Bonus Tolerance on an Out-of-Round Hole
A machined aluminum housing carries a Ø10.00 ±0.10 through hole located by position Ø0.30 Ⓜ to A|B|C. The hole is drilled and reamed in one pass. The CMM reports: Part 1 - every cross-section measures 10.04 and the hole axis is 0.20 mm from true position; Part 2 - cross-sections measure 10.04 and 9.94 at 90° to each other, axis 0.20 mm from true position; Part 3 - every cross-section measures 9.92, axis 0.22 mm from true position. The shop inspector awards bonus using the largest diameter found anywhere in the hole, so he passes Parts 1 and 2 and rejects Part 3. The customer's functional gauge is a Ø9.60 pin perpendicular to a datum A plate, with B and C simulated by fixed pins.
1. State MMC, LMC and virtual condition for the hole, and tabulate the maximum permissible axis offset for a perfectly round hole produced at 9.90, 9.94, 10.00 and 10.10 mm.
Read the worked answer
Feature limits
- MMC (internal feature, smallest hole) = 9.90
- LMC = 10.10
- Virtual condition = MMC − geometric tolerance = 9.90 − 0.30 = Ø9.60
Available tolerance at any size
Total diametral tolerance = 0.30 + bonus, where bonus = (actual mating envelope − MMC). Because the zone is diametral and centered on true position, the permitted radial offset is half of that.
| Actual mating size | Bonus | Total Ø tolerance | Max radial offset |
|---|---|---|---|
| 9.90 | 0.00 | 0.30 | 0.15 |
| 9.94 | 0.04 | 0.34 | 0.17 |
| 10.00 | 0.10 | 0.40 | 0.20 |
| 10.10 | 0.20 | 0.50 | 0.25 |
Note the pattern: the radial limit equals (size − 9.60)/2 throughout. That is the same statement as the virtual-condition boundary, which is why the functional gauge and the calculated bonus must agree.
2. Evaluate Parts 1, 2 and 3. Explain how bonus is determined and why the inspector's 'largest diameter' rule gives the wrong result for Part 2.
Read the worked answer
Part 1 - round at 10.04, so the actual mating envelope (AME) is 10.04. Bonus = 0.14, total tolerance = 0.44, radial limit = 0.22. Measured 0.20 ≤ 0.22. Conforms.
Part 2 - oval 10.04 × 9.94. The largest inscribed cylinder in an oval of those dimensions is Ø9.94, so the AME is 9.94, not 10.04. Bonus = 9.94 − 9.90 = 0.04, total = 0.34, radial limit = 0.17. Measured 0.20 > 0.17. Rejects.
Part 3 - round at 9.92. Bonus = 0.02, total = 0.32, radial limit = 0.16. Measured 0.22 > 0.16. Rejects (the inspector reached the right answer for the wrong reason if he used largest diameter, which here happens to equal the AME).
Why the inspector's rule fails: with the Ⓜ modifier, the tolerance grows from the departure of the actual mating envelope from MMC, because the mating envelope is what a mating part actually feels. In an oval hole the wide direction buys nothing - the pin is stopped by the narrow direction. Taking the largest local size overstates the bonus and lets nonfunctional parts through.
3. What does the Ø9.60 functional gauge actually verify, does it agree with your Task 2 conclusions, and what does it not verify?
Read the worked answer
The gauge verifies the virtual condition boundary: no material of the hole may intrude inside a Ø9.60 cylinder perpendicular to A and located at the basic position from B and C. Equivalently, the hole must clear the pin - a clearance test of (AME/2 − axis offset) ≥ 4.80 mm.
- Part 1: 10.04/2 − 0.20 = 4.82 ≥ 4.80 → pin enters, accept. Matches.
- Part 2: 9.94/2 − 0.20 = 4.77 < 4.80 → pin will not enter, reject. Matches.
- Part 3: 9.92/2 − 0.22 = 4.74 < 4.80 → reject. Matches.
The gauge is the arbiter of the MMC assembly condition, and it automatically accounts for bonus because a smaller hole leaves less clearance.
What it does not verify:
- Size limits. A hole reamed to 10.15 would pass the gauge yet violate LMC.
- Form or surface texture beyond the envelope.
- The position of the hole at sizes other than those the drawing permits - the gauge cannot tell you why a part failed, only that it did.
Practical rule: a gauge proves acceptance, never rejection, unless the gauge itself is certified to the correct virtual condition and the datum features are simulated at their stated boundary (MMB here for B and C).