20.1 SECTION 20 GEAR DESIGN AND APPLICATION Analyzing Gears for Dynamic Loads 20.1 Helical-Gear Layout Analysis 20.12 Analyzing Shaft Speed in Epicyclic Gear Trains 20.14 Speeds of Gears and Gear Trains 20.17 Selection of Gear Size and Type 20.18 Gear Selection for Light Loads 20.21 Selection of Gear Dimensions 20.25 Horsepower Rating of Gears 20.26 Moment of Inertia of a Gear Drive 20.28 Bearing Loads in Geared Drives 20.29 Force Ratio of Geared Drives 20.30 Determination of Gear Bore Diameter 20.31 Transmission Gear Ratio for a Geared Drive 20.32 Epicyclic Gear Train Speeds 20.33 Planetary-Gear-System Speed Ratio 20.34 ANALYZING GEARS FOR DYNAMIC LOADS A two-stage, step-up gearbox drives a compressor and has a lubrication pump mounted on one of the gear shafts, Fig. 1. Tables 1, 2, and 3 show the spur-gear data, tolerances for tooth errors, and polar moments of inertia for the masses in the compressor drive. All gears in the drive have 20 Њ pressure angles. The gears are made of steel; the compressor is made of cast iron. A 50-hp (37.3-kW) motor drives the gearbox at 3550 rpm. What are the dynamic loads on this gearbox? Calculation Procedure: 1. Determine the pitchline velocity, V, applied load, W, at each mesh, and shaft speed, n In the equations that follow, subscripts identify the shaft, gear, or mesh under con- sideration. For example, the subscript A denotes Shafts A 1 and A 2 ; B, Shaft B; and C, Shaft C. Also, subscripts 1, 2, 3, 4 refer respectively to Gears 1, 2, 3, and 4 in the gearbox. The subscripts a and b denote the mesh of Gears 1 and 2 and the mesh of Gears 3 and 4, respectively. Finally, subscripts r and n refer to the driver gear and driven gear. Full nomenclature appears at the end of this procedure. For Shaft A 1 and A 2 , Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 20.2 DESIGN ENGINEERING SI Values 0.95 in (22.7 mm) 4.5 in (114.3 mm) 3.2 in (81.3 mm) 1.0 in (25.4 mm) 5.8 in (147.3 mm) 8 in (203.2 mm) 1.25 in (31.8 mm) 6.65 in (168.9 mm) 1.5 in (38.1 mm) 14 in (355.6 mm) 1.57 in (39.9 mm) 16 in (406.4 mm) 2 in (50.8 mm) 2.35 in (59.7 mm) 2.5 in (63.5 mm) FIGURE 1 Two-stage step-up gearbox driving a compressor (Machine Design.) TABLE 1 Spur Gear Data Number of teeth, N Pitch radius R (in) mm Face width F (in) mm Lewis form factor, y Gear 1 133 3.325 84.5 1.25 31.8 0.276 Gear 2 47 1.175 29.8 1.25 31.8 0.256 Gear 3 116 2.90 73.7 1.00 25.4 0.275 Gear 4 64 1.60 40.6 1.00 25.4 0.264 n ϭ input speed A ϭ 3550 r/min P ϭ input power A ϭ 50 hp (37.3 kW) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION GEAR DESIGN AND APPLICATION 20.3 TABLE 2 Tolerances for Tooth Errors Tolerance for tooth profile error e p in mm Tolerance for tooth spacing error e s in mm Gear 1 0.0010 0.0254 0.0007 0.0178 Gear 2 0.0009 0.0229 0.00065 0.0165 Gear 3 0.0010 0.0254 0.0007 0.0178 Gear 4 0.0009 0.0229 0.00065 0.0165 TABLE 3 Polar Moment of Inertia of Cylindrical Masses Element Diameter, D in mm Length, L in mm Polar moment of inertia, I o in 2 ⅐ lb ⅐ s 2 /ft cm 2 ⅐ kg ⅐ s 2 /m Motor 14 355.6 16 406.4 534.751 5138.78 Shaft A 1 0.95 24.1 2.5 63.5 0.002 0.0192 Shaft A 2 1.57 39.88 4.0 101.6 0.021 0.2018 Gear 1 6.65 168.9 1.25 31.75 2.127 20.439 I oA ϭ ͚ I o ϭ 536.901 5159.44 Gear 2 2.35 59.69 1.25 31.75 0.033 0.3171 Shaft B 2.0 50.8 12.543 318.59 0.175 1.6816 Gear 3 5.8 147.3 1.0 25.4 0.985 9.465 I oB ϭ ͚ I o ϭ 1.193 11.463 Gear 4 3.2 81.28 1.0 25.4 0.091 0.8744 Shaft C 1.5 38.1 4.5 114.3 0.020 0.1921 Compressor 8.0 203.2 1.5 