采用微分法确定曲柄滑块机构最大行程速比系数  

Determination on speed ratio coefficient of maximum travel of crank and slider mechanism by the differential method

作　　者：孟兆明[1] 唐仁刚[2] 赵海霞[1] 赵红滨[1] 

机构地区：[1]青岛科技大学机电工程学院,山东青岛266061 [2]海军航空工程学院青岛分院,山东青岛266041

出　　处：《机械设计》2007年第9期42-44,共3页Journal of Machine Design

摘　　要：曲柄滑块机构,瞬间传动角愈大,其运动性能愈好;单向工作的曲柄滑块机构,行程速比系数愈大,急回特性愈强,工作效率愈高。采用微分法,以机构允许的最小传动角[γ]为设计依据,以机构行程速比系数为函数,对设计变量求导联立,导出机构能够实现最大行程速比系数km ax的函数表达式,绘制出函数曲线,并得出结论:机构最大行程速比系数km ax只与许用传动角[γ]有关,且随着[γ]增大而减小。The larger the instantaneous transmission angle is in the crank-slider mechanism the better its moving performance is. The larger the speed ratio coefficient of travel and the stronger the quick return characteristics in the unidirectional operated crank-sli- der mechanism are the higher its working efficiency is. Adopting the differential method, taking the allowable minimum transmission angle[γ] of the mechanism as the foundation of design and using the travel speed ratio coefficient of mechanism as the function, the differentiation and simultaneity were made on the designing variables, and the functional expression of maximum travel speed ratio coefficient kmax that could be achieved by the mechanism was derived. The functional curves were protracted and the result was obtained that the maximum travel speed ratio coefficient kmax is only related to the allowable transmission angle [γ] and decreases along with the increase of [γ].

关 键 词：行程速比系数 急回特性 最小传动角 曲柄滑块机构 

分 类 号：TH112[机械工程—机械设计及理论]