Uncertainty

Dividing the circumference of a circle by its diameter always gives the same value of π=3.14159\pi = 3.14159\ldots In theory, we should be able to measure π\pi ourselves. Wrap a measuring tape around a mug, place a ruler across it, and divide our two results.

Trouble shows up the moment you read the tape or ruler: the edges never lands exactly on a mark. A measurement is always a best estimate together with an uncertainty, written as

measurement=(best estimate±uncertainty)×unit.\text{measurement} = (\text{best estimate} \pm \text{uncertainty}) \times \text{unit}.

When you read a single mark off a ruler, the edge falls somewhere inside the smallest division. Half of that smallest division is a standard choice of uncertainty.

Drag the reading along the ruler to measure the object.

object012345678910cmreading
Your measurement
L = 1.0 ± 0.5cm
relative uncertainty = 50.0%
reading = keep adjusting

Something like ±0.5 mm\pm 0.5\ \text{mm} expresses the absolute uncertainty, which carries units. Dividing it by the measurement gives the relative (percent) uncertainty: 0.5 mm0.5\ \text{mm} on a 200 mm200\ \text{mm} length is only 0.25%0.25\%, but on a 5 mm5\ \text{mm} length it is a hefty 10%10\%. It can be difficult to measure small things.

Furthermore, the number of digits used implies something about the uncertainty as well. Reporting 24.8 cm24.8\ \text{cm} implies the uncertainty is on the order of 0.1 cm0.1 \ \text{cm}. We often round the uncertainty to one significant figure, then round the measured value to the same decimal place: 3.097±0.0833.097 \pm 0.083 is reported as 3.10±0.083.10 \pm 0.08.

Propagation of uncertainty

To get π\pi we have to divide one uncertain number by another. So how do we quantify the uncertainty of the resulting quotient?

The simplest method is to just try the extremes. Take the smallest and largest plausible value of each input, recompute the result every which way, and see how far apart the answers land. The range of values quantifies the uncertainty; the half-width of this range is the new ±\pm.

Combine two measurements. The bracket spans every plausible result — its half-width is the uncertainty you carry forward.

low 15.68high 20.48best 18.00
5.60 × 2.80 = 15.68 ← low
5.60 × 3.20 = 17.92
6.40 × 2.80 = 17.92
6.40 × 3.20 = 20.48 ← high

Once you’ve done this procedure a few times, you may start to notice a pattern:

Addition/Subtraction: absolute uncertainties add. If R=A±BR = A \pm B, then δR=δA+δB\delta R = \delta A + \delta B.

Multiplication/Division: relative uncertainties (approximately) add. If R=A×BR = A \times B or R=A/BR = A / B, then

δRR=δAA+δBB.\frac{\delta R}{|R|} = \frac{\delta A}{|A|} + \frac{\delta B}{|B|}.

Note that when you subtract two measurements, the absolute uncertainties still add. If the two measurements are nearly equal, the result itself is tiny — so the relative uncertainty can blow up. Measuring a thin gap as the difference of two lengths is a good way to get a useless answer.

Comparing your result to theory

If you’ve followed these procedures, each measurement isn’t a single value. Instead, it’s a range around your best estimate of the value: from estimateuncertainty\text{estimate} - \text{uncertainty} to estimate+uncertainty\text{estimate} + \text{uncertainty}. Drawn on a graph, that range is an error bar.

This is a good time to clarify the scientific meaning of two similar words: error and uncertainty. Error is the difference between a measured value and the true value. Since the true value of something is rarely known, we don’t know the error in our measurements (if we did, we’d just take better measurements). The best we can do is estimate the error, i.e. express uncertainty.

Rather than compare results to some true value, we compare them to a theoretical prediction. In this case, theory predicts that π3.14\pi \approx 3.14 (nevermind that this is a rare instance in which we know the true value). Theory agrees with your measurement when it falls within your uncertainty range.

Each object's error bar is its measured π give-or-take the propagated uncertainty.

3.003.103.203.30π = 3.14159mug3.10 ± 0.08plate3.16 ± 0.05coin3.22 ± 0.03bucket3.13 ± 0.06crosses π — agreesmisses π — disagrees
Verdict
consistent with π = 3 of 4
coin discrepancy = 2.6×its own ±

By random chance, it’s common to have some measurements in disagreement with the theoretical value. If the magnitude of this disagreement is improbable (statistics can help quantify this probability), either you made a mistake or there’s a systematic error — a consistent bias, like a tape that was read off-center every time — that your uncertainty didn’t account for.

