being radius.
Whence, at the directions of 15' or 75°, the amplitude is equal to the impetus ; for, from what has been said, half the im petus being radius, a fourth part of the amplitude is the sine of twice the angle of elevation; but the sine of twice 15°, that is, the sine of 30°, is always equal to half the radius ; or in this case a fourth part of the impetus is equal to a fourth part of the amplitude. From this and the preceding proposition, there are two easy practical methods for finding the im petus of any piece of ordnance. The fourth part of the amplitude is a mean pro portional between the impetus at the curve's principal vertex and its altitude. ForMN:Na::NA=slis_-_-vD ' The altitudes are as the versed sines of double the angles of elevation, the impe tus remaining the same. For making half the impetus radius, A N the altitude is the versed sine of the angle A C s twice angle a A F. And also, radius : tangent angle elevation :: one-fourth amplitude : altitude: that is, It :tangent angle a A f D v.