The compound gear system of a reduction gearbox is shown in FIGURE 2 and consists of an input gear A with 68 teeth rotating at 850 r/min and an intermediate shaft on which two gears B and C are mounted having 136 and 70 teeth respectively and D having 100 teeth.

Question

If this gear system has a module of 6 mm, calculate the following:
5.1
The rotational speed of B, C and D
The addendum and dedendum of the gear teeth
5.2
5.3
The centre distance X
5.4
The centre distance Y

Answer 1

To analyze the compound gear system, we first need to apply the fundamental concepts of gear design. Given data includes the number of teeth on each gear, their module, and the input speed. Here's the analysis:

Given Data:

• Gear A: \(N_A = 68\) teeth
• Gear B: \(N_B = 136\) teeth
• Gear C: \(N_C = 70\) teeth
• Gear D: \(N_D = 100\) teeth
• Input speed (RPM) of Gear A: \(N_A = 850\) r/min
• Module \(m = 6\) mm

5.1 Rotational Speeds of Gears B, C, and D

The speed ratio between two gears in mesh is inversely proportional to the ratio of their teeth:

\[ \text{Speed Ratio} = \frac{N_1}{N_2} \]

1. Finding the speed of Gear B (driven by Gear A): \[ N_B = N_A \cdot \frac{N_A}{N_B} \] \[ N_B = 850 \cdot \frac{68}{136} = 850 \cdot 0.5 = 425 \text{ r/min} \]

2. Finding the speed of Gear C (driven by Gear B): \[ N_C = N_B \cdot \frac{N_B}{N_C} \] \[ N_C = 425 \cdot \frac{136}{70} = 425 \cdot 1.942857 \approx 826.07 \text{ r/min} \]

3. Finding the speed of Gear D (driven by Gear C): \[ N_D = N_C \cdot \frac{N_C}{N_D} \] \[ N_D = 826.07 \cdot \frac{70}{100} = 826.07 \cdot 0.7 \approx 577.34 \text{ r/min} \]

5.2 Addendum and Dedendum of the Gear Teeth

The addendum and dedendum can be calculated using the module:

• Addendum (ha): \[ ha = m = 6 \text{ mm} \]

• Dedendum (hf): \[ hf = m \times (1.25) = 6 \times 1.25 = 7.5 \text{ mm} \]

5.3 Centre Distance \(X\) (Between Gears A and B)

The center distance \(X\) between two gears can be calculated as:

\[ X = \frac{(N_A + N_B) \cdot m}{2} \]

Applying values:

\[ X = \frac{(68 + 136) \cdot 6}{2} = \frac{204 \cdot 6}{2} = 612 \text{ mm} \]

5.4 Centre Distance \(Y\) (Between Gears C and D)

Similarly, the center distance \(Y\) between gears C and D is:

\[ Y = \frac{(N_C + N_D) \cdot m}{2} \]

Applying values:

\[ Y = \frac{(70 + 100) \cdot 6}{2} = \frac{170 \cdot 6}{2} = 510 \text{ mm} \]

Summary of Results

• Rotational Speed of Gear B: \(425\) r/min
• Rotational Speed of Gear C: \(826.07\) r/min
• Rotational Speed of Gear D: \(577.34\) r/min
• Addendum: \(6\) mm
• Dedendum: \(7.5\) mm
• Centre Distance \(X\): \(612\) mm
• Centre Distance \(Y\): \(510\) mm

These should cover all the specified metrics related to the gearbox. If you have any other requirements, feel free to ask!