已知一对渐开线标准外啮合圆柱齿轮传动[1],其模数m=10mm,压力角a=20°,中心距a=350mm,传动比i1=9/5,试计算这对齿轮传动的几何尺寸

步骤 1：计算齿轮的齿数
根据已知条件，中心距 \(a = 350mm\)，模数 \(m = 10mm\)，传动比 \(i_{12} = \frac{9}{5}\)。首先，我们需要计算小齿轮和大齿轮的齿数 \(z_1\) 和 \(z_2\)。
\[ z = \frac{2a}{m(1+i_{12})} = \frac{2 \times 350}{10(1+\frac{9}{5})} = \frac{700}{10 \times \frac{14}{5}} = \frac{700}{28} = 25 \]
\[ z_2 = i_{12} \cdot z_1 = \frac{9}{5} \times 25 = 45 \]

步骤 2：计算齿轮的几何尺寸
根据计算出的齿数，我们可以计算出齿轮的分度圆直径、齿顶圆直径、齿根圆直径、基圆直径、全齿高、齿顶高、齿根高、齿距、齿厚和齿槽宽。
- 分度圆直径：\(d_1 = mz_1 = 10 \times 25 = 250mm\), \(d_2 = mz_2 = 10 \times 45 = 450mm\)
- 齿顶圆直径：\(d_{a1} = d_1 + 2h_{a}m = 250 + 2 \times 10 = 270mm\), \(d_{a2} = d_2 + 2h_{a}m = 450 + 2 \times 10 = 470mm\)
- 齿根圆直径：\(d_{f1} = d_1 - 2(h_{a} + c)m = 250 - 2(1 + 0.25) \times 10 = 225mm\), \(d_{f2} = d_2 - 2(h_{a} + c)m = 450 - 2(1 + 0.25) \times 10 = 445mm\)
- 基圆直径：\(d_{b1} = d_1 \cos \alpha = 250 \cos 20^\circ = 234.923mm\), \(d_{b2} = d_2 \cos \alpha = 450 \cos 20^\circ = 422.862mm\)
- 全齿高：\(h = (2h_{a} + c)m = 2(1 + 0.25) \times 10 = 22.5mm\)
- 齿顶高、齿根高：\(h_{a} = h_{a}m = 10mm\), \(h_{f} = (h_{a} + c)m = 12.5mm\)
- 齿距：\(p = m\pi = 10 \times \pi = 31.416mm\)
- 齿厚、齿槽宽：\(s = \frac{1}{2}m\pi = 15.708mm\), \(e = \frac{1}{2}m\pi = 15.708mm\)