一种摆线齿轮齿形曲线的变步长离散方法

摘要

本发明公开了一种摆线齿轮齿形曲线的变步长离散方法，包括以下步骤：S1.设定摆线齿轮齿廓曲线参数，S2.推导出摆线齿轮齿廓曲线方程，S3.推导出摆线齿轮实际齿廓曲线的曲率半径ρ的公式，S4.推导出ψ关于压力角α的方程，S5.推导出摆线齿轮齿廓曲线拐点方程，S6.给出曲线线性插补初始步长的离散方法。本发明通过将离散误差与给定误差进行比较而自动调整逼近步长，从而获得小于逼近误差的最大步长，在满足精度要求的前提下，减少了走刀次数，提高了加工效率；本发明的摆线齿轮齿形曲线的变步长离散方法在逼近精度、离散效率和计算速度方面具有良好的综合指标。

主权项

一种摆线齿轮齿形曲线的变步长离散方法，其特征在于，包括如下步骤：S1.设定摆线齿轮齿廓曲线参数：针齿分布圆半径为R_z、针齿半径为r_z、偏心距为A、摆线齿轮齿数为Z_a、针销数为Z_b，短幅系数为K₁，K₁＝AZ_b/R_z；S2.推导出摆线齿轮齿廓曲线方程如下：$\left\{\begin{array}{c}x=R_{z}sin\psi -A\sin Z_b\psi +r_z\frac{K_1{\mathrm{sinZ}}_b\psi -sin\psi }{\sqrt{1+K_1^2-2K_1{\mathrm{cosZ}}_a\psi }}\\ y=R_{z}cos\psi -{\mathrm{AcosZ}}_b\psi +r_z\frac{K_1{\mathrm{cosZ}}_b\psi -cos\psi }{\sqrt{1+K_1^2-2K_1{\mathrm{cosZ}}_a\psi }}\end{array}\right.$其中，ψ为滚圆中心绕基圆中心转过的角度；S3.推导出摆线齿轮实际齿廓曲线的曲率半径ρ的公式如下：$\rho ={\rho }_0+r_z=\frac{{(1+K_1^2-2K_1{\mathrm{cosZ}}_a\psi )}^{\frac{3}{2}}{R}_z}{K_1(1+Z_b){\mathrm{cosZ}}_a\psi -(1+Z_b{K}_1^2)}+r_z$其中ρ₀为摆线齿轮理论齿廓曲线的曲率半径；S4.推导出ψ关于压力角α的方程如下：$\alpha =arccos\frac{K_1{\mathrm{sinZ}}_a\psi }{\sqrt{1+K_1^2-2K_1{\mathrm{cosZ}}_a\psi }}$S5.根据S2中方程推导出摆线齿轮齿廓曲线拐点方程如下：$\begin{array}{c}\left({\mathrm{AZ}}_b\sin \right(Z_b\psi )-R_z\sin (\psi )+r_z\frac{\sin \left(\psi \right)-K_1{Z}_b\sin \left(Z_b\psi \right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1}}+\frac{K_1{Z}_a{r}_z\left(\cos \right(\psi )-K_1\cos \left(Z_b\psi \right))\sin \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}})\\ \left(\begin{array}{c}{R}_z\sin \left(\psi \right)-{\mathrm{AZ}}_b^2\sin \left(Z_b\psi \right)+r_z\frac{\sin \left(\psi \right)-K_1{Z}_b^2\sin \left(Z_b\psi \right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1}}-\frac{K_1{Z}_a^2{R}_z\left(\sin \right(\psi )-K_1\sin \left(Z_b\psi \right))\cos \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}}\\ +\frac{2K_1{Z}_a{R}_z\left(\cos \right(\psi )-K_1{Z}_b\cos \left(Z_b\psi \right))\sin \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}}+\frac{3K_1^2{Z}_a^2{R}_z\left(\sin \right(\psi )-K_1{\mathrm{sinZ}}_b\psi ){\left(\sin \right(Z_a\psi \left)\right)}^2}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{5}{2}}}\end{array}\right)\\ +\left(R_z\cos \right(\psi )-{\mathrm{AZ}}_b\cos (Z_b\psi )+r_z\frac{\cos \left(\psi \right)-K_1{Z}_b\cos \left(Z_b\psi \right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1}}+\frac{K_1{Z}_a{r}_z\left(\sin \right(\psi )-K_1\sin \left(Z_b\psi \right))\sin \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}})\end{array}$$\frac{d^{2}y}{{\mathrm{dx}}^2}=-\frac{\left(\begin{array}{c}{R}_z\cos \left(\psi \right)-{\mathrm{AZ}}_b^2\cos \left(Z_b\psi \right)+r_z\frac{\cos \left(\psi \right)-K_1{Z}_b^2\cos \left(Z_b\psi \right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1}}-\frac{K_1{Z}_a^2{r}_z\left(\cos \right(\psi )-K_1\cos \left(Z_b\psi \right))\cos \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}}\\ +\frac{2K_1{Z}_a{R}_z\left(\sin \right(\psi )-K_1{Z}_b\sin \left(Z_b\psi \right))\sin \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}}+\frac{3K_1^2{Z}_a^2{R}_z\left(\cos \right(\psi )-K_1{\mathrm{cosZ}}_b\psi ){\left(\sin \right(Z_a\psi \left)\right)}^2}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{5}{2}}}\end{array}\right)}{{\left(R_z\cos \right(\psi )-{\mathrm{AZ}}_b\cos (Z_b\psi )-r_z\frac{\cos \left(\psi \right)-K_1{Z}_b\cos \left(Z_b\psi \right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1}}+\frac{K_1{Z}_a{r}_z\left(\sin \right(\psi )-K_1\sin \left(Z_b\psi \right))\sin \left(Z_a\psi \right)}{{(K_1^2-2K_1{\mathrm{cosZ}}_a\psi +1)}^{\frac{3}{2}}})}^3}$S6.给出曲线线性插补初始步长的离散方法，具体包括如下步骤：(1)根据上述步骤S4、S5确定ψ的拐点；(2)确定初始离散点步长α₀设定α₀为初始离散点步长，η为前进系数，λ为后退系数，ε₀为允许的逼近误差，ε‘为实际逼近误差，ε为近似逼近误差，建立近似逼近误差ε方程如下：$ϵ=\sqrt{\begin{array}{c}{(\frac{A\sin \left(Z_b{\alpha }_0\right)}{2}-\sin (\frac{Z_b{\alpha }_0}{2})-R_z\frac{\sin \left({\alpha }_0\right)}{2}+R_z\sin (\frac{{\alpha }_0}{2})-\frac{r_z}{2}\frac{\sin \left({\alpha }_0\right)-K_1\sin \left(Z_b{\alpha }_0\right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a{\alpha }_0+1}}-r_z\frac{\sin \left(\frac{{\alpha }_0}{2}\right)-K_1\sin \left(\frac{Z_b{\alpha }_0}{2}\right)}{\sqrt{K_1^2-2K_1\cos \frac{Z_a{\alpha }_0}{2}+1)}})}^2+\\ {(\frac{A}{2}-\frac{R_z}{2}+\frac{r_z}{2}+\frac{A\cos \left(Z_b{\alpha }_0\right)}{2}-A\cos (\frac{Z_b{\alpha }_0}{2})-\frac{R_z}{2}\cos ({\alpha }_0)+R_z\cos (\frac{{\alpha }_0}{2})+\frac{r_z}{2}\frac{\cos \left({\alpha }_0\right)-K_1\cos \left(Z_b{\alpha }_0\right)}{\sqrt{K_1^2-2K_1{\mathrm{cosZ}}_a{\alpha }_0+1}}-r_z\frac{\cos \left(\frac{{\alpha }_0}{2}\right)-K_1\cos \left(\frac{Z_b{\alpha }_0}{2}\right)}{\sqrt{K_1^2-2K_1\cos \frac{Z_a{\alpha }_0}{2}+1)}})}^2\end{array}}$(3)给出摆线齿轮齿形曲线的变步长离散方法设定α₁＝α₀+α₀，采用步骤(2)中方程计算摆线齿轮齿廓曲线上(x(α₀),y(α₀))、(x(2α₀),y(2α₀))两点之间的近似逼近误差ε；若则输出α₁参量离散点；若将初始离散点步长α₀乘以前进系数η重新计算近似逼近误差ε；若将初始离散点步长α₀乘以后退系数λ重新计算近似逼近误差ε；判断α₁是否到达曲线终点，未到达则重复以上步骤；否则，结束计算近似逼近误差ε。