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

The outside diameter (OD) of the big gear is given by the formula:
$OD = m(T + 2)$
Using the values from part 5.1.1:
$OD = 3(30 + 2) = 3 imes 32 = 96 mm$
Therefore, the outside diameter of the big gear is 96 mm.

The PCD of the big gear can be calculated using the formula:
$PCD = m \times T$
Substituting the known values:
$PCD = 3 \times 30 = 90 mm$
Thus, the PCD of the big gear is 90 mm.

The dedendum (d) can be found using the formula:
$d = 1.25 \times m$
For the big gear:
$d = 1.25 \times 3 = 3.75 mm$
Therefore, the dedendum of the big gear is 3.75 mm.

The centre distance (Y) between the two gears is calculated as:
$Y = \frac{PCD_A + PCD_B}{2}$
Substituting the respective values:
$Y = \frac{90 + 90}{2} = 180 mm$
Thus, the centre distance is 180 mm.

The indexing can be calculated using the formula:
$Index = \frac{40}{n}$
where n is the number of teeth on the gear.
For a gear with 33 teeth:
$Index = \frac{40}{33} \approx 1.21$
This means that approximately 1.21 full turns are required for 33 teeth, which can be rounded to 1 full turn and 14 holes in a plate with 66 holes.

The width of the key can be calculated by:
$Width = \frac{Diameter}{4}$
Using the diameter of 92 mm:
$Width = \frac{92}{4} = 23 mm$
Therefore, the width of the key is 23 mm.

The length of the key is determined using the equation:
$Length = 1.5 \times Diameter$
Substituting the diameter of 92 mm:
$Length = 1.5 \times 92 = 138 mm$
Hence, the length of the key is 138 mm.

The thickness of the key at the bigger end is calculated as:
$Thickness = \frac{Diameter}{6}$
Using the diameter of 92 mm:
$Thickness = \frac{92}{6} \approx 15.33 mm$
Thus, the thickness at the bigger end is approximately 15.33 mm.

To calculate the thickness of the key at the smaller end, we use:
$T_s = T_b - 0.5 \times \frac{T_b - T_s}{100}$
Assuming the thickness at the bigger end is 15.33 mm, the smaller end will be 15.33 mm - 1.38 mm, giving us:
$Thickness = 15.33 - 1.38 \approx 13.95 mm$
Therefore, the thickness at the smaller end is approximately 13.95 mm.

An index plate is a component on the dividing head that has holes enabling it to subdivide one full turn of the crank into fractions. This allows for precise positioning and indexing during machining.

Sector arms are used to determine the spacing of required holes. They allow the operator to set up the dividing head without recounting the number of holes needed, facilitating efficient task execution.

Two methods that may be used on a centre lathe to cut external V-screw threads are:

1. Compound Slide Method: This involves using the compound rest of the lathe to adjust the angle for threading.
2. Cross Slide Method: This technique uses the cross slide to control the depth of cut, allowing for accurate threading by moving the tool in a transverse direction.