Machinery's Handbook, 31st Edition
HELICAL GEARING
2279
To find: 1) D = . 2) d = . 3) O = D +
. N 0 70711 0 70711 10 27 n # = n 0 70711 0 70711 10 18 n # = P . P
= 3.818 inches
= 2.545 inches
. P 2 3818
10 2
= 4.018 inches
= +
n
. P 2 2545
10 2
4) o = d +
= 2.745 inches
= +
n
. N 0353 0353 = 27 = 76.5, say 76 teeth
5) T = . 6) t = .
. n 0353 0353 = 18 = 51 teeth
7) L = π D = 3.1416 × 3.818 = 12 inches 8) l = π d = 3.1416 × 2.545 = 8 inches 9) C = . . D d 2 2
+ = 3818+2545 = 3.182 inches 4A. Shafts at Right Angles, Center Distance Exact.— Gears have same direction of helix. Sum of the helix angles will equal 90 degrees. Given or assumed: Driven
1) Position of gear having right- or left-hand helix depending on rotation and direction in which thrust is to be received 2) P n = normal diametral pitch (pitch of cutter) 3) R = ratio of number of teeth in large gear to number of teeth in small gear 4) α a = approximate helix angle of large gear 5) C = exact center distance
R.H.
R.H.
To find: 1) n = number of teeth in small gear nearest = 2 CP n sin α a ÷ 1 + R tan α a 2) N = number of teeth in large gear = Rn 3) α = exact helix angle of large gear, found by trial from R sec α + cosec α = 2 CP n ÷ n
4) β = exact helix angle of small gear = 90 ° − α 5) D = pitch diameter of large gear = cos P N n α 6) d = pitch diameter of small gear = cos P n n β 7) O = outside diameter of large gear = D + P 2 n
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