The required number of teeth in wheel A.

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

$$T_A = 30$$

The required number of teeth in wheel A is 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 $T_A$:

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

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 write:

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

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):

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

So, the sun gear B has 6 teeth, and the planet gear C has 12 teeth.

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

$$T_D = T_C = 12$$

The required number of teeth in wheel D is 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 4.2, we found $T_D = 12$. The module $m$ is not provided in the problem. Therefore, 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 $\boxed{\text{12m}}$.