# Step-by-Step Solution: Cylindrical Worm Gear Mesh Design Let's break down the worm gear design process for your scenario. --- ## **Given Data** - **Motor speed, \( N_1 \):** 1125 rev/min (rpm) - **Velocity ratio, \( i \):** 10:1 (i.e., \( N_1 / N_2 = 10 \)) - **Output power, \( P_{out} \):** 25 hp - **Shaft axes:** 90° (standard for worm gears) - **Overload factor, \( K_o \):** 1.25 - **Application factor, \( K_a \):** 1.25 - **Design factor, \( n_d \):** 1.1 - **Worm:** Triple start, steel, HRC 58 - **Gear:** Sand-cast bronze - **Ambient temperature:** 70°F --- ## **Step 1: Calculate Output Shaft Speed** Given the velocity ratio (\( i = 10 \)), the output speed (\( N_2 \)) is: \[ N_2 = \frac{N_1}{i} = \frac{1125}{10} = 112.5\ \text{rpm} \] --- ## **Step 2: Convert Output Power to Watts** \[ P_{out} = 25\ \text{hp} \times 746\ \frac{\text{W}}{\text{hp}} = 18,650\ \text{W} \] --- ## **Step 3: Determine Required Input Power** Assume a typical efficiency (\( \eta \)) for a steel-bronze worm gear: \[ \eta \approx .85 \quad (\text{typical range .8 to .9, use .85 for design}) \] \[ P_{in} = \frac{P_{out}}{\eta} = \frac{18,650}{.85} \approx 21,941\ \text{W} \] --- ## **Step 4: Calculate Output Torque** \[ T_{out} = \frac{P_{out} \times 60}{2\pi N_2} \] \[ T_{out} = \frac{18,650 \times 60}{2\pi \times 112.5} = \frac{1,119,000}{706.858} \approx 1,584\ \text{Nm} \] --- ## **Step 5: Apply Load Factors** The **design tangential load** on the teeth: \[ W_{t,design} = \frac{2 T_{out}}{d_g} \cdot K_o \cdot K_a \cdot n_d \] We **don't yet know the gear pitch diameter \( d_g \)**, so we'll need to estimate it. First, let's calculate the **total design factor**: \[ \text{Total Factor} = K_o \cdot K_a \cdot n_d = 1.25 \times 1.25 \times 1.1 = 1.71875 \] --- ## **Step 6: Select Gear Materials and Allowable Stresses** - **Worm:** Steel, HRC 58 (very hard, strong) - **Gear:** Sand-cast bronze (softer, but good wear resistance) Typical **allowable bending stress** for sand-cast bronze: \[ \sigma_{allow} \approx 55\ \text{MPa} = 7,977\ \text{psi} \] --- ## **Step 7: Choose Number of Teeth** Let: - **Worm (pinion):** \( z_1 \) - **Gear (wheel):** \( z_2 \) For triple-start worm: \[ z_1 = \text{number of starts} = 3 \] \[ \text{Velocity ratio} = \frac{z_2}{z_1} \implies z_2 = i \cdot z_1 = 10 \times 3 = 30 \] --- ## **Step 8: Estimate Gear Dimensions** For a **preliminary design**, select a **module** (\( m \)) or **diametral pitch** (\( P \)). Let's use the **metric module system**: \[ d_g = m \cdot z_2 \] Let’s estimate module \( m \): ### **Lewis Equation for Tangential Load** \[ W_{t,design} = \frac{2 T_{out}}{d_g} \cdot \text{Total Factor} \] But Lewis equation for gear tooth bending: \[ W_t = \sigma_{allow} \cdot b \cdot m \cdot Y \] where: - \( b \): face width (typically \( 8m \) for worm gears) - \( Y \): Lewis form factor (\( \approx .1 \) for 30 teeth) Set \( W_t = W_{t,design} \): \[ \frac{2 T_{out}}{d_g} \cdot \text{Total Factor} = \sigma_{allow} \cdot b \cdot m \cdot Y \] But \( d_g = m \cdot z_2 \), so: \[ \frac{2 T_{out}}{m z_2} \cdot \text{Total Factor} = \sigma_{allow} \cdot b \cdot m \cdot Y \] \[ \frac{2 T_{out} \cdot \text{Total Factor}}{m z_2} = \sigma_{allow} \cdot b \cdot m \cdot Y \] \[ \frac{2 T_{out} \cdot \text{Total Factor}}{z_2} = \sigma_{allow} \cdot b \cdot m^2 \cdot Y \] Substitute \( b = 8m \): \[ \frac{2 T_{out} \cdot \text{Total Factor}}{z_2} = \sigma_{allow} \cdot 8m \cdot m^2 \cdot Y = 8 \sigma_{allow} Y m^2 \] \[ m^2 = \frac{2 T_{out} \cdot \text{Total Factor}}{8 \sigma_{allow} Y z_2} \] \[ m = \sqrt{\frac{2 T_{out} \cdot \text{Total Factor}}{8 \sigma_{allow} Y z_2}} \] Plug in values: - \( T_{out} = 1,584\ \text{Nm} = 1,584,000\ \text{Nmm} \) - \( \text{Total Factor} = 1.71875 \) - \( \sigma_{allow} = 55\ \text{MPa} = 55\ \text{N/mm}^2 \) - \( Y = .1 \) - \( z_2 = 30 \) \[ m = \sqrt{\frac{2 \times 1,584,000 \times 1.71875}{8 \times 55 \times .1 \times 30}} \] Calculate the denominator: \[ 8 \times 55 \times .1 \times 30 = 8 \times 55 \times 3 = 8 \times 165 = 1,320 \] Calculate the numerator: \[ 2 \times 1,584,000 \times 1.71875 = 3,168,000 \times 1.71875 ≈ 5,445,000 \] Now, \[ m = \sqrt{\frac{5,445,000}{1,320}} = \sqrt{4,126.5} ≈ 64.2\ \text{mm} \] This is **very large** (unrealistically so). This suggests: - The torque is very high for the output speed - The allowable stress is conservative - The face width is perhaps too narrow ### **Check calculation** Let’s try face width \( b = 10m \) (sometimes used for worm gears): \[ 8 \rightarrow 10, \quad 10 \times 55 \times .1 \times 30 = 1,650 \] \[ m = \sqrt{\frac{5,445,000}{1,650}} = \sqrt{3,300} ≈ 57.4\ \text{mm} \] Still large. #### **Possible Issue** The torque at the output is very high (1,584 Nm). For practical gears, you might use a **higher allowable stress** (say, 80 MPa), or increase the number of teeth. Let’s assume \( \sigma_{allow} = 80\ \text{MPa} \): For \( b = 10m \): \[ 10 \times 80 \times .1 \times 30 = 2,400 \] \[ m = \sqrt{\frac{5,445,000}{2,400}} = \sqrt{2,268.75} ≈ 47.6\ \text{mm} \] Still, the module is very large. For high-torque, low-speed applications, worm gears must be large. --- ## **Step 9: Final Gear Dimensions (Summary Table)** | Parameter | Value | |-------------------|--------------------------| | Number of starts (worm) | 3 | | Number of teeth (gear) | 30 | | Module (approximate) | 48–65 mm | | Pitch diameter (gear) | \( m \times z_2 \) ≈ 1,440 mm | | Face width | \( 10m \) ≈ 480–650 mm | --- ## **Step 10: Check and Refine** - If the calculated gear seems too large, consider using a **multistage reduction** (not just a single worm gear) or increase allowable stresses if material allows. - For **final design**, detailed AGMA calculations and manufacturer consultation are necessary. --- ## **Summary** 1. **Motor at 1125 rpm** drives worm (triple start, steel); the worm meshes with a **bronze gear** (30 teeth) at 112.5 rpm, transmitting 25 hp (18,650 W). 2. **Total torque at output**: 1,584 Nm. 3. **Design factors applied**: Overload (1.25), application (1.25), and design (1.1). 4. **Calculated gear size is large** (module 48–65 mm, gear pitch diameter ≈ 1.5 m). 5. **Face width**: around .5–.65 m. 6. **Materials**: Steel worm (hardened), sand-cast bronze gear. --- ## **Conclusion** - Worm gear must be **very robust** for this torque and speed reduction. - Consider **multistage gearing** if physical size is a constraint. - Use AGMA standards and manufacturer data for final detail design. --- ## **References** - Shigley, J.E., Mischke, C.R., & Budynas, R.G. "Mechanical Engineering Design", 9th Edition. - AGMA Standards for Worm Gear Design --- **If you have constraints on size or want a more compact design, let me know!**