Example 2 — Pressure transducer, quadratic fit (auto degree selection)

Exercises: automatic polynomial-degree selection (AIC), non-linearity detection, and the t-distribution / Welch-Satterthwaite coverage-factor path.

2.0 Setup

One asset, one channel:

Leave measurement_uncertainty blank (not used in this example).

2.1 Calibration wizard settings

Type external (no reference asset needed), Regression degree: Auto (the point of this example), distribution t-distribution, confidence 95% (the t-distribution path derives its own coverage factor from the confidence level and the fit's effective degrees of freedom), decision rule Simple acceptance.

Data points — a transducer with a small but real quadratic non-linearity across its range:

The math

Open Gauge tries degree 1 through 5 and picks the lowest degree whose AIC isn't beaten by more than 2 points by a higher degree (the parsimony rule). For this data it selects degree 2:

$$\text{reference} = 0.00040 \cdot \text{measured}^2 + 0.98014 \cdot \text{measured} + 0.50917$$

(Try forcing "Regression degree: 2" explicitly — you should get the same coefficients as the auto-selected result, confirming AIC picked the same degree you'd pick by eye from the data.)

$n = 8$, $k = 3$ (degree 2 → 3 coefficients) → residual dof $= 5$. Max error = 0.169881 (point 2); span $= 882.38 - 0.6 = 881.78$ → %FS error $= 0.169881/881.78 \times 100\% = \mathbf{0.0193\%}$.

Non-linearity = 3.17% FS — the deviation of the quadratic fit from its own best-fit straight line. This is the number confirming "yes, a straight line would have been a noticeably worse fit" — that degree 2 was the right call, not degree 1.

Uncertainty budget:

• Combined: $u_c = \sqrt{0.119470^2 + 0.144338^2} = \mathbf{0.187367\ kPa}$.
• $\nu_{\text{eff}} = u_c^4 / (u_A^4/5) \approx \mathbf{30.2}$ (only fit_residuals has finite dof).
• Expanded (t-distribution): $k = t_{0.975,\,\nu=30.2} \approx 2.042 \Rightarrow U = 0.187367 \times 2.042 = 0.382523 \approx \mathbf{0.38\ kPa}$ (2 sig figs).

Decision rule (simple acceptance, spec = ±1.0% FS = ±8.818 kPa):

$$\text{max error} = 0.169881 \le 8.818 \implies \textbf{CONFORMS}$$

(comfortably — this is a "good" calibration example).

Expected results