the King's gunners. To every scholar he administers an oath, not to serve, without leave, any other prince or state; or teach any one the art of gunnery, but such as have taken the said oath.

GUNNERA, in botany, so named in honour of J. E. Gunnerus, Bishop of Drontheim, in Norway, a genus of the Gynandria Diandria class and order. Natural order of Urtica, Jussieu. Essential character: ament with one-flowered scales; calyx and corolla none; germ two toothed; styles two; seed one. There is but one species, viz. G. Perpensa, marsh ma rygold-leaved gunnera. Native of the Cape.

GUNNERY, is the art of determining the course and directing the motion of bodies shot from artillery, or other warlike engines.

The great importance of this art is the reason it is distinguished from the doctrine of projectiles in general; for it is no more than an application of those laws which all bodies observe, when cast into the air, to such as are put in motion by the explosion of guns, or other engines of that sort. And it is the same thing, whether it is treated in the manner of projectiles in general, or of such only as be long to gunnery; for from the moment the force is impressed, all distinction with regard to the power which put the body first in motion is lost, and it can only be considered as a simple projectile. See PROJECTILES.

Prob. I. The impetus of a ball, and the horizontal distance of an object aimed at, with its perpendicular height or depression, 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 representing the horizontal distance, and Bm the perpendicular height of the object aimed at bisect A m in H, and A H inf; on H and ƒ erect H T, ƒ F perpendicular 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 AM 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 radius C A, describe the circle A G M, cutting, if possible, the line F S in S, s, points equally distant from G ; lines drawn from A through S, 8, will be the tangents or directions required.

Continue A S, A 8 to T, t; bisect D T, Dt, in V, v; and draw lines from M to

=

S, s; then the angle A SF: = = angle MAS angle A M s = angle s A F; and for the same reason angle A & F angle MAS angle A MS= angle SAF; wherefore the triangles M AS, SA F, s A Fare similar, and AM: A s :: As: s F = tv; consequently A T 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 H, 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 Tv H is perpendicular to the horizon.

When the mark can be hit with two directions (the triangles S A M, 8 A F being similar) the angle which the lowest directtion makes with the plane of projection is equal to that which the highest makes with the perpendicular A M, or angle 8 AF angle S A M. And the angle SA s, 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, & coincide with G, or when the directions A S, A s 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 coinciding 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 AB, c A C are similar, because the angle of obliquity mAB=cA C; wherefore m Am B:: one-half impetus : c C, and m A: A B :: A c: A C.

Horizontal Projections (ibid. fig. 10, 11.)

When the impetus is greater than half the amplitude, there are two directions, TAH and AH, for that amplitude; when equal to it, only one; and when less, none at all; and conversely. For in the first case the line FS cuts the circle in two points S, s, in 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 m, the angle of elevation is 45°; and when the angle of elevation is of 45°, A m 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 conversely. For drawing s N (fig. 10.) parallel and equal to A F, a fourth part of the amplitude, and supposing lines drawn from s to the points C and M, the angle AC & = 2 A M 8 = 2 s A F; therefore

Ns, the sine of A Cs, is the sine of twice the angle & 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 impetus 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 easy practical methods for finding the impetus of any piece of ordnance. The fourth part of the amplitude is a mean proportional between the impetus at the curve's principal vertex and its altitude. For MN: Ns:: N As F = v D

The altitudes are as the versed sines of

=

double the angles of elevation, the impetus remaining the same. For making half the impetus radius, A N the altitude is the versed sine of the angle ACs twice angle & A F. And also, radius: tangent angle elevation: one-fourth amplitude: altitude: that is, R: tangent angle & Af :: Af:f8D v.

Projections on Ascents and Descents, fig. 8, 9.

If the mark can be hit only with one direction A G, the impetus in ascents will be equal to the sum of half the inclined plane and half the perpendicular height, and in descents it will be equal to their difference; but if the mark can be reach. ed with two directions, the impetus will be greater than that sum or difference. For when A G is the line of direction, the angle g G A being: MAGGA g;

-

Gg ed from both, makes G z half the impetus equal to the sum or difference of A g, a fourth part of the inclined plane, and g z a fourth part of the perpendicular height. In any other direction F P is greater than FoAF; and Ff, added to or subtracted from both, makes fP half the impetus greater than the sum or difference of A F, & fourth part of the inclined plane, and Ff a fourth part of the perpendicular height. Whence, if in ascents the impetus be equal to the sum of half the inclined plane and half the perpendicular height, or if in descents it be equal to their difference, the mark can be reached only with one direction; if the impetus is greater than that sum or difference, it may be hit with two directions; and if the impetus is less, the mark can be hit with

A g and g z added to or substract

none at all.

