Length — Question 5

Question 5

A fine micrometer scale does not settle a tolerance decision

A micrometer screw advances $0.500\ \mathrm{mm}$ per full revolution. Its thimble has 50 equal divisions. The sleeve reading gives the completed half-millimetres of travel; the thimble reading adds the fractional revolution beyond that sleeve reading.

For one measurement the last exposed sleeve mark is $6.500\ \mathrm{mm}$ and the reference line aligns with thimble division 24. A closed-jaw test gives a signed indication of $-0.030\ \mathrm{mm}$. Treat that zero correction as known exactly. After it is applied, the manufacturer supplies a total hard error bound of $\pm0.020\ \mathrm{mm}$ under the stated use conditions; this bound already includes all remaining reading effects.

For this exercise the part can be certified only if its entire allowed diameter interval lies in the inclusive specification $[6.760,6.780]\ \mathrm{mm}$.

1. Derive the travel per thimble division and find the raw indicated diameter.

2. Apply the signed zero correction and give the allowed true-diameter interval. Do not add a separate resolution bound to the stated total bound.

3. Can the part be certified from this result? Does failing that certification test establish that the part is defective?

4. Explain why recording extra decimal places or repeatedly observing division 24 cannot, by itself, justify replacing the total error bound by half a division.

Question 5 - Solution

A fine micrometer scale does not settle a tolerance decision

(a) Divide screw travel by the number of divisions.

$$\text{travel per division}=\frac{0.500\ \mathrm{mm}}{50} =\boxed{0.010\ \mathrm{mm}}.$$ Division 24 adds $24(0.010\ \mathrm{mm})=0.240\ \mathrm{mm}$ of travel. Thus the raw reading is $(6.500+0.240)\ \mathrm{mm}=\boxed{6.740\ \mathrm{mm}}$.

(b) Subtract the signed indication at true zero.

For a constant zero error $z$, $L_{\mathrm{corrected}}=L_{\mathrm{raw}}-z$. Here $$L_{\mathrm{corrected}}=6.740\ \mathrm{mm}-(-0.030\ \mathrm{mm}) =\boxed{6.770\ \mathrm{mm}}.$$ The given total hard bound yields $$\boxed{L\in[6.750,\ 6.790]\ \mathrm{mm}}.$$ No additional half-division allowance is added: the stated total bound already includes residual reading effects.

(c) Distinguish non-certification from demonstrated failure.

$$[6.750,6.790]\not\subseteq[6.760,6.780]\ \mathrm{mm};\qquad 0.020\ \mathrm{mm}>0.010\ \mathrm{mm}.$$

The corrected centre lies inside the specification, but its full allowed interval extends outside both limits. Therefore the part cannot be certified by the stipulated rule using this measurement.

That does not prove the part is defective. Allowed values include both conforming diameters, such as $6.770\ \mathrm{mm}$, and nonconforming diameters, such as $6.755\ \mathrm{mm}$. A sufficiently tighter valid measurement bound could resolve the decision.

(d) Resolution is not a bound on all errors.

The division size describes the readout increment. The supplied total error bound describes how far the corrected reading may differ from the true diameter under the use conditions. Repeated identical readings can preserve an unrecognized systematic effect; extra printed digits do not establish a smaller bound. Reducing the bound requires supporting calibration or measurement evidence, not a change in formatting.