Thabo is a machinist and is tasked to cut a spur gear with a pitch-circle diameter of 136 mm and a module of 4 - NSC Mechanical Technology Fitting and Machining - Question 6 - 2021 - Paper 1

The dedendum (d) can be calculated using the formula:

$d = 1.157 \times m$

Substituting the module:

$d = 1.157 \times 4 = 4.63 \text{ mm}$

Therefore, the dedendum is 4.63 mm.

The outside diameter (BD) can be calculated using the formula:

$BD = m \times (T + 2)$

Substituting the values:

$BD = 4 \times (34 + 2) = 4 \times 36 = 144 \text{ mm}$

Thus, the outside diameter is 144 mm.

The circular pitch (SS) can be calculated using the formula:

$SS = m \times \pi$

Substituting the module value:

$SS = 4 \times \pi \approx 12.57 \text{ mm}$

Therefore, the circular pitch is approximately 12.57 mm.

To calculate the minimum width (w), use the relationship:

$w = 190 - 2(DE)$

We need to calculate DE first:

Using the tangent relationship:

$\tan(\theta) = \frac{DE}{ED}$

Substituting DE with determined values:

• Calculate DE as:

$DE = 38 \tan(30^{\circ}) \approx 21.94 \text{ mm}$

Then substituting back for w:

$w = 190 - 2(21.94) = 190 - 43.88 = 146.12 \text{ mm}$

Thus, the minimum width of the dovetail is approximately 146.12 mm.

The distance over the precision rollers (M) can be calculated using:

$M = w + 2(AC) + 2(R)$

From previous calculations:

• w = 146.12 mm
• AC (after calculation) = 25.98 mm and R = 15 mm

Thus,

$M = 146.12 + 2(25.98) + 2(15) \approx 228.08 \text{ mm}$

Therefore, the distance over the precision rollers is approximately 228.08 mm.

For simple indexing, the indexing can be calculated as:

$\text{Indexing} = \frac{40}{N}$

where N is the number of divisions: N = 40.

Substituting the value:

$\text{Indexing} = \frac{40}{40} = 1 \text{ (direct ratio)}$

Thus, the indexing needed is 1.

To calculate the change gears needed with 163 teeth:

Use the format:

$D_{BR} = \left(\frac{D_{G}}{A - n}\right) \times 40$

1. For the first set:

   • $D_{G} = 40$, $A = 40$, $n = 0$: $D_{BR} = \left(\frac{40}{40 - 0}\right) \times 40 = 40$

2. For the second set:

   • $D_{G} = 40$, $A = 40$, $n = 163$: $D_{BR} = \left(\frac{(160-163)}{40}\right) \times 40 = 24$

Thus, the change gears needed are 24.