1. Load the sweep. Three cooling rates, three trial counts, three objective values, all from one paper's own table. Cell 1 prints them beside accuracy, taken as one over the objective.
2. Read the headline ratio. 279,927 over 2,989 is 93.7 times the compute. 0.022276 over 0.000451 is 49.4 times the accuracy. In decades that is 1.97 of compute for 1.69 of accuracy, and the trial counts themselves span three successive decades.
3. Fit the exponent. Take log base ten of trials and of accuracy and run an ordinary least squares fit. You should get about 0.859, so accuracy grows roughly as the compute to the power 0.86.
4. Check the legs before you trust the fit. The cell prints each segment separately: 0.742 for the first, 0.976 for the second. A three-point fit hides its own curvature, and the residuals tell you how much. Report the legs whenever you report the slope.
5. Extrapolate once, and label it. At that exponent, one further decade of accuracy costs 14.6 times more trials, roughly 4.1 million. That number is an extrapolation of a fit to three points and is not a measurement.
6. Now price a run in joules. Cell 2 takes an 80 W controller draw from one arm's datasheet and a 60 s excitation trajectory from a Chinese identification method, and multiplies. 4,800 J, so 4.8 kJ.
7. Check the instrument that could have done this properly. The German record gives a worked production cycle at 2,446 W over 14.64 s. Multiply and you get 9.95 Wh, which is what the source states, so the instrument is consistent. It has never been pointed at an identification run.
8. Read the floor as a floor. Machine power over controller power is 30.6, so the same 60 s at the machine figure is 146.8 kJ. One identification run brackets 4.8 kJ to 146.8 kJ and neither end was measured.
9. Try to price the Japanese sweep and fail on purpose. The last table prints each region's unit beside whether it converts to joules. Only one does, and to convert 279,927 trials you would need seconds per trial, which this review did not locate.