Smallest gap: 0.30 mm
49.90 − 30.05 − 19.55 = 0.30 mm.
The smallest envelope meets the largest components. The result is 0.10 mm above the assumed lower acceptance limit.
ORIGINAL EDUCATIONAL EXAMPLE / NOT A CLIENT PROJECT
Will the assembly retain its gap?
A positive nominal gap is a starting point. This example checks whether the gap stays within an agreed range when every component reaches its dimensional limit.
The review brief
Two components sit inside a 50 mm envelope. For this illustrative exercise, the required final gap is 0.20–0.80 mm.
- Envelope: 50.00 ± 0.10 mm
- Component A: 30.00 ± 0.05 mm
- Component B: 19.50 ± 0.05 mm
- Functional loop: gap = envelope − A − B
Rigid parts, one dimension and one consistent unit system. Coatings, deformation, thermal growth and assembly effects are excluded from this example.
CALCULATION NOTE / 001Worked example
A small gap. A clear answer.
One envelope. Two components. Worst-case fit.
Minimum gap0.30 mm
Maximum gap0.70 mm
g = Envelope − A − B
g min = 49.90 − 30.05 − 19.55 = 0.30 mm
Check both extremes.
Largest gap: 0.70 mm
50.10 − 29.95 − 19.45 = 0.70 mm.
The largest envelope meets the smallest components. The result is 0.10 mm below the assumed upper acceptance limit.
Conclusion for this example: The 0.30–0.70 mm worst-case range falls within the assumed 0.20–0.80 mm requirement. This establishes dimensional fit only under the stated assumptions. It does not establish seal performance or equipment qualification.