Figure 5.27 shows a simple gear system consisting of a gearwheel A and pinion B of which the centre distance is approximately 700 mm. If the circular pitch of the gears is 40 mm and the gears have a velocity ratio of 4:1, calculate: 5.1 The number of teeth on pinion B.

The velocity ratio ($VR$) for a gear system is the ratio of the number of teeth of the driven gear to the number of teeth of the driving gear. Since A is a gearwheel and B is a pinion, A is typically larger than B, meaning $T_A > T_B$. A velocity ratio of 4:1 means that the larger gear has 4 times the teeth of the smaller gear, or the speed of the smaller gear is 4 times the speed of the larger gear.
Given $VR = 4:1$, we assume this means $\frac{T_A}{T_B} = 4$.
Therefore, $T_A = 4 T_B$.

The circular pitch ($P_c$) is given as 40 mm.
The pitch diameter ($D$) of a gear is related to its number of teeth ($T$) and circular pitch ($P_c$) by the formula:

So, for gear A: $D_A = \frac{P_c T_A}{\pi}$
And for pinion B: $D_B = \frac{P_c T_B}{\pi}$

The centre distance ($C$) between two meshing gears is half the sum of their pitch diameters:

Substitute the expressions for $D_A$ and $D_B$:

Given:
$C = 700$ mm (approximately)
$P_c = 40$ mm
$T_A = 4 T_B$

Now, solve for $T_B$:

Using $\pi \approx 3.14159$:

Since the number of teeth must be an integer and the centre distance is given as "approximately 700 mm", we round $T_B$ to the nearest whole number.

The number of teeth on pinion B is $\text{22}$.