To Inverse
The Steinhart Equation models the relationship between the temperature and resistance of a thermistor. It’s defined like this:
\[\frac{1}{T} = A + B \ln R + C\ln^3R\]\(A\), \(B\) and \(C\) are the coefficients used to determine the thermistor.
\[\frac{1}{T} - A = B \ln R + C\ln^3R\] \[\frac{\frac{1}{T} - A}{C} = \frac{B \ln R}{C} + \ln^3R\]By using Cardano’s method of the cubic equation we can determine \(\ln R\). To solve the depressed cubic we define the following equations.
\[\frac{B}{C} = 3st \rightarrow s = \frac{B}{3tC}\] \[\frac{\frac{1}{T} - A}{C} = s^3 - t^3\]Substituting for \(s\) makes:
\[\left(\frac{B}{3tC}\right)^3 - t^3 = \frac{\frac{1}{T} - A}{C}\] \[\frac{B^3}{27t^3C^3} - t^3 = \frac{\frac{1}{T} - A}{C}\] \[t^6 + t^3 * \frac{\frac{1}{T} - A}{C} - \frac{B^3}{27C^3} = 0\]Substituting \(t^3 = z\) allows us to solve the equation.
\[z^2 + z * \frac{\frac{1}{T} - A}{C} + \left(\frac{\frac{1}{T} - A}{2C}\right)^2 - \left(\frac{\frac{1}{T} - A}{2C}\right)^2 - \frac{B^3}{27C^3} = 0\] \[\left(z + \frac{\frac{1}{T} - A}{2C}\right)^2 = \left(\frac{\frac{1}{T} - A}{2C}\right)^2 + \left(\frac{B}{3C}\right)^3\]To shorten up we can define \(x = \frac{\frac{1}{T} - A}{2C}\) and \(y = \sqrt{x^2 + \left(\frac{B}{3C}\right)^3}\).
\[z = \sqrt{x^2 + \left(\frac{B}{3C}\right)^3} - x = y - x\]We substitute \(t\) back in.
\[t = \sqrt[3]{y - x}\]Now that we have \(t\) we can substitute it to get \(s\). We continue using \(s^3 - t^3 = 2x\).
\[s^3 - (y - x) = 2x\] \[s^3 = x + y = y + x\] \[s = \sqrt[3]{y + x}\]Cardano’s method states that \(y = s - t\). Thus we get the solution:
\[\ln R = \sqrt[3]{y + x} - \sqrt[3]{y - x}\]\(R = e^{\sqrt[3]{y + x} - \sqrt[3]{y - x}}\) with \(x = \frac{\frac{1}{T} - A}{2C}\), \(y = \sqrt{x^2 + \left(\frac{B}{3C}\right)^3}\)
From Inverse
To bring back the equation to its original form we define a simpler version:
\(c = a - b\) with \(a = \sqrt[3]{y + x}\), \(b = \sqrt[3]{y - x}\), \(c = \ln R\)
Thus:
\[a^3 = y + x \qquad b^3 = y - x\] \[a^3b^3 = y^2 - x^2 = (ab)^3\] \[ab = \sqrt[3]{y^2 - x^2}\]Expressing \(c\) as a cubic binominal enables us to get \(\ln R\).
\[c^3 = (a - b)^3 = a^3 - b^3 + 3ab^2 - 3a^2b = a^3 - b^3 - 3ab(a - b)\]Notice we can substitute for \(ab\), \(a - b\), \(a^3\) and \(b^3\) which leads us to the solution.
\[c^3 = y + x - (y - x) - 3\sqrt[3]{y^2 - x^2} * c\] \[\ln^3R = 2x - 3\sqrt[3]{y^2 - x^2} * \ln R\] \[\ln^3R = \frac{\frac{1}{T} - A}{C} - 3\sqrt[3]{\left(\sqrt{x^2 + \left(\frac{B}{3C}\right)^3}\right)^2 - x^2} * \ln R\] \[\ln^3R = \frac{\frac{1}{T} - A}{C} - \frac{B}{C} * \ln R\] \[C\ln^3R + B\ln R + A = \frac{1}{T}\]