按许用传动角反求曲柄滑块机构最大行程速比系数的新方法  

A New Method with the Minimum Transmission Angle to search for the Maximum Time Ratio of slider-Crank Mechanism to the Contrary

作　　者：孟兆明[1] 赵红滨[1] 邹玉静[1] 

机构地区：[1]青岛科技大学机电工程学院,青岛266061

出　　处：《现代制造技术与装备》2007年第5期20-21,68,共3页Modern Manufacturing Technology and Equipment

摘　　要：单向工作的曲柄滑块机构,机构的行程速比系数愈大,滑块回程用时愈短,机构的工作效率愈高。但同时机构的最小传动角会减小,机构的运动性能会趋劣。本文利用机构的几何关系,应用最优化技术,针对机构许用传动角[g],采用0.618法一次寻优,快速确定出机构具有最大行程速比系数Kmax。例举了设计实例,并通过实例对比,指出前期一些文献计算方法和计算结果的不妥。For a crank-slider mechanism working in a sin- gle-track, the larger the velocity ratio of its stroke is, the shorter its return time work is and the higher its work efficiency is, but at the same time the minimum transmission angle of the mechanism de- crease, its character will go worse. This paper uses the geometry of the mechanism with 0.618 and the optimization method; is directed against minimum transmission angle [ g] to define the maximum velocity ratio . This paper gives an example and contrast with living example; prints out the shortcoming in the method and result of some previous papers.

关 键 词：0.618法 曲柄滑块机构 行程速比系数 最小传动角 

分 类 号：TH112[机械工程—机械设计及理论]