38.1 4.915 47.231 I oC ϭ ͚ I o ϭ 5.026 48.297 Tabulated values do not include the polar moment of inertia for the lubrication pump. A material mass factor B ϭ 0.00087 lb ⅐ s 2 /in 3 ⅐ ft was assumed for steel; B ϭ 0.00080 lb ⅐ sec 2 /in 3 ⅐ ft (0.000073 kg ⅐ s 2 /cm 3 ⅐ m) cast iron. The polar moment of inertia for cylindrical masses was calcualted as, I o ϭ BD 4 L. T ϭ 63,025 P /n AAA ϭ 63,025 (50)/3550 ϭ 888 lb ⅐ in (100.3 N ⅐ m) W ϭ T / R aA1 ϭ 888/3.325 ϭ 267 lb (121.2 kg) V ϭ 0.5236 Rn a 1 A ϭ 0.5236 (3.325) (3550) ϭ 6180 ft/min (1883.7 m/min) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION 20.4 DESIGN ENGINEERING Therefore, for Shafts B and C, the following values for speed, torque, load and pitchline velocity are obtained. n ϭ n (N /N ) BA12 ϭ 3550 (133/47) ϭ 10,046 r/min T ϭ T (N /N ) BA21 ϭ 888 (47/133) ϭ 314 lb ⅐ in (35.5 N ⅐ m) n ϭ n (N /N ) CB34 ϭ 10,046 (116/64) ϭ 18,208 r/min T ϭ T (N /N ) CB43 ϭ 314 (64/116) ϭ 173 lb ⅐ in (19.5 N ⅐ m) W ϭ T /R bB3 ϭ 314/2.90 ϭ 108 lb (49 kg) V ϭ 0.5236 Rn b 3 B ϭ 0.5236 (2.9) (10,046) ϭ 15,254 ft/min (4649.4 m/min) From these values, dynamic loads can be calculated. Find the Total Effective Mass at Mesh a and Mesh b Total effective mass m t must be determined at each gear mesh. The parameter m t can be calculated from 1/m ϭ 1/m ϩ 1/m trn where m r is the effective mass at the driver gear and m n is the effective mass at the driven gear. For rotating components such as gears, an effective mass m at the pitch radius R can be calculated from Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION GEAR DESIGN AND APPLICATION 20.5 2 m ϭ ͚ I /R o The gears in the compressor drive mesh at two points, for which the total ef- fective mass m t must be determined. At Mesh a and Mesh b, m t is 1/m ϭ 1/m ϩ 1/m ta ra na 1/m ϭ 1/m ϩ 1/m tb rb nb However, a lubrication pump is directly coupled to Shaft A 2 . For Mesh a, the effective mass at the driver gear m ra is infinite because of the incompressible fluid in the pump that prevents instantaneous accelerations. Also at Mesh b, the effective mass at the driver gear m rb is infinite because the reflected inertia of the pump is also infinite. Consequently, the equations for m t become 1/m ϭ 1/ ϱϩ 1/m ta na m ϭ m ta na and 1/m ϭ 1/ ϱϩ 1/m tb nb m ϭ m tb nb Finally, the total effective mass m t at Mesh a and Mesh b can be calculated as 2 I ϩ I (R / R ) oB oC 34 m ϭ m ϭ ta na 2 R 2 2 1.193 ϩ 5.026(2.90/160) ϭ 2 1.175 22 ϭ 12.82 lb ⅐ s /ft (19.1 kg ⅐ s/m) 2 m ϭ m ϭ I / R tb nb oC 4 2 ϭ 5.026/1.6 22 ϭ 1.963 lb ⅐ s /ft (2.924 kg ⅐ s/m) 3. Compute the acceleration force for the spur gears For spur gears, the acceleration force ƒ 1 is 2 ƒ ϭ Hm V 1 t where H ϭ A 1 (1/R r ϩ 1/R n ). The calculation factor A 1 ϭ 0.00086 for 14.5 Њ teeth, 0.00120 for 20 Њ teeth, and 0.00153 for 25 Њ teeth. The parameters R r and R n are the pitch radii of driver and driven gears, respectively. Therefore at Mesh a, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION 20.6 DESIGN ENGINEERING H ϭ 0.00120(1R ϩ 1/R ) arn ϭ 0.00120(1/3.325 ϩ 1/1.175) ϭ 0.00138 V ϭ 6180 ft/min (1883.7 m/min) a 22 m ϭ 12.82 lb ⅐ s /ft (19.1 kg ⅐ s/m) ta 2 ƒ ϭ Hm V 1a a ta a 2 ϭ (0.00138)(12.82)(6180) ϭ 675,685 lb (306,761 kg) And at Mesh b, H ϭ 0.00120(1/R ϩ 1/R ) brn ϭ 0.00120(1/2.90 ϩ 1/1.60) ϭ 0.00116 V ϭ 15,254 ft ⅐ min (4649.4 m/min) b 22 m ϭ 1.963 lb ⅐ s /ft (2.924 kg ⅐ s/m) tb 2 ƒ ϭ Hm V 1b b tb b 2 ϭ (0.00116)(1.963)(15,254) ϭ 529,841 lb (240,548 kg) 4. Calculate the deflection