To solve the problem systematically, we'll break down each part of the question related to the gear train with a pinion (A) and a gear wheel (B).
Given:
- Module (m) = 3.5 mm
- Gear ratio (A:B) = 3:1
- Pitch circle diameter (PCD) of pinion A (d_A) = 66.5 mm
5.1: Number of teeth on each gear wheel
The pitch circle diameter (PCD) is related to the module and number of teeth (N) according to the formula:
\[ d = m \times N \]
For pinion A:
\[ d_A = m \times N_A \]
Rearranging gives us the number of teeth on pinion A:
\[ N_A = \frac{d_A}{m} = \frac{66.5 \text{ mm}}{3.5 \text{ mm}} \approx 19 \]
For the gear wheel B, the gear ratio \( \frac{N_A}{N_B} = \frac{3}{1} \).
Let \( N_B \) be the number of teeth on gear B:
\[ \frac{19}{N_B} = \frac{3}{1} \implies N_B = \frac{19}{3} \approx 6.33 \]
Since the number of teeth must be a whole number, we round down \( N_B = 6 \).
Now, using the gear ratio:
\[ \frac{N_A}{N_B} = 3 \implies N_B = N_A / 3 \approx 19 / 3 = 6.33 \]
On deciding \( N_B = 6 \): Adjusting \( N_A \) to \( 18 \), gives a proper integer.
Thus, the adjusted values are:
- Number of teeth on pinion A (N_A) = 18
- Number of teeth on gear B (N_B) = 6
5.2: Addendum and Dedendum of the gear teeth
The addendum (a) and dedendum (b) can be calculated using the following formulas:
-
Addendum (a): \[ a = m = 3.5 \text{ mm} \]
-
Dedendum (b): \[ b = 1.25 \times m = 1.25 \times 3.5 \text{ mm} = 4.375 \text{ mm} \]
5.3: Outside diameters of both gears
The outside diameter (OD) is calculated using the formula:
\[ \text{OD} = PCD + 2 \times \text{addendum} \]
For pinion A:
\[ \text{OD}_A = d_A + 2 \times a = 66.5 \text{ mm} + 2 \times 3.5 \text{ mm} = 66.5 \text{ mm} + 7 \text{ mm} = 73.5 \text{ mm} \]
For gear B, first we need its PCD, which is given by:
\[ d_B = m \times N_B = 3.5 \text{ mm} \times 6 = 21 \text{ mm} \]
Now we calculate its OD:
\[ \text{OD}_B = d_B + 2 \times a = 21 \text{ mm} + 2 \times 3.5 \text{ mm} = 21 \text{ mm} + 7 \text{ mm} = 28 \text{ mm} \]
5.4: The pitch circle diameter of the gear wheel
From the earlier calculation:
- Pitch Circle Diameter of Gear B (d_B) = 21 mm
5.5: The tooth thickness at the pitch circle
The tooth thickness at the pitch circle (t) can be calculated with:
\[ t = \frac{m}{2} = \frac{3.5 \text{ mm}}{2} = 1.75 \text{ mm} \]
Summary of Results
- 5.1 Number of teeth: Pinion A (N_A) = 18, Gear B (N_B) = 6
- 5.2 Addendum: 3.5 mm, Dedendum: 4.375 mm
- 5.3 Outside Diameters: Pinion A (OD_A) = 73.5 mm, Gear B (OD_B) = 28 mm
- 5.4 Pitch Circle Diameter of Gear B: 21 mm
- 5.5 Tooth Thickness at Pitch Circle: 1.75 mm