Tolerance Stack-up Analysis

Parts are made with some variation. A single part’s tolerance is easy to read; whether an assembly will always go together depends on how the variations of several parts add up. A tolerance stack-up adds up the variation of every dimension that decides a gap, a clearance or a fit, so you can predict how the assembly will behave before anything is made. This page follows the method in the Machinery’s Handbook (“Tolerance Analysis and Assignment”), including its worked example, and extends it with geometric tolerances and design guidance. The basic one-dimensional worst-case and RSS examples are also on the GD&T Inspection and Tolerance Stack-up page. It is part of the Fits & Tolerances guide.

What a stack-up is

The Handbook describes two complementary jobs. Tolerance analysis works bottom-up: given the tolerances on each part, what variation will the assembly have? Tolerance assignment works top-down: given the variation the assembly can accept, what tolerance can each part have? Analysis is done on preliminary or final part tolerances to find needed design changes; assignment is done at the design stage to meet the assembly and function requirements. In practice you iterate between them.

The Handbook adds a warning that applies to all of it: part tolerances are not assigned arbitrarily, but based on manufacturing process capability, cost limits, and assembly and functional requirements. Overly restrictive tolerances limit the choice of processes and drive up cost, and small differences in tolerance can change cost dramatically.

The stack-up chain

A stack-up starts with a chain diagram: a picture of the assembly that identifies the dimension you want to solve (the assembly dimension, A) and every part dimension that contributes to it. Then write the chain as an equation, with a sign convention that gives each dimension a direction: positive for right or up, negative for left or down.

A = − B − C − D + E + F + G

That is the equation for the Handbook’s example (its Figure 5), a small bearing assembly with six contributing dimensions. The sizes below are in inches; the table also gives the mean and standard deviation of each dimension from manufacturing data.

Dimension	Sign	Max	Min	Mean (µ), manufacturing data	Std. deviation (σ), manufacturing data
B	−	0.3760	0.3750	0.375464	0.000106
C	−	0.5906	0.5858	0.588126	0.001004
D	−	1.7510	1.7490	1.750128	0.000203
E	+	0.1875	0.1825	0.185147	0.000874
F	+	1.7550	1.7450	1.750748	0.001200
G	+	0.9395	0.9355	0.937331	0.000517

• Clear part definition is essential. The Handbook stresses that unambiguous definition, as provided by geometric dimensioning and tolerancing, is required for accurate analysis, and that any variation allowed by reference drawings must be included in the dimensions of the stack.
• Use the loop that matters. Follow the chain from one side of the gap, through each part and each contact, to the other side. A missing dimension or a wrongly signed one is the commonest source of a wrong answer.

Four methods

Method	Idea	Use it when	Result
Worst case	Every dimension at its extreme, in the worst direction	You must guarantee fit; few dimensions (about seven or fewer) in the chain; parts are inspected so none are out of tolerance	Guarantees 100% fit; tends to ask for tight, costly tolerances
Statistical	Each dimension is a normal distribution with a mean and a standard deviation from real data; the assembly dimension is another normal distribution	You have data from a controlled process (SPC); many dimensions contribute	Likely values and ranges; allows larger tolerances; predicts conformance and defect rates; does not guarantee conformance
Root sum of squares (RSS)	A statistical method with simplifying assumptions: each dimension’s mean is the middle of its tolerance and σ = (max − min) ÷ 6	You have no manufacturing data but expect centered, normally distributed parts	The assembly is predicted to fall within ±3σ of its mean
Monte Carlo	A computer picks random values for every dimension, from its real distribution (normal, uniform or other), and combines them through the chain thousands of times	Distributions are not normal, the equation is nonlinear, or you want defect rates	A probability distribution of the assembly dimension, including defect rates

Worst case

A worst-case calculation is done twice: once for the maximum assembly dimension and once for the minimum. For a positive contributor use its maximum in the maximum equation and its minimum in the minimum equation; for a negative contributor the opposite:

A_max = − B_min − C_min − D_min + E_max + F_max + G_max
A_min = − B_max − C_max − D_max + E_min + F_min + G_min

Statistical

Assembly mean: µ_A = − µ_B − µ_C − µ_D + µ_E + µ_F + µ_G
Assembly standard deviation: σ_A = √( σ_B² + σ_C² + σ_D² + σ_E² + σ_F² + σ_G² )

The signs matter for the mean but not for σ, because variations add as squares. The normal distribution has about 68.26% of its area within ±1σ, 95.44% within ±2σ, and 99.74% within ±3σ (the Handbook’s statistical chapter). For 99% of assemblies, use ±2.5758σ.

