周轉輪系,若輪2轉速為N2=+6,旋臂轉速為m=-4
T2=100,T3=80,T4=120,T5=50
(N5-m)/(N2-m)=+[(T2*T4)/(T3*T5)]
(N5-(-4))/(6-(-4))=+[(100*120)/(80*50)]
N5=26 rpm
N4=N3
(N3-m)/(N2-m)=-(T2/T4)
(N3-(-4))/(6-(-4))=-(100/80)]
N3=N4=-16.5 rpm
e2→5 = T2×T4T3×T5 = 100×12080×50 =3
e2→5 = N5-(-4)6-(-4)
N5=26 rpm 順時計
880.如圖所示之周轉輪系,若輪2轉速為+6,旋臂轉速為-4,試求輪5、輪4之轉..-阿摩線上測驗