Let's break down the problem: - **Given:** - **Pinion teeth, \( N_p \) = 15** - **Full depth teeth** (Standard, so addendum = 1 module) - **Pressure angle, \( \phi \) = \( 20^\circ \)** - **We want:** The **largest number of teeth for the gear (\( N_g \))** such that an **interference-free** gear set is obtained. ## Step 1: **Interference Condition** For a standard full-depth involute spur gear (20° pressure angle), **interference** occurs if the gear ratio is too low (i.e., the gear is not large enough compared to the pinion). The limiting condition is: The **minimum number of teeth on a gear (\( N_g \))** to avoid interference with a 15-tooth pinion: \[ N_g \geq \frac{2N_p \sin^2 \phi}{1 - 2N_p \left( \frac{\sin^2 \phi}{N_p} \right)} \] But it's easier to recall the standard limiting formula for **no interference** for full-depth 20° involute gears: \[ N_g \geq \frac{N_p}{\left[2 \sin^2 \phi\right]} - N_p \] But let's use the more standard approach: For two standard gears (full-depth, 20°) with teeth \( N_p \) and \( N_g \): \[ N_p \geq \frac{2a}{\sin^2 \phi} \] and \[ N_g \geq \frac{2a}{\sin^2 \phi} \] where \( a \) is the addendum (full-depth = 1 module). But usually, **minimum number of teeth on pinion to avoid interference** is: \[ N_{min} = \frac{2}{\sin^2 \phi} \] For 20°, \( \sin 20^\circ = 0.3420 \), so: \[ N_{min} = \frac{2}{(0.3420)^2} = \frac{2}{0.116} \approx 17.24 \] But our pinion has only 15 teeth, so to avoid interference, the **gear must be big enough**. The limiting number of teeth on the gear to avoid interference is: \[ N_{g,min} = \frac{N_p}{\left[2 \sin^2 \phi\right]} - N_p \] But the more precise and general formula for the **minimum number of teeth on the gear to avoid interference** is: \[ N_g \geq \frac{N_p}{\left[2 \sin^2 \phi - \frac{N_p}{N_g} (1 + 2\cos^2 \phi)\right]} \] However, this is an implicit equation, but there's a standard formula: \[ N_g = \frac{N_p}{\sin^2 \phi} - 2N_p \] But let's use the **standard equation for minimum number of teeth on gear to avoid interference with a given pinion:** \[ N_{g,min} = \frac{N_p}{\sin^2 \phi} - 2N_p \] Now, plug in the numbers: - \( N_p = 15 \) - \( \sin 20^\circ = 0.3420 \) - \( \sin^2 20^\circ = 0.116 \) \[ N_{g,min} = \frac{15}{0.116} - 2 \times 15 = 129.31 - 30 = 99.31 \] Therefore, the **minimum number of teeth on the gear** to avoid interference with a 15-tooth pinion is about **99.31**. ## Step 2: **Largest Number of Teeth for Interference-Free Gear** But the question asks for the **largest number of teeth for the gear that is interference-free**. In practice, **larger gears are always interference-free with a small pinion**; the interference risk is with small gears. So, with a 15-tooth pinion and standard 20° full-depth teeth, **any gear with 100 or more teeth** (since teeth must be integers) will be interference-free. ### **Final Answer:** > **The largest number of teeth for the spur gear that is interference-free is “no upper limit” (it can be as large as you like), as long as it is greater than or equal to 100.** > > **The minimum number of teeth for the gear to avoid interference with a 15-tooth pinion (full-depth, 20°) is 100 teeth.** ### **Summary Table:** | Pinion Teeth (\( N_p \)) | Pressure Angle | Gear Teeth (\( N_g \)), Minimum for No Interference | |--------------------------|----------------|------------------------------------------------------| | 15 | 20° | 100 (rounded up) | - **Any gear with ≥ 100 teeth will be interference-free with a 15-tooth pinion.** - **There is no upper limit to the gear teeth for interference-free operation.** --- **If you meant "maximum" instead of "largest", for a **given center distance**, you would need to specify that constraint.** Otherwise, as above, the answer is: **no upper limit, as long as \( N_g \geq 100 \).**