Help: Cannot solve this equation

The way it is. Where does this equation come from? How is it derived? Is it a solution to a differntial equation? If so then maybe that equation could be solved for resistance.

It can be solved graphically, I reckon. You would have to plot the final element (c*logR^3) first (just substitute it for 'y' for example), to obtain a trace for the cubic element. Plotting the first 2 elements together would then give you a second trace. Where these traces intersect on the graph would then give you a solution for 'R'. I can't see how it could be solved by transposing the equation to give a solution for 'R'. If there's a prize I'll try harder! :-)

The only prize I can offer is the satisfaction you would have knowing that you have solved for the elusive ln(R). Through manipulation of values, we have worked a way to get around solving the equation. This equation, however, has eluded us here for months and undoubtedly for years to come. It has gotten to the point where the ln(R) has become a common joke among us. We have multiple mech. engineers, and electrical, computer engineers and computer programers at this facility, all of which are stumped on how to algebraicly solve it. One of them suggesting that we solve for the cubed root which would yield 3 posible values for "R", but I don't think that would work because it is not the "R" that is cubed, but rather the (ln(R)) quantity that is cubed at the end of the equation. Please give it another shot algebraicly if at all possible.

Apparently the equation is called the Steinhart-Hart equation. The solution for Resistance can be found at the following website http://www.atpsensor.com/ntc/steinhart/steinhart.h tml?=steinhart_main.html

Some more interesting information as to where the equation comes from can be found here. http://www.betatherm.com/stein.htm

Of course, I still can't derive it yet. I'll try to work it backwards the next time I've got time.

Sorry, I still have trouble putting links into comments. The links are there above, but the important one that I think answers your question was this one. http://www.atpsensor.com/ntc/steinhart/steinhart.h tml?=steinhart_main.html

"R" is the actual resistance of the thermocouple being used for your equation. Two dissimular metals combined at single juction, create a thermocouple. Thus, they create a current as their electrons flow across the juction. This electron flow can be measured in millivolts. The actual resistance of the thermocouple wire is your base number for the equation. It is measured in OHMS. At 0*C with the use of an ice bath, you will find the base value of the TC. Then you 212*F, boiling point of water, is also your top end of a known value. Once you have your output number for your TC, that is the millivolt reading at both these tempuratures, you can actually creat the linear curve of temperature point per millivolt. But you must have the resistance of the wire to be known.

The eqution is for a negative temperature coefficient thermistor -- not a thermocouple.

I'm not a Math expert but using standard Math-techniques I got this solution 1. rewrite equation to the form z^3 + p*z + q = 0 (which can be solved thanks to Girolamo Cardano) > z = ln(R) (A) > c*z^3 + b*z + a - (1/T) = 0 > z^3 + (b/c)*z + (a/c) - (1/(T*c)) = 0 (1) > p = b/c, q = (a/c) - (1/(T*c)) 2. Girolamo Cardano solution > z = u + v (C) > (1) u^3 + v^3 + q + (3*u*v + p) * (u + v) = 0 > u^3 + v^3 + q = 0 (2) AND 3*u*v + p = 0 (3) > (3) u = - (p/(3*v)) (D) > (2) - p^3 / (27*v^3) + v^3 + q = 0 > - p^3 + 27*(v^3)^2 + 27*q*(v^3) = 0 > x = v^3 (E) > 27*x^2 + 27*q*x - p^3 = 0 (4) 2. (4) is a standard 2nd grade equation with these solutions: > x(1,2) = ( - 27*q +/- SQRT( (27*q)^2 + 108*p^3 ) ) / 54 "+/-"; "+" for x1, "-" for x2 3. Reverse substitution of x1 and x2 > (E) gives v1 and v2 > (D) gives u1 and u2 > (C) gives z1 and z2 > (A) gives R1 = exp (z1) and R2 = exp (z2) If you work down the last equations with the basic values (a,b,c,T) you get 2 large equations which gives the R values directly.