393. Exam. 2. To find the altitude, when the state of the barometers and thermometers is as follows, viz.

ON THE RESISTANCE OF FLUIDS, WITH THEIR FORCES AND ACTIONS ON BODIES.

PROPOSITION LXXVII.

394. If any Body Move through a Fluid at Rest, or the Fluid Move against the Body at Rest; the Force or Resistance of the Fluid against the Body, will be as the Square of the Velocity and the Density of the Fluid. That is, R a dv2.

FaR, the force or resistance is as the quantity of matter or particles struck, and the velocity with which they are struck. But the quantity or number of particles struck in any time, are as the velocity and the density of the fluid. Therefore the resistance, or force of the fluid, is as the density and square of the velocity.

395. Corol. 1. The resistance to any plane, is also more or less, as the plane is greater or less; and therefore the resistance on any plane, is as the area of the plane a, the density of the medium, and the square of the velocity. That, is, R α adv3.

396. Corol. 2. If the motion be not perpendicular, but oblique to the plane, or to the face of the body; then the resistance, in the direction of motion, will be diminished in the triplicate ratio of radius to the sine of the angle of inclination of the plane to the direction of the motion, or as the cube of radius to the cube of the sine of that angle. So that R∞ adv3, putting 1 = radius, and s=sine of the angle of inclination CAB.

For, if AB be the plane, AC the direction of motion, and вс perpendicular to ac; then no more particles meet the plane than what meet the perpendicular Be, and therefore their number is diminished as AB to Bc or as 1 to 8. But the force of each par

B

ticlė, striking the plane obliquely in the direction CA, is also diminished as AB to BC, or as 1 to 8; therefore the resistance, which is perpendicular to the face of the plane by art. 52, is as 12 to 2. But again, this resistance in the direction perpendicular to the face of the plane, is to that in the direction AC, by art. 51, as AB to BC, or as 1 to s. Consequently, on all these accounts, the resistance to the plane when moving perpendicular to its face, is to that when moving obliquely, as 13 to s3, or 1 to s3. That is, the resistance in the direction of the motion, is diminished as I to, or in the triplicate ratio of radius to the sine of inclination.

PROPOSITION LXXVIII.

397. The Real Resistance to a Plane, by a Fluid acting in a Direction perpendicular to its Face, is equal to the Weight · of a Column of the Fluid, whose Base is the Plane, and Altitude equal to that which is due to the Velocity of the Motion, or through which a Heavy Body must fall to acquire that Velocity.

THE resistance to the plane moving through a fluid, is the same as the force of the fluid in motion with the same velocity, on the plane at rest. But the force of the fluid in motion, is equal to the weight or pressure which generates that motion; and this is equal to the weight or pressure of a column of the fluid, whose base is the area of the plane, and its altitude that which is due to the velocity.

398. Corol. 1. If a denote the area of the plane, u the velocity, n the density or specific gravity of the fluid, and g= 16,2 feet, or 193 inches. Then the altitude due to the velocity being therefore a xnx 4g'

02

be the whole resistance, or motive force R.

anr 2

45

will

399. Corol. 2. If the direction of motion be not perpendicular to the face of the plane, but oblique to it, in any angle, whose sine is s. Then the resistance to the plane will

be

and 283

48

400. Corol. 3. Also, if w denote the weight of the body, whose plane face a is resisted by the absolute force R; then the retarding force f, or will be

R

and $3

4gro

401. Corol. 4. And if the body be a cylinder, whose face

or

?

or end is a, and radius r, moving in the direction of its axis; because then s = 1, and a = pr2, where p =

pnv 2 will be the resisting force R, and

pno 2,2 4g force f.

4gw

3.1416; then the retarding

402. Corol. 5. This is the value of the resistance when the end of the cylinder is a plane perpendicular to its axis, or to the direction of motion. But were its face an elliptic section, or a conical surface, or any other figure everywhere equally inclined to the axis, or direction of motion, the sine or inclination being : then, the number of particles of the fluid striking the face being still the same, but the force of each, opposed to the direction of motion, diminished in the duplicate ratio of radius to the sine of inclination, the resisting force R would be pur2282

4g

PROPOSITION LXXIX.

