found in the first column of the table of ranges, the opposite number in the 2d col is 37° 15' for the elevation of the piece, and in the 3d column 2.14, multiplied by 7014, gives 15010 feet, or nearely 3 miles. So that our 13 inch shells, discharged at an elevation of about 374 degrees, would range nearly the distance mentioned by the French, when filled with lead, if they can be projected with so much as 2000 feet velocity, or upwards. This however it is thought cannot possibly be effected by our mortars; and that it is therefore probable the French, to give such a velocity to those shells, must have contrived some new kind of large cannon on the occasion.

4. Having shown in the preceding articles and problems, how, from our theory of the air's resistance, can be found, first the initial or projectile velocity of shot and shells; 2dly, the terminal velocity, or the greatest velocity a ball can acquire by descending by its own weight in the air; 3dly, the height a ball will ascend to in the air, being projected verti cally with a given velocity, also the time of that ascent; 4thly, the greatest horizontal ranges of given shot, projected with a given velocity; as also the particular angle of elevation of the piece, to produce that greatest range. It remains then now to inquire, what laws and regulations can be given respecting the ranges, and times of flight, of projects made at other angles of elevation.

Relating to this inquiry, the Encyclopædia Britannica mentions the two following rules: 1st. "Balls of equal density projected with the same elevation, and with velocities which are as the square roots of their diameters, will describe similar curves. This is evident, because, in this case, the resistance will be in the ratio of their quantities of motion; therefore all the homologous lines of the motion will be in the proportion of the diameters." But though this may be nearly correct, yet it can hardly ever be of any use in practice, since it is usual and proper to project small balls, not with a less, but with a greater velocity, than the larger ones. 2dly, the other rule is, "If the initial velocities of balls, projected with the same elevation, be in the inverse subduplicate ratio of the whole resistance, the ranges, and all the home. logous lines in their track, will be inversely as those resistances." This rule will come to the same thing, as having the initial velocities in the inverse ratio of the diameters, as distant perhaps from fitness as the former. Two tables are next given in the same place, for the comparison of ranges and projectile velocities, the numbers in which appear to be much wide of the truth, as depending on very erroneous effects of the resistance. Most of the accompanying remarks, however,

however, are very ingenious, judicious, and philosophical, and very justly recommending the making and recording of good experiments on the ranges and times of flight of projects, of various, sizes, made with different velocities, and at various angles of elevation.

Besides the above, we find rules laid down by Mr. Robins and Mr. Simpson, for computing the circumstances relating to projectiles as affected by the resistance of the air Those of the former respectable author, in his ingenious Tracts on Gunnery, being founded on a quantity which he calls F, (answering to our letter a in the foregoing pages), I find to be almost uniformly double of what it ought to be, owing to his improper measures of the air's resistance; and therefore the conclusions derived by means of those rules must needs be very erroneous. Those of the very ingenious Mr. Simpson, contained in his Select Exercises, being partly founded on experiment, may bring out conclusions in some of the cases not very incorrect; while some of them particularly those relating to the impetus and the time of flight, must be very wide of the truth. We must therefore refer the student, for more satisfaction, to our rules and examples before given in pa. 142 this vol &c. especially for the circumstances of different ranges and elevations, &c. after having determined, as. above, those for the greatest ranges, founded on the real measure of the resistances.

PROMISCUOUS

PROMISCUOUS PROBLEMS, AS EXERCISES IN MECHANICS, STATICS, DYNAMICS, HYDROSTATICS, HYDRAULICS, PROJECTILES, &c. &c.

PROBLEM I

Let AB and AC be two inclined planes, whose common altitude AD is given =64 feet; and their lengths such, that a heavy body is 2 seconds of time longer in descending through AB than through AC, by the force of gravity; and if two balls, the one weighing 3 and the other 2lb, be connected by a thread and laid on the planes, the thread sliding freely over the vertex A, they will mutually sustain each other. Quere the lengths of the two planes.

THE lengths of the planes of the same height being as the times of descent down them (art 133 this vol.), and also as the weights of bodies mutually sustaining each other on them (art. 122), therefore the times must be as the weights; hence as 1, the difference of the weights, is to 2 sec. the diff. of S3: 6 sec. times, :: the times of descending down the two 22 : 4 sec. planes. And as √ 16: √64: 1 sec. : 2 sec. the time of descent down the perpendicular height (art 70,).

laws of descents (art. 132), as 2 sec.: 64 feet feet, the lengths of the planes.