It’s helpful to quantify any discrepancy as a multiple of the uncertainty:

experimenttheoryuncertainty.\frac{|\,\text{experiment} - \text{theory}\,|}{\text{uncertainty}}.

If this quantity is about 11, the uncertainty is likely a good estimate of the error; very close to 00 or much greater than 22 typically means something is wrong. This is also where precision and accuracy differ: small values of the numerator correspond to accurate measurements, while small values of the denominator are precise. As an experimentalist, the goal is usually to make both as small as possible while keeping a ratio close to 1.

Problem Solving

Area of a tabletop

Problem

A tabletop measures L=120.0±0.5 cmL = 120.0 \pm 0.5\ \text{cm} by W=80.0±0.5 cmW = 80.0 \pm 0.5\ \text{cm}. Find its area and the uncertainty in that area.

Use

Area is a product, so relative uncertainties add:

δAA=δLL+δWW.\frac{\delta A}{A} = \frac{\delta L}{L} + \frac{\delta W}{W}.

Solve

A=120.0×80.0=9600 cm2,A = 120.0 \times 80.0 = 9600\ \text{cm}^2,δAA=0.5120.0+0.580.0=0.42%+0.63%=1.04%.\frac{\delta A}{A} = \frac{0.5}{120.0} + \frac{0.5}{80.0} = 0.42\% + 0.63\% = 1.04\%.δA=9600×0.0104100 cm2    A=9600±100 cm2.\delta A = 9600 \times 0.0104 \approx 100\ \text{cm}^2 \;\Rightarrow\; A = 9600 \pm 100\ \text{cm}^2.

Check

The bigger side carries the smaller relative uncertainty, and the percentages simply add — exactly what the high–low machine shows for a product.

A thin gap

Problem

Two scratches sit at x2=12.4±0.1 cmx_2 = 12.4 \pm 0.1\ \text{cm} and x1=11.9±0.1 cmx_1 = 11.9 \pm 0.1\ \text{cm}. How wide is the gap between them?

Use

The gap is a difference, so absolute uncertainties add:

δ(Δx)=δx2+δx1.\delta(\Delta x) = \delta x_2 + \delta x_1.

Solve

Δx=12.411.9=0.5 cm,δ(Δx)=0.1+0.1=0.2 cm.\Delta x = 12.4 - 11.9 = 0.5\ \text{cm}, \qquad \delta(\Delta x) = 0.1 + 0.1 = 0.2\ \text{cm}.

So Δx=0.5±0.2 cm\Delta x = 0.5 \pm 0.2\ \text{cm} — a relative uncertainty of 40%40\%.

Check

Each length was known to under 1%1\%, yet their difference is uncertain to 40%40\%. Subtracting nearly equal numbers throws precision away; measure the gap directly if you can.

Does this circle give π?

Problem

A plate gives a circumference C=69.5±0.2 cmC = 69.5 \pm 0.2\ \text{cm} and a diameter d=22.0±0.1 cmd = 22.0 \pm 0.1\ \text{cm}. Find C/dC/d with its uncertainty, and decide whether it agrees with π=3.14159\pi = 3.14159.

Solve

Cd=69.522.0=3.159,δ(C/d)C/d=0.269.5+0.122.0=0.29%+0.45%=0.74%.\frac{C}{d} = \frac{69.5}{22.0} = 3.159, \qquad \frac{\delta(C/d)}{C/d} = \frac{0.2}{69.5} + \frac{0.1}{22.0} = 0.29\% + 0.45\% = 0.74\%.δ(C/d)=3.159×0.00740.02    Cd=3.16±0.02.\delta(C/d) = 3.159 \times 0.0074 \approx 0.02 \;\Rightarrow\; \frac{C}{d} = 3.16 \pm 0.02.

Verdict

The range is [3.14, 3.18][3.14,\ 3.18], and π=3.14159\pi = 3.14159 lands just inside it — consistent, though only barely. The discrepancy is about 3.163.1420.020.9\tfrac{|3.16 - 3.142|}{0.02} \approx 0.9 of an uncertainty. A finer tape would tighten the bar and make the test sharper.

Uncertainty Checkpoint

Question 1 of 4

You compute a speed $v = d/t$ from a measured distance and time. How do you get the uncertainty in $v$?

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