RSS

µ = (max + min) ÷ 2
σ = (max − min) ÷ 6
A_max = µ_A + 3σ_A, A_min = µ_A − 3σ_A

The dividing-by-6 assumes a process whose ±3σ spread just fits the tolerance (a capability index of 1.0; see Statistical Tolerancing and Process Capability). Real processes that are better than that give a smaller assembly spread; worse processes, a larger one.

Modified RSS

Many practitioners add a safety factor to RSS (commonly 1.5, from Bender’s work, discussed in Fischer’s Mechanical Tolerance Stackup and Analysis) to allow for drift and non-centered processes: T_assembly = 1.5 × √ΣT². It sits between RSS and worst case and is a convenient compromise when there is little data. It is an industry convention, not a standard.

The Handbook’s example, step by step

Modified RSS

Summing the ± tolerances gives 0.0134 (worst case). Their RSS is 0.0065. With the 1.5 factor: ± 0.0098, about the middle of the assembly range 0.1588: 0.1490 to 0.1686 inch.

Geometric tolerances in a stack

A stack of plus-minus size dimensions is only part of the story. Form, orientation, location and runout tolerances also move features, and they have to be put in the chain as numbers. The Handbook warns that care must be taken to ensure the effects of geometric variation are included in the dimensions of the stack-up.

Tolerance in the loop	How it enters the stack
Position (⌀t)	A feature’s axis may be up to t ÷ 2 from true position in any direction. In a one-dimensional stack the contribution is ± t ÷ 2; add the bonus tolerance if the size is at a modifier condition. Datum shift adds more (see the Position and Location page)
Size at MMC or LMC	The virtual condition (size plus geometric tolerance) is the worst-case boundary for fits and gaps; use it in a worst-case loop
Flatness, perpendicularity, parallelism	Add the full zone width on the surface involved (a flat surface that sits across a gap can be out by its flatness)
Profile	Equal bilateral profile t contributes ± t ÷ 2; a unilateral zone contributes t on one side only
Runout	Contributes its full indicator movement in the direction of the gap, and half of it to the centerline offset
Fits	A bearing or press fit between two parts adds its clearance to the loop wherever the loop passes through the fit (see How Fits Work)

Angles and two dimensions

When parts act at an angle, a dimension’s effect on the assembly is its component in the direction of the gap: ΔA = ΔX × cos θ, where θ is the angle between the dimension and the gap. The factor cos θ is the sensitivity of the assembly to that dimension. A stack-up of 2D or 3D features is a vector loop: write each dimension as a vector, resolve it along the direction you care about, and add the contributions with their sensitivities. Worst case sums |cos θ| × T; RSS sums (cos θ × T)². An angle tolerance contributes the length of the part times the angle in radians.

Assigning tolerances

Assignment spreads a “tolerance budget” among the contributing dimensions so that the assembly meets its target. The Handbook lists the common ways to do it:

Method	How it works
Even assignment	Part feature tolerances are assigned equally among the dimensions
Proportional scaling	Preliminary tolerances are assigned and analyzed. If the assembly tolerance exceeds the allowed amount, the part tolerances are scaled down proportionally
Weight factors	A weight factor, based on manufacturing difficulty, is given to each tolerance; the assembly tolerance is divided in proportion to the weights
Cost minimization	A cost-against-tolerance relationship is determined for each dimension, and an optimization finds the combination of tolerances with the lowest total cost
Total cost minimization	Loosens tolerances to cut cost while also counting the cost of lower yield from defective parts

There are infinitely many ways to assign part tolerances to reach a given assembly tolerance, so analytical solutions are usually not possible; the methods above are rules for choosing a good one.

More worked examples

Example 1: assigning a budget three ways

The assembly gap may vary by ± 0.30 mm in total (the stack is four dimensions, n = 4).

• Even assignment, worst case: T ÷ n = 0.30 ÷ 4 = ± 0.075 mm on each part.
• Even assignment, RSS: T ÷ √n = 0.30 ÷ 2 = ± 0.150 mm on each part: twice as generous, if you accept the statistical risk.
• Weight factors (worst case), with weights 1, 1, 2, 4 for the four parts in order of how hard they are to hold: the budget is split in proportion to the weights:

Part	Weight	Tolerance (± mm)
Part 1	1	0.038
Part 2	1	0.038
Part 3	2	0.075
Part 4	4	0.150

The tolerances add to ± 0.30 mm, as they must, but the hard part gets the largest share.

