Question:medium
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).
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).
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).
Show 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.
Updated On: Aug 5, 2026
Show Solution
Correct Answer: 1
Solution and Explanation
Step 1: An alternate route through the graph length:
Instead of applying both magnifications together in one shot, let us first work out how long the trace is on paper, then how tall it is on average, and only then scale down to the real surface.
Step 2: Length of the trace on the graph paper:
The real sampling length is 0.8 mm, and it was stretched by the horizontal magnification of 100:
\[ L_{graph} = 0.8 \times 100 = 80 \text{ mm} \]
Step 3: Average height on the graph, then scale down:
Total enclosed area on the graph, above and below the line together:
\[ \sum A = (140+60+150+50) + (70+50+120+160) = 400 + 400 = 800 \text{ mm}^2 \]
Average height on the graph is area divided by graph length:
\[ h_{graph} = \frac{800}{80} = 10 \text{ mm} \]
This 10 mm is the roughness height as blown up by the vertical magnification alone, so it must be divided back down:
\[ R_a = \frac{h_{graph}}{V_{mag}} = \frac{10}{10000} = 0.001 \text{ mm} \]
Final Answer:
$0.001$ mm equals $1.00$ $\mu m$, matching the direct formula method.
\[ \boxed{1.00 \ \mu m} \]
Instead of applying both magnifications together in one shot, let us first work out how long the trace is on paper, then how tall it is on average, and only then scale down to the real surface.
Step 2: Length of the trace on the graph paper:
The real sampling length is 0.8 mm, and it was stretched by the horizontal magnification of 100:
\[ L_{graph} = 0.8 \times 100 = 80 \text{ mm} \]
Step 3: Average height on the graph, then scale down:
Total enclosed area on the graph, above and below the line together:
\[ \sum A = (140+60+150+50) + (70+50+120+160) = 400 + 400 = 800 \text{ mm}^2 \]
Average height on the graph is area divided by graph length:
\[ h_{graph} = \frac{800}{80} = 10 \text{ mm} \]
This 10 mm is the roughness height as blown up by the vertical magnification alone, so it must be divided back down:
\[ R_a = \frac{h_{graph}}{V_{mag}} = \frac{10}{10000} = 0.001 \text{ mm} \]
Final Answer:
$0.001$ mm equals $1.00$ $\mu m$, matching the direct formula method.
\[ \boxed{1.00 \ \mu m} \]
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