= Pt100 local linear model
{c}
{title2=$R=R_0(1+\alpha t)$}
Choose the illustrative local model
$$
R(t)=R_0(1+\alpha t),\qquad
R_0=100\,\Omega,\qquad
\alpha=0.00385\,({}^\circ\mathrm C)^{-1}.
$$
Here $t$ is the Celsius <temperature>, measured relative to $0\,{}^\circ\mathrm C$, so $\alpha t$ is dimensionless. For this chosen <affine function>, the <derivative> and inverse relation are
$$
\frac{dR}{dt}=R_0\alpha=0.385\,\Omega/{}^\circ\mathrm C,\qquad
t=\frac{R/R_0-1}{\alpha},\qquad
\Delta t=\frac{\Delta R}{0.385\,\Omega/{}^\circ\mathrm C}.
$$
Calculated values are $R(0\,{}^\circ\mathrm C)=100\,\Omega$, $R(20\,{}^\circ\mathrm C)=107.70\,\Omega$, $R(25\,{}^\circ\mathrm C)=109.625\,\Omega$, and $R(50\,{}^\circ\mathrm C)=119.25\,\Omega$. These are model outputs, not measurements or an assertion that any real <Pt100> follows this line exactly. The number $0.00385$ here is the chosen slope parameter; it is not a universal local <derivative> for all <platinum resistance thermometers>. Curvature, manufacturing differences, and the fitted calibration interval matter. A wider-range calibration uses a validated response curve rather than extrapolating this line indefinitely.
\Image[/platinum-resistance-temperature-calibration.png]
{title=Chosen Pt100 line and an illustrative calibration with an independent check}
{id=platinum-resistance-temperature-calibration}
{description=The left panel plots the <Pt100 local linear model>. The right panel uses stipulated values for a different illustrative element: two points set a line, while a third point tests it. Every point is synthetic; none is measured calibration data or a tolerance band.}
{source=platinum-resistance-temperature-calibration.py}
For a separate synthetic <thermometer calibration>, stipulate $R_a=100.20\,\Omega$ at $t_a=0\,{}^\circ\mathrm C$ and $R_b=119.60\,\Omega$ at $t_b=50\,{}^\circ\mathrm C$. A two-point <affine function> has slope $0.388\,\Omega/{}^\circ\mathrm C$ and gives
$$
t=\frac{R-100.20\,\Omega}{0.388\,\Omega/{}^\circ\mathrm C}.
$$
Thus $R=109.90\,\Omega$ would indicate $25\,{}^\circ\mathrm C$. If an additional synthetic check at reference $25\,{}^\circ\mathrm C$ instead supplied $110.10\,\Omega$, the line would indicate $25.515\ldots\,{}^\circ\mathrm C$. The $0.20\,\Omega$ residual exposes disagreement that two fitted points cannot reveal. This separate example changes both offset and slope; it is not a replacement specification for the nominal <Pt100 local linear model>.
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