One reference for the Steinhard-Hart equation gives:Not too comfortable. But if I get the formula(s) and where to find the starting numbers, I'd like to try.
\(
\frac{1}{T} = A\;+\;B\ln(R)\;+\;C\ln^2(R)\;+\;D\ln^3(R)
\)
Now, I hate this equation because the units aren't consistent, specifically, you can't take the logarithm of anything with units. We could normalize this away pretty easily by simply dividing each of the R's by Ro=1Ω, but my understanding is that if you normalize the equation around the center of your range of interest (or at least something that is close to your range), then you will tend to get something that is a better fit over that range.
\(
\frac{1}{T}\;=\;A\;+\;B\ln \left(\frac{R}{R_0}\right)\;+\;C\ln^2\left(\frac{R}{R_0}\right)\;+\;D\ln^3\left(\frac{R}{R_0}\right)
\)
So, pick four temperatures, T_0 through T_3, and measure the resistances, R_0 through R_3, at those temperatures. Chose T_0 to be the one you are normalizing to. You then have the following four equations:
\(
\frac{1}{T_0}\;=\;A\;+\;B\ln \left( \frac{R_0}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_0}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_0}{R_0} \right)
\;
\frac{1}{T_1}\;=\;A\;+\;B\ln \left( \frac{R_1}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_1}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_1}{R_0} \right)
\;
\frac{1}{T_2}\;=\;A\;+\;B\ln \left( \frac{R_2}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_2}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_2}{R_0} \right)
\;
\frac{1}{T_3}\;=\;A\;+\;B\ln \left( \frac{R_3}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_3}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_3}{R_0} \right)
\)
Now, note that by choosing this normalization, you get 'A' for free since Ro/Ro=1 and ln(1)=0. So A = 1/To.
\(
\frac{1}{T_0}\;=\;A
\;
\frac{1}{T_1}\;-\;\frac{1}{T_0}\;=\;B\ln \left( \frac{R_1}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_1}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_1}{R_0} \right)
\;
\frac{1}{T_2}\;-\;\frac{1}{T_0}\;=\;B\ln \left( \frac{R_2}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_2}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_2}{R_0} \right)
\;
\frac{1}{T_3}\;-\;\frac{1}{T_0}\;=\;B\ln \left( \frac{R_3}{R_0} \right) \;+\;C\ln^2 \left( \frac{R_3}{R_0} \right) \;+\;D\ln^3 \left( \frac{R_3}{R_0} \right)
\)
Everything above, except the B, C, and D, are just numbers that you can calculate so that you have three equations of the form:
\(
b_1B\;+\;c_1C\;+\;d_1D=a_1
\;
b_2B\;+\;c_2C\;+\;d_3D=a_2
\;
b_3B\;+\;c_3C\;+\;d_2D=a_3
\)
which are three linear equations in B, C, and D.