Suppose, for example, that a pulley of 7 in. diameter drives one of 14 in. diameter, then if there is no slip on either pulley both pulleys will run at the same velocity as the belt, and this velocity must be equal to the velocity of the driver, because the belt is moved by the driver. Now, suppose the driver which is of 7 in. diameter makes one revolution in a minute, and as it is only one-half the diameter of the driven wheel, its circumference will also be half that of the driven, so that it must make two revolutions to carry around length of belt enough to pass once around the driven pulley. The revolutions of the two are, therefore, in the same proportions as are their diameters, which in this case is two to one. As the driven pulley is the largest diameter, it will make one revolution in the same time that the driver makes two. But suppose the driving pulley was 14 and the driven was 7 inches in diameter, then the proportion would still be two to one, and the driven would make two revolutions to every revolution of the driver.
Fig. 2654.
If we are given the number of revolutions a driving pulley makes and the diameter or circumference of both pulleys, and require to find the number of revolutions the driven pulley will make to one or to any given number of the driver, we may consider as follows: Suppose the circumference of the driver to be 24 inches and that of the driven to be 18 inches, then in [Fig. 2654] let circle a represent the driver, and circle b the driven pulley. If we divide the circumference of a into four equal divisions, as at 1, 2, 3, and 4, each of these divisions will equal 6 inches, because the whole circumference being 24 inches, one quarter of it will be 6. If we divide the circumference of b into six-inch divisions there will be but three of them as marked, because one-third of 18 (its circumference) is 6. Now three of the divisions at a will move a a full revolution, and the remaining division on a will move b through another one-third of a revolution, hence, each revolution of a equals 11⁄3 revolutions on b. The proportions of the circumference are, therefore, as 11⁄3 to 1, or as 133 is to 100, taking a as the driver, and, therefore, as the basis of the proportion. But suppose we take b as the basis of the proportion, and one revolution of b will cause a to make three quarters of a revolution, or during 100 revolutions of b, a will make 75. But nevertheless during the period that a is making 100 revolutions b will have made one-third more, or 1331⁄3, because b makes 11⁄3 revolutions to cause a to make one revolution. From this it will be seen that the proportion is as the greater is to the lesser, and not as the lesser is to the greater, or, in other words, it is in this case as 24 is to 18, which is one and one-third times, for one-third of 18 is 6, and 18 + 6 = 24.
Suppose, now, we take the four divisions on a and the three on b to consider their proportions, and we may say 4 is 11⁄3 times 3, or we may with equal propriety say 3 is 3⁄4 of 4, hence 4 is not in the same proportion to 3 that 3 is to 4. Let it now be supposed that a driven pulley b is 18 inches in diameter, and requires to be driven one quarter faster than the driver, what then must be the diameter of that driver? As the revolutions require to be increased one-fourth the pulley diameter must be increased one-fourth. Thus one quarter of 18 = 41⁄2, and this added to 18 is 221⁄2, which is therefore the diameter of the driving pulley, as may be proved as follows: Suppose the circumferences instead of the pulley diameters to be 221⁄2 and 18 respectively, and that the largest pulley makes 100 revolutions, then it will pass 2,250 (221⁄2 × 100 = 2,250) inches of belt over its circumference, and every 18 inches of this belt will cause the small pulley to make one revolution; hence we divide 2,250 by 18, which gives us 125 as the revolutions made by the small pulley, while the large one makes 100. Thus it appears that we obtain the same result whether we take the circumferences or the diameters of the pulleys, because it is their relative proportions or relative revolutions that we are considering, and their actual diameters do not affect their proportions one to the other. Thus, if a 10-inch pulley drives a 30-inch one, the proportions being three to one, the revolutions will be three to one, and the driven being three times the largest, will make one revolution to every three of the driver. If the driver was 3 inches in diameter and the driven 9, the revolutions would be precisely the same as before, but with equal revolutions the velocities would be different, because in each revolution of the driver it will move a length of belt equal to its circumference; hence, the greater the circumference the greater length of belt it will move per revolution. To take the velocity into account, we must take into consideration the number of revolutions made in a given time by the driver. Suppose, for example, that the driver being 3 inches in diameter makes one revolution in a minute, then it will move in that minute a length of belt equal to its circumference, so that the circumference of the driver, multiplied by the number of its revolutions per minute, gives its velocity per minute. Thus, if a pulley has a circumference of 50 inches, and makes 120 revolutions per minute, then its velocity will be 6,000 inches per minute, because 50 × 120 = 6,000. The velocity of the belt, and therefore that of the driven wheel, will also be 6,000 inches per minute, as has already been shown. From this train of reasoning the following rules will be obvious:—
To find the diameter of the driving pulley when the diameter of the driven pulley and the revolutions per minute of each are given:
Rule.—Multiply the diameter of the driven by the number of its revolutions, and divide the product by the number of revolutions of the driver, and the quotient will be the diameter of the driver.
The diameter and revolutions of the driver in a given time being known, to find the diameter of a driven wheel that shall make a given number of revolutions in the same time:
Rule.—Multiply the diameter of the driver by its number of revolutions, and divide the product by the number of revolutions of the driven. The quotient will be the diameter of the driven.