Machinery's Handbook, 31st Edition
2282
HELICAL GEARING
To find: 1) D =
. 8 086603 132 # . 8 086603 33 #
cos P N n
= 19.052 inches
=
α
cos P n n
2) d =
= 4.763 inches
=
β
. P 2 19052 =
8 2
3) O = D +
+ = 19.302 inches
n
. P 2 4763
8 2
4) o = d +
n = + = 5.013 inches
. 065 132
N
= 203 teeth
5) T =
3 α =
cos
. 065 33
cos n
6) t = = 51 teeth 7) L = π D cot α = π × 19.052 × 1.732 = 103.66 inches 8) l = π d cot β = π × 4.763 × 1.732 = 25.92 inches 9) C = . . D d 2 2 + = 19052+4763 = 11.9075 inches 3 β =
6. Shafts at Any Angle, Center Distance Exact.— The sum of the helix angles of the two gears equals the shaft angle, and the gears are of the same hand, if each angle is less than the shaft angle. The difference between the helix angles equals the shaft angle, and the gears are of opposite hand, if either angle is greater than the shaft angle. Given or assumed: 1) Hand of helix, depending on rotation and direction in which thrust is to be received 2) C = center distance 3) P n = normal diametral pitch (pitch of cutter) Driven L.H.
4) α a = approximate helix angle of gear 5) β a = approximate helix angle of pinion 6) R = ratio of gear to pinion size = n N 7) n = number of pinion teeth nearest cos R CP 2 n
cos cos cos a a β
α α β +
a
a
8) N = number of gear teeth = Rn To find: 1) α and β , exact helix angles, found by trial from R sec α + sec β = n CP 2 n
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