force The deflection force ƒ 2 is 1/ƒ ϭ 1/C ϩ (1/C ϩ 1/C ϩ ⅐⅐⅐ 1/C ) 212x where C accounts for deflection from bending and compressive loads in the gear teeth, and C x accounts for deflection from torsional loads in shafts, flexible cou- plings, and other components. C is the load required to deflect gear teeth by an amount equal to the error in action. C x is the load required at the pitch radius to deflect a shaft or coupling by an amount equal to the error in action. Previously, calculations of ƒ 2 only considered the parameter C. With the addition of C x to the equation for ƒ 2 , the value of ƒ 2 will always be less than the smallest value of C, C 1 , C 2 .C x . The parameter C is C ϭ W ϩ 1000 eFA where A is the load required to deflect teeth by 0.001 in (0.0254 mm). If the Lewis form factor y is known, A can be calculated from Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION GEAR DESIGN AND APPLICATION 20.7 EZEZ rrnn A ϭ 1000(EZ ϩ EZ) rr nn where for the driver and driven gears, E is the modulus of elasticity, and Z is the calculation factor for A. Thus calculation factor Z is Z ϭ y/(0.242 ϩ 7.25y) The table below summarizes the results of calculations for A. For both Mesh a and Mesh b, A ϭ 1837 lb (834 kg). Summary of Calculations for Deflection Load Lewis form Calculation Deflection factor, y factor, Z load, A (lb) Gear 1 0.276 0.123 At Mesh a, Gear 2 0.256 0.122 A ϭ 1837 (834 kg) Gear 3 0.275 0.123 At Mesh b, Gear 4 0.264 0.122 A ϭ 1837 (834 kg) Previously, the error in action e had to be assumed for a given class of gears, based on recommendations tabulated in handbooks. However, this error can be approximated as 22 e ϭ ͙ (e ϩ e ) ϩ (e ϩ e ) pr sr pn sn where for the driver and driven gears, e p is the tooth profile error and e s is the tooth spacing error. If actual measurements of e p and e s are not available, then the tol- erances for allowable e p and e s can be used for calculations instead. Therefore, for both Mesh a and Mesh b, the approximate error in action e is 22 e ϭ ͙ (0.0010 ϩ 0.0007) ϩ (0.0009 ϩ 0.00065) ϭ ͙ 0.0000053 ϭ 0.0023 in (0.058 mm) Finally, C can be calculated for both meshes. At Mesh a, C ϭ W ϩ 1000eFA aa aaa ϭ 267 ϩ 1000(0.0023)(1.25)(1837) ϭ 5548 lb (2519 kg) and at Mesh b, C ϭ W ϩ 1000eFA bb bbb ϭ 108 ϩ 1000(0.0023)(1.00)(1837) ϭ 4333 lb (1967 kg) For steel shafts, the parameter C x is calculated from Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION 20.8 DESIGN ENGINEERING 42 C ϭ 1,080,000eD /RL xss where D s is the shaft diameter, and L s is the shaft length. Generally, for other components such as flexible couplings, C x has to be determined experimentally or from information supplied by the component manufacturer. In the gear box, the portion of Shaft A 2 between the motor and Gear 1 deflects interdependently with Shaft A 1 . On the other hand, the portion of Shaft A 2 between the lubrication pump and Gear 1 deflects independently of the shaft elements to the right of Gear 1. The table below summarizes the calculations of C x for all shaft elements affecting each mesh. Summary of Calculations for Shaft Deflection Load Shaft elements for Mesh a Shaft diameter, D s (in) mm Shaft length, L s (in) mm Pitch radius, R (in) mm Shaft deflection load, C x (lb) kg Shaft A 1 0.95 24.1 2.5 63.5 3.325 84.5 73 33.1 Shaft A 2 1.57 39.9 2.5 63.5 3.325 84.5 546 247.9 Shaft A 2 1.57 39.9 1.5 38.1 3.325 84.5 910 413.1 Shaft B 2.00 50.8 8.543 216.9 1.175 29.8 3370 1529.9 Shaft elements for Mesh