Prob. II. The angles of elevation, the horizontal distance, and perpendicular height, being given, to find the impetus. Fig. 8, 9.

From these data you have the angle of obliquity, and length of the inclined plane; then as

A&: AM:: S. angle A Ma: S. angle A's M:: S. angle & AF: S. angle MA F, and AF: As: S. angle M AS: S. angle MAF; whence, by the ratio of equality, A FAM: S. angle s A F× S. angle MA & S. angle M A F ̈× S. angle M A F, which gives this rule.

:

logarithmic sine of the angle M A F; from Add the logarithm of A F to twice the their sum subtract the logarithmic sines of the angles s A F and MA s, and the remainder will give the logarithm of A M the impetus.

tion are given, and the length of the inWhen the impetus and angles of eleva. clined plane is required, this is the rule. Add the logarithm of A M to the logarith from their sum subtract twice the logamic sines of the angles s A Fand MA *; rithmic sine of angle M AF, and the remainder will give the logarithm of A F, the fourth part of the length of the inclined plane.

If the angle of elevation t A H, and its amplitude A B (fig. 11,) and any other angle of elevation A H is given; to find the amplitude Ab for that other angle, DA H remaining the same. the impetus AM and angle of obliquity

Describe the circle A G M, take A Fa fourth part of A B, and A fa fourth part of A b; from the points F, f, draw the lines F 8, and fp parallel to A M, and cutting the circle in the points s,p; then AF AMS. angle s AFX S. angle M As: S. angle M AFX S. angle MAF ; and AM: Af: S. angle MAF X S. angle MAF: S. angle p Af× S. angle p A M ; whence by the ratio of equality,

:

AF Af S. angle s AFX S angle M As: S. angle p AƒX S. angle p A M, which gives this rule.

Add the logarithm of A F to the logarithmic sines of the angles pAf. pAM; from their sum subtract the logarithmic sines of the angles s A F, s A M, and the remainder will give the logarithm of Aƒ, a fourth part of the amplitude required.

Prob. Ilk To find the force or velocity of a ball or projectile at any point of the curve, having the perpendicular height of that point, and the impetus at the point of projection given. From these two data find out the impetus at that point; then

2 x 16 feet 1 inch is the velocity acquir ed by the descent of a body in a second of time; the square of which (4 × the square of 16 feet 1 inch) is to the square of the velocity required, as 16 feet 1 inch is to the impetus at the point given; wherefore multiplying that impetus by four times the square of 16 feet 1 inch, and dividing the product by 16 feet 1 inch, the quotient will be the square of the required velocity: whence this rule. Multiply the impetus by four times 16 feet 1 inch, or 64 feet and the square root of the product is the velocity.

Thus suppose the impetus at the point of projection to be 3,000, and the perpendicular height of the other point 100; the impetus at that point will be 2,900. Then 2,900 feet multiplied by 644 feet gives 186,566 feet, the square of 432 nearly, the space which a body would run through in one second, if it moved uniformly.

And to determine the impetus or height, from which a body must descend, so as at the end of the descent it may acquire a given velocity, this is the rule:

Divide the square of the given velocity (expressed in feet run through in a second) by 644 feet, and the quotient will be the impetus.

The duration of a projection made perpendicularly upwards is to that of a projection in any other direction whose impetus is the same, as the sine complement of the inclination of the plane of projection (which in horizontal projections is radius) is to the sine of the angle contained between the line of direction and that plane.

Draw out A t (fig. 8,) till it meets m B continued in E, the body will reach the mark B in the same time it would have moved uniformly through the line A E; but the time of its fall through MA the impetus, is to the time of its uniform motion through A E, as twice the impetus is to A E. And therefore the duration of the perpendicular projection being double the time of its fall, will be to the time of its uniform motion through A E, as four times the impetus is to A E; or as A E is to EB; that is, as A t is to tD; which is as the sine of the angle t D A (or MA B its complement to a semicircle) is the sine of the angle t A D.

