推荐回答(5个)
实际啮合线段长度渐开线标准圆柱齿轮的啮合传动一对渐开线齿轮正确啮合的条件是两个齿轮的法向齿距相等。
即是下面公式成立:pb1=pb2pb1=m1πcosα1pb2=m2πcosα2m1=m2=mα1=α2=α。
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扩展资料:
满足渐开线齿轮的正确啮合条件的注意事项:
但是仅满足正确啮合条件有时还不能保证齿轮能够连续传动。
齿轮的传动是依靠两轮的轮齿依次接触来实现的,为了使传动不中断,当前一对轮齿尚未脱离啮合时,后—对轮齿就应进人啮合,这样齿轮传动才具有连续性。
要了解齿轮机构能否实现连续传动,就必须研究齿轮的啮合过程。当两轮的一对齿开始啮合时,一定是主动轮的齿根推动从动轮的齿顶,这时从动轮齿顶圆与啮合线N1N2的交点B2是一对轮齿啮合的起始点。
参考资料来源:百度百科——实际啮合线段长度
齿轮啮合的条件:一对齿轮基圆齿距必须相等。因为,渐开线齿轮加工刀具标准化,所以,齿轮啮合条件,是(分度圆)模数、压力角必须分别相等。
齿轮连续啮合条件,是重合度必须大于1 。
一对渐开线齿轮的工作一侧齿廓的啮合点必须同时都在啮合线N1N2上,若有两对轮齿同时参加啮合,则两对齿工作一侧齿廓的啮合点必须同时都在啮合线上,如图所示:
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有两个齿轮相邻两齿轮同侧齿廓沿法线的距离相等, 即满足
时齿轮才能连续传动。又根据渐开线性质,上述条件可化为pb1=pb2=pb , 即要求两齿轮的基圆齿距相等。
对于渐开线齿廓,基圆齿距pb=πmcosα ,由于分度圆上的模数m和压力角α都必须取标准值,故只有使两齿轮的模数m和压力角α分别相等,即:
m1=m2=m
α1=α2=α
这是齿轮正确啮合的必要条件之一
一对齿廓开始啮合时,应是主动轮的齿根的齿廓与从动轮的齿顶接触,所以开始啮合点是从动轮的齿顶圆与啮合线N1N2的交点A。当两轮继续转动时,啮合点的位置沿啮合线向左下方移动,从动轮2齿廓上的接触点由齿顶向齿根移动,而主动轮1齿廓上的接触点则由齿根向齿顶移动。终止啮合点是主动轮的齿顶圆与啮合线N1N2的交点E。线段AE为啮合点的实际轨迹,称为实际啮合线.
当两轮齿顶圆加大时,点A和E分别趋近于点N1和N2,因基圆以内无渐开线,故线段N1N2为理论上可能的最长啮合线,称为理论啮合线。
如上所述,满足正确啮合条件的一对齿轮可保证每对齿都在啮合线上啮合。但是,若前后两对齿的啮合点之间的距离大于实际啮合线AB的长度,则前一对齿终止啮合而分离时,后一对齿尚未开始啮合,连续传动不能实现,也就是说满足正确啮合条件只是匀速连续传动的必要条件,还不是充分条件。因此,要保证连续传动,必须在前一对齿在E点终止啮合以前,后一对齿已在A点进入啮合。显然要做到这点,必须使AE大于基圆齿距pb。根据渐开线性质,线段AE应等于齿廓从
开始啮合到终止啮合在基圆上转过的弧长ae。
我们把实际啮合线长度与基圆齿距pb的比值称为重合度,用ε表示,它也等于齿廓从开始啮合到终止啮合在基圆上经过的弧长与基圆齿距之比,因此为保证连续传动,应使ε=
>1
综上所述,渐开线齿轮匀速连续传动的条件是两齿轮的模数和压力角必须分别相等,且啮合的重合度大于1。
渐开线齿轮的正确啮合条件是:两轮的模数和压力角分别相等
补充:
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渐开线齿轮啮合条件模数和压力角相等,斜齿轮是法面模数相等,齿形角相等,螺旋角相等且螺旋角方向相反……
两轮的模数相等,等于国家标准值;两轮分度圆上的压力角相等=20°
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