Calculating a tolerance stack-up by the worst-case method
In the design office of a gearbox manufacturer, the design engineer is sizing the axial clearance of the input pinion. If the clearance drops to zero, the ball bearing seizes when hot and wrecks the housing. If it exceeds 0.4 mm, the gear teeth run out of true and the noise level becomes unacceptable. An arithmetic tolerance stack-up is the only rigorous way to guarantee assembly without risk.
Principles and formula of a worst-case tolerance stack-up
To calculate a stack-up by the worst-case method, you write the vector closing equation on the functional clearance condition, then add up the extreme limits to find the largest and smallest possible clearances. This deterministic approach guarantees 100% interchangeability of parts, by assuming that every contributing dimension can reach its least favourable tolerance limit at the same time.
| Calculation parameter | Formula | Metrological meaning |
|---|---|---|
| Closing equation | J = Σ A+ - Σ A- | Sum of positive dimensions minus sum of negative dimensions |
| Nominal clearance, J nom | Σ A+ nom - Σ A- nom | Theoretical centre value of the functional clearance |
| Maximum clearance, J max | Σ A+ max - Σ A- min | Case where the components open the clearance as much as possible |
| Minimum clearance, J min | Σ A+ min - Σ A- max | Extreme condition at maximum material |
| Clearance tolerance, TJ | J max - J min, i.e. Σ TA | Straight arithmetic sum of all the tolerances |
This method offers complete safety compared with statistical methods (such as the root sum square, RSS), which accept a small percentage of scrap at assembly.
Drawing the loop diagram and orienting the contributing dimensions
The first step is to draw the functional condition vector J between the two end surfaces that define the clearance to be preserved. By convention, this vector points from the reference face to the face resting against it.
Then the closed loop is built by linking the adjacent parts through their respective manufacturing dimensions A1, A2, up to An.
Dimensions that run in the same direction as the clearance vector get a positive sign. They enlarge the available space when their value increases.
Dimensions whose vector runs opposite to the clearance vector get a negative sign. They reduce the useful space and eat into the assembly clearance.
To check the related dimensional specifications, see the guide on ISO 286 fits and numerical limits.
Writing the minimum and maximum limit equations
Once the signs are assigned, the method gives the algebraic expressions of the maximum and minimum clearance.
For the maximum clearance, give every positive dimension its upper limit and every negative dimension its lower limit:
J max = Σ A+ max - Σ A- min
For the minimum clearance, swap the limits: the lower limit for positive dimensions and the upper limit for negative ones:
J min = Σ A+ min - Σ A- max
The resulting tolerance on the functional condition is the difference between the maximum and minimum clearance, which is the same as the plain sum of the individual tolerances:
TJ = TA1 + TA2 + … + TAn
A full worked example with three components
Consider a stack made of a housing bore A1, a ground spacer A2 and a shaft shoulder A3. The axial clearance J must stay positive to avoid any mechanical jamming.
The specifications of the three contributing dimensions are as follows.
| Dimension | Nominal | Limits |
|---|---|---|
| Housing bore, A1 | 60.0 ± 0.1 mm | 59.90 to 60.10 mm |
| Spacer width, A2 | 15.0 ± 0.05 mm | 14.95 to 15.05 mm |
| Shaft length, A3 | 40.0 ± 0.1 mm | 39.90 to 40.10 mm |
Closing equation of the loop: J = A1 - A2 - A3.
| Quantity | Calculation | Result |
|---|---|---|
| Nominal clearance | 60.0 - 15.0 - 40.0 | 5.00 mm |
| Maximum clearance | 60.10 - 14.95 - 39.90 | 5.25 mm |
| Minimum clearance | 59.90 - 15.05 - 40.10 | 4.75 mm |
| Clearance tolerance | 5.25 - 4.75 | 0.50 mm |
Note that 0.50 mm is indeed the sum of the three individual intervals, 0.20 + 0.10 + 0.20. It’s the quickest consistency check on a worst-case stack.
Checking the physical assembly: after building a sub-assembly whose measured parts give A1 = 59.88 mm, A2 = 15.04 mm and A3 = 40.16 mm, the actual clearance read with a feeler gauge is:
J actual = 59.88 - 15.04 - 40.16 = 4.68 mm
Verdict: the measured clearance, 4.68 mm, falls below the required minimum of 4.75 mm. Part A1, at 59.88 mm, was 0.02 mm below its lower limit, which made the whole assembly non-conforming.
Classic pitfalls in tolerance stack-ups
The first pitfall is leaving the geometric tolerances of form and orientation out of the stack. A parallelism error of 0.03 mm on a spacer face eats 0.03 mm of the axial clearance, exactly like a pure dimensional deviation.
The second pitfall is over-tolerancing the individual parts. Guaranteeing a very tight clearance by the worst-case method forces very narrow tolerances on every component, and machining costs soar. When the loop has more than six dimensions, it becomes better to switch to a statistical root-sum-square (RSS) analysis, accepting a controlled risk of parts not assembling.
The third pitfall is misidentifying the closing link. If an intermediate dimension can’t be measured directly on the manufacturing drawing, the inspector can’t validate the stack at goods-in inspection.
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