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Gage R&R Calculator: Formula, Example & Instant Result

A Gage R&R study shows how much of the variation you see in measurements comes from the measuring system itself rather than from the parts. Two appraisers measure the same parts several times; the study separates repeatability (the gauge) from reproducibility (the people) and compares their combined spread with the part-to-part variation and the tolerance.

↳ Values are pre-filled with an example — replace them with your own data.

Your data

Appraiser A
T1T2
#1
#2
#3
#4
#5
Appraiser B
T1T2
#1
#2
#3
#4
#5

Leave empty to judge against study variation only.

Paste readings (CSV / spreadsheet)

One row per part: first all parts of appraiser A, then all parts of appraiser B. Separate the trials with tabs, semicolons or spaces.

Result

%GRR of toleranceMarginal29.6 %

Share of the tolerance used up by measurement error (6 × GRR).

%GRR of study variationMarginal26.3 %

Share of the observed variation that comes from measuring.

Distinct categories (ndc)5

The system can separate the parts into this many groups — enough for process control.

Repeatability (EV)0.0266

Spread from the gauge: same person, same part, repeated.

Reproducibility (AV)0.0416

Extra spread because different people measure.

Gage R&R (GRR)0.0494

Combined measurement-system spread (one standard deviation).

Part variation (PV)0.1813

Spread of the parts themselves.

14.1EV22.1AV26.3GRR96.5PV
Components of variation as a share of total variation: the measurement system (GRR) takes 26.3 %, part-to-part variation 96.5 %.

Formula

EV (repeatability) = R̄ × K1    (K1 = 0.8862 for 2 trials, 0.5908 for 3)
AV (reproducibility) = √( (X̄diff × K2)² − EV² / (n × m) )    (K2 = 0.7071 for 2 appraisers; 0 if negative)
GRR = √(EV² + AV²)
PV (part variation) = Rp × K3    (K3 by number of parts)
TV (total variation) = √(GRR² + PV²)
%GRR of study variation = 100 × GRR / TV
%GRR of tolerance = 100 × 6 × GRR / tolerance
ndc = 1.41 × PV / GRR, rounded down

Worked example

Two inspectors measure five turned shafts twice each with the same micrometer. The diameter tolerance is 1.00 mm. Is the micrometer good enough?

Data

  • • Average range of repeat readings R̄ = 0.030 mm (2 trials → K1 = 0.8862)
  • • Difference between the two inspectors' averages X̄diff = 0.060 mm
  • • Range of the five part averages Rp = 0.450 mm (5 parts → K3 = 0.4030)
  • • Tolerance = 1.00 mm; n = 5 parts, m = 2 trials

Step by step

  1. 1EV = 0.030 × 0.8862 = 0.0266 mm
  2. 2AV = √((0.060 × 0.7071)² − 0.0266² / 10) = √(0.001800 − 0.000071) = 0.0416 mm
  3. 3GRR = √(0.0266² + 0.0416²) = 0.0494 mm
  4. 4PV = 0.450 × 0.4030 = 0.1813 mm; TV = √(0.0494² + 0.1813²) = 0.1880 mm
  5. 5%GRR of study variation = 0.0494 / 0.1880 = 26.3 %
  6. 6%GRR of tolerance = 6 × 0.0494 / 1.00 = 29.6 %
  7. 7ndc = 1.41 × 0.1813 / 0.0494 = 5.2 → 5

What it means

Both views land in the marginal band (10–30 %): the micrometer may be acceptable for this application, but only just. Reproducibility (0.0416) is larger than repeatability (0.0266), so the inspectors measure differently — align the method (where on the shaft, how much pressure) before buying a better gauge. With a tolerance, the tolerance view is usually the one to decide on; ndc = 5 means the system can just separate the parts into five groups.

Tips & common mistakes

  • Pick parts that span the real process range; ten nearly identical parts make any gauge look bad on the study-variation view.
  • Measure in random order and hide part numbers so appraisers cannot remember earlier readings.
  • Use the same procedure and fixture the operators use in production, not the lab's best practice.
  • Check resolution first: the gauge should read in steps of a tenth of the tolerance or finer.
  • AI assistants often drop the EV²/(n·m) correction in the AV formula or confuse %study with %tolerance; check their numbers against this calculator.

AI prompt for this calculation

Paste it into your AI assistant to be walked through the calculation with your own data.

Act as a measurement systems engineer and help me evaluate a Gage R&R study (average and range method, 2 appraisers).

Use exactly these formulas:
EV = R̄ × K1 (K1 = 0.8862 for 2 trials, 0.5908 for 3)
AV = √((X̄diff × K2)² − EV²/(n × m)), K2 = 0.7071, set AV to 0 if the value under the root is negative
GRR = √(EV² + AV²)
PV = Rp × K3 (K3 for n parts: 2→0.7071, 3→0.5231, 4→0.4467, 5→0.4030, 6→0.3742, 7→0.3534, 8→0.3375, 9→0.3249, 10→0.3146)
TV = √(GRR² + PV²)
%GRR of study variation = 100 × GRR / TV; %GRR of tolerance = 100 × 6 × GRR / tolerance
ndc = 1.41 × PV / GRR, rounded down

Rules:
1. Work step by step and show every intermediate value.
2. Restate my readings first and check the decimal separators.
3. If any reading, the number of trials or the tolerance is missing or unclear, ask me — never assume.
4. Judge both percentages against <10 % acceptable, 10–30 % marginal, >30 % unacceptable, and explain if they disagree.
5. Tell me whether repeatability or reproducibility dominates and what to fix first.

My study:
Number of parts = 
Trials per part = 
Tolerance (USL − LSL) = 
Appraiser A readings (one row per part) = 
Appraiser B readings (one row per part) = 

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Frequently asked questions

What is the difference between repeatability and reproducibility?

Repeatability (EV, equipment variation) is the spread when the same person measures the same part several times with the same gauge. Reproducibility (AV, appraiser variation) is the extra spread caused by different people measuring. A large EV points at the gauge or fixture; a large AV points at method and training.

Why can %GRR against study variation and against tolerance disagree?

They compare the same measurement error with different things. The study view compares it with the variation of the parts you picked, so a set of very similar parts makes it look large. The tolerance view compares it with what the specification allows. When a tolerance exists, it is usually the index to decide on; if you also use the gauge for process control, the study view matters too.

What does ndc (number of distinct categories) mean?

It estimates how many groups of parts the measurement system can reliably tell apart within the part variation. Fewer than 5 means the gauge cannot distinguish parts well enough for process control, even if the tolerance view looks acceptable.

What should I do when a study fails?

Look at which component dominates. High repeatability: check gauge resolution, wear, fixturing and the measuring point. High reproducibility: standardize the method, train appraisers and use the same fixture. Also check for bias and linearity with a reference part before investing in new equipment.

Last tested: October 2026 · Tested with: automated math tests (3+ cases) · Reviewed by: Quality for Everyone editorial team