Determine: 4.1 The required number of teeth in wheel A. 4.2 The required number of teeth in wheel D. 4.3 The pitch circle diameter (PCD) of D.

The diagram explicitly labels the ring gear A with "30 teeth".

$$T_A = 30$$

The required number of teeth in wheel A is $\boxed{\text{30 teeth}}$.

• b) 4.2 The required number of teeth in wheel D.

For an epicyclic gear train with a ring gear (A), sun gear (B), and planet gear (C), the relationship between their number of teeth is:

$$T_A = T_B + 2T_C$$

Substitute the value of $T_A = 30$:

$$30 = T_B + 2T_C \quad (1)$$

By visual inspection of Figure 5.45, the diameter of the planet gear C appears to be twice the diameter of the sun gear B. Since the number of teeth is proportional to the diameter for gears with the same module ($D = m T$), we can infer:

$$D_C = 2D_B \implies T_C = 2T_B \quad (2)$$

Substitute equation (2) into equation (1):

$$30 = T_B + 2(2T_B)$$ $$30 = T_B + 4T_B$$ $$30 = 5T_B$$ $$T_B = \frac{30}{5} = 6$$

Now, calculate $T_C$ using equation (2):

$$T_C = 2T_B = 2 \times 6 = 12$$

Since "wheel D" refers to the planet gear C:

$$T_D = T_C = 12$$

The required number of teeth in wheel D is $\boxed{\text{12 teeth}}$.

• c) 4.3 The pitch circle diameter (PCD) of D.

The PCD of a gear is given by the formula:

$$\text{PCD} = m \times T$$

where $m$ is the module and $T$ is the number of teeth.

From part 4.2, we found that the number of teeth on wheel D is $T_D = 12$. The module $m$ is not provided in the problem, so the PCD of D must be expressed in terms of $m$.

$$\text{PCD}_D = m \times T_D = m \times 12 = 12m$$

The pitch circle diameter (PCD) of D is $\boxed{\text{12m}}$.