1 .017452 19.325568 37 .601815 55 .819152 73 .956305 74 .961262 57.838671 75 .965926 .970296 .974370 60.866025 78 .978148 79 .981627 .984808 .987688 .990268 .992546 .994522 7.121869 25.422618 43 .681998 61 .874620 8.139173 26.438371 44.694658 62.882948 80 9.156434 27 .453990 45.707107 63.891007 81 10.173648 28.469472 46.719340 64.898794 82 11.190809 29.484810 47.731354 65.906308 83 12 .207912 30.500000 48.743145 66 .913545 84 13 .224951 31.515038 49 .754710 67.920505 85 14 .241922 32.529919 50.766044 68 .927184 86 15 .258819 33.544639 51.777146 69 .933580 87 16.275637 34.559193 52.788011 70 .939693 88 17.292372 35.573576 53.798636 71.945519 89 .999848 18 .309017 36 .587785 54.809017 72.951057 90 1.000000 .996195 .997564 .998637 .999391 The natural sines for degrees and parts of a degree, as that for 10 degrees and 20 minutes, may be found very nearly as follows:-Subtract the natural sine of the given number of degrees from the next greater; and say:-As 60 is to the remainder, so is the number of minutes to the part to be added to the sine of the given number of degrees. Thus, .190809– .173648 = .017161. Then, as 60: .017161 :: 20 to the part to be added to the sine of 10 degrees, viz. 0.005720; consequently, the sine of 10 degrees and 20 minutes = .179368, nearly. 6. A cannon ball is fired with a velocity of 480 feet per second, at an elevation of 75 degrees, and another is fired with the same velocity at an angle of 15 degrees; required the ranges. Ans. 3,600 ft., nearly. 7. If a shell be fired with a velocity of projection sufficient to give it a range of 3,600 feet at an elevation of 45 degrees; required its range and time of flight, when discharged with the same velocity at an angle of 32 degrees. The natural sine of twice the angle of elevation, (viz. 64 degrees,) is .898794, which multiplied by 3600, gives the required range 3235.65 feet; and the velocity of projection equals the square root of 3600 × 32=339.4 feet per second; and said velocity, viz. 339.4÷16=21.212, which multiplied by .529919, the natural sine of 32 degrees, gives 11.24 seconds, the time of flight; and half the range at 45 degrees, viz. 1800, multiplied by (.529919), the square of the sine of the angle of elevation, gives 672.3 feet, the greatest altitude of the projectile. The ranges are directly proportional to the charges of powder. 8. If a shell range 1,000 yards at an elevation of 45 degrees, how far will it range at an elevation of 301 degrees, the charge being double? Ans. 1,742 yards. All ranges at the same elevation are directly proportional to the squares of the velocities. 9. If a shell ranges 4,000 feet, the velocity of projection being 480, how far will it range if the velocity be 600 feet? Ans. 6,250 feet. Although the above rules give results nearly correct, when the velocity of projection does not exceed 1,000 feet per second; yet the results are far from the truth for greater velocities. This difference is occasioned by the increased pressure of the air on the anterior surface of the projectile, and the greatly diminished or entire removal of a corresponding pressure on the posterior surface; the consequence of which is a retardation of the velocity of the projectile, which greatly diminishes the range. 74. MEASUREMENT OF DISTANCES BY THE VELOCITY OF SOUND. It has been proved by careful experiments, that sound is propagated through dry air at the freezing temperature (32 de grees Fahrenheit) at the rate of 1,090 feet per second; and through dry air at the temperature of 60 degrees at the rate of 1,125 feet per second, the velocity increasing with the temperature very nearly at the rate of 1.25 feet for every degree of heat. And since light moves with the astonishing rapidity of nearly 200,000 miles in a second of time, its transmission through short spaces becomes quite insensible, and may be regarded as instantaneous. It is therefore very plain, that when the time elapsed between a flash of lightning, or of the powder of a gun, and the perception of the sound is known, the distance at which the sound was produced may be easily determined. EXAMPLES. 