Example 2: how many parts is “too many” for worst case?

Parts in the chain (each ± 0.05)	Worst case	RSS	Worst case ÷ RSS
3	± 0.150	± 0.087	1.73×
5	± 0.250	± 0.112	2.24×
7	± 0.350	± 0.132	2.65×
10	± 0.500	± 0.158	3.16×
15	± 0.750	± 0.194	3.87×

Worst case grows with n and RSS with √n, so the gap between them widens as the chain gets longer. The Handbook’s “seven or fewer” guideline for worst case reflects this: at seven dimensions worst case is already 2.6 times the RSS result.

Example 3: a vector loop component

A pin is toleranced ± 0.10 mm along a line at 30° to the gap direction. Its effect on the gap is 0.10 × cos 30° = ± 0.0866 mm. At 90° it contributes nothing, at 0° all of it.

Troubleshooting

Symptom	Likely cause	Try
Assemblies fail although the stack-up passed	A contributor was missing from the loop, a sign was wrong, or a geometric tolerance was left out	Rebuild the chain from the drawing, check each sign, and add position, form and fit contributions
Statistical result looks too good	Parts are not centered or not normally distributed; the process has drifted	Use measured means, or Monte Carlo, or modified RSS; check the process with SPC
Worst case demands impossible tolerances	Too many dimensions in the chain, or a loose assembly requirement	Shorten the chain (datum from one feature), or accept a small defect rate with RSS
The same assembly dimension depends on a fit	Clearance of a bearing or pin was not counted	Add the fit’s clearance (for example H7/g6) as a contributor
An angle’s effect is much larger than expected	A small angle error over a long arm	Use the leg length times the angle in radians, or tolerance the end position directly

Related

• Fits & Tolerances: the overview and quick reference.
• Dimensional Tolerancing Basics: chain and baseline dimensioning.
• Statistical Tolerancing and Process Capability: Cp, Cpk and defect rates.
• Inspection and Tolerance Stack-up: a basic one-dimensional stack-up and the fastener formulas.
• Tolerance Stack-up Calculator: worst case, RSS, modified RSS and Monte Carlo.

FAQ

Which method should I use?

Worst case if every part must fit and the chain is short; statistical or RSS when there are many dimensions and the processes are capable and centered; Monte Carlo when the data are not normal or you need a defect rate. Many designs check worst case first, then relax.

Why is the RSS range so much narrower than worst case?

Because it is very unlikely that every part will be at its extreme in the same direction at once. RSS exploits that, which is why it needs the assumptions of independent, centered, roughly normal dimensions.

Where does σ = (max − min) ÷ 6 come from?

It assumes the tolerance band is ±3σ, which gives 99.73% of parts inside the limits for a centered normal distribution (capability index 1.0). If your process is better than that, σ is smaller.

How do I handle a dimension with a bilateral and a unilateral tolerance?

Convert every dimension to a mean (the middle of its range) and an equal ± tolerance before you add. A dimension 25.00 +0.10 −0.00 becomes 25.05 ± 0.05.

Should I include a safety factor?

On RSS, many companies do (the modified RSS factor of 1.5); on worst case, no, because it already assumes the worst. Choose a factor your quality system can justify with data.

Can a stack-up replace a gage or a test?

No. It predicts the assembly from the drawing; it does not know the actual parts. Verify critical assemblies with measured parts or a functional test.

References

• Machinery’s Handbook, 31st edition, Industrial Press: “Tolerance Analysis and Assignment” (the tolerance stack-up chain; worst-case, statistical, RSS and Monte Carlo methods; tolerance assignment) and “Statistical Analysis of Manufacturing Data”.
• Fischer, B. R., Mechanical Tolerance Stackup and Analysis, CRC Press (modified RSS, geometric tolerances in stack-ups, vector loops).
• Bender, A., “Statistical Tolerancing as it Relates to Quality Control and the Designer”, Automotive Division Newsletter of ASQC, 1968 (the 1.5 factor).
• ASME Y14.5-2018, Dimensioning and Tolerancing.

Last reviewed: October 2026. The worst-case, statistical and RSS results reproduce the Handbook’s worked example.