b Shaft B 2.00 50.8 8.543 216.9 2.90 73.7 553 251 Shaft C 1.5 38.1 4.5 114.3 1.60 40.6 1091 495.3 When shafts have steps of different diameter, the loads required to deflect each step torsionally by the error in action must be determined. Then the loads for each step must be combined as an inverse of the sum of the reciprocals of the loads. For example, Shaft A 2 has a different diameter than Shaft A 1 has. Consequently, the effective value for C x between the motor and pump is C ϭ C ϭ C ϩ 1/(1C ϩ 1/C ) x 43 2 1 ϭ 910 ϩ 1/(1/546 ϩ 1/73) ϭ 974 lb (442.2 kg) Note that values of C x for each shaft element in the table are differentiated from each other by the use of a subscript x, enumerated as x ϭ 1, 2, 3, 4 . . . This subscript should not be confused with numerical subscripts used to denote Gears 1, 2, 3, 4. For Mesh a,ƒ 2 is calculated as 1/ƒ ϭ 1/C ϩ 1/C ϩ 1/C 2aa45 ϭ 1/5548 ϩ 1/974 ϩ 1/3370 F ϭ 665 lb (301.9 kg) 2a and for Mesh b,ƒ 2 is Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION GEAR DESIGN AND APPLICATION 20.9 1/ƒ ϭ 1/C ϩ 1/C ϩ 1/C 2bb67 ϭ 1/4333 ϩ 1/553 ϩ 1/1091 ƒ ϭ 338 lb (153.5 kg) 2b 5. Find the resultant force and dynamic load at each mesh At Mesh a, 1/ƒ ϭ 1/ƒ ϩ 1/ƒ aa 1a 2a ϭ 1/675,685 ϩ 1/665 ƒ ϭ 664 lb aa W ϭ W ϩ ͙ ƒ (2ƒ Ϫ ƒ ) da a aa 2aaa ϭ 267 ϩ ͙ 664[2(665) Ϫ 664] ϭ 932 lb (423.1 kg) and at Mesh b, 1/ƒ ϭ 1/ƒ ϩ 1/ƒ ab 1b 2b ϭ 1/529,841 ϩ 1/338 ƒ ϭ 338 lb ab W ϭ W ϩ ͙ ƒ (2ƒ Ϫ ƒ ) db b ab 2bab ϭ 108 ϩ ͙ 338[2(338) Ϫ 338] ϭ 446 lb (202.5 kg) At Mesh a and Mesh b, W d /W ϭ 3.49 and 4.13 respectively. Because the W d /W ratio for both meshes is greater than two, there will be free impact between the gear teeth. Also, the drive will be noisy and may wear rapidly. The dynamic load W d can de decreased by placing a flexible coupling between the motor and Shaft A 1 . This will effectively isolate the motor mass. Also, the lubrication pump can be isolated by driving it with a quill shaft. Finally, the diameter of Shaft B can be minimized to meet the required torque capacity while increasing shaft resilience to lower the magnitude of W d . Related Calculations. Gears in mesh never operate under a smooth, continuous load. Factors such as manufacturing errors in tooth profile and spacing, tooth de- flections under load, and imbalance all interact to create a dynamic load on gear teeth. The resulting action is similar to that of a variable load superimposed on a steady load. Consider, for example, how tooth loads can fluctuate from manufacturing errors. The maximum, instantaneous tooth load occurs at the maximum error in action. The average load is the applied load on the teeth at the pitchline of the gears. As each pair of teeth moves through its duration of contact, errors in action create periods of sudden acceleration that momentarily separate mating gear teeth. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION 20.10 DESIGN ENGINEERING Impact occurs as each pair of teeth returns to mesh to complete its contact duration. The impact loads can be significantly greater than the applied load at the pitchline. The maximum magnitude of the dynamic load depends on gear and pinion masses, connected component masses, operating speed, and material elasticity. Elastic de- flections in gear teeth and drive components, such as shafts and flexible couplings, help reduce dynamic loads. Generally, gears should be designed so that their bending and wear capacities are equal to or greater than the maximum, instantaneous dynamic