FIGURE 9.1 below shows a gear-drive system. Driver gear A on the shaft of the electric motor has 18 teeth that mesh with gear B, with 36 teeth, on a countershaft. There is another driver gear, C, with 16 teeth, on this countershaft, which meshes with a gear D, with 46 teeth, on the output shaft. Calculate the: 9.1.1 Rotation frequency of the output shaft if the electric motor rotates at 1 660 r/min 9.1.2 Velocity ratio between the input shaft and output shaft.

NSC Mechanical Technology Welding and Metalwork Basic calculations: FIGURE 9.1 below shows a gear-dr

FIGURE 9.1 below shows a gear-drive system - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

Question 9

FIGURE 9.1 below shows a gear-drive system. Driver gear A on the shaft of the electric motor has 18 teeth that mesh with gear B, with 36 teeth, on a countershaft. Th... show full transcript

Worked Solution & Example Answer:FIGURE 9.1 below shows a gear-drive system - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

Step 1

9.1.1 Rotation frequency of the output shaft if the electric motor rotates at 1 660 r/min

96%

114 rated

Answer

To calculate the rotation frequency of the output shaft, we can apply the gear formula relating the input and output speeds based on gear teeth:

$N_{OUTPUT} = N_{INPUT} \times \frac{T_A \times T_D}{T_B \times T_C}$

Where:

$N_{INPUT} = 1660$ r/min
$T_A = 18$ teeth (Driver gear A)
$T_B = 36$ teeth (Gear B)
$T_C = 16$ teeth (Gear C)
$T_D = 46$ teeth (Gear D)

Substituting the values:

$N_{OUTPUT} = 1660 \times \frac{18 \times 46}{36 \times 16}$

Calculating:

$N_{OUTPUT} = 1660 \times \frac{828}{576} = 1660 \times 1.4375 \approx 2389.06$

Thus, the rotation frequency of the output shaft is approximately 288.70 r/min.

Step 2

9.1.2 Velocity ratio between the input shaft and output shaft

99%

104 rated

Answer

The velocity ratio (VR) between the input shaft and the output shaft can be calculated using:

$VR = \frac{N_{INPUT}}{N_{OUTPUT}}$

Using the values:

$N_{INPUT} = 1660$ r/min
$N_{OUTPUT} \approx 288.70$ r/min

So:

$VR = \frac{1660}{288.70} \approx 5.75 : 1$

This indicates that the input shaft speed is 5.75 times greater than that of the output shaft.

FIGURE 9.1 below shows a gear-drive system - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

Worked Solution & Example Answer:FIGURE 9.1 below shows a gear-drive system - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

9.1.1 Rotation frequency of the output shaft if the electric motor rotates at 1 660 r/min

9.1.2 Velocity ratio between the input shaft and output shaft

Join the NSC students using SimpleStudy...