Hence the time a projection will take to arrive at any point in the curve may be found from the following data, viz. the impetus, the angle of direction, and the inclination of the plane of projection, which in this case is the angle the horizon make

VOL. VI.

with a line drawn from the point of projection to that point.

Hence also, in horizontal cases, the durations of projections in different directions with the same impetus are as the sines of the angles of elevation. But in ascents or descents, their durations are as the sines of the angles which the lines of direction make with the inclined plane. Thus, suppose the impetus of any projection were 4,500 feet; then 16 feet 1 inch: 1": 4,500 feet: 275", the square of the time a body will take to fall perpendicularly through 4,500 feet, the square root of which is 16" nearly, and that doubled gives 32", the duration of the projection made perpendicularly upwards Then, to find the duration of a horizontal projection at any elevation, as 20°; say R: S. angle 200 :: 32": duration of a projection at that elevation with the impetus 4,500. the direction of 35° was projected on a Or if with the same impetus a body at plane inclined to the horizon 17°, say as sine 73° sine 18° :: 32′′: duration required.

:

The tables in the next leaf, at one view, give all the necessary cases, as well for shooting at objects on the plane of the horizon, with proportions for their solutions, as for shooting on ascents and descents. We shall in this place mention some of the more important maxims laid down by Mr. Robins, as of use in prac tice. 1 In any piece of artillery, the greater quantity of powder with which it is charged, the greater will be the velocity of the bullet. 2. If two pieces of the same bore, but of different lengths, are fired with the same charge of powder, the longer will impel the bullet with a greater celerity than the shorter. 3. The ranges of pieces at a given elevation are no just measures of the velocity of the shot: for the same piece fired successively at an invariable elevation, with the powder, bullet, and every other circumstance, as nearly the same as possible, will yet range to very different distances. 5. The greatest part of the uncertainty in the ranges of pieces arises from the resistance of the air. 6. The resistance of the air acts upon projectiles by opposing their motion, and diminishing celerity; and it also diverts them from the regular track which they would otherwise follow. 7. If the same piece of cannon be successively fired at an invariable elevation, but with va rious charges of powder, the greatest charge being the whole weight of the ball in powder, and the least not less than the fifth part of that weight; then, if the elevation be not less than eight or ten de

M

GUNPOWDER, a composition of nitre, sulphur, and charcoal, mixed together, and usually granulated. This easily takes fire, and when fired it rarifies or expands with great vehemence, by means of its elastic force. It is to this powder that we owe all the effect and action of guns, and ordnance of all sorts, so that fortification, with the modern military art, &c. in a great measure depends upon it.

The invention of gunpowder is ascribed by Polydore Virgil to a chemist, who having accidentally put some of his composition in a mortar, and covered it with a stone, it happened to take fire, and blew up the stone. Thevet says, that the person here spoken of was a monk of Fribourg, named Constantine Anelzen; but Belleforet, and other authors, with more probability, hold it to be Bartholdus Schwartz, or the black, who discovered it, as some say, about the year 1320; and the first use of it is ascribed to the Vene

S. angle p A f+Log. S. angle PAM- Log. S angle s A F Log. S. angle MA 8.

Fig. 5, 6.

T. angle GA z: Sec. angle g Az: Gz: A g.

tians in the year 1380, during the war with the Genoese. But there are earlier accounts of its use, after the accident of Schwartz, as well as before it for Peter Mexia, in his "Various Readings," mentions, that the Moors being besieged, in 1343, by Alphonsus the Eleventh, King of Castile, discharged a kind of iron mortars upon them, which made a noise like thunder: and this is seconded by what is related by Don Pedro, Bishop of Leon, in his Chronicle of King Alphonsus, who reduced Toledo, viz. that in a sea combat, between the King of Tunis and the Moorish King of Seville, about that time, those of Tunis had certain iron tubs or barrels, with which they threw thunderbolts of fire.

Du Cange adds, that there is mention made of gunpowder in the registers of the chambers of accounts in France, as early as the year 1338. But it appears that Roger Bacon knew of gunpowder near