Example 1 — Temperature RTD, linear fit, full uncertainty budget

Exercises: linear regression, all four Type A/B budget contributions at once (fit residuals, reference standard, resolution, sensor nominal accuracy), the reference-standard-uncertainty auto-fetch feature, and the three decision rules diverging on identical data.

Every number below was produced by running Open Gauge's real calculation engine against the input data shown, then independently re-derived by hand from the underlying formulas to confirm they match. If you enter the same setup and data through the UI, you should see the same numbers (up to the display rounding described at the end).

This example has two calibrations: first you calibrate a reference thermometer (so it has an expanded uncertainty on file), then you calibrate the working sensor against that reference — this is what makes Open Gauge auto-populate the "reference standard" budget row.

1.0 Setup — two assets

Asset A — reference standard (Assets → New Asset):

One channel:

Leave accuracy/resolution/uncertainty blank on this asset — only its calibration record's expanded uncertainty matters here (computed in 1.1).

Asset B — working sensor:

One channel:

Leave "Output signal unit" blank on both channels so the measured-value unit defaults to °C — this avoids unit-conversion complexity for this example.

1.1 Calibrate the reference standard first

On Asset A, start a calibration: type internal (or external — doesn't matter here, no reference asset needed), distribution Normal, confidence 95%, no accuracy spec needed (skip channel accuracy — it isn't set on this channel).

Data points (reference = a higher-tier standard's value; measured = this reference thermometer's reading):

The math

Fitting $\text{reference} = a \cdot \text{measured} + b$ (degree 1) to these 3 points gives $a = 1.00025$, $b = -0.01750$.

Residuals (reference − calculated):

$n = 3$ points, $k = 2$ parameters → residual degrees of freedom $= 1$.

• Type A standard uncertainty = sample std dev of residuals ($\text{ddof}=1$) = 0.012994 °C.
• No Type B contributions configured on this channel, so the budget has exactly one row:

• Combined uncertainty: $u_c = \sqrt{0.012994^2} = \mathbf{0.012994\ °C}$.
• Coverage factor: for a normal distribution at 95% confidence, $k = \Phi^{-1}(0.975) \approx \mathbf{1.96}$.
• Expanded uncertainty: $U = k \cdot u_c = 1.96 \times 0.012994 = 0.025467 \approx \mathbf{0.025\ °C}$ (2 sig figs).

Expected results in the wizard (Step 3)

Save this calibration. Write down the Expanded uncertainty (0.025 °C) — Asset B's calibration fetches this automatically in the next step.

1.2 Calibrate the working sensor against the reference standard

On Asset B, start a calibration: type internal, reference asset Reference PT100 Standard (Asset A) — as soon as you pick it, the wizard should show "Ref. standard U: 0.025 °C (last calibration of Reference PT100 Standard)." If it shows a manual-entry field instead, the fetch didn't find a calibration on Asset A — go back and confirm 1.1 was saved.

Distribution Normal, confidence 95%, decision rule Guard band (tolerance − U), and check "Incl. sensor nominal accuracy" (folds in the channel's 0.3 °C manufacturer spec).

Data points:

The math

Fit: $a = 1.00038$, $b = -0.04739$.

$n = 6$, $k = 2$ → residual dof $= 4$. Max error = 0.116956 (point 5) → %FS error = $0.116956/100 \times 100\% = \mathbf{0.117\%}$.

Uncertainty budget (four rows this time):

• Combined: $u_c = \sqrt{0.083509^2 + 0.012734^2 + 0.028868^2 + 0.150000^2} = \mathbf{0.174554\ °C}$.

• Effective degrees of freedom (Welch-Satterthwaite — only fit_residuals has finite dof; the three Type B rows are exactly-known and drop out of the sum):

  $$\nu_{\text{eff}} = \frac{u_c^4}{u_A^4 / \text{dof}_A} = \frac{0.174554^4}{0.083509^4/4} \approx 76.4$$

• Expanded (normal distribution — $\nu_{\text{eff}}$ isn't used for "normal," only for "t"/"chi_squared"): $k = \Phi^{-1}(0.975) \approx 1.96$, $U = 1.96 \times 0.174554 = 0.342120 \approx \mathbf{0.34\ °C}$ (2 sig figs).

Decision rule (guard band, spec = ±0.5 °C absolute):

$$\text{guard} = U = 0.342120, \quad \text{max error} + \text{guard} = 0.116956 + 0.342120 = 0.459076 \le 0.5 \implies \textbf{CONFORMS}$$

Expected results

1.3 Bonus check — watch the three decision rules disagree

Repeat 1.2 with the channel's Accuracy value changed to 0.4 (edit the channel, save, then re-run the same 6 points through Step 3 of a new calibration) and try all three decision rules with "Incl. sensor nominal accuracy" still checked. Everything above is unchanged (same fit, same budget, same $U = 0.342120$) — only the pass/fail flips:

This is the cleanest way to confirm the decision-rule feature is wired correctly: identical data and identical spec, three different verdicts depending only on which rule is selected.