PROBLEM III. To lay out a given area from an irregular field. EXAMPLE I. Let it be required to cut off i acres towards the north fide of the irregular field ABCD. Ftg. 4. 82 80 2)3200 50 UIO 2)65602 1440|2)42002)3000 2)88202)5500 98 yo Ex. 2. It is required to lay off 2 roods towards the south side of the same field, and to know how far up the lines AC, BD, the march-line must be struck, Ans: 921, links. GUNNERI. GUNNERY. t ples of GUNNERY UNNERY is the art of charging, dirècting, and exploding fire-arms, such as cannons, mortars, &c. to the best advantage. To this art belongs the knowledge of the force of gunpowder, the dimensions of cannon, the proportion of powder and ball they carry. From experiment and observation alone the history of nature can be collected, or her phenomena described. By the princi geometry and mechanics we are enabled to carry on the analysis from the phenomena to the powers or causes that produce them. The same power which renders bodies heavy when at rest, accelerates their motion when they descend in the direction of their gravity; and, if poojected in any other direction, bends their motion into a curve line, which, from its properties and flexure, is known to be a parabola. For every body, projected into the air, moves under the influence of two distinct forces, viz. its projectile force, and that of gravity. By the first, it is carried forward with an equal motion, and describes equal Spaces in equal times. By the latter, it is drawn downwards in lines perpendicular to the horizon, with a motion incessantly accelerated. If either of these forces were destroyed, the body would move for ever in the direction of the remaining force alone, (if its motion was not hindered by the interposition of other bodies ;) but, as both continue to act, the course of a projectile must be determined by a power compounded of these two forces. DEFINITIONS. DEFINITIONS. 1. The impetus of a piece is the perpendicular height to which it would shoot a ball with its ordinary charge of powder; or the height from which it must fall perpendicularly to acquire the velocity with which it was projected. Thus, BA is the impetus. Fig. 1. 2. The diameter, or axis to any point of the curve, is a line drawn from that point perpendicular to the horizon. Thus, HQ is the diameter to the point H. 3. The point H is called the vertex. 4. The ordinates to any diameter are lines drawn parallel to the tangent, where the diameter cuts the curve. Thus GK is an ordinate to the axis HQ. 5. The absciss is that part of the diameter intercepted between the ordinate and the curve. Thus, HQ is Thus, HQ is an absciss of the diameter HF. 6. The altitude of the curve is the perpendicular height of the vertex above the horizontal plane. Thus, HQ is the altitude of the curve AHK. 7. The amplitude is the distance between the object aimed at and the piece, and is sometimes called the random, or range. Thus, AK is the amplitude of the curve ABK. - 8. The elevation of the piece is the angle its direction makes with the horizontal plane. 9. The inclination of a plane is the angle it makes with the horizon, and is either elevated or depressed. 10. The directrix is a line parallel to the horizon, and whose distance from the horizon is the impetus. N. B. The vertex is equidistant from the directrix and focus. The focus may be found by various methods. These following are most commonly used. PROBLEM. PROBLEM. Fig. 1. To describe the path of a projectile. Draw AL the horizontal plane, and, from a scale of equal párts, lay off the amplitude AX, and through the point A erect a perpendicular AB cqual to the impetus taken from the same {cale ; through B draw the directrix parallel to AK; then bisect AK in Q, and draw QN at right angles to AK; upon A, as centre with the distance AB, describe the semicircle BFFR, and the point F is the focus. Or, If the direction AD is given, upon AB, as diameter, describe a semicircle BDA; and through the point of intersection D draw BD, and produce it to F; fo fhall BD and DF be equal, and F will be the focus. Or, Through the point D draw PD parallel to the horizon ; then Thall PD=DH, and NH=HF, and H will be the vertex. Cor. 4. times PD is equal to the amplitude. PROBLEM I. The impetus of a piece and the angle of elevation being given, to find the amplitude. EXAMPLE I. How far will a cannon, whose impetus is 1200 feet, carry, at an elevation of 30°? Geometrically. Let AB represent the impetus of the piece, or the velocity a heavy body would acquire in falling from B to A. Through the point A draw the horizontal line AL, and make the angle O. LAM |