PULLEYS. Proportions of Pulleys. (See also Fly-wheels, pages 820 to 823.)Let n = number of arms, D = diameter of pulley, S = thickness of belt, t = thickness of rim at edge, T thickness in middle, B = width of rim, ß width of belt, h = breadth of arm at hub, h1 = breadth of arm at rim, e = thickness of arm at hub e1 = thickness of arm at rim, c = amount of crowning; dimensions in inches. Reuleaux. 9/88 to 5/48 (thick, of rim.) 1/5h to 1/4h BD 4 14" B D 201 n e thickness of arm at hub. (not less than 2.5S, B for sin.-arm pulleys. M= thickness of metal in hub....... c = crowning of pulley........ B. 1/24B .... The number of arms is really arbitrary, and may be altered if necessary. (Unwin.) Pulleys with two or three sets of arms may be considered as two or three separate pulleys combined in one, except that the proportions of the arms should be 0,8 or 0.7 time that of single-arm pulleys. (Reuleaux.) EXAMPLE.-Dimensions of a pulley 60" diam., 16" face, for double belt 1⁄2" thick. Solution by.... n h hi e e1 t T L M с The following proportions are given in an article in the Amer. Machinist, authority not stated: h = .0825D+.5 in., h1 = .04D +3125 in., e = .025D+.2 in., e1 = .016D+ .125 in. These give for the above example: h = 4.25 in., h1 = 2.71 in., e = 1.7 in. e1 = 1.09 in. The section of the arms in all cases is taken as elliptical. The following solution for breadth of arm is proposed by the author: Assume a belt pull of 45 lbs. per inch of width of a single belt, that the whole strain is taken in equal proportions on one half of the arms, and that the arm is a beam loaded at one end and fixed at the other. We have the Rbd2 formula for a beam of elliptical section ƒP = .0982 in which P = the load, R= the modulus of rupture of the cast iron, b = breadth, d depth, and = length of the beam, and f = factor of safety. Assume a modulus of rupture of 36,000 lbs., a factor of safety of 10, and an additional allowance for safety in taking l = 1⁄2 the diameter of the pulley instead of 1⁄2D less the radius of the hub. Take dh, the breadth of the arm at the hub, and be = B 3535 X 0.4h3 1⁄2D reached by Unwin from a different set of assumptions. Convexity of Pulleys.-Authorities differ. Morin gives a rise equal to 1/10 of the face; Molesworth, 1/24; others from 1 to 1/96. Scott A. Smith says the crown should not be over 1 inch for a 24-inch face. Pulleys for shifting belts should be "straight," that is, without crowning. CONE OR STEP PULLEYS. = To find the diameters for the several steps of a pair of cone-pulleys: 1. Crossed Belts.-Let D and d be the diameters of two pulleys con nected by a crossed belt, L = the distance between their centres, and ẞ the angle either half of the belt makes with a line joining the centres of the πρ pulleys then total length of belt (D+ d) + (D+ d); +2L cos B. 180 angle whose sine is D+d. Cos B =, B = 2L = π L2 (+). The length of 2 the belt is constant when D + d is constant; that is, in a pair of steppulleys the belt tension will be uniform when the sum of the diameters of each opposite pair of steps is constant. Crossed belts are seldom used for cone-pulleys, on account of the friction between the rubbing parts of the belt. To design a pair of tapering speed-cones, so that the belt may fit equally tight in all positions: When the belt is crossed, use a pair of equal and similar cones tapering opposite ways. 2. Open Belts.-When the belt is uncrossed, use a pair of equal and similar conoids tapering opposite ways, and bulging in the middle, according to the following formula: Let L denote the distance between the axes of the conoids; R the radius of the larger end of each; r the radius of the smaller end; then the radius in the middle, ro, is found as follows: If D. the diameter of equal steps of a pair of cone-pulleys, D and d = the diameters of unequal opposite steps, and L distance between the D+d (D-d)2 axes, D1 = + 2 12.566L If a series of differences of radii of the steps, R - r, be assumed, then for each pair of steps R+r = 10 (R − 1)2 and the radii of each may be computed from their half sum and half difference, as follows: A. J. Frith (Trans. A. S. M. E., x. 298) shows the following application of Rankine's method: If we had a set of cones to design, the extreme diameters of which, including thickness of belt, were 40" and 10", and the ratio desired 4, 3, 2, and 1, we would make & table as follows, L being 100": The above formulæ are approximate, and they do not give satisfactory results when the difference of diameters of opposite steps is large and when the axes of the pulleys are near together, giving a large belt-angle. The following more accurate solution of the problem is given by C. A. Smith (Trans. A. S. M. E., x. 269) (Fig 152): Lay off the centre distance C or EF, and draw the circles D, and d, equal to the first pair of pulleys, which are always previously determined by known conditions. Draw HI tangent to the circles D, and d,. From B midway between E and F, erect the perpendicular BG, making the length BG .314C. With G as a centre, draw a circle tangent to HI. Generally this circle will be outside of the belt-line, as in the cut, but when C is short and the first pulleys D, and d, are large, it will fall on the inside of the beltline. The belt-line of any other pair of pulleys must be tangent to the circle G; hence any line, as JK or LM, drawn tangent to the circle G, will give the diameters Da, da or D3, ds of the pulleys drawn tangent to these lines from the centres E and F. The above method is to be used when the belt-angle A does not exceed 18°. When it is between 18° and 30° a slight modification is made. In that case, in addition to the point G, locate another point m on the line BG .298 C above B. Draw a tangent line to the circle G, making an angle of 18° to the line of centres EF, and from the point m draw an arc tangent to this tangent line. All belt-lines with angles greater than 1° are tangent to this arc. The following is the summary of Mr. Smith's mathematical method: A = angle in degrees between the centre line and the belt of any pair of pulleys; a = .314 for belt-angles less than 18°, and .298 for angles between 18° B° C and 30°; an angle depending on the velocity ratio; the centre distance of the two pulleys; D, d = diameters of the larger and smaller of the pair of pulleys; E an angle depending on B"; L= the length of the belt when drawn tight around the pulleys; r = D÷d, or the velocity ratio (larger divided by smaller). (7) L = 2C cos A+.01745d[180+(1)(90+ 4)]. Equation (1) is used only once for any pair of cones to obtain the constant cos A, by the aid of tables of sines and cosines, for use in equation (3). BELTING. Theory of Belts and Bands.