Graphing measurements with uncertainty

Every measured value carries an uncertainty $u$ set by the instrument resolution and the scatter of repeated readings. On a graph the value is drawn as a point and the uncertainty as a bar ('whisker') of total length $2u$: the true value most likely lies inside that interval. A linear model $y = kx + b$ is admissible if a single straight line can pass through all the uncertainty intervals at once.

The slope of that line gives $k$, and the extreme admissible lines bound a corridor whose width is $2\Delta k = k_{\max} - k_{\min}$; half of that width is reported as the slope uncertainty $\Delta k$. Points at the ends of the range constrain $k$ most strongly, so the $x$ values should be spread as widely as possible. If no straight line can be drawn through all the bars, the linear model is rejected — and that too is a genuine measurement result.

Measurements of a spring extension gave the table:

Uncertainties: $u_x = 0.1$ cm, $u_y = 0.2$ N. Plot these points with bars and find the slope $k \pm \Delta k$.

For the same data ($x$ from 1.0 to 3.0 cm, $u_y = 0.2$ N) find the value of $y$ at $x = 0$ (the intercept $b$) together with its uncertainty. Why is $\Delta b$ larger than a single $u_y$?

A model predicts a slope of $2.2$ N/cm. Dataset A (the data above) gives $k = 1.85 \pm 0.20$ N/cm. Dataset B, taken with a coarser scale, gives $k = 2.0 \pm 0.6$ N/cm. For each dataset decide whether the model is compatible with the data, and explain which dataset is more informative.

A computational study in a spreadsheet or in Python — safe, no heating and no hazardous substances are required.

1. On the interval $x = 0.5\ldots 4.0$ cm generate 8 values from the model $y = 1.85x + 0.35$ and add a small random noise to each.
2. Set $u_x = 0.1$ cm, $u_y = 0.2$ N and make a scatter plot with cross-shaped bars.
3. Draw the central line and the two extreme admissible lines; record $k \pm \Delta k$ and $b \pm \Delta b$.
4. Repeat everything with $u_y$ doubled. Note by what factor $\Delta k$ grew and whether the verdict on compatibility with the model $y = 2.2x$ changed.

Save both graphs side by side and compare the widths of the corridors.

Show how the uncertainty of a single point turns into a corridor of admissible lines, and how $\Delta k$ and the interval for $b$ are read off from that corridor.

A labelled sketch-graph: $x$ and $y$ axes; three points with rectangular $2u_x \times 2u_y$ bars; a solid central line; two dashed extreme lines forming a shaded corridor; horizontal brackets labelled $2\Delta k$ (full corridor width in slope; $\Delta k$ is half of it) and $2\Delta b$ (full corridor width at $x = 0$).

A student films a classmate running $60$ m, using a phone in slow-motion mode at $240$ frames per second. Marks every $10$ m are measured with a tape to $\pm 0{,}02$ m, and the moment the runner passes a mark can be read to one frame, that is to $\pm 0{,}004$ s. The data are: $s = 0$ m at $t = 0$ s; $10$ m at $1{,}55$ s; $20$ m at $2{,}95$ s; $30$ m at $4{,}38$ s; $40$ m at $5{,}81$ s.

Every pair of numbers becomes a point with an error cross: a horizontal bar of $\pm 0{,}004$ s and a vertical bar of $\pm 0{,}02$ m. A best-fit straight line over the $20$–$40$ m section has a slope of $7{,}0$ m/s; the corridor allowed by the crosses is $6{,}95$ to $7{,}05$ m/s. Over the first $20$ m, however, the points lie systematically above the line and spread wider than the crosses, so a realistic speed estimate is $(7{,}0 \pm 0{,}2)$ m/s. A graph with uncertainties marks the boundary between instrument scatter and an imperfect model.

A cyclist brakes on dry asphalt from several different speeds: the bike computer reads speed to $\pm 0{,}5$ km/h, and the braking distance is measured with a tape to $\pm 0{,}1$ m. The data are: $20$ km/h — $2{,}3$ m; $25$ km/h — $3{,}5$ m; $30$ km/h — $5{,}1$ m; $35$ km/h — $6{,}9$ m.

The dependence of $d$ on $v$ is a parabola, so the data are moved to a graph of $d$ against $v^2$, where the points fall on a straight line. Its slope is $0{,}073$ s$^2$/m; the error crosses (a $\pm 0{,}5$ km/h uncertainty in $v$, which at $30$ km/h is about $\pm 2$ m$^2$/s$^2$ along the $v^2$ axis) widen the admissible slope to $0{,}066$–$0{,}080$ s$^2$/m. Since $d = v^2/(2\mu g)$ makes the slope equal to $1/(2\mu g)$, this gives $\mu = 0{,}70 \pm 0{,}07$: the size of the crosses on the graph fixes how accurately the friction coefficient can be known at all.

A mug of tea at $85{,}0\ ^\circ$C is left in a room at $22\ ^\circ$C. The thermometer has a scale division of $0{,}5\ ^\circ$C and the stopwatch reads to $\pm 1$ s. The recorded temperatures are: $85{,}0\ ^\circ$C at $t = 0$; $72{,}0\ ^\circ$C after $5$ min; $62{,}5\ ^\circ$C after $10$ min; $55{,}0\ ^\circ$C after $15$ min; $49{,}0\ ^\circ$C after $20$ min.

The excess over room temperature, $\Delta T = T - 22\ ^\circ$C, falls as $\Delta T = \Delta T_0 e^{-t/\tau}$, so one plots $\ln(\Delta T)$ against $t$ and gets a straight line with slope $-0{,}042$ min$^{-1}$. The vertical cross of each point is the thermometer division $\pm 0{,}5\ ^\circ$C; at $\Delta T \approx 30\ ^\circ$C that is only $\pm 0{,}017$ in the logarithm, and the lines still passing through all crosses give slopes from $-0{,}038$ to $-0{,}046$ min$^{-1}$. Hence the time constant is $22$–$26$ min, that is $(24 \pm 2)$ min. Turning the exponential into a straight line makes the uncertainty visible.