fhall therefore proceed in this place to fhew a practical method of determining the most useful of thofe particulars, when the angle of elevation, and the greatest range the cannon is capable of, are known. This greatest range, viz. the distance to which the cannon when charged with the ufual quantity of powder, and elevated to an angle of 45°. is capable of throwing the shot, must be afcertained by means of actual experiment and menfuration in every piece of artillery, especially with large cannons, and mortars; fince the balls from thofe pieces deflect much more from the ftraight direction. The angle of elevation is afcertained by means of a graduated circular inftrument, and a plummet, or a level. Let la is the fame as it would acquire by defcending perpendicularly through a space equal to the fourth part of the parameter belonging to that point as a vertex; fuppofing that the force of gravity is uniform, and that it acts in directions perpendicular to the horizontal plane; also that the air offers no refiftance to the motion of projectiles. Let a body be projected from A, fig. 5. Flate IX. in the direction AE, and let AE represent the space, through which the projecting force alone would carry it with an equable motion in the time T. Alfo let AB represent the fpace through which the force of gravity alone would cause it to defcend in the fame time T. Complete the parallelo gram ABEC, and it is evident that the body, being impelled both by the projecting force, and by the force of gravity, Let the greatest horizontal range of a cannon, or mortar, be 6750 yards, and let the actual angle of elevation be 25°; the other particulars may be found by delineating this cafe upon paper; for which purpose the inftruments that are generally put gravity, must, at the end of the time T, be found at C. Now AE is as the time T, because it represents the space defcribed uniformly; but AB is as the fquare of the time T; therefore AE or its equal BC, is as the fquare of AB. And the fame reafoning may be applied to any other contemporary distances, as AH, AF, or FG, AF. But AE is a tangent to the curve at the point A, AF is a diameter at the point A, and BC, FG, &c. being parallel to the tangent AE, are ordinates to the diameter AF; and fince the squares of those ordinates have been demonftrated to be as the respective abfciffas AB, AF, &c.; therefore the curve ACGD is the parabola. The velocity of the projectile at any point, as A, in the curve is fuch, that the space AE would be defcribed uniformly by it, in the fame time that the body would employ in descending perpendicularly by the force of gravity from A to B. Also (see p. 66.) the velocity acquired by the perpendicular defcent AB is fuch as would carry the body equably through twice AB in the fame time, (that is, in the fame time that AE is defcribed;) therefore the velocity which is acquired by the perpendicular descent AB, is to the velocity with which AE is described, as twice AB is to AE. But the velocity acquired by the perpendicular defcent through AB, is to the velocity acquired by the perpendicular descent through a quarter of the parameter be. longing to the vertex A of the parabola ACG, alfo as put in a common cafe of drawing inftruments, are quite fufficient, viz. a pair of compaffes, a ruler with a scale of equal parts, and a protractor. Draw an indefinite right line AK, fig. 4. Plate IX. to represent an horizontal plane, paffing through twice AB is to AE; for that parameter is (by conics) laws of gravity, the velocity acquired by the perpendicular descent AB, is to the velocity acquired by the perpendicular as AB is to as the square roots of those spaces; viz. AE ; or as twice AB is to AE. Therefore, fince the like reasoning may be applied to any other point of the parabolic path, we conclude that, univerfally, the velocity of the projectile in any point of its path is the fame as would be acquired by a perpendicular descent through a space equal to the fourth part of the parameter belonging to that point as a vertex. Corollary 1. It is evident that the projectile must move in the plane of the two forces, viz. in the plane which paffes through AE, AB, and is, of courfe, perpendicular to the horizon. Cor. 2. It follows from the laws of compound motion, that the projectile will defcribe the arch AC, in the fame. time in which it would descend by the force of gravity from A to B, or in which it would defcribe uniformly the space AE. Cor. 3. When a body is projected from A in the direction AE, if the parameter which belongs to the vertex A, through the point of projection A. Make AB perpendicular to it, and equal to twice the greatest horizontal range, viz. equal to 13500 yards; which is done by making it equal to 13500 divifions of the be equal to point C. AE the parabola must pass through the Cor. 4. Either in the fame, or in different parabolas, the parameters belonging to different points are to each other as the fquares of the velocities of the projectile at those points (fee p. 65); whence it follows, that at the vertex of the parabola the velocity or the momentum of the projectile is the leaft, and at equal diftances from that vertex the velocities or the momentums are equal. Propofition II. The initial velocity being given, to find the direction in which a body must be projected in order to bit a given point. Let A, fig. 6, Plate IX, be the projecting point, and C the object, or point which is required to be hit. The velocity of projection being given, the parameter of the parabola which must pass through the point C will eafily be found by means of the preceding propofition; viz. by finding the space, through which a body muft fall from reft, in order to acquire the given velocity; for that space is equal to the fourth part of the parameter belonging to the point A. Join AC, draw the horizontal line AL, and at A erect AP perpendicular to the horizontal line AL, and equal to the above-mentioned parameter. Divide AP into two equal parts at G, and through G draw an indefinite right line KGH parallel to the horizontal line AL. Through A draw the fcale of equal parts, for those parts must reprefent yards. Upon AB, as a diameter, defcribe the femicircle AFB. At A, by means of the protractor, draw the line of projection AF, making an angle of 25° with the horizontal line AK. Through draw AK perpendicular to the direction AC of the object, which AK will meet KH in a point K. With the centre K and radius KA draw the circular arch PHEA. Through the point or object C draw BCI perpendicular to the horizon, and if this perpendicular meets the circular arch, as at E and I, draw AE, AI; and either of thofe directions will answer the defired purpose. Join PL and PE; and the triangles PAE, EAC are fimilar; the angle PAE being equal to the angle AEC (Eucl. p. 29, B. I.) and the Angle APE equal to the angle EAC (Eucl. p. 32, B. III.) Hence PA: AE AE: EC; A E2 EC therefore PA = Farther, the triangles PAI, AIC are alfo fimilar; the angle PAI being equal to the angle AIC (Eucl. prop. 29, B. I.) and the angle API equal to IAC (Eucl. p. 32, B. III.) Hence PA: AI:: AI: IC; and PA = A I IC Therefore fince PA is the pa rameter belonging to the point A of the parabola, which is to be defcribed by the projectile, &c. the faid parabola (by cor. 3 of the preceding prop.) muft pafs through the point C. Corollary 1. The angular diftance CAP between the object and the zenith, is divided into two equal parts by the line AH; for KH being equal to KA, the angle AHK is equal to the angle HAK, and likewife equal (on account of |