Chapter 13, Problem 1P

A 17-tooth spur pinion has a diametral pitch of 8 teeth/in, runs at 1120 rev/min, and drives a gear at a speed of 544 rev/min. Find the number of teeth on the gear and the theoretical center-to-center distance.

To determine

The number of teeth on the gear.

The theoretical center to-center distance.

Answer to Problem 1P

The number of teeth on the gear is $35$.

The theoretical center to-center distance is $82.6\text{mm}$.

Explanation of Solution

Write the expression for the diametrical pitch for the pinion.

$p_d=\frac{t}{d}$ . (I)

Here, the number of teeth on pinion is $t$ and pitch circle diameter of pinion is $d$.

Write the expression of velocity ratio in terms of the number tooth.

$\frac{N_G}{N_P}=\frac{t}{T}$ (II)

Here, the rotational speed of the gear is $N_G$, the rotational speed of the pinion is $N_P$, and the number of teeth on gear is $T$.

Write the relation between number of tooth on pinion and gear and diameters of the pinion and gear.

$\frac{T}{t}=\frac{D}{d}$                                              (III)

Here, the pitch circle diameter of the gear is $D$.

Write the expression for the centre distance between gear and pinion.

$CD=\frac{d+D}{2}$ (IV)

Conclusion:

Substitute $17\text{teeth}$ for $t$ and $8\text{teeth/in}$ for $p_d$ in Equation (I).

$\begin{array}{l}\text{8}\text{teeth}/\text{in}\text{=}\frac{\text{17}\text{teeth}}{d}\\ \left(\text{88}\text{teeth}/\text{in}\right)\left(\frac{1\text{in}}{\text{25}\text{.4}\text{mm}}\right)=\frac{\text{17}\text{teeth}}{d}\\ d=\text{53}\text{.975}\text{mm}\\ d\approx \text{54}\text{mm}\end{array}$

Substitute $17$ for $t$, $544\text{rpm}$ for $N_G$ and $1120\text{rpm}$ for $N_P$ in Equation (II).

$\begin{array}{l}\frac{544\text{rpm}}{1120\text{rpm}}=\frac{17}{T}\\ T=17\left(\frac{1120}{544}\right)\\ T=35\end{array}$

Thus, the number of teeth on the gear is $35$.

Substitute $17$ for $t$,$35$ for $T$ and $\text{54}\text{mm}$ for $d$ in Equation (III).

$\begin{array}{l}\frac{35}{17}=\frac{D}{54\text{mm}}\\ D=\left(54\text{mm}\right)\left(2.06\right)\\ D=111.2\text{mm}\end{array}$

Substitute $\text{111}\text{.2}\text{mm}$ for $D$ and $\text{54}\text{mm}$ for $d$ in Equation (IV).

$\begin{array}{c}CD=\frac{54\text{mm}+111.2\text{mm}}{2}\\ =\frac{165.2\text{mm}}{2}\\ =82.6\text{mm}\end{array}$

Thus, the centre distance is $82.6\text{mm}$.