1. Find the distance of a thunder-cloud, the time elapsed between the lightning and the thunder being 6 seconds, and the temperature of the air 60 degrees. Ans. 6,750 feet. 2. An echo of sound was reflected from a rock in 4 seconds after the sound, the temperature of the air being 60 degrees; required the distance of the rock. Ans. 2,250 feet. 3. The flash of a cannon is seen at a distance, and 10 seconds after the report is heard; required the distance, the temperature of the air being 42 degrees. Ans. 11,025 feet. The velocity of sound is slightly affected by the variable density of the atmosphere, by the quantity of vapor which it contains, and by the direction of the wind; but none of these causes, nor all of them combined, materially affect the results arrived at by the above data. ¶ 75. PROPERTIES OF THE ATMOSPHERE AND THE The atmosphere which we breathe is a permanently elastic fluid surrounding our globe to the height of about 45 miles. It is almost wholly composed of two gases, oxygen and nitrogen, there being in its composition 4 parts of nitrogen to 1 of oxygen, by volume, and 75 parts of nitrogen to 23 of oxygen, by weight. Oxygen gas supports respiration, and its existence. is necessary to the maintenance of both animal and vegetable life. This gas is highly inflammable, and when a current of it is directed upon burning charcoal, the heat thus developed is so intense as to consume or evaporate iron, tin, copper, and other metals. Nearly half the solid content of the globe is composed of oxygen; and it enters largely into the composition of a great number of bodies. Nitrogen gas is a nonsupporter of combustion; and an animal immersed in this gas is killed very shortly by suffocation. The mean pressure of the atmosphere on a square inch, at the surface of the ocean, is 14.6 pounds avoirdupois; consequently, the pressure on a square foot would be 2102.4 pounds, or 2102 pounds, nearly; and on a circular inch 11.467 pounds, or 111⁄2 pounds, nearly, and on a circular foot, 1651.22 pounds. This pressure is sufficient to support a column of mercury whose height is 28.6 inches, and a column of water whose height is 33 feet. The atmospheric pressure continually changes from various causes, as may be shown by the rise and fall of the mercury in the tube of the barometer, its height varying between 27 and 31.8 inches at the level of the sea. The density of the atmosphere decreases in the duplicate ratio of the altitude; that is to say, if at a certain altitude above the earth's surface the weight be half what it is at the surface, then at twice that altitude the density or weight will be only one-fourth of what it is at the surface. If at the surface of the ocean the height of the mercury in the barometric tube be 30 inches, at an altitude of 90 feet it will be about 29.9, or one-tenth of an inch less, and at the height of twice 90, or 180 feet, it will stand at the height of 29.8 inches, nearly, or one-fifth of an inch less, and so on. It is necessary, however, to take into the account several other particulars in order to render this method of measuring heights by the fall of mercury in the barometer, accurate; for which information we refer the student to Chambers' Mathematics, Part 1, and to Bonnycastle's Trigonometry. WIND is air in motion. The velocity of wind varies from ( to more than 100 miles per hour. A gentle breeze moves with a velocity of 6.8 miles per hour, or 10 feet per second, and exerts a force equal to 0.229 of a pound on a square foot. A very high wind moves 47.7 miles per hour, or 70 feet per second, and exerts a force of 11.207 pounds on a square foot. The wind in a tempest moves 54.3 miles per hour, or 80 feet per second, and exerts a force of 14.638 pounds on a square foot. A hurricane moves 81.8 miles an hour, or 120 feet per second, and exerts a force of 32.926 pounds on a square foot. A violent hurricane moves 102.3 miles per hour, or 150 feet per second, with a force equal to 51.426 pounds on a square foot. To find the force of wind, its velocity being given :— Divide the velocity per second by 10, and multiply the square of the quotient by 0.229, and the product will be the force exerted on a square foot in pounds. To find the velocity of wind, its force being given :— Divide the force exerted on one square foot in pounds by 0.229, and extract the square root of the quotient; multiply this root by 10, and the product will be the velocity per second in feet. EXAMPLES. 1. What is the force exerted on a square foot of surface by the wind, when its velocity is 36 miles per hour, or 52.8 feet per second? Ans. 6.384 lbs. 2. If the velocity of wind be 40 miles per hour, what force will it exert on a square foot? Ans. 7.882 lbs. 3. If a horse-power be equal to 400 lbs., to how many horse-power will the wind be equal when exerted on 40 square yards of canvas, its velocity being 60 miles per hour? Ans. 16, very nearly. 4. If the force of the wind on the surface of one square foot be equal to 20 pounds; required its velocity. Ans. 93.453 feet per second. 5. If the pressure of the atmosphere be 14.6 pounds on a |