Machinery's Handbook, 31st Edition
2280
HELICAL GEARING
8) o = outside diameter of small gear = d + P 2 n 9) N ′ and n ′ = numbers of teeth marked on cutters for large and small gears (see page 2283) 10) L = lead of helix on large gear = π D cot α 11) l = lead of helix on small gear = π d cot β Example: Given or assumed: 1) See illustration; 2) P n = 8; 3) R = 3; 4) α a = 45 degrees; and 5) C = 10 in. To find: 1) n = . tan sin R CP 1 2 1 3 2 10 8 0 70711 a n a # # # α α + = + = 28.25, say 28 teeth
2) N = Rn = 3 × 28 = 84 teeth 3) R sec α + cosec α = n CP 2
28 2 10 8
n # # = = 5.714, or α = 46 ° 6 ′
4) β = 90 ° − α = 90 ° − 46 ° 6 ′ = 43 ° 54 ′ 5) D = . cos P N 8 06934 84 n # α =
= 15.143 inches
. 8 072055 28 #
cos P n n
6) d =
= 4.857 inches
=
β
7) O = D + P 2 n
= 15.143 + 0.25 = 15.393 inches
8) o = d + P 2 n
= 4.857 + 0.25 = 5.107 inches
9) N ′ = 275; n ′ = 94 (see page 2283) 10) L = π D cot α = 3.1416 × 15.143 × 0.96232 = 45.78 inches 11) l = π d cot β = 3.1416 × 4.857 × 1.0392 = 15.857 inches
4B. Shafts at Right Angles, Any Ratio, Helix Angle for Minimum Center Distance.— Diagram similar to 4A. Gears have same direction of helix. The sum of the helix angles will equal 90 degrees. For any given ratio of gearing R there is a helix angle α for the larger gear and a helix angle β = 90 ° − α for the smaller gear that will make the center distance C a minimum. Helix angle α is found from the formula cot α = R 1 ∕ 3 . As an example, using the data found in Case 4A, helix angles α and β for minimum center distance would be: cot α = R 1 ∕ 3 = 1.4422; α = 34 ° 44 ′ and β = 90 ° − 34 ° 44 ′ = 55 ° 16 ′ . Using these helix angles, D = 12.777; d = 6.143; and C = 9.460 from the formulas for D and d given under Case 4A.
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