I am given a graph of a motor's torque vs RPM values at two different current draws, 730A and 300A (see graph). I need to obtain the 3D function to find current as a function of torque and RPM by linearly interpolating and extrapolating the two slices I have available. I need to use the current function in Matlab to determine power draw at different torques and RPMs.
I have obtained the equations for the 2D slices using a curve fitting software and found the 730A slice to be best fit with the Richards curve and the 300A slice to be best fit with the logistic power curve.
The slice at 730A has the equation:
$$
f(x)=\frac{501.451}{[1+\exp(-25.544+0.006604\cdot x)]^{\frac{1}{31.698}}}
$$
The slice at 300A has the equation:
$$
f(x)=\frac{231.650}{1+\left(\dfrac{x}{8566.519}\right)^{4.196}}
$$
How do I take these two functions and find the current given a torque and RPM?
Best Answer
This can be done by a simple homotopy between curves: if we put $$ \begin{align} f(x,730\mathrm{A}) &= \frac{501.451}{[1+\exp(-25.544+0.006604\cdot x)]^{\frac{1}{31.698}}}\\ \\ f(x,300\mathrm{A}) &= \frac{231.650}{1+\left(\dfrac{x}{8566.519}\right)^{4.196}}\\ \end{align} $$ then $$ \begin{split} f(x, \mathrm{I}_M)& =\frac{\mathrm{I}_M-300\mathrm{A}}{730\mathrm{A}-300\mathrm{A}} f(x,730\mathrm{A}) \\ &\quad+ \frac{730\mathrm{A} -\mathrm{I}_M}{730\mathrm{A}-300\mathrm{A}} f(x,300\mathrm{A})\\ & = \frac{\mathrm{I}_M-300\mathrm{A}}{430\mathrm{A}} f(x,730\mathrm{A}) + \frac{730\mathrm{A} -\mathrm{I}_M}{430\mathrm{A}} f(x,300\mathrm{A}) \end{split}\quad 300\mathrm{A}\le \mathrm{I}_M \le 730\mathrm{A} $$ where $\mathrm{I}_M$ is the motor current.
I think that with only two curves this, which is really equivalent to a linear interpolation, is the best we can hope: however, you could try to take more "slices" of the torque vs. (RPM, $\mathrm{I}_M$) surface, and use higher order interpolation methods