- #1
DevonZA
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Homework Statement
Homework Equations
The Attempt at a Solution
1. A gearbox is required to transmit 5.5 kW from a shaft rotating at 2000 rpm. The desired output speed is approximately 350 rpm. The output must rotate in the same direction as the input and along the same axis. Select suitable gears.
2. Lewis Formula: ##F= \frac{Wt}{(Kv)(m)(Y)}##
3.
I get an overall ratio of 2000rpm/350rpm = 5.71
Then I choose Np=18 and NG=100 (SG1-100) because 100/18=5.5 (these teeth numbers are from standard tables provided in the textbook - I unfortunately do not have a copy with me but I can post later)
m=1 according to table 6.7 where I have chosen SG1-100
d=mN
d=(1)(Np)
= 1x18
= 18
d=(1)(NG)
=1x100
=100
V=(d/2)XnX(2pi/60)
= (18/2)X2000X(2pi/60)
= 1884.96m/s
V=(d/2)XnX(2pi/60)
= (100/2)X350X(2pi/60)
= 1832.60m/s
= (18/2)X2000X(2pi/60)
= 1884.96m/s
V=(d/2)XnX(2pi/60)
= (100/2)X350X(2pi/60)
= 1832.60m/s
Kv=6/(6 +V)
= 6/(6+1884.96)
= 3.172991496x10^-3
Kv=6/(6 +V)
= 6/(6+1832.6)
=3.263352551x10^-3
= 6/(6+1884.96)
= 3.172991496x10^-3
Kv=6/(6 +V)
= 6/(6+1832.6)
=3.263352551x10^-3
Wt= Power/V
= 5500W/1884.96
= 2.92N
Wt= Power/V
= 5500W/1832.6
= 3N
= 5500W/1884.96
= 2.92N
Wt= Power/V
= 5500W/1832.6
= 3N
F= Wt/(Kv)(m)(Y)
= 2.92/(3.172991496x10^-3)(1)(0.39502)
= 2329.67
F= Wt/(Kv)(m)(Y)
= 3/(3.263352551x10^-3)(1)(0.51321)
= 1791.27
= 2.92/(3.172991496x10^-3)(1)(0.39502)
= 2329.67
F= Wt/(Kv)(m)(Y)
= 3/(3.263352551x10^-3)(1)(0.51321)
= 1791.27
I'm not sure what to do from here. This question actually appears in Mechanical Design by K.Maekawa, T. Obikawa, Y. Yamane, T.H.C Childs but no solution is provided.
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