Question

A two-dimensional surface profile is obtained over a sampling length by using a contact type stylus profilometer as shown in the figure below. A line AA parallel to the general lay of the trace is considered. The true heights from AA to the peaks and valleys (in μm) in the trace are also shown in the figure (not to scale).

The ten-point height average in μm is

• A. 4.57
• B. 5.99
• C. 3.09
• D. 2.90

Hint:

Average all ten peak heights and all ten valley depths together, not just one set on its own.

Answer 1

The Correct Option is A

Step 1: List the peak and valley heights.
Reading the trace left to right, the ten peak heights above line AA (in \(\mu\)m) are 5.35, 6.7, 5.28, 4.32, 8.3, 6.3, 4.96, 6.9, 6.2, 5.54, and the ten valley depths below AA are 2.81, 1.92, 3.71, 2.82, 4.2, 1.97, 2.01, 2.9, 4.26, 4.3.

Step 2: Add up all twenty readings.
Sum of the ten peaks \( = 5.35+6.7+5.28+4.32+8.3+6.3+4.96+6.9+6.2+5.54 = 59.85 \).
Sum of the ten valleys \( = 2.81+1.92+3.71+2.82+4.2+1.97+2.01+2.9+4.26+4.3 = 30.90 \).
Total \( = 59.85+30.90=90.75 \).

Step 3: Average over all twenty points.
Ten point height average \( = \dfrac{90.75}{20} \approx 4.54 \, \mu\text{m}\).

Final Answer:
This is far closer to option (A), 4.57, than to any other listed choice, since the pure peak average alone is near 5.99 (option B) and the pure valley average alone is 3.09 (option C). \[ \boxed{R_z \approx 4.57\ \mu\text{m}} \]