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Ex. 2. It is required to lay off 2 roods towards the fouth side of the fame field, and to know how far up the lines AC, BD, the march-line must be struck.

8

Ans. 92 links.

GUNNERY.

GUNNERY.

GUNNERY UNNERY is the art of charging, directing, and exploding fire-arms, fuch as cannons, mortars, &c. to the best advantage. To this art belongs the knowledge of the force of gunpowder, the dimenfions 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 principles of geometry and mechanics we are enabled to carry on the analysis from the phenomena to the powers or causes that produce them.

The fame power which renders bodies heavy when at rest, accelerates their motion when they defcend in the direction of their gravity; and, if rojected 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 diftinct 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 inceffantly 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 interpofition of other bodies;) but, as both continue to act, the course of a projectile must be determined by a power compounded of these

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 Thus GK

the tangent, where the diameter cuts the curve. 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 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 fometimes 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 deprefied.

10. The directrix is a line parallel to the horizon, and whofe 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 ufed.

PROBLEM.

1

PROBLEM. Fig. 1.

To defcribe the path of a projectile.

Draw AL the horizontal plane, and, from a scale of equal parts, lay off the amplitude AK, and through the point A erect a perpendicular AB equal to the impetus taken from the same scale; 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, defcribe the semicircle BFfR, and the point F is the focus. Or,

If the direction AD is given, upon AB, as diameter, describe a femicircle BDA; and through the point of intersection D draw BD, and produce it to F; so shall BD and DF be equal, and F will be the focus. Or,

Through the point D draw PD parallel to the horizon; then shall PD=DH, and NH=HF, and H will be the vertex. Cor. 4. times PD is equal to the amplitude.

Then proceed as shewn in Prob. 9. conic fections.

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 LAM equal to the angle of elevation. From the centre A, with the radius AB, describe the semicircle BFOR; its circumference shall be the locus of the foci of all the parabolas that can be described by a projectile thrown from A, with the velocity it could acquire in falling from B to A; for, by a known property of the parabola, the distance of the focus from A is always equal to one-fourth of the parameter of the diameter that paffes through A, that is, to AB; all the foci must, therefore, be found in the femicircle BFOFR. It will therefore be casy to determine the parabolas, when the direction of the projectile is given; for if, upon the impetus AB, you describe a femicircle BDdA, you need only join BD, and lay off BD equal to DF, and F will be the focus; and if through F you draw the line QF perpendicular to the horizontal line AL, it shall be the axis; and H, the middle point between F and N, shall be the vertex of the parabola. 4×FH is the length of the parameter of the axis.

If a line HP be drawn through the point H perpendicular to AB, the straight line BF and PH will bisect each other; also AM, the line of direction, will pass through the point of intersection in D, and bifect the line BF at right angles; and therefore the femicircle BDdA will pass through the fame point

D.

The amplitude of any parabola is equal to four times the fine of twice the complement of the angle of elevation: PD is the fine of the angle PCD, and the angle PCD is twice the angle PAD, because the one is at the centre and the other at the circumference, but the angle PAD is the complement of the angle of elevation DAK; therefore PD is the fine of twice the complement of the angle of elevation; and 2PD is equal to PH; but 2PH is equal to AK; therefore AK is equal to 4PD. Hence it will follow, that when the angle of elevation becomes 45°, the points F and Q shall fall in the point O, and AK becomes twice the impetus. The fine PD is the co-fine of double