Question

3) (15 pts] Number of teeth and theoretical center to center distance distance from given data A 17-tooth spur pinion has a diametral pitch of 8 teethlin, runs at 1386 rev/min, and drives a gear at a speed of 462 rev/min. Find the number of teeth on the gear and the theoretical center-to-center distance. The number of teeth on the gear is The theoretical center-to-center distance is in. 4) [15 pts] Gear Train Analysis The tooth numbers for the gear train illustrated are N2 = 12, N3 = 16, N4 = 30, N6 = 36, and N7 = 38. Gear 7 is fixed. If shaft a is turned through 16 revolutions, how many turns will shaft b make? 4) 6) 5 F3 27 -7- -b The shaft b turns revolutions.

Answer 1

Problem 3 is solved here.

Given:

• Number of teeth on the pinion,
  T,
  = 17
• Diametral pitch of the pinion= 8 teeth/in
• Speed of the pinion, $N_{p}$ = 1386 rev/min
• Speed of the gear, $N_{g}$ = 462 rev/min

Find:

• Number of teeth on the gear, $T_{g}$
• Center to center distance, C

Solution:

We know that, by law of gearing, $\frac{N_{p}}{N_{g}}= \frac{d_{g}}{d_{p}}$ , where,

• $N_{p}$ is speed of pinion.
• $N_{g}$ is speed of gear.
• $d_{g}$ is pitch circle diameter of gear.
• $d_{p}$​​​​​​​ is pitch circle diameter of pinion.

We know that, module, m= d/T, where, d is pitch circle diameter and T is number of teeth.

We know that, Module for pinion and gear are same. Hence, , where, $m_{p}$ is module of pinion and $m_{g}$ is module of gear.

• 
• 
  dg dp To 9
  
• 
  Tg T dp
  ​​​​​​​

From law of gearing and by equating module, we can make a equation,

• $\frac{N_{p}}{N_{g}}= \frac{d_{g}}{d_{p}}=\frac{T_{g}}{T_{p}}$​​​​​​​

i) Number of teeth on the gear, $T_{g}$ :

From the above equation,

• $\frac{T_{g}}{T_{p}}=\frac{N_{p}}{N_{g}}$
• $T_{g}=\frac{N_{p}}{N_{g}}(T_{p})$
• $T_{g}=\frac{1386}{462}(17)$
• $T_{g}=51$
• Number of teeth on the gear, $T_{g}=51$ ​​​​​​​

ii) Center to center distance, C:

We know that, diametral pitch= T/d, where, T is number of teeth and d= pitch circle diameter.

Given that, diametral pitch of pinion= 8 teeth/in

• $\frac{T_{p}}{d_{p}}=8$
• $\frac{17}{d_{p}}=8$
• 
• Pitch circle diameter of pinion, ​​​​​​​

From the equations,

• $\frac{d_{g}}{d_{p}}=\frac{T_{g}}{T_{p}}$
• $\frac{d_{g}}{2.125}=\frac{51}{17}$
• 
• Pitch circle diameter of gear, ​​​​​​​

We know that, center to center distance, C = $\frac{d_{p}+d_{g}}{2}$

• $C=\frac{d_{p}+d_{g}}{2}$
• $C=\frac{2.125+6.375}{2}$
• 
• Center to center distance,