Home / Courses / Manufacturing / Metrology, Tolerances, and Quality

Manufacturing · Chapter 9 of 10 · Advanced

Metrology, Tolerances, and Quality

No part is made exactly to size, so engineering runs on tolerances. Specify them well, measure them honestly, and prove with statistics that the process can actually hold them.

01

Readiness check

This chapter mixes dimensions and statistics. Tick only what you can do closed-notes.

  • Add and subtract dimensions with plus and minus tolerances.
  • Recall the mean and standard deviation.
  • Picture a normal (bell) distribution.
  • Compare a value against limits.
  • Compute a simple ratio.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRefresh mean and standard deviation in statistics.
3 or more weak itemsReview basic statistics before continuing.
02

The core idea

Parts are made to a tolerance, not a number; a fit is the relationship between two toleranced parts, and process capability measures whether production can stay inside the tolerance.

clearance = hole − shaftCp = (USL − LSL)/6σCpk = min[(USL−μ), (μ−LSL)]/3σ

Every dimension carries a tolerance band. When two parts mate, the combination of their bands defines the fit: clearance (always a gap), interference (always a squeeze), or transition (either). Measuring confirms the parts; statistics confirms the process. Capability indices compare the tolerance width with the process spread (Cp) and account for whether the process is centred (Cpk).

The skill works when: you compute fits from the worst-case limits and judge capability with both Cp and Cpk.
The skill breaks down when: tolerances are specified tighter than needed, or capability is read from Cp alone while the process is off-centre.
The concept. Each part's tolerance is a band around the basic size. A hole zone above and a shaft zone below leave a clearance; how the two zones overlap (or not) defines the type of fit.
03

The skills, taught in order

Quality is dimensions plus statistics. Five skills cover tolerances, fits, GD&T, measurement, and process control.

9.1 Tolerances

A tolerance is the allowed variation on a dimension, the difference between its upper and lower limits. Tighter tolerances cost more (finer processes, more inspection, more scrap), so the rule is to specify the loosest tolerance the function allows.

9.2 Fits

A fit describes how a shaft and hole mate, found from their limit dimensions.

FitConditionUse
Clearancealways a gap (hole > shaft)free rotation or sliding
Interferencealways a squeeze (shaft > hole)permanent press-fit assembly
Transitionmay clear or interfereaccurate location, light press

9.3 Geometric dimensioning and tolerancing

Plus-minus limits do not control shape or relationship; GD&T does. It specifies form (flatness, roundness), orientation (perpendicularity), and position relative to datums, so a feature is toleranced for how it must function, not just its size. It also enables bonus tolerance at the material condition.

9.4 Measurement and metrology

Dimensions are checked with calipers, micrometers, gauges, and coordinate measuring machines. Accuracy (closeness to truth) differs from precision (repeatability); a good instrument and procedure need both, and must be calibrated against traceable standards.

9.5 Statistical process control and capability

Control charts track a process over time to separate normal variation from a real shift. Capability compares the tolerance width with the spread: Cp = (USL − LSL)/6σ assumes a centred process, while Cpk penalises off-centre operation. A process needs Cpk ≥ 1.33 to be considered capable.

Index valueInterpretation
Cpk < 1.0not capable; defects expected
1.0 to 1.33marginal; little margin
≥ 1.33capable
Cpk < Cpprocess is off-centre

Engineering connection: tolerances and capability decide whether parts assemble and function, and they feed the cost and yield models of Chapter 10.

04

Worked example 1: clearance of a fit

A hole is 25.000 +0.021/0 mm and a shaft is 25.000 −0.020/−0.041 mm. Find the maximum and minimum clearance, and classify the fit.

Figure 1. The hole zone sits entirely above the shaft zone, so there is always a gap. The extreme limits give the maximum and minimum clearance.
  1. ProblemFind the clearance range and fit type for the parts in Figure 1.
  2. Given / findHole 25.000 to 25.021 mm; shaft 24.959 to 24.980 mm. Find max and min clearance.
  3. AssumptionsWorst-case (limit) analysis of two mating cylindrical parts.
  4. ModelMaximum clearance pairs the largest hole with the smallest shaft; minimum clearance pairs the smallest hole with the largest shaft.
  5. EquationsCmax = holemax − shaftmin Cmin = holemin − shaftmax
  6. SolveCmax = 25.021 − 24.959 = 0.062 mm. Cmin = 25.000 − 24.980 = 0.020 mm. Both positive, so it is a clearance fit with an allowance (minimum clearance) of 0.020 mm.
  7. CheckBoth extremes give a gap, confirming a clearance fit (no chance of interference). The 0.020 to 0.062 mm range is typical of a running fit, loose enough to turn freely but not sloppy.
  8. ConclusionThe fit, not either part alone, governs function. Worst-case limit analysis guarantees the parts always assemble as intended, the safe way to specify a fit.
Result. Clearance 0.020 to 0.062 mm; a clearance (running) fit.
05

Worked example 2: process capability

A dimension is specified 50.00 ± 0.15 mm. The process runs at a mean of 50.04 mm with a standard deviation of 0.04 mm. Find Cp and Cpk, and judge whether the process is capable.

