Zero Error, Least Count and a Lot of 200 Pins
In workshop practice, 200 mild-steel pins of nominal diameter 12.00 mm must be inspected before assembly; the drawing allows ±0.05 mm. The store holds only two instruments: a steel rule with a least count of 0.5 mm, and a vernier caliper with a least count of 0.02 mm whose jaws, when fully closed, show a vernier reading of 0.06 mm. One group measures with the steel rule, finds every pin 'about 12 mm' and stamps the whole lot accepted. A second group uses the vernier but ignores the closed-jaw reading entirely, obtains values between 11.92 mm and 12.10 mm, rejects the 60 pins reading below 11.95 mm as undersize, and passes the rest. No micrometer is available in the store today, the assembly line wants the pins within the hour, and the instructor must decide which readings can be trusted, how many pins actually fail, and whether any pin may leave the workshop.
1. Correct the vernier observations and decide how many pins genuinely lie outside 11.95–12.05 mm. Was the second group right to reject 60?
Read the worked answer
A positive zero error of +0.06 mm means the instrument reads 0.06 mm too large, so true diameter = observed − 0.06 mm.
The observed spread 11.92 to 12.10 mm therefore corresponds to true diameters of 11.86 to 12.04 mm.
Acceptance band is 12.00 ± 0.05, i.e. 11.95 to 12.05 mm. Every pin with true diameter below 11.95 mm fails, which means every observed reading below 11.95 + 0.06 = 12.01 mm fails; and since the largest true size is 12.04 mm, none is oversize.
The group was pointing the right way but undercounted badly: after correction, everyone between 11.95 and 12.01 mm on their sheet also fails, so the true rejection is well above 60. Their error was to apply the acceptance band to uncorrected readings. The steel-rule group's blanket acceptance is worthless.
2. Is either instrument fit for verifying a ±0.05 mm tolerance? Explain using least count.
Read the worked answer
Total tolerance band = 0.10 mm. The working rule is that the least count should be of the order of one-tenth of the tolerance band, i.e. about 0.01 mm, so that measurement uncertainty does not consume the tolerance itself.
The vernier's least count of 0.02 mm is one-fifth of the band. It is usable for sorting only if the zero error is corrected, jaws and pin are wiped clean, one operator measures the whole lot, and readings are taken at three positions 120° apart. Even then each reading carries roughly ±0.01–0.02 mm uncertainty, which is 10–20 per cent of the band, so borderline pins cannot be finally judged.
The steel rule's least count of 0.5 mm is five times the entire tolerance band. Its 'pass' says only that the pin is near the nearest half millimetre. It cannot be used for inspection at all, only for rough checking.
3. Set out what you would do in the remaining hour before releasing any pins.
Read the worked answer
- Record and display the zero error of the vernier at the start of the session, and re-check it at the end - a knock changes it.
- Discard the steel-rule verdict; that group re-measures with the corrected vernier.
- Sort with the vernier into three bins: clear accept (true 11.96–12.04 mm), clear reject (true outside 11.94–12.06 mm), and borderline (about 11.93–11.96 mm and 12.04–12.06 mm). Release only the clear-accept bin today.
- Tag and quarantine the borderline pins and ask the store to buy a 0–25 mm micrometer (least count 0.01 mm) before the next session.
- Keep an inspection sheet with operator name, instrument number and readings, so any dispute can be traced.
- Report upwards that the process, not only the inspection, is off-centre: the corrected mean sits about 0.06 mm below nominal, so the turning setting needs adjusting.