Then, by the 6 sec. : 1922 24 sec. : 128

Note. In this solution we have considered 16 feet as the space freely descended by bodies in the 1st second of time, and 32 feet as the velocity acquired in that time, omitting the fractions and, to render the numeral calculations simpler, as was done in the preceding chapter on projectiles, and as we shall do also in solving the following questions, wherever such numbers occur.

Another Solution by means of Algebra.

Put the time of decent down the less plane; then will x + 2 be that of the greater, by the question. Now the weights being as the lengths of the planes, and these again as the times, therefore as 2 : 3 :: * : x + 2; hence

2x + 4

x

3x, and = 4 sec. Then the lengths of the planes are found as in the last proportion of the former solution.

PROBLEM 2.

If an elastic ball fall from the height of 50 feet above the plane of the horizon, and impinge on the hard surface of a: filane inclined to it in an angle of 15 degrees; it is required tofind what part of the plane it must strike, so that after reflection, it may fall on the horizontal plane, at the greatest distance possible beyond the bottom of the inclined plane?

Here it is manifest that the ball must strike the oblique plane continued on a point somewhere below the horizontal plane; for otherwise there could be no maximum. Therefore let Bc be

the inclined plane, CDG the horizontal one, в the point on which the ball impinges after falling from the point A, BEGI the parabolic path, E its vertex, BH a tangent at B, being the direction in which the ball is reflected; and the other lines as are evident in the figure. Now, by the laws of reflection, the angle of incidence ABC, is equal to the angle of reflection HBм, and therefore this latter, as well as the former, is equal to the complement of the Ze the inclination of the two planes; but the part 1BM is = c, therefore the angle of projection HBI is the comp, of double the Ze, and being the comp. of HBK, theref. HBK 2 c. Now, put a = 5) = AD the height above the horizontal line, tang, LDBC or 75° the complement of the plane's inclination, tang HBI or H=60° the comp. of 2 Lc, s = sine of 24HBX = 120° the double elevation, or sine of 42c; also x=AB the impetus or height failen through. Then,

and

==

B14KH 2.x, by the projectiles prop. 21,
SBK = 7X KH = F*X

CD = t X BD = t (x − a). Š

also, KD BK — BD =

then, by the parabola,

N

by trigonometry;

: √ DK ::

STX -x+a, and RE =
K

B = 8X; KE: FG = EE X 28

=

2

- XD

T

2b √ (ax-b2x2), putting b = sine of 2 c sine of 30°. Hence CGCD + DF FG= 1x-ta + sx25、/(ux − b2x2) a maximum, the fluxion of which made == 0, and the equa

+t, and the double sign answers to the two roots or values of x, or to the two points G, G, where the parabolic path cuts the horizontal line CG, the one in ascending and the other in descending.

Now, in the present case, when the c 75° = 2 + √ 3, tan. 60°=√√3, 8 = sin

sin. 30°ns + 1 = 2 + 2√3; then,

n2

52

262

150,= tang. 60°=} √ 3, b=

2a=100, and

= 100 x (1 ± √ 41+63)=100×(1±99414)-199-414

13

or 586; but the former must be taken. Hence the body must strike the inclined plane at 149:414 feet below the horizontal line; and its path after reflection will cut the said line in two points; or it will touch it when =

Hence

also the greatest distance ce required is 826-9915 feet. Corol If it were required to find CG or tx 2b√(ax-b2x2)g a given quantity, this equation would give the value of x by solving a quadratic.

PROBLEM 3.

Suppose a ship to sail from the Orkney Islands, in latitude 59° 3' north, on a N. N. E. course, at the rate of 10 miles an hour; it is required to determine how long it will be before she arrives at the pole, the distance she will have sailed, and the difference of longitude she will have made when she arrives there?

Let ABC represent part of the equator; P the pole; AmrP a loxodromic or rhumb line, or the path of the ship continued to the equator; PB, PC, any two meridians indefinitely near each other; nr, or mt, the part of a parallel of latitude intercepted between them.

Put c for the cosine, and for the tangent of the course, or angle nmr to the radius r;

Am, any variable part of the rhumb from the equator,=v; the latitude Bmw; its sine x, and cosine y; and AB, the dif. of longitude from A, = z. Then, since the elementary triangle mnr may be considered as a right-angled plane triangle, it is, as rad. r: c = sin. mrn: mri w = mn :::; theref. cv rw, or v= by putting & for the secant of the

טדיו

1

nmr the ship's course. In like man

ner