Figure 2. The spread fits comfortably between the limits (Cp = 1.25), but the mean sits high, so the right tail crowds the upper limit and Cpk drops below one.
  1. ProblemFind Cp and Cpk for the process in Figure 2 and judge capability.
  2. Given / findUSL = 50.15, LSL = 49.85, μ = 50.04, σ = 0.04 mm. Find Cp, Cpk.
  3. AssumptionsApproximately normal output, σ estimated from a stable process.
  4. ModelCp compares tolerance width with 6σ; Cpk uses the nearer spec limit to capture centring.
  5. EquationsCp = (USL − LSL)/6σ Cpk = min[(USL − μ), (μ − LSL)]/3σ
  6. SolveCp = 0.30/(6 × 0.04) = 1.25. The nearer limit is the USL: (50.15 − 50.04)/(3 × 0.04) = 0.11/0.12 = 0.92, so Cpk = 0.92.
  7. CheckCpk < Cp confirms the process is off-centre, and Cpk = 0.92 < 1.33 means it is not capable: some parts will exceed the upper limit. Re-centring the mean to 50.00 would raise Cpk to match Cp = 1.25.
  8. ConclusionSpread alone (Cp) is not enough; centring matters, which is why Cpk is the index quoted. Here the fix is to shift the mean, not to tighten the process.
Result. Cp = 1.25 but Cpk = 0.92: not capable, because the process is off-centre.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Tighter tolerance is always saferCost and scrap balloon"Does the function need this tolerance?"Specify the loosest tolerance that works.
Judging capability by Cp aloneOff-centre process passes on paper"Is Cpk below Cp?"Use Cpk; it accounts for centring.
Confusing accuracy and precisionRepeatable but biased readings trusted"Is it close to true, or just consistent?"Calibrate for accuracy; repeatability is precision.
Size controls shapeBent or out-of-round part within size limits"Is form or position toleranced?"Use GD&T for form, orientation, and position.
07

Practice ladder

Level 1 · Direct skill

A hole is 40.000 to 40.025 mm and a shaft 40.000 to 39.984 mm. Find the minimum clearance.

Show answer

Cmin = holemin − shaftmax = 40.000 − 40.000 = 0. The minimum clearance is zero (line-to-line), the boundary of a clearance fit.

Level 2 · Mixed concept

For the Worked Example 2 process, what does Cpk become if the mean is re-centred to 50.00 mm?

Show answer

Centred, both limits are 0.15 away: Cpk = 0.15/(3 × 0.04) = 1.25, equal to Cp. Centring alone lifts the process from not-capable to the same 1.25, showing centring is free capability.

Level 3 · Independent problem

A press-fit needs the shaft always larger than the hole by 0.01 to 0.04 mm. If the hole is 30.000 to 30.021 mm, what shaft limits achieve it?

Show answer

For minimum interference 0.01 at worst case, shaftmin = holemax + 0.01 = 30.031. For maximum interference 0.04, shaftmax = holemin + 0.04 = 30.040. So the shaft is 30.031 to 30.040 mm, an interference fit.

Level 4 · Transfer to real engineering

Find a real assembly with a fit (a bearing on a shaft, a dowel in a hole). Identify the fit type and estimate the tolerances, and say what Cpk the process would need.

What good work looks like

The fit classified from the limits, a clearance or interference range computed, and a capability target (Cpk ≥ 1.33) tied to the consequence of a bad fit.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I judged capability with Cpk, not just Cp."
"Give me five mating pairs; I will classify each fit from the limits."
"Compute the clearance." Pairing the worst-case limits yourself is the skill.
"Is this process capable?" Reasoning from Cpk and centring is the point.

Portfolio task

Analyse one fit and one process: classify the fit from limit dimensions, and compute Cp and Cpk from process data, recommending any re-centring.

Must include: a worst-case clearance or interference, Cp and Cpk, and a capability verdict.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is a tolerance, and why not specify it as tight as possible?

The allowed variation on a dimension; tighter tolerances raise cost and scrap, so specify the loosest that works.

2. Name the three fit types.

Clearance (always a gap), interference (always a squeeze), transition (either).

3. Write Cp and Cpk.

Cp = (USL − LSL)/6σ; Cpk = min[(USL − μ), (μ − LSL)]/3σ.

4. What does Cpk < Cp tell you?

The process is off-centre; re-centring would raise Cpk.

5. Accuracy versus precision?

Accuracy is closeness to the true value; precision is repeatability.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a fit and a capability index from a blank page.
+3 daysOne fit and one Cpk problem.
+7 daysCarry yield and quality into economics, Chapter 10.
+30 daysRelate tolerances back to the processes that can hold them.
10

Textbook mapping

ItemMapping
Primary sourceKalpakjian and Schmid, Manufacturing Engineering and Technology, Chapters 35 and 36 (engineering metrology, quality assurance)
Cross-referenceGroover, Ch. 5 and 42 · Montgomery, Statistical Quality Control
Core topics9.1 Tolerances · 9.2 Fits · 9.3 GD&T · 9.4 Measurement · 9.5 SPC and capability
Engineering connectionWhether parts assemble and function, and the yield that drives cost.
Read nextChapter 10: Automation, CNC, and Economics.