Machinery's Handbook, 31st Edition
2276 HELICAL GEARING 1. Shafts Parallel, Center Distance Approximate.— 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 a = approximate center distance 3) P n = normal diametral pitch
L.H.
4) N = number of teeth in large gear 5) n = number of teeth in small gear 6) α = angle of helix
R.H.
Driver
To find: 1) D = pitch diameter of large gear = cos P N n α 2) d = pitch diameter of small gear = cos P n n α 3) O = outside diameter of large gear = D + P 2 n 4) o = outside diameter of small gear = d + P 2 n 5) T = number of teeth marked on formed milling cutter (large gear) = N cos 3 a 6) t = number of teeth marked on formed milling cutter (small gear) = n cos 3 a 2 ( D + d ) Example: Given or assumed: 1) See illustration; 2) C a = 17 inches; 3) P n = 2; 4) N = 48; 5) n = 20; and 6) α = 20. To find: 1) D = . cos P N 2 09397 48 n # α = = 25.541 inches 2) d = . cos P n 2 09397 20 n # α = = 10.642 inches 3) O = . P 2 25541 7) L = lead of helix on large gear = π D cot α 8) l = lead of helix on small gear = π d cot α 9) C = center distance (if not right, vary α ) = 1 ∕ 2 2 = + = 26.541 inches
n
. P 2 10642 =
2 2
+ = 11.642 inches
4) o = d +
n
. 09397 48
N
= 57.8, say 58 teeth
5) T =
=
3
cos 3 a
h
^
. 09397 20
n
6) t = = 24.1, say 24 teeth 7) L = π D cot α = 3.1416 × 25.541 × 2.747 = 220.42 inches 8) l = π d cot α = 3.1416 × 10.642 × 2.747 = 91.84 inches 9) C = 1 ∕ 2 ( D + d ) = 1 ∕ 2 (25.541 + 10.642) = 18.091 inches 3 = ^ h cos 3 a
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