## LINE SHAFT PULLEYS and BELTINGBrought to you by Harry's Old Engine...Everything you've ever wanted to know about a lineshaft and then some... Click for color photo.## Kinematics of Flexible Gearing
For transmitting considerable power between shafts at some distance apart, ropes are frequently used instead of belts. The advantages of rope driving are lightness, low first cost, smooth running, flexibility, and case of repair. With rope drives, the shafts connected are generally run at high speeds. Chain gearing is used when positive driving is essential, and also for the transmission of large powers. Ordinarily, with chain drives the motion is slow, although recently chains of special forms have been devised for highspeed transmission, and these are being used quite extensively in place of belts. ## VELOCITY RATIO IN BELT AND ROPE DRIVES
Then v = pi Da Na and v = pi Db Nb; hence, pi Da Na = pi Db Nb and (1) Da Na=Db Nb or (2) Na/Nb = Db/DaThat is, the speeds or numbers of revolutions of two pulleys connected by belt are inversely proportional to their diameters.In formulas 1 and 2, the diameters may be taken in either feet or inches, but the same unit must in every case be used for both diameters. EXAMPLE 1. A pulley 30 inches in diameter, making 210 revolutions per minute, drives a second pulley 14 inches in diameter; how many revolutions per minute does the latter pulley make? SOLUTION. From formula 1, 30 x 210 = 14 x Nb; whence Nb = 30 x 210 / 14 = 450 R. P. M. Ans. EXAMPLE 2. The driving pulley of a machine is 1 foot in diameter and must make 750 revolutions in 5 minutes; what size pulley should be used on the driving shaft, if its speed is 143 revolutions per minute? SOLUTION. In all examples of this kind, the speeds and diameters must be reduced to the same units. 750 revolutions in 5 minutes = 750 / 5 = 150 R. P. M.; 1 ft. = 12 in. Hence, 12 X 150 = Db X 143, whence Db = 12 X 150 / 143 = 12.6 in., nearly. Ans.
(1) Na (Da + t) = Nb (Db + t)and formula 2, Art. 2, becomes (2) Na/Nb = Db + t / Da + tEXAMPLE. If, in the first example of Art. 2, the belt thickness is 3/8 inch, how many revolutions per minute does the 14 inch pulley make, taking account of this thickness? SOLUTION. From formula 1,Na (Da + t) = Nb (Db +t), whence,210 X (30 + 3/8) = Nb X (14 + 3/8) and Nb = 210 X (30 3/8 / 14 3/8) = 444 R. P. M., nearly. Ans. Hence, by taking belt thickness into account, the speed of the driven pulley is reduced from 450 to 444 revolutions per minute, or about 1.3 percent. In practice, however, the thickness of the belt is usually neglected.
In Fig. 2 is shown a compound system of belts in which there are six pulleys, Now, pulleys b and c are fixed to the same shaft and make the same number of revolutions per minute; hence, Considering separately the three pair connected by the three belts, pullevs a, c, and e may be taken as the drivers,
The thickness of the belt may be taken into account by increasing all pulley diameters by an amount equal to that thickness. EXAMPLE. An emery grinder Is to be set up to run at 1,200 revolutions per minute. The countershaft has pulleys 20
and 8 inches in diameter. The pulley on the grinder is 6 inches In diameter and is belted to the 20-inch pulley on the
countershaft. The line shaft runs at 180 revolutions per minute and carries a pulley that is belted to the 8-inch pulley
on the countershaft. Calculate the diameter of tile line shaft pulley. SOLUTION. From formula
EXAMPLE. It is required to run a machine at 1,600 revolutions per minute, the driving shaft making 320 revolutions
per minute; what size pulleys are required: (a) when two pulleys are used; (b) when four pulleys are used? SOLUTION. (a) 1,600 / 320 = 5. The two pulleys must, therefore, be in the ratio of 5 to 1, the driving pulley being five
times as large as the driven pulley, since the latter has the greater speed. Then assume diameters of 30 and 6 in. Ans.
## EXAMPLES FOR PRACTICE1. A driving pulley is 54 inches in diameter, and the driven pulley, which runs at 112 revolutions per minute,
is 2 1/2 feet in diameter; what is the speed of the driving shaft? 2. The flywheel of an engine running at 180 revolutions per minute is 8 feet 5 inches in diameter; what should
be the diameter of the pulley that it drives if the required speed of the latter is 600 revolutions per minute? 3. A machine is to be belted through a countershaft, so as to run at 1,200 revolutions per minute, the speed of
the driving shaft being 120 revolutions per minute; find three ratios that could be used for two pair of pulleys. 4. An emery grinder is to be set to run at 1,400 revolutions per minute; the countershaft has pulleys 30 and 8 inches
in diameter; the pulley on the grinder is 7 inches in diameter; what size pulley should be used on the line shaft, its
speed being 185 revolutions per minute? ## LENGTHS OF OPEN AND CROSSED BELTS
In Fig. 1, In the figure, the radii The circumference of the larger pulley is 2 Substituting these values of Since EXAMPLE. In Fig. 1, let the pulleys have diameters of 20 and 12 inches, respectively, and let the distance between the centers
of the shafts be 4 feet; calculate the length of the open belt required. SOLUTION. R = 10, r = 6, d = 10 - 6 = 4, s = 10 + 6 = 16, and h = 48 in.
