Machinery's Handbook, 31st Edition
2286 HELICAL GEARING To eliminate the need for making the calculations indicated in Formulas (3) and (4), the accompanying graph may be used to obtain the value of K h directly for full-depth pinions of 20-degree normal pressure angle. (1) (2) (3) (4) (5) Example: Find the outside diameter of a helical pinion having 12 teeth, 32 normal diametral pitch, 20-degree pressure angle, and 18-degree helix angle. P t = P n cos ψ = 32 cos 18 ° = 32 × 0.95106 = 30.4339 d = n ÷ P t = 12 ÷ 30.4339 = 0.3943 inch K h = 0.851 (from graph) d o = 0.3943 + . 32 2+ 0851 = 0.4834 Center Distance at which Modified Mating Helical Gears Mesh Without Backlash.— If the helical pinion in the previous example on page 2286 had been made to standard dimensions, that is, not enlarged, and was in tight mesh with a standard 24-tooth mat ing gear, the center distance for tight mesh could be calculated from the formula on page 2214 : (1) However, if the pinion is enlarged as in the example and meshed with the same standard 24-tooth gear, then the center distance for tight mesh will be increased. To calculate the new center distance, the following formulas and calculations are required: First, calculate the transverse pressure angle φ t using Formula (2): (2) and from a calculator the angle φ t is found to be 20 ° 56 ′ 30 ″ . In the table on page 112 , inv φ t is found to be 0.017196, and the cosine from a calculator as 0.93394. Using Formula (3), calculate the pressure angle φ at which the gears are in tight mesh: (3) In this formula, the value for t nP for 1 diametral pitch is that found in Table 9c on page 2231 , for a 12-tooth pinion, in the fourth column: 1.94703. The value of t nG for 1 diametral pitch for a standard gear is always 1.5708. . . . . 0017196 12 24 1 94703 1 5708 0027647 inv φ π = + + + − = ^ h From the table on page 113 , or a calculator, 0.027647 is the involute of 24 ° 22 ′ 7 ″ and the cosine corresponding to this angle is 0.91091. Finally, using Formula (4), the center distance for tight mesh, C ′ is found: (4) . tan 20 cos 18 0 38270 tan tan cos n ' φ φ = t ' ψ = = ° ° n N + + − ^ h t t inv inv t nP nG φ φ π = + . 091091 0 5914 0 93394 0 606 inch . . . cos cos C C t # z z = = = l P t = P n cos ψ d = n ÷ P t tan ψ t = tan φ n ÷ cos ψ K h = 2.1 − (sin φ t − cos φ t tan 5 ° ) sin φ t cos n ψ d o = d + P K 2 n h + . 05914inch cos ψ = + = n N 2 cos C P 2 32 18 12 24 + n # # = °
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