Measuring Things — Question 6

Question 6

Measure many to infer one

Identical rigid sheets are stacked without gaps or compression. A height instrument adds the same unknown offset $b$ to every stack reading. Its model is $$H_N=Nt+b,$$ where $t$ is one sheet’s thickness and $N$ is an exact count. Each displayed height also has a worst-case reading-error bound of $\pm0.10$ mm. The measured pairs $(N,H_N)$ are $(20,2.80\ \mathrm{mm})$, $(50,6.40\ \mathrm{mm})$, and $(80,10.00\ \mathrm{mm})$.

1. Use the 20-sheet and 80-sheet central readings to determine $t$ and $b$. Does the 50-sheet central reading support the model?

2. Bound $t$ using the two endpoint readings. Do not divide the offset by the number of sheets or assume random-error cancellation.

3. Evaluate the shortcut $t=H_{80}/80$. Explain its error here and one physical condition whose failure would invalidate the stacking model even after the offset is removed.

Question 6 - Solution

Measure many to infer one

(a) Differences isolate the thickness.

Subtracting $H_{20}=20t+b$ from $H_{80}=80t+b$ removes $b$: $$t=\frac{10.00-2.80}{80-20}=\boxed{0.120\ \mathrm{mm}},\qquad b=2.80-20(0.120)=\boxed{0.40\ \mathrm{mm}}.$$ The predicted middle reading is $50(0.120)+0.40=6.40$ mm, matching the data. This is a consistency check, not proof that the model holds for every stack.

(b) Propagate worst-case bounds.

$$\delta t_{\max}=\frac{0.10+0.10}{80-20}\ \mathrm{mm}=\frac{1}{300}\ \mathrm{mm}.$$

The difference $H_{80}-H_{20}$ lies in $[7.00,7.40]$ mm because its two reading bounds can oppose each other. Dividing by the exact count difference, 60, $$\boxed{t\in[0.116\overline6,\,0.123\overline3]\ \mathrm{mm}},\qquad \delta t=\frac{0.20}{60}=0.003\overline3\ \mathrm{mm}.$$ This is a bounded-error interval, not a statistical standard deviation. The third reading is compatible with both extreme slopes; it does not tighten this endpoint-derived interval here.

(c) Diagnose the shortcut.

$$\frac{H_N}{N}=t+\frac bN,\qquad\frac{b}{80}=\frac{0.40}{80}\ \mathrm{mm}=0.005\ \mathrm{mm}.$$

$H_{80}/80=0.125$ mm is too large by $0.005$ mm because it includes $b/80$. Using more sheets reduces this offset contribution but does not justify ignoring it. Gaps, unequal sheet thicknesses, or compression that depends on stack size could violate $H_N=Nt+b$; these require separate physical controls.