Machinery's Handbook, 31st Edition
HELICAL GEARING
2277
2. Shafts Parallel, Center Distance Exact.— Given or assumed:
Driven
1) Position of gear having right- or left-hand helix, depending upon rota tion and direction in which thrust is to be received 2) C = exact center distance 3) P n = normal diametral pitch (pitch of cutter)
L.H.
4) N = number of teeth in large gear 5) n = number of teeth in small gear To find: 1) cos α = P C N n 2 n +
R.H.
Driver
cos P N n
2) D = pitch diameter of large gear =
α 3) d = pitch diameter of small gear = cos P n n α 4) O = outside diameter of large gear = D + P 2 n 5) o = outside diameter of small gear = d + P 2 n 6) T = number of teeth marked on formed milling cutter (large gear) = N cos 3 a 7) t = number of teeth marked on formed milling cutter (small gear) = n cos 3 a 8) L = lead of helix (large gear) = π D cot α 9) l = lead of helix (small gear) = π d cot α Example: Given or assumed: 1) See illustration; 2) C = 18.75 inches; 3) P n = 4; 4) N = 96; and 5) n = 48. 1) cos α = . P C N n 2 2 4 1875 96 48 n # # + = + = 0.96, or α = 16 ° 16 ′ 2) D = . cos P N 4 096 96 n # α = = 25 inches 3) d = . cos P n 4 096 48 n # α = = 12.5 inches 4) O = D + P 2 25 4 2
n = + = 25.5 inches n = + = 13 inches . P 2 125 4 2
5) o = d +
. 096 96 ^ h
cos N
= 108 teeth
6) T =
a =
3
3
. 096 48 ^ h
cos n
7) t = = 54 teeth 8) L = π D cot α = 3.1416 × 25 × 3.427 = 269.15 inches 9) l = π d cot α = 3.1416 × 12.5 × 3.427 = 134.57 inches 3 3 α =
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