EXAMPLE.-Required the velocity a ball will acquire in descending through 201 feet. √64.33 × 201 = 113.7 feet. To find the Space through which a Body will fall in any given time. RULE.-Multiply the square of the time in seconds by 16.083, and the product will be the space in feet. EXAMPLE.-Required the space fallen through in 7 seconds. 16.083 x 49788.067 feet. NOTE. The velocity acquired by a body in falling from rest, through a given height, is the same whether it fall freely or descend through a plane any way inclined. The diameter of a circle perpendicular to the horizon, and any chord terminating at either extremity of that diameter, are fallen through in the same time. And the velocities which bodies acquire by descending along chords of the same circle are as the lengths of those chords. ON PENDULUMS. A pendulum that vibrates seconds, or 60, in the latitude of London, is 39.1393 inches long; and /39.1393 x 60375.36, which serves as a constant number for other pendulums: thus, 375.36 divided by the square root of the pendulum's length, gives the number of vibrations per minute; and divided by the vibrations per minute, gives the square root of the length of pendulum. EXAMPLE 1.-Required the number of vibrations a pendulum of 25 inches long will make per minute. 375.36 25 = 75.072 vibrations per minute. EXAMPLE 2.-Required the length of a pendulum to make 80 vibrations per minute. 375.36 80 = 4.6922 = 22.014864 inches long. ON THE VELOCITY OF WHEELS, DRUMS, PULLEYS, &c. When wheels are applied to communicate motion from one part of a machine to another, their teeth act alternately on each other, consequently, if one wheel contain 60 teeth, and another 20, the one containing 20 teeth will make three revolutions while the other makes but one; and if drums or pulleys are taken in place of wheels, the result will be the same; because their circumferences, describing equal spaces, render their revolutions unequal: from this the rule is derived, namely, Multiply the velocity of the driver by the number of teeth it contains, and divide by the velocity of the driven; the quotient will be the number of teeth it ought to contain.-Or, Multiply the velocity of the driver by its diameter, and divide by the velocity of the driven; the quotient will be the diameter of the driven. EXAMPLE 1.-If a wheel that contains 75 teeth makes 16 revolutions per minute, required the number of teeth in another to work in it, and make 24 revolutions in the same time. EXAMPLE 2.-A wheel 64 inches diameter, and making 42 revolutions per minute, is to give motion to a shaft at the rate of 77 revolutions in the same time; required the diameter of a wheel suitable for that purpose. EXAMPLE 3.-Required the number of revolutions per minute made by a wheel or pulley 20 inches diameter, when driven by another of 4 feet diameter, and making 46 revolutions per minute. EXAMPLE 4.-A shaft at the rate of 22 revolutions per minute is to give motion, by a pair of wheels, to another shaft at the rate of 15, the distance of the shafts from centre to centre is 45 inches; the diameters of the wheels at the pitch lines is required. And 45.5 x 15.5 45.5 x 22 = 18.81 radius of the driving wheel. 26.69 radius of the driven wheel. EXAMPLE 5.-Suppose a drum making 20 revolutions per minute, required the diameter of another to make 58 revolutions in the same time. 58 20 2.9, that is, their diameters must be as 2.9 to 1; thus, if the one making 20 revolutions be called 30 inches, the other will be 302.9 10.345 inches diameter nearly. EXAMPLE 6-Required the diameter of a pulley, to make 12 revolutions in the same time as one of 32 inches making 26. EXAMPLE 7.-A shaft at the rate of 16 revolutions per minute, is to give motion to a piece of machinery at the rate of 81 revolutions in the same time; the motion is to be communicated by means of two wheels and two pulleys with an intermediate shaft; the driving wheel contains 54 teeth, and the driving pulley is 25 inches diameter; required the number of teeth in the other wheel, and the diameter of the other pulley. √81 × 16=36, the mean velocity between 16 and 81;-then 81 24 teeth;-and- EXAMPLE 8.-Suppose in the last example the revolutions of one of the wheels being given, the number of teeth in both, and likewise the diameter of each pulley, to find the revolutions of the last pulley. 16 x 54 24 = 36, velocity of the intermediate shaft; and, 36 x 25 = 81 the velocity of the machine. A TABLE the radius is given, teeth, and any pitch For finding the radius of a wheel when the pitch is given, or the pitch of a wheel when that shall contain from 10 to 150 required. 31 32 27 4.307 63 10.031 98 15.600 133 21.169 10.190 99 15.759 134 21.328 135 21.488 136 21.647 16.236 137 21.806 138 21.965 16.554 139 22.124 140 22.283 |