403. The Resistance to a Sphere moving through a Fluid, is but Half the Resistance to its Great Circle, or to the End of a Cylinder of the same Diameter, moving with an Equal Velocity.

A

H

C

E

B

LET AFEB be half the sphere, moving in the direction CEG. Describe the paraboloid AIEKB on the same base. Let any particle of the medium meet the semicircle in F, to which draw the tangent FG, the radius FC, and the ordinate FIH. Then the force of any particle on the surface at F, is to its force on the base at н, as the square of the sine of the angle G, or its equal the angle FCH, to the square of radius, that is, as HF2 to CF2. Therefore the force of all the particles, or the whole fluid, on the whole surface, is to its force on the circle of the base, as all the HF to as many times cr2. But cr2 is = CA2 = AC. Cв, and нF2 = AH Hв by the nature of the circle: also, AH. HB AC. CB HI: CE by the nature of the parabola; consequently the force on the spherical surface, is to the force on its circular base, as all the HI's to as many cE's, that is, as the content of the paraboloid to the content of its circumscribed cylinder, namely, as 1 to 2.

404. Corol. Hence, the resistance to the sphere is R = being the half of that of a cylinder of the same

diameter.

diameter. For example. a 9lb iron ball, whose diameter is 4 inches, when moving through the air with a velocity of 1600 feet per second, would meet a resistance which is equal to a weight of 1324lb, over and above the pressure of the atmosphere, for want of the counterpoise behind the wall.

PRACTICAL EXERCISES CONCERNING SPECIFIC

GRAVITY.

The Specific Gravities of Bodies are their relative weights contained under the same given magnitude; as a cubic foot, or a cubic inch, &c.

The specific gravities of several sorts of matter, are expressed by the numbers annexed to their names in the Table of Specific Gravities, at page 211; from which the numbers are to be taken, when wanted.

Note. The several sorts of wood. are supposed to be dry. Also, as a cubic foot of water weighs just 1000 ounces avoirdupois, the numbers in the table express, not only the specific gravities of the several bodies, but also the weight of a cubic foot of each in avoirdupois ounces; and hence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known, as in the following problems.

PROBLEM I.

To find the Magnitude of any Body, from its Weight.

As the tabular specific gravity of the body,

Is to its weight in avoirdupois ounces,

So is one cubic foot, or 1728 cubic inches,
To its content in feet, or inches, respectively.

EXAMPLES.

EXAM. 1. Required the content of an irregular block of common stone, which weighs lcwt. or 112lb.

Ans. 1228 cubic inches.

EXAM. 2. How many cubic inches of gunpowder are there in 1lb weight? Ans. 29 cubic inches nearly.

EXAM. 3. How many cubic feet are there in a ton weight of dry oak? Ans. 381 cubic feet.

PROBLEM

PROBLEM II.

To find the Weight of a Body from its Magnitude.

As one cubic foot, or 1728 cubic inches,
Is to the content of the body,
So is its tabular specific gravity,
To the weight of the body.

EXAMPLES.

EXAM. 1. Required the weight of a block of marble, whose length is 63 feet, and breadth and thickness each 12 feet; being the dimensions of one of the stones in the walls of Balbeck ?

Ans. 6837 ton, which is nearly equal to the burden of an East-India ship.

EXAM. 2. What is the weight of 1 pint, ale measure, of gunpowder? Ans. 19 oz. nearly. EXAM. 3. What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet broad, and 24 feet deep? Ans. 4335 lb.

PROBLEM III.

To find the Specific Gravity of a Body.

CASE 1. When the body is heavier than water, weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then say,

As the weight lost in water,

Is to the whole weight,

So is the specific gravity of water,
To the specific gravity of the body.

EXAMPLE.

A piece of stone weighed 10lb, but in water only 6lb, required its specific gravity? Ans. 2.609.

CASE 2. When the body is lighter than water, so that it will not quite sink, affix to it a piece of another body, heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass separately, both in water and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say,

As