load. However, the exact magnitude of the maximum dynamic load is seldom known. Although many gear studies have been conducted, there is no full agreement on the single best method for determining dynamic loads. One of the most widely accepted methods for calculating dynamic loads con- siders the maximum dynamic load W d , resulting from elastic impact, to be W ϭ W ϩ W di The term W is the applied load at the pitchline of a gear. For gears, the incremental load W i is W ϭ ͙ ƒ (2ƒ Ϫ ƒ ) ia2 a In the above equation, ƒ a is the resultant force required to accelerate masses in a system of elastic bodies. The resultant acceleration force ƒ a is 1/ƒ ϭ 1/ƒ ϩ 1/ƒ a 12 The term ƒ 1 is the force required to accelerate masses in a system of rigid bodies. The term ƒ 2 is the force required to deflect the system elastically by the amount of error in action. As mentioned previously, free impact loads occur in gears when the teeth separate and return to mesh suddenly during the contact interval. For free impact, the value of the ratio W d /W will always be greater than two. If forces ƒ 1 , ƒ 2 , and ƒ a are plotted as functions of pitchline velocity, force ƒ 2 is seen to be an asymptote of the incremental load W i . The equations that define ƒ 1 and ƒ 2 depend on the type of gears being analyzed for dynamic loads. The example presented in this procedure provides equations defining ƒ 1 and ƒ 2 for spur gear applications. Sometimes this method for calculating W d gives values for dynamic load that are conservative. These high dynamic load estimates can lead to overdesigned gears. Conservative calculations of W d can be minimized if ƒ 2 can be calcualted more accurately. A less conservative calculation of ƒ 2 effectively reduces W d because ƒ 2 is an asymptote for the incremental load W i . The method presented here has been refined to give more accurate values for ƒ 2 . Previously, only the elastic deflection in the gear teeth was considered in the calculation of ƒ 2 . Now, elastic deflections in other mechanical components, such as flexible couplings and shafts, are also considered for their effects on ƒ 2 through inclusion of a parameter C x . Also, the calculation of ƒ 2 is further refined by a more accurate approximation of the error in action. Previously, the error in action was assumed for a given class of gears, based on recommendations commonly tabulated in gear design handbooks. Now the error in action can be calculated from tooth profile errors and tooth-to- tooth spacing errors. The equation for approximating error in action results from tests where gears were measured for error in action, profile error, spacing error, and runout error. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. GEAR DESIGN AND APPLICATION [...]... shaft What is the moment of inertia of the high-speed and low-speed assemblies of this gearset? Calculation Procedure: 1 Compute the moment of inertia of each gear The moment of inertia of a cylindrical body about its longitudinal axis is Ii ϭ WR2, where Ii ϭ moment of inertia of a cylindrical body, in4 / in of length; W ϭ weight of cylindrical material, lb / in3; R ϭ radius of cylinder to its outside... One of the gears is driven by another gear; the second gear of the compound set drives another gear In a compound gear train, the product of the number of teeth of the driving gears and the rpm of the first driver equals the product of the number of teeth of the driven gears and the rpm of the last driven