Let's break down the problem step by step: ### **Given Data** - Number of teeth on gear (\( N_g \)) = **57** - Number of teeth on pinion (\( N_p \)) = **23** - Helix angle (\( \psi \)) = **10°** - Normal diametral pitch (\( p_d \)) = **6** - Pressure angle (\( \phi \)) = **25°** We are to find: - **Transverse contact ratio (\( \epsilon_t \))** - **Axial contact ratio (\( \epsilon_a \))** --- ## **1. Calculate Required Parameters** ### **a. Pitch Circle Diameters** \[ d = \frac{N}{p_d \cos \psi} \] - For **gear**: \( d_g = \frac{57}{6 \cos 10^\circ} \) - For **pinion**: \( d_p = \frac{23}{6 \cos 10^\circ} \) \[ \cos 10^\circ \approx .9848 \] \[ d_g = \frac{57}{6 \times .9848} \approx \frac{57}{5.9088} \approx 9.65~\text{in} \] \[ d_p = \frac{23}{6 \times .9848} \approx \frac{23}{5.9088} \approx 3.89~\text{in} \] --- ### **b. Base Circle Diameters** \[ d_b = d \cos \phi \] \[ \cos 25^\circ \approx .9063 \] - For **gear**: \( d_{bg} = 9.65 \times .9063 \approx 8.75~\text{in} \) - For **pinion**: \( d_{bp} = 3.89 \times .9063 \approx 3.52~\text{in} \) --- ### **c. Circular Pitch in the Transverse Plane (\( p_t \))** \[ p_t = \frac{\pi}{p_d} \] \[ p_t = \frac{\pi}{6} \approx .5236~\text{in} \] --- ### **d. Addendum (Standard Gear)** \[ a = \frac{1}{p_d} = \frac{1}{6} = .1667~\text{in} \] --- ### **e. Path of Action (Transverse Contact Length, \( L_t \))** For a standard gear, the **length of action** in the **transverse plane** is: \[ L_t = \sqrt{(r_{a,g})^2-(r_{b,g})^2} + \sqrt{(r_{a,p})^2-(r_{b,p})^2} - (r_{o,g} + r_{o,p}) \sin \phi \] Where: - \( r_{a,g} \), \( r_{a,p} \): Addendum circle radii of gear and pinion - \( r_{b,g} \), \( r_{b,p} \): Base circle radii - \( r_{o,g} \), \( r_{o,p} \): Pitch circle radii **Calculate:** - \( r_{a,g} = \frac{d_g}{2} + a = \frac{9.65}{2} + .1667 = 4.825 + .1667 = 4.9917~\text{in} \) - \( r_{a,p} = \frac{d_p}{2} + a = \frac{3.89}{2} + .1667 = 1.945 + .1667 = 2.1117~\text{in} \) - \( r_{b,g} = d_{bg}/2 = 8.75/2 = 4.375~\text{in} \) - \( r_{b,p} = d_{bp}/2 = 3.52/2 = 1.76~\text{in} \) - \( r_{o,g} = d_g/2 = 4.825~\text{in} \) - \( r_{o,p} = d_p/2 = 1.945~\text{in} \) Now, \[ L_t = \sqrt{4.9917^2 - 4.375^2} + \sqrt{2.1117^2 - 1.76^2} - (4.825 + 1.945)\sin 25^\circ \] Calculate each term: - \( 4.9917^2 - 4.375^2 = 24.917 - 19.141 = 5.776 \Rightarrow \sqrt{5.776} = 2.404 \) - \( 2.1117^2 - 1.76^2 = 4.461 - 3.098 = 1.363 \Rightarrow \sqrt{1.363} = 1.168 \) - \( (4.825 + 1.945) = 6.77 \) - \( \sin 25^\circ \approx .4226 \) - \( 6.77 \times .4226 = 2.862 \) So, \[ L_t = 2.404 + 1.168 - 2.862 = 3.572 - 2.862 = .710~\text{in} \] --- ### **f. Transverse Contact Ratio (\( \epsilon_t \))** \[ \epsilon_t = \frac{L_t}{p_t} = \frac{.710}{.5236} \approx 1.36 \] --- ### **g. Axial (Face) Contact Ratio (\( \epsilon_a \))** \[ \epsilon_a = \frac{b \tan \psi}{p_t} \] But **face width (\( b \))** is not given in the problem. If the face width is not specified, we can't calculate the exact value. **Assume a typical face width** for helical gears, e.g., **1 inch** (if not specified, leave the result as a function of \( b \)): \[ \epsilon_a = \frac{b \tan 10^\circ}{.5236} \] \[ \tan 10^\circ \approx .1763 \] \[ \epsilon_a = \frac{b \times .1763}{.5236} = .337 \times b \] - For \( b = 1 \) inch: \( \epsilon_a = .337 \) - For \( b = 2 \) inch: \( \epsilon_a = .674 \) - **Leave as \( .337 \times b \) (in inches)** if face width is unspecified. --- ## **Final Answers** ### **Transverse Contact Ratio** \[ \boxed{\epsilon_t \approx 1.36} \] ### **Axial Contact Ratio** \[ \boxed{\epsilon_a = .337 \times b} \] Where \( b \) is the face width in inches. --- **Summary Table:** | Parameter | Value | |-----------------------------|----------------------| | Transverse Contact Ratio | \( \epsilon_t \approx 1.36 \) | | Axial Contact Ratio | \( \epsilon_a = .337\,b \) | If you know the face width, substitute to get the actual axial contact ratio.