下圖中之周轉輪系,A輪有54 齒,B輪有14 齒,C輪有28 齒,D輪有12 齒,若主動輪D設計以800 rpm之轉速順轉,求搖臂E 之轉速為多少rpm? 求公式和算法?

2015-12-06 7:33 am
更新1:

答案是80 RPM

回答 (1)

2015-12-07 10:45 am
✔ 最佳答案
Ta/Tc * Tb/Ta

= 12/28 * 14/54

= 1/9

= (Na - Ne)/(Nd - Ne)

= (0 - Ne)/(800 - Ne)

= -Ne/(Ne - 800)



=> Ne/800 = 1/10

=> Ne = 80 rpm

=> Answer = 80 rpm


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