gear In this gearset, the first driver has 24 teeth and the second driver has 72 teeth The rpm of. .. ENGINEERING the idler gear An idler gear is generally used to reduce the required diameter of the driving and driven gears on two widely separated shafts 3 Determine the effect of two idler gears The effect of more than one idler is the same as that of a single idler—i.e., the speed of the driving and driven gears remains the same, regardless of the number of idlers used 4 Determine the direction of. .. (7104.7 cm4) The total moment of inertia of the gear shaft equals the sum of the individual moments, or It ϭ 5.33 ϩ 170.69 ϭ 176.02 in4 (7326.5 cm4) 3 Compute the high-speed-assembly moment of inertia The effective moment of inertia at the high-speed assembly input ϭ Ithi ϭ Ith ϩ Itl / (Rh / Rl)2, where Ith ϭ moment of inertia of high-speed assembly, in4; Itl ϭ moment of inertia of low-speed assembly, in4;... force Related Calculations Use these procedures to compute the bearing loads in any type of geared drive—open, closed, or semiclosed—serving any type of load Computation of the bearing load is necessary step in bearing selection FORCE RATIO OF GEARED DRIVES A geared hoist will lift a maximum load of 1000 lb (4448.2 N) The hoist is estimated to have friction and mechanical losses of 5 percent of the maximum... Calculation Procedure: 1 Choose the type of gear to use Use Table 4 of the previous calculation procedure as a general guide to the type of gear to use Make a tentative choice of the gear type 2 Select the pitch diameter of the pinion and gear Compute the pitch diameter of the pinion from dp ϭ 2c / (R ϩ 2), where dp ϭ pitch diameter, in, of the pinion, which is the smaller of the two gears in mesh; c ϭ center... Procedure: 1 Compute the number of teeth on the gear For any gearset, RD / Rd, ϭ Nd / ND, where RD ϭ rpm of driver; Rd ϭ rpm of driven gear; Nd ϭ number of teeth on the driven gear; ND ϭ number of teeth on driving gear Thus, 900 / 300 ϭ Nd / 20; Nd ϭ 60 teeth 2 Compute the diametral pitch of the gear The diametral pitch of the gear must be the same as the diametral pitch of the pinion if the gears are... number of teeth on pinion gear; Ng ϭ number of teeth on gear Thus, dc ϭ (20 ϩ 60) / [2(6)] ϭ 6.66 in (16.9 cm) 5 Compute the dimensions of the gear teeth Use AGMA Standards, Dudley—Gear Handbook, or the engineering tables published by gear manufacturers Each of these sources provides a list of factors by which either the circular or diametral pitch can be multiplied to obtain the various dimensions of. .. diameteral pitch number of teeth in gear or worm wheel number of teeth in pinion or threads in worm ratio, n / N helix angle of gear or worm wheel, deg Find the value of the lefthand portion of this equation for this gear layout from: 2(4)(20) / 64 ϭ 2.5 Then, R ϭ 32 / 64 ϭ 0.5 2 Find the suitable helix angle or angles Knowing the value of the lefthand side of the equation and the value of R we can assume... Determine the effect of one idler gear An idler gear has no effect on the speed of the driving or driven gear Thus, the speed of each gear would remain the same, regardless of the number of teeth in Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at . is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 20.2 DESIGN ENGINEERING SI Values 0.95. number of teeth of the driving gears and the rpm of the first driver equals the product of the number of teeth of the driven gears and the rpm of the last driven