Instrument resolution and measurement uncertainty

Every instrument has a resolution — the smallest change in the measured quantity that produces a noticeable change in the reading.

For an analog scale the resolution equals the value of the smallest division; for a digital display it is one unit in the last digit. A measurement result is always written as $x \pm \Delta x$, and resolution sets only the lower bound of the uncertainty: random scatter and systematic errors add to it. For a single reading on an analog scale one takes $\Delta x = \frac{1}{2}\,\delta$, where $\delta$ is the smallest division; for a digital instrument $\Delta x = \delta$ (one unit in the last digit). The relative uncertainty $\varepsilon = \frac{\Delta x}{x}\cdot 100\%$ shows whether the instrument suits the task: when $x$ is comparable to $\delta$, $\varepsilon$ becomes enormous. In multiplication and division relative uncertainties add, so the accuracy of the result is set by the crudest of the measured quantities.

A ruler with millimetre divisions is used to measure the length of a bar; the reading is $L = 24.6$ cm. Write the result with its uncertainty and find the relative uncertainty.

The mass of a sample on a digital balance is $m = 12.34$ g (one unit in the last digit is $0.01$ g). The volume was measured with a graduated cylinder: $V = 5.0 \pm 0.5$ cm$^3$. Find the density of the sample and its uncertainty.

You must measure the thickness of a single sheet of paper with a relative uncertainty of no more than $1\%$. The only available tool is a ruler with a resolution of $0.5$ mm. A direct measurement of one sheet gives about $0.1$ mm. Propose a method and estimate the achievable uncertainty.

Self-study with two instruments of different resolution.

1. Take a ruler with millimetre divisions and a digital balance (or graph paper and a kitchen scale). Identify and write down the resolution of each instrument.
2. Measure the thickness of a single sheet of a book directly with the ruler; record $t \pm \Delta t$ and $\varepsilon$.
3. Measure the thickness of a stack of 100 sheets and obtain the thickness of one sheet by division; again record $t \pm \Delta t$ and $\varepsilon$.
4. Compare the two values of $\varepsilon$ and explain why dividing by a large number $N$ reduces the relative uncertainty.
5. Extra: weigh an object, then weigh several identical objects together and repeat the reasoning for the case of addition.

Safety. Work only with ordinary safe objects (sheets of paper, a book, coins, dry food). Do not measure hot, live, sharp, or fragile objects. If no instruments are available, carry out the steps in a simulation (for example a virtual ruler or PhET "Measurement") — it fully replaces the experiment.

Show that any reading on a scale lies inside an interval one division wide and that this interval sets the minimum uncertainty of the result.

Sketch. A magnified section of a ruler: marks every $0.1$ cm, labelled ticks. The reading $L = 24.6$ cm is shown by a vertical arrow standing exactly on the $24.6$ cm tick (a reading truly between ticks would be, for instance, $24.65$ cm between the $24.6$ cm and $24.7$ cm marks). Around the reading a band is shaded whose full width is $\delta = 0.1$ cm, that is from $L - \Delta L$ to $L + \Delta L$ with $\Delta L = 0.05$ cm: its half-width is $\Delta L = 0.05$ cm, while its full width equals the smallest division $\delta$ rather than being narrower. Beside it, a bar chart for Example 2: two bars $\varepsilon_m \approx 0.08\%$ and $\varepsilon_V = 10\%$, making it obvious which quantity determines the final uncertainty.

You time the winner of a school $100\ \text{m}$ race with a phone stopwatch: the screen shows $t = 13.42\ \text{s}$. The display changes in steps of $0.01\ \text{s}$, so the instrument resolution is $\Delta t_{\text{res}} = 0.01\ \text{s}$, and reading the display gives an uncertainty of $\pm 0.005\ \text{s}$. But you press the button yourself: the reaction delay at start and finish adds at least $\pm 0.15\ \text{s}$. The honest result is $t = (13.42 \pm 0.15)\ \text{s}$, the relative uncertainty is $0.15/13.42 \approx 1\,\%$, and the runner's average speed $v = 100/13.42 \approx 7.45\ \text{m/s}$ is known no better than to $\pm 0.08\ \text{m/s}$.

The distinction between resolution and uncertainty is at work here: resolution fixes only the last digit the display can print, while the measurement uncertainty is set by the largest contribution — here $\pm 0.15\ \text{s}$ of reaction time, 15 times the display step. So the hundredths ($13.42$) are not significant, and the result is rounded to tenths: $t = (13.4 \pm 0.2)\ \text{s}$, with the speed computed from the rounded values.

You need $250\ \text{mL}$ of water for porridge, and the jug at hand is a $500\ \text{mL}$ one with graduations every $25\ \text{mL}$. Its resolution is $\Delta V_{\text{res}} = 25\ \text{mL}$, so reading the water level gives an uncertainty of $\pm 12.5\ \text{mL}$, i.e. $V = (250 \pm 12)\ \text{mL}$. The relative uncertainty of the volume is $12.5/250 = 5\,\%$. With a beaker graduated every $5\ \text{mL}$ you would get $\pm 2.5\ \text{mL}$, that is $1\,\%$.

Half a scale division is the standard estimate of the uncertainty of a single reading from a scale, and dividing it by the measured value gives the relative uncertainty. That relative uncertainty then carries over to any computed quantity: if the recipe uses $125\ \text{g}$ of grain per $250\ \text{mL}$ of water, the water-to-grain ratio $= 2.00$ is known to the same $5\,\%$, that is $2.0 \pm 0.1$.

A phone in a holder shows a travelled distance $s = 1.20\ \text{km}$ (resolution $0.01\ \text{km}$, reading uncertainty $\pm 0.005\ \text{km}$) and a travel time $t = 72.0\ \text{s}$ (resolution $0.1\ \text{s}$, reading uncertainty $\pm 0.05\ \text{s}$). The average speed is $v = s/t = 1200/72.0 = 16.7\ \text{m/s} = 60.0\ \text{km/h}$. The relative uncertainties are $0.005/1.20 = 0.42\,\%$ and $0.05/72.0 = 0.07\,\%$; added together they give about $0.4\,\%$, that is $v = (16.7 \pm 0.07)\ \text{m/s}$.

This shows how two independent reading uncertainties, both born from instrument resolution, flow into the uncertainty of a computed quantity: under division the relative uncertainties add, and the answer is rounded to the digit that is still significant. Note that the distance term dominates ($0.42\,\%$), so halving the uncertainty in the time reading would be pointless.