The arithmetic mean of the absolute deviations of the profile from the mean line

The root mean square of the deviations

The distance from highest peak to lowest valley

The sum of the signed deviations

Which is correct for Rq with sum y^2 = 54.55?

Rq = sqrt(54.55/10) = 2.34 micrometre

Rq = 54.55/10 = 5.455 micrometre

Rq = 54.55/22.5 = 2.42 micrometre

Rq = sqrt(54.55) = 7.39 micrometre

What are Ry and Ra in micro-inches (1 micrometre = 39.37 micro-inch)?

Ry = 6.2 micrometre (3.2 + 3.0); Ra = 88.6 micro-inch

Ry = 0.2 micrometre (3.2 - 3.0); Ra = 88.6 micro-inch

Ry = 6.2 micrometre; Ra = 0.057 micro-inch

Ry = 3.2 micrometre; Ra = 22.5 micro-inch

Practice · 6 marks

Roughness parameters from a surface trace

Short answer

Ra = arithmetic mean of absolute deviations; Rq = RMS; Rz is peak-to-valley height. From given ordinates: Ra = 2.25 µm (88.6 µ in), Rq = 2.34 µm, maximum peak-to-valley height = 6.2 µm.

Define Ra, Rq (RMS) and Rz/Ry. The heights of ten equally spaced ordinates of a surface profile from the mean line are (in micrometres): +2.1, -1.8, +3.2, -2.4, +1.5, -3.0, +2.6, -1.2, +1.9, -2.8. Calculate Ra, Rq and the maximum peak-to-valley height. Convert Ra to micro-inches.

Definitions

Ra (centre-line average, CLA): the arithmetic mean of the absolute deviations of the profile from the mean line over the sampling length:

R_a = \frac{1}{n}\sum_{i=1}^{n}|y_i|

Rq (RMS): the root mean square deviation, $R_q = \sqrt{\dfrac{1}{n}\sum y_i^2}$ . It is about 1.11 times Ra for a sine-like profile.
Rz / Ry (peak to valley): the vertical distance between the highest peak and the lowest valley within the sampling length (Rz in ISO is the mean of five such heights in five sampling lengths).

Calculation

$i$	$y_i$ ( $\mu$ m)	$\lvert y_i\rvert$	$y_i^2$
1	+2.1	2.1	4.41
2	-1.8	1.8	3.24
3	+3.2	3.2	10.24
4	-2.4	2.4	5.76
5	+1.5	1.5	2.25
6	-3.0	3.0	9.00
7	+2.6	2.6	6.76
8	-1.2	1.2	1.44
9	+1.9	1.9	3.61
10	-2.8	2.8	7.84
Sum		22.5	54.55

R_a = \frac{22.5}{10} = 2.25\ \mu\text{m}

R_q = \sqrt{\frac{54.55}{10}} = \sqrt{5.455} = 2.34\ \mu\text{m}

Highest peak $= +3.2$ , deepest valley $= -3.0$ :

R_y = 3.2 - (-3.0) = 6.2\ \mu\text{m}

In micro-inches, 1 $\mu$ m $= 39.37\ \mu$ in: $R_a = 2.25 \times 39.37 = 88.6\ \mu$ in.

Answer: $R_a$ = 2.25 $\mu$ m (88.6 $\mu$ in), $R_q$ = 2.34 $\mu$ m, $R_y$ = 6.2 $\mu$ m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Roughness parameters from a surface trace

Definitions

Calculation

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