Gear angle calculation

Acceleration a=3.2m/s2, the acceleration thrust of the moving part Fa=ma=2 800×3.2=8 960N, the friction force of the moving part f=mgµ=2 800×10×0.005=140N, the total thrust of the moving part F = Fa + f = 8 960 + 140 = 9 100N, fast forward speed v fast = 48m/min = 48/60 = 0.8m/s, the maximum speed of output helical gear n tooth = v fast / (3.14 × D) = 0.80/( 3.14×111.4×0.001)=2.29r/s, the maximum angular velocity of the output helical gear ω tooth=n·2π=2.29×2×3.14=14.38rad/s.

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