Measuring Temperature — Question 9

Question 9

Digit size competes with calibration uncertainty

Thermometer A has slope $.20\,\mathrm{V/K}$, rounds voltage to the nearest $.002\,\mathrm V$, and has an additional independent hard calibration error of $.030\,\mathrm K$. Thermometer B has slope $.050\,\mathrm{V/K}$, rounds to the nearest $.0002\,\mathrm V$, and has calibration error $.010\,\mathrm K$. Use closed half-bin bounds and add worst-case contributions.

1. Find each total temperature half-width. Which instrument meets a requirement of at most $.015\,\mathrm K$?

2. If A displays ten times more voltage digits without changing its calibration error, can it meet that requirement? Quantify its best resulting bound.

Question 9 - Solution

Digit size competes with calibration uncertainty

(a) Method and calculation

$$\delta T_A=\frac{.002/2}{.20}+.030=.035\,\mathrm K.$$$$\delta T_B=\frac{.0002/2}{.050}+.010=.012\,\mathrm K.$$Only B meets $.015$ K despite its smaller voltage sensitivity.

(b) Interpretation and further calculation

$$\delta T_{A,\mathrm{finer}}=\frac{.0002/2}{.20}+.030=.0305\,\mathrm K.$$$$\lim_{\text{digit size}\to0}\delta T_A=.030\,\mathrm K>.015\,\mathrm K.$$Calibration dominates, so finer display increments alone cannot meet the specification.

Physical check

The contributions are stipulated hard bounds; statistical independence does not justify replacing the worst-case sum by a root-sum-square.