I was wondering it's possible to calculate a formula from the graph that I have attached below. (the purple line 85% RH can be ignored)

  • coefficient = 1.7 at -10 degrees and 33 RH
  • coefficient = 1.4 at 0 degrees and 33 RH
  • coefficient = 1 at 20 degrees and 33 RH
  • coefficient = 0.9 at 50 degrees and 33 RH

I would need the formula so that I could calculate the coefficient at any temperature knowing only Ro, Rs and the temperature.

E.g. how much will the coefficient be at -5 degrees? or how much at +15 degrees?

this is the image of the graph

  • $\begingroup$ Do you know if it should be linear? quadratic or exponential? $\endgroup$ – Jean-Sébastien Sep 24 '12 at 15:36
  • 2
    $\begingroup$ unfortunately I do not have this information, the only datasheet has that piece of informartion from -10 to +50. The good thing is that I won't need to use any values that are above +50 or below -10 $\endgroup$ – martin Sep 24 '12 at 15:50

Without further knowledge about the underlying physics, I'd say that $f(x) = \frac75 - \frac 3{100} \cdot x$ is a good approximation for $-10\le x \le 5$ and $f(x) = \frac{16}{15} - \frac1{300} \cdot x$ is a good approximation for $20\le x \le 50$. How to fill the gap is not so clear. An exponential or quadratic interpolation might look fine. More data would help.


Without more information about the functional form you expect, we can't generate a function that works for every possible value of the dependent variable. If you're only interested in interpolating that data (i.e. you don't care about values outside of [-10,50]), it looks like your best bet is linear interpolation. $\textbf{EDIT}:$ I see from your recent comment that you're only interested in values in$[-10,50]$, so this would be an appropriate method for these data.

So, for example, to find the value at -5, we'd find the equation of the line connecting -10 and 0. If we let $y$ be the 'coefficent' and $x$ be degrees, that equation is: $$\frac{1.7-1.4}{-10-0}x + 1.4 = -.03x + 1.4$$ Then, plugging in -5 for $x$, we get: $$.03(-5)+1.4 = 1.25$$ This kind of interpolation only works because the lines between the points look roughly linear. You need to be cautious and look for when it would over- or underestimate the actual line. For example, interpolating 0 to 20 would probably overestimate as the curve is more pronounced in that region.

Also, interpolation only works for data between the two points we're interpolating. So, we would not expect the above equation to give reasonable answers for 15 degrees. To estimate that, we would interpolate for 0 to 20 (and from 20 to 50 if we wanted e.g. 30 and so on).


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