Question

In the measurement of surface roughness using a 2D stylus profilometer, a surface profile measured over a length of 0.8 mm was recorded on a graph paper. During recording, vertical magnification of 10,000 and horizontal magnification of 100 were used. The areas in the as-recorded graph above and below the datum line are as follows:

Above (mm\(^2\)): 140, 60, 150, 50
Below (mm\(^2\)): 70, 50, 120, 160

The average surface roughness (\(R_a\)) of the surface is ______ \(\mu m\) (rounded off to two decimal places).

Hint:

Add all the areas above and below the datum line, then divide by the true length times both magnifications to get Ra directly in mm.

Answer 1

Correct Answer: 1

Step 1: Understanding the Question:
The centre line average roughness \(R_a\) is found from the total area enclosed between the profile trace and the datum line, divided by the actual length of the surface, then corrected for the magnifications used while plotting.


Step 2: Key Formula:
\[ R_a = \frac{\sum A}{L \times V_{mag} \times H_{mag}} \]
where \(\sum A\) is the total area (above plus below the datum) on the recorded graph, \(L\) is the true sampling length, \(V_{mag}\) is the vertical magnification and \(H_{mag}\) is the horizontal magnification.


Step 3: Plugging in the numbers:
Sum of areas above the datum: \(140 + 60 + 150 + 50 = 400 \text{ mm}^2\)
Sum of areas below the datum: \(70 + 50 + 120 + 160 = 400 \text{ mm}^2\)
Total area: \(\sum A = 400 + 400 = 800 \text{ mm}^2\)
\[ R_a = \frac{800}{0.8 \times 10000 \times 100} = \frac{800}{800000} = 0.001 \text{ mm} \]


Final Answer:
Converting to micrometres, \(0.001 \text{ mm} = 1.00 \ \mu m\).
\[ \boxed{1.00 \ \mu m} \]