FIGURE 5.1 shows two spur gears that mesh - NSC Mechanical Technology Automotive - Question 5 - 2016 - Paper 1

The outside diameter (OD) of the big gear can be calculated using the formula:

$OD = m(T + 2)$

Substituting values: $OD = 3(30 + 2) = 3 \times 32 = 96 mm$

Therefore, the outside diameter of the big gear is 96 mm.

The PCD for the big gear can be calculated using the module and the number of teeth:

$PCD = m \times T$

Here:

• m = 3
• T (Number of Teeth) = 30

So: $PCD = 3 \times 30 = 90 mm$

The PCD of the big gear is 90 mm.

The dedendum can be calculated using:

$Dedendum = 1.25 m$

Using m = 3: $Dedendum = 1.25 \times 3 = 3.75 mm$

Thus, the dedendum of the big gear is 3.75 mm.

The center distance (Y) between the two gears can be calculated using:

$Y = \frac{PCD_A + PCD_B}{2}$

Since PCD_A = 90 mm and PCD_B = 90 mm: $Y = \frac{90 + 90}{2} = \frac{180}{2} = 90 mm$

The center distance between the two gears is 90 mm.

The indexing can be calculated using:

$Indexing = \frac{40}{n}$

where n is the number of teeth, which is 33: $Indexing = \frac{40}{33} \approx 1.21$

This means that in one full turn of the crank, 1.21 holes are indexed. It can also be represented as:

• One full turn of the crank and 14 holes in a 66 hole plate.