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DQ = h = 10; AL = c=10}=3 ; a=log. of 7=·8450980;

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log. of 2.562070408591; cy + a = •0408591y + •8450980 =log. of n. From the general equation then, viz.

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equal to 1, 2, 3, 4, &c., thence finding the corresponding values of cy+ or 0408591y+8450980, and to these, as common logs, taking out the corre sponding natural numbers, which will be the values of n; then the above theorem will give the several values of w or ci, as they are here arranged in the annexed table, from which the figure of the curve is to be constructed, by thus finding so many points in it.

Otherwise. Instead of making n the number of the log. cy+a, if we put in the natural number of the log. w+√(w2— a2)

cy only; then m=

or by squaring, &c., a2m2.
m2+1
2m

w=

a

-

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and am―w = √ (w2—a2),

2amw + w2 = w2-a2 and hence

Xa to which the numbers being applied, the

very same conclusions result as in the foregoing calculation and table.

PROBLEM VIII.

To construct Powder Magazines with a Parabolical Arch.

It has been shown, in my tract on the Principles of Arches of Bridges, that a parabolic arch is an arch of equilibration, when its extrados, or form of its exterior covering, is the very same parabola as the lower or inside curve. Hence then a parabolic arch, both for the inside and outer form, will be very proper for the structure of a powder magazine. For, the inside parabolic shape will be very convenient as to room for stowage : 2dly, the exterior parabola, every where parallel to the inner one, will be proper enough to carry off the rain water: 3dly, the structure will be in perfect equilibrium: and 4thly, the parabolic curve is easily constructed, and the fabric erected.

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tity a, what it is at the vertex; that is, ci is everywhere equal to KD.

Consequently KR is DP; and since RI is = PC, it is evi. dent that KI is the same parabolic curve with DC, and may be placed any height above it, always producing an arch of equilibration, and very commodious for powder magazines.

PROMISCUOUS PROBLEMS, AS EXERCISES IN MECHANICS, STATICS, DYNAMICS, HYDROSTA. TICS, 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 planes of the same height being as the times of descent down them (art. 195, Dynam.), and also as the weights of bodies mutually sustaining each other on them (art. 193), therefore the times must be as the weights: hence as 1, the difference of the weights, is to 2 sec. the diff. of times, :: 3:6 sec. the times of descending down the two

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2:4 sec.

planes. And as 16:64: 1 sec. : 2 sec. the time of descent down the perpendicular height. Then, by the laws of

descents as 2 sec. : 64 feet ::

{

6 sec. : 192
4 sec. : 128

feet, the lengths

of the planes.

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 often do in solving the following questions, wherever great accuracy is not required.

Another Solution by means of Algebra.

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

PROBLEM II.

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?

A

H

K

F

B

M

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 ABH, 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 Lc, therefore the angle of projection HBI is the comp. of double the ▲ c, and being the comp. of HBK, theref. HBK = 2 c. Now, put

=

a = 50 = AD the height above the horizontal line, t = tang. DBC or 75° the complement of the plane's inclination, = tang. HBI or н=60° the comp. of 2c, s=sine of 2▲Het 120° the double elevation, or sine of 4 c; also x=AB the impetus or height fallen through. Then,

and

BI=4Kн=2sx, by the projectiles prop. 176, p. 213.
BKXKH=187x

{

CD=t XBD=(a) by trigonometry.

also, KD = BK BD= = 18Tx

x+a, and кE = ¦BI = 82 ;

then, by the parabola,✓ BK: ✔✅ DK :: KE: FG = KE X

28

KD

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28

=√

= v[

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KB

T

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Hence CGCD + DF FG=1x

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·la + sx

a maximum, the fluxion of which made

tion reduced, gives a =

a

262

=

c sine of 30°. ± 26 √(ax —b2x2) = 0, and the equa.

=

na

×(1± √ (n2 +46*)*

where n=s

+t, and the double sine 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 descending,

Now, in the present case, when the c= 15°, t=tang. 75°=2+ √/3, r= tan. 60°=√√3, s=sin. 60°=√3, b ÷ sin. 30°, n=s+1=2+}√3; then

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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

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

line in two points; or it will touch it when x = bb'

PROBLEM III.

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 be fore 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 póle; Amrr a loxodromic or rhumb line, or the path of the ship continued to the equator; PB, PC, any two meridians in. definitely near each other; nr, or mt, the part of a parallel of latitude intercepted

between them.

P

201

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, 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 rightangled plane triangle, it is, as rad. r : c = sin. ▲ mrn ; ; v

rw

= mr : w = mn::v: w; theref. cv = rw, or v = с

sw

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

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ponding length of the rhumb; then v == ; and hence v

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rd

с

or D- by putting D v-v the distance,

с

and dw-w the dif. of latitude; which is the common rule.

mrn

The same is evident without fluxions: for since the is the same in whatever point of the path Amre the point m is taken, each indefinitely small particle of amrr, 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 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. yż = tio =

tri

ri because i =

y

y

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by the

; the gene

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