9.4 Coordinate Conversion and Appendix B Position Calculations

Coordinate Conversion and Appendix B Position Calculations

Quick Answer: Square coordinate zone to diametral position zone: position Ø = 2t√2 ≈ 2.83t for a ±t square zone. Position back to worst-case coordinate: ±(Ø/2)/√2 ≈ ±0.354Ø. Actual deviation from measured coordinates: Ø_actual = 2√(Δx² + Δy²). Accept when Ø_actual ≤ stated tolerance + bonus.

The Senior examination competencies include performing calculations associated with GD&T to determine geometric requirements, including formulas from Nonmandatory Appendix B. Appendix B is the positional-tolerancing formula annex, and it is named explicitly in the body of knowledge for every revision. Section 9.2 covered the fixed and floating fastener formulas that live there. This section covers the other half: converting between coordinate and positional systems, and evaluating measured results.

Why coordinate tolerancing fails

A hole dimensioned 25 ± 0.2 in x and 40 ± 0.2 in y sits inside a square tolerance zone 0.4 on a side. Two problems follow:

1. The zone shape does not match the function. A pin either fits the hole or it does not, and the clearance available is the same in every direction. A square zone permits 0.283 of deviation on the diagonal but only 0.2 along an axis — it is stingy where the function does not care and generous where the function does.
2. The zone is not tied to datums. Plus-minus dimensions carry no explicit datum reference frame, so different inspectors set up differently and get different answers.

Position tolerancing fixes both. The zone becomes a cylinder whose axis is at true position, established by basic dimensions from an explicit datum reference frame.

Converting coordinate to position

Take the square zone and circumscribe a circle around it. Any point the square permits, the circle also permits; the circle additionally permits the corners' worth of extra area.

For a square zone of ±t in each direction (total width 2t per side):

• Half-diagonal = t√2
• Diametral position zone = 2 × t√2 = 2t√2 ≈ 2.83t

Worked example. A hole is dimensioned ±0.2 in x and ±0.2 in y. So t = 0.2.

• Half-diagonal = 0.2 × 1.414 = 0.283
• Position tolerance = 2 × 0.283 = Ø0.566, rounded in practice to Ø0.56

Note what happened: the designer converted to position, and the hole gained permissible deviation in every direction except exactly on the diagonals, where it broke even. That is not a giveaway — it is the removal of an artificial restriction the square zone imposed for no functional reason.

The 57% figure

Compare areas. The square of side 2t has area 4t². The circumscribed circle of diameter 2t√2 has radius t√2 and area π(t√2)² = 2πt² ≈ 6.283t².

6.283t² ÷ 4t² = 1.571

The cylindrical zone offers about 57% more area than the square zone it replaces. Memorize the number; it appears in exam questions about why position tolerancing increases yield without loosening the functional requirement.

Converting position back to coordinate

Sometimes you need the reverse: a machine tool programmed in coordinates, or a legacy inspection routine that reports x and y only. Inscribe the square inside the circle.

For a position tolerance of ØT:

• Radius = T/2
• Half-side of the inscribed square = (T/2)/√2 = 0.354T
• Equivalent coordinate tolerance = ±0.354T in each direction

Worked example. Position Ø0.5 converts to approximately ±0.177 in x and y.

This conversion is conservative — every part that passes the coordinate check will pass the position check, but some parts that pass position will fail the coordinate check. Never use the inscribed conversion to reject a part that was specified with position; use it only to program or to screen.

Computing actual position deviation

An inspector reports a hole's centre as deviating Δx = 0.06 and Δy = 0.08 from true position. What is the actual position deviation?

The deviation of the axis from true position is the radial distance √(Δx² + Δy²). Because position tolerance zones are stated as diameters, the reported actual position must also be a diameter:

Ø_actual = 2 × √(Δx² + Δy²)

• √(0.06² + 0.08²) = √(0.0036 + 0.0064) = √0.01 = 0.1
• Ø_actual = 2 × 0.1 = Ø0.20

Forgetting the factor of two is the most common arithmetic error on the Senior exam. A radial deviation of 0.1 is a Ø0.2 position deviation.

Deciding acceptance with bonus

Acceptance is a single comparison:

Accept when Ø_actual ≤ T_stated + bonus

where bonus is the departure from the applicable material condition, and is zero when the callout is RFS.

Worked example. A hole is Ø10.0 +0.3/−0 with position Ø0.2 at MMC to A|B|C. The hole is produced at Ø10.2, with measured deviations Δx = 0.08, Δy = 0.15.

1. MMC of the hole = 10.0 (smallest). Bonus = 10.2 − 10.0 = 0.2.
2. Total available position tolerance = 0.2 + 0.2 = Ø0.4.
3. Actual deviation = 2√(0.08² + 0.15²) = 2√(0.0064 + 0.0225) = 2√0.0289 = 2 × 0.17 = Ø0.34.
4. 0.34 ≤ 0.40, so the hole passes.

Had the callout been RFS, bonus would be zero, the available tolerance would be Ø0.2, and the same hole at Ø0.34 actual would fail. Same part, same measurements, different answer — which is the whole reason material condition modifiers exist.

Quick-reference table

Traps

1. Dropping the factor of 2 when converting measured coordinates to an actual position diameter.
2. Using the inscribed conversion to reject a part specified with position; it is conservative, not equivalent.
3. Adding bonus on an RFS callout. No modifier means RFS by default, and RFS earns no bonus.
4. Mixing radial and diametral values inside one calculation — decide which you are in and stay there.
5. Converting the coordinate tolerance instead of the coordinate deviation. The 2.83 factor converts zones; the factor of 2 converts measured deviations.