The greater the difference in radii and the smaller the distance between the pulleys, the less exact is the formula. EXAMPLE. Solve the example of Art. 7 by the approximate method.
The angles The angle EXAMPLE. In the example of Art. 7, calculate the length of the belt if it is a crossed belt. SOLUTION. R - 10; r = 6; d = 10 - 6=4; s =10 + 6 = 16; and h = 48. By formula 4, ## EXAMPLES FOR PRACTICE1. Two pulleys have diameters of 36 inches and 20 inches, respectively, and the distance between their centers is
128 Inches; find the length of an open belt to connect them. Ans. 344.5 in. 2. What length of crossed belt is required to connect two pulleys of 42 inches and 18 inches diameter, respectively, if the distance
between shaft centers is 8 feet? Ans. 295.7 in. 3. Using the formula of Art. 8, find the length of an open belt to connect two pulleys 30 inches and 36 inches in diameter,
respectively, the distance between centers being 12 feet. Ans. 391.7 In. ## CONE PULLEYS
Since From formula 1, Art. 2, That is, N must equal the square root of the product of Having determined EXAMPLE. Two continuous speed cones are to be designed to give a range of speed between 100 and 700 revolutions per minute; they are to be alike in all respects. What must be the speed of the driving shaft and the large diameter of the cones, assuming the small diameter to be 4 inches? SOLUTION. From formula 1,
EXAMPLE. The steps of a cone pulley have diameters of 7, 8 3/8, 10, and 12 inches, respectively, and the diameter of the step on the other cone corresponding to the 10-inch step is 8 3/4 inches; calculate the diameters of the remaining steps for a crossed belt. SOLUTION. SUM of diameters = 10 + 8 3/4 = 18 3/4 In. Since this Is to be the same for all pairs, the diameters of the remaining steps are as follows:
Let N1, N2, N3, etc., to the right of 0, and from these draw the vertical lines N1, P1 N2 P2, etc., to cut the line X X in the points P1 P2, P3, etc., and the line A W in the points M1, M2, M3 etc. Now assume that P1 N1 represents the radius of a step to the same scale that A B represents the distance h; then 0 P1, or the equal length P1 M1, represents to the same scale the radius of the step that is the mate to the step P1 N1. Similarly, P2 N2 and 0 P2, P3 N3 and 0 P3, P4 N4 and 0 P4 are corresponding radii of steps.If The actual sizes of the cones depend on the location of the axis Usually the radii of one pair of steps are given or assumed, and in this case the axis
For an open belt, the other steps are found by the following method: In Fig. 10, make The lines
For a crossed belt; the problem is easily solved by a simple calculation. The sum of the radii of the equal middle steps is R3 + r3 = 8 + 8 = 16 inches, and this sum is constant. That is, R1 + r1 = 16. But R1 = 4 r1 for this pair of pulleys. Then 4 r1 + r, = 16, 5 r1 = 16, or r1 = 3.2 inches, and R1 = 16 - 3.2 = 12.8 inches. Also, R2 + r2 = 16. But R2 = 2 r2 for this pair. Hence, 2 r2 + r2 = 16, 3 r2 = 16, or r2 = 5.33 inches, and R2 = 16 - 5.33 = 10.67 inches. If the distance between, the axes of the pulleys to be connected by open belt is great, or if, as is sometimes the case, one of the axes is adjustable, the diameters can be calculated as though the belt were crossed. Otherwise, when designed for open belts, they should be laid out as described in Arts. 14 and 15. ## EXAMPLES FOR PRACTICE1. Two continuous speed cones are required to give a range of speed between 100 and 600 revolutions per minute; assuming the large diameters of the cones to be 14 inches, what must be: (a) the small diameters; and (b) the speed of the driving shaft? Both cones are to be alike. Ans. (a) 5.71 in. (b) 244.95 R. P. M. 2. In example 1, if the speed of the driving shaft were 260 revolutions per minute, and the slowest speed of the driven cone 140 revolutions: (a) what would be the greatest speed of the driven cone? (b) what would be the ratio of the large and small diameters of the cones? Ans. a) 482.86 R. P. M. (b) 1. 86 : 1 3.The radii of one of two speed cones are 3, 5 1/2, 8, and 10 inches and the radius of the step on the second cone corresponding to the 8-inch step on the first cone is 6 inches. Find the radii of the second cone, the distance between shafts being 30 inches and an open belt being used. Ans. 10 7/16, 8 1/16, 6, and 3 5/8 in. 4. The distance between the axes of two cones is 36 inches and an open belt is used. The two cones are to be alike, and are required to give velocity ratios as follows: The radii of the first pair are, respectively, 15 Inches and 5 inches. Find, by the graphic method, the radii of the other pairs.
5. With the data of example 4, calculate the radii for a pair of cones with a crossed belt. Ans. ## POWER TRANSMISSION BY BELT |