True Position Calculator (GD&T): Positional Deviation, MMC/LMC Bonus Tolerance and Pass/Fail
Find the GD&T true position deviation 2√(ΔX² + ΔY²) from measured coordinates, add MMC or LMC bonus tolerance and see if the hole or pin passes inspection.
How it works
True position (Ø) = 2 × √(ΔX² + ΔY²), or 2 × √(ΔX² + ΔY² + ΔZ²) for a spherical zone · ΔX = measured X − basic X · Bonus at MMC = |actual size − MMC size| · Bonus at LMC = |actual size − LMC size| · Allowed tolerance = position tolerance + bonus · Pass when true position ≤ allowed tolerance and the size is within its limits
A position tolerance in GD&T is a cylinder (a circle in the drawing view) centered on the true position, the exact location given by the basic dimensions. The feature axis must sit inside it. Because the zone is a diameter, the measured radial offset √(ΔX² + ΔY²) is doubled before it is compared with the tolerance. A part that is 0.004 off in X and 0.003 off in Y is 0.005 from true position, which is a Ø0.010 deviation. An Ⓜ (MMC) modifier gives bonus tolerance as the feature departs from its maximum material size: a hole larger than its smallest limit, or a pin smaller than its largest limit, can move further and still assemble. An Ⓛ (LMC) modifier works the other way and is used to protect wall thickness. With no modifier (RFS) there is no bonus. The calculator also shows the virtual condition, the worst-case boundary a functional gauge pin or hole checks, and the ± per-axis tolerance of the square that fits inside the round zone, to show how much more a round zone allows than coordinate tolerancing. The tolerance stack-up calculator on this site adds up linear tolerances along a chain of parts; this one checks a single feature against its position callout. It does not include datum shift from datum features of size, or projected tolerance zones.
Worked example
With the defaults, a hole with basic location X 2.000, Y 1.500 measures X 2.004, Y 1.497, so ΔX = +0.0040 and ΔY = −0.0030 in. True position = 2 × √(0.004² + 0.003²) = 2 × 0.00500 = Ø0.0100 in. The callout is Ø0.010 at MMC, and the hole size is 0.502–0.508 in, so the MMC size is 0.502. The hole measures 0.505, giving a bonus of 0.505 − 0.502 = 0.0030 in and an allowed tolerance of Ø0.0130 in. The deviation uses 76.9% of the tolerance with 0.0030 in to spare, so the hole passes. The virtual condition (gauge pin) is 0.502 − 0.010 = Ø0.4920 in.
Sources: Omni Calculator: True position (PV = 2√(ΔB² + ΔC²), MMC and LMC bonus examples) · GD&T Basics: True position (diametral deviation, bonus tolerance, gauge formulas, .003/.002 example)
Frequently asked questions
What is the formula for true position?
True position = 2 × √(ΔX² + ΔY²), where ΔX and ΔY are the measured minus the basic (true) coordinates. The factor 2 converts the radial offset into a diameter so it can be compared directly with the Ø position tolerance.
How do you calculate bonus tolerance at MMC?
Take the difference between the actual feature size and its MMC size. For a hole the MMC is the smallest allowed size, so bonus = actual − smallest size; for a pin the MMC is the largest size, so bonus = largest size − actual. Add the bonus to the position tolerance.
Why is true position multiplied by 2?
Position tolerances are stated as a diameter (Ø), but √(ΔX² + ΔY²) is the radius of the offset. Doubling it lets you compare like with like. Forgetting the 2 makes a part look twice as good as it is.
How much more tolerance does a round position zone give than ± coordinates?
The largest square that fits inside a round zone has a half-width of (Ø ÷ 2) × 0.707. A Ø0.010 zone therefore matches ±0.0035 in X and Y, but it also accepts diagonal offsets up to 0.005, about 57% more area than the square.
What happens if the hole size is out of tolerance?
The part is rejected for size regardless of its position. Bonus tolerance is only available while the feature is within its size limits.
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This calculator is for educational purposes. Double-check important results before acting on them.