-A pulley is driven by a belt by means of the friction between the surfaces in contact. Let 7, be the tension on the driving side of the belt, T, the tension on the loose side; then S, = T1 T2, is the total friction between the band and the pulley, which is equal to the tractive or driving force. Let f = the coefficient of friction, the ratio of the length of the arc of contact to the length of the radius, a = the angle of the arc of contact in degrees, e = the base of the Naperian logarithms = 2.71828, m = the modulus of the common logarithms = 0.434295. The following formulæ are derived by calculus (Rankine's Mach'y & Millwork, p. 351; Carpenter's Exper. Eng'g, p. 173): T1 − T1 = T1(1 − e −ƒ0) = T1(1 − 10−50m) = T1(1 — 10−.00758fa); If the arc of contact between the band and the pulley expressed in turns and fractions of a turn = n, 0 = 2n; e = 102.7288; that is, efe is the natural number corresponding to the common logarithm 2.7288fn. The The value of the coefficient of friction f depends on the state and material of the rubbing surfaces. For leather belts on iron pulleys, Morin found f = .56 when dry, .36 when wet, .23 when greasy, and .15 when oily. In calculating the proper mean tension for a belt, the smallest value, f= .15, is to be taken if there is a probability of the belt becoming wet with oil. experiments of Henry R. Towne and Robert Briggs, however (Jour. Frank. Inst., 1868), show that such a state of lubrication is not of ordinary occurrence; and that in designing machinery we may in most cases safely take f 0.42. Reuleaux takes f = 0.25. The following table shows the values of the coefficient 2.7288f, by which n is multiplied in the last equation, corresponding to different values of f; also the corresponding values of various ratios among the forces, when the arc of contact is half a circumference: In ordinary practice it is usual to assume T2 = S; T1 = 28; T1 + T2+ 2S 1.5. This corresponds to f = 0.22 nearly. For a wire rope on cast iron f may be taken as 0.15 nearly; and if the groove of the pulley is bottomed with gutta-percha, 0.25. (Rankine ) Centrifugal Tension of Belts.-When a belt or band runs at a high velocity, centrifugal force produces a tension in addition to that existing when the belt is at rest or moving at a low velocity. This centrifugal tension diminishes the effective driving force. Rankine says: If an endless band, of any figure whatsoever, runs at a given speed, the centrifugal force produces a uniform tension at each crosssection of the band, equal to the weight of a piece of the band whose length is twice the height from which a heavy body must fall, in order to acquire the velocity of the band. (See Cooper on Belting, p. 101.) g acceleration due to gravity 32.2; W = weight of a piece of the belt 1 ft. long and 1 sq. in. sectional area,— Leather weighing 56 lbs. per cubic foot gives W = 56 +144 = .888. Belting Practice. Handy Formulæ for Belting. Since in the practical application of the above formulæ the value of the coefficient of friction must be assumed, its actual value varying within wide limits (15% to 135%), and since the values of T, and T, also are fixed arbitrarily, it is customary in practice to substitute for these theoretical formulæ more simple empirical formulæ and rules, some of which are given below. Let d diam. of pulley in inches; d= circumference; V = velocity of belt in ft. per second; v = vel. in ft. per minute; a = angle of the arc of contact; L= length of arc of contact in feet da + (12 × 360); F= tractive force per square inch of sectional area of belt; w = width in inches; t = thickness; S tractive force per inch of width = F÷t; rpm. = revs. per minute; rps. = revs. per second = rpm. + 60. If F= working tension per square inch = 275 lbs., and t = 7/32 inch, S= 60 lbs. nearly, then If F 180 lbs. per square inch, and t = 1/6 inch, S= vw 1100 H.P. = = .055 Vw = .000238wd X rpm. = (2) If the working strain is 60 lbs. per inch of width, a belt 1 inch wide travelling 550 ft. per minute will transmit 1 horse-power. If the working strain is 30 lbs. per inch of width, a belt 1 inch wide, travelling 1100 ft. per minute, will transmit 1 horse-power. Numerous rules are given by different writers on belting which vary between these extremes. A rule commonly used is : 1 inch wide travelling 1000 ft. per min. I.H.P. This corresponds to a working strain of 33 lbs. per inch of width. Many writers give as safe practice for single belts in good condition a working tension of 45 lbs. per inch of width. This gives H.P.= =.0818w= .000357wd X rpm. = wv wd X rpm. (4) 2800 For double belts of average thickness, some writers say that the trans mitting efficiency is to that of single belts as 10 to 7, which would give H.P. of double belts = = .1169 Vw = .00051wd rpm. = wv Other authorities, however, make the transmitting-power of double belts twice that of single belts, on the assumption that the thickness of a doublebelt is twice that of a single belt. Rules for horse-power of belts are sometimes based on the number of square feet of surface of the belt which pass over the pulley in a minute. Sq. ft. per min. = wv ÷ 12. The above formulæ translated into this form give: |