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as depending on very erroneous effects of the resistance. Most of the accompanying remarks, 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 page 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.





Let AB and AC be two inclined planes, whose common altitude ap 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
§ 3: 6 sec.
times, ::
the times of descending down the two
2: 4 sec.
planes. And as ✅✔✅ 16: ✓✓ 64 : : 1 sec. : 2 sec. the time of
descent down the perpendicular height (art 70.) Then by the
laws of descents (art. 132), as 2 sec. : 64 feet (4 sec.: 123

feet, the lengths of the planes.


6 192

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, whereever such numbers occur.


Another Solution by means of Algebra.

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Put the time of descent down the less plane; then will x2 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:x + 2; hence 2x+4=3x, and x 4 sec. Then the lengths of the planes are found as in the last proportion of the former solution.



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 plane inclined to it in an angle of 15 degrees; it is required to find 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?


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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 вс 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 в, 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


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HBм, and therefore this latter, as well as the former, is equal
to the complement of the c, the inclination of the two planes;
but the part IBM is c, therefore the angle of projection
HBI is
the comp. of double the c, and being the comp. of
HBK, theref. HBK 24c. Now, put a=50=AD the height
above the horizontal line, t-tang. DBC or 75o the comple-
ment of the plane's inclination, tang. HBI or H=60° the
comp. of 2c, s-sine of 24HBI=120° the double elevation,
or sine of 4 c; also x = AB the impetus or height fallen
through. Then,

and {

BI = 4KH=2sx, by the projectiles prop. 21,
BK = 7X KH=8TX

CD=tXBD=(-a). by trigonometry;

also, KD = BK
BK BD 1 ST X x + α, and KE — BI = SX;
then, by the parabola, BK: ✓ DK :: KE : FG = KE × √
Ts2x2 - 2sx2 +2αsx



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√[ax- - — s2 ) x2] =2b√

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(ax-b2x2), putting b = sine of 2 c=sine of 30°. Hence CG=CD+DFFG=tx-ta+sx±2b✓ (ax -- b2x2) a maximum, the fluxion of which made =0, and the equation reduced,


gives x= ×(1± √


double sign


where nst, and the (n2 +-464)3 answers to the two roots or values of x, or to the two points G, G, where the parabolic path cuts the horizontal line ce, the one in ascending and the other in descend

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•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 =



est distance co required is 826.9915 feet.

Hence also the great

Corol. If it were required to find ca or tx-ta + sx±2b√√ (ax-b2x2)=g a given quantity, this equation would give the value of x by solving a quadratic.


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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; Amre 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.



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Put c for the cosine, and t for the tangent of the course, or angle nmr to the radius r; am, any variable part of the rhumb from the equator,

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the latitude вm w; its sine x, and cosine y; and AB, the dif. of longitude from A, z. Then, since the elementary triangle mar may be considered as a right-angled plane triangle, it is, as rad. resin. ▲mrn :: v=mr : w = mn :::; theref. cv=rw, or v=

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by putting s for the secant of the nmr the ship's course. In like manner, if w be any other latitude, and v its corresponding length

of the rhumb; then v = ; and hence v v = r X

or D=






by putting D=v- the distance, and d
v dw-w

the dif. of latitude; which is the common rule.

The same is evident without fluxions: for since the Zmrn is the same in whatever point of the path amre the point m is taken, each indefinitely small particle of Amre, must be to the corresponding indefinitely small part of Bm, in the constant ratio of radius to the cosine of the course: and therefore the whole lines, or any corresponding parts of them, must be in the same ratio also, as above determined. In the same manner it is proved that radius: sine of the course :: distance the departure.

Again, as the radius r: t tang nmr :: w=mn: nr or mt, and as ry:: PB ; PM :: ¿➡BC: mt; hence, as the extremes of these proportions are the same, the rectangles of the means must be equal, viz, yz=tw= because by the pro



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fluents of these are z=tX hyp. log. ✔✅✔✅

+c; which

corrected by supposing 2=0 when x = a, are z tx (hyp.

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r - X

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- hyp. log. ✔+) is the meridional parts of the dif. of the

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latitudes whose sines are x and a, which call b; then is z



the same as it is by Mercator's sailing.

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Further, putting m=2-71828 the number whose hyp. log. r+x


is 1, and n = ; then, when z begins at a, m2 = and




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2r : hence it appears that mr +1 as m2, or rather n or z increases (since m is constant), that x

approximates to an equality with r, because

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or converges to 0, which is its limit; consequently r is the limit or ultimate value of x: but when xr, the ship will be at the pole; theref. the pole must be the limit, or evanescent state, or the rhumb or course: so that the ship may be said to arrive at the pole after making an infinite number 2r

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of revolutions round it; for the above expression
nishes when n, and consequently z, is infinite, in which case x
is = r..

rd: ·sd


Now, from the equation D = it is found, that when d≈ 30° 57', the comp. of the given lat. 59° 3′, and c= sine of 67o, 30′ the comp. of the course, p will be = 2010 geographical miles, the required ultimate distance; which, at the rate of 10 miles an hour, will be passed over in 201 hours, or 83 days. The dif. of long. is shown above to be infinite. When the ship has made one revolution, she will be but about a yard from the pole, considering her as a point.

When the ship has arrived infinitely near the pole, she will go round in the manner of a top, with an infinite velocity; which at once accounts for this paradox, viz. that though she make an infinite number of revolutions round the pole, yet her distance run will have an ultimate and definite value, as above determined: for it is evident that however great the number of revolutions of a top may be, the space passed over


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