Combining Uncertainties

Spec mapping: OCR H556 Module 2.2 — Making measurements and analysing data, specifically the propagation of uncertainty through addition, subtraction, multiplication, division and powers. These are the highest-yielding rules in the entire Practical Endorsement and recur in every PAG analysis question. (Refer to the official OCR H556 specification document for exact wording.)

In practice you rarely measure a single quantity in isolation. Resistance is calculated from $V$ and $I$ ; density from $m$ and $V$ ; the Young modulus $E$ from $F$ , $L$ , $A$ and $x$ . In each case you must start with uncertainties on the raw measurements and end with an uncertainty on the final result. The process of doing this is called propagating uncertainty, and it is governed by a small set of simple rules.

These three rules appear on every H556 paper that involves practical analysis. You must know them fluently — they are among the highest-yielding pieces of knowledge in the specification.

Key definition. Propagation of uncertainty is the procedure by which the uncertainty on a derived quantity is calculated from the uncertainties on the measured quantities that go into it. At A-Level the procedure is rule-based; at undergraduate level it is replaced by the calculus-based partial-derivative formula.

The Three Rules

The A-Level rules (OCR and AQA both) are simplifications of the rigorous statistical formulas used at university. They are slightly conservative — they tend to give an uncertainty marginally larger than the strict quadrature combination — but they are easy to apply by hand and match the OCR mark-scheme expectations.

Rule 1 — Addition and subtraction: add absolute uncertainties

If $Z = X + Y$ or $Z = X - Y$ , then

$\Delta Z = \Delta X + \Delta Y.$

Absolute uncertainties add — never subtract, even when the quantities are subtracted. The uncertainty of a subtraction is the worst-case spread of possible results, which is the sum.

Rule 2 — Multiplication and division: add percentage uncertainties

If $Z = X \times Y$ or $Z = X / Y$ , then

$\frac{\Delta Z}{Z}\times 100\% = \frac{\Delta X}{X}\times 100\% + \frac{\Delta Y}{Y}\times 100\%,$

or in shorthand $\%Z = \%X + \%Y$ . The rule holds regardless of whether the operation is multiplication or division — division is just multiplication by the reciprocal, and reciprocals have the same percentage uncertainty as the original.

Rule 3 — Powers: multiply percentage uncertainty by the power

If $Z = X^{n}$ , then

$\%Z = |n|\times \%X.$

This follows from Rule 2 applied repeatedly: squaring is $X \cdot X$ , so $\%X + \%X = 2\%X$ . Cubes give $3\%X$ . Reciprocals ( $n = -1$ ) and square roots ( $n = 1/2$ ) follow the same rule with $|n|$ .

Combining the rules

For a general formula

$Z = k\, X^{a}\, Y^{b}\, W^{c}$

where $k$ is a dimensionless constant and $a, b, c$ are powers, the rules combine cleanly:

$\%Z = |a|\,\%X + |b|\,\%Y + |c|\,\%W.$

Dimensionless constants (such as $\tfrac{1}{2}$ , $\pi$ , $g$ treated as exact for OCR purposes) carry no uncertainty and do not appear in the propagation formula. Their job is to scale the value, not its uncertainty.

Which Rule to Use? — A Decision Flowchart

graph TD
    A[Start: Z depends on measured X, Y, ...] --> B{What operation joins them?}
    B -->|+ or −| C["Add ABSOLUTE uncertainties<br/>ΔZ = ΔX + ΔY"]
    B -->|× or ÷| D["Add PERCENTAGE uncertainties<br/>%Z = %X + %Y"]
    B -->|power X^n| E["Multiply percentage by |n|<br/>%Z = |n| × %X"]
    B -->|mixed e.g. X − Y / W| F["Stage 1: combine X − Y in absolute form<br/>Stage 2: convert to %<br/>Stage 3: add % with W"]
    C --> G["Convert to percentage if needed<br/>for downstream operations"]
    D --> H["Convert to absolute<br/>for the final quoted answer"]
    E --> H
    F --> H
    G --> H
    H --> I["Quote value ± uncertainty<br/>uncertainty to 1 s.f."]

This flowchart is the heart of the lesson — refer back to it as you work through the examples below.

Worked Example 1 — Adding two lengths

A student measures two pieces of string and lays them end-to-end.

$L_1 = (245 \pm 2)\,\text{mm}$ .
$L_2 = (130 \pm 1)\,\text{mm}$ .

Calculate the total length and its uncertainty.

Solution.

$L = L_1 + L_2 = 245 + 130 = 375\,\text{mm}.$

$\Delta L = \Delta L_1 + \Delta L_2 = 2 + 1 = 3\,\text{mm}.$

$L = (375 \pm 3)\,\text{mm}.$

Worked Example 2 — subtraction can magnify percentage uncertainty

Two masses are weighed on the same digital balance:

$m_1 = (52.34 \pm 0.01)\,\text{g}$ .
$m_2 = (52.18 \pm 0.01)\,\text{g}$ .

Calculate $m_1 - m_2$ and its percentage uncertainty.

Solution.

$m_1 - m_2 = 52.34 - 52.18 = 0.16\,\text{g}.$

$\Delta(m_1 - m_2) = 0.01 + 0.01 = 0.02\,\text{g}.$

Percentage uncertainty:

$\frac{0.02}{0.16}\times 100\% = 12.5\%.$

Quote $(0.16 \pm 0.02)\,\text{g}$ , $\approx 13\%$ .

Compare with the individual measurements: each was $0.02\%$ . The subtraction has inflated the percentage uncertainty by a factor of $\sim 600$ . This is the subtraction catastrophe: subtracting two nearly equal numbers throws away precision. The absolute uncertainty is unchanged ( $0.02\,\text{g}$ ), but the value has shrunk to $0.16\,\text{g}$ , so $\Delta/x$ explodes. Avoid experimental designs that rely on this kind of subtraction wherever possible. If you must subtract, increase the absolute resolution of the measurement (analytical balance instead of digital tabletop).

Worked Example 3 — multiplication for area

A rectangular plate has width $w = (30.0 \pm 0.5)\,\text{mm}$ and length $l = (80.0 \pm 0.5)\,\text{mm}$ . Calculate the area and its uncertainty.

Solution.

$A = w \times l = 30.0 \times 80.0 = 2400\,\text{mm}^{2}.$

Percentage uncertainties:

$\%w = \frac{0.5}{30.0}\times 100 = 1.67\%, \qquad \%l = \frac{0.5}{80.0}\times 100 = 0.625\%.$

Add:

$\%A = 1.67 + 0.625 = 2.29\%.$

Absolute uncertainty:

$\Delta A = \frac{2.29}{100}\times 2400 = 55\,\text{mm}^{2}.$

Quote $A = (2400 \pm 60)\,\text{mm}^{2}$ , or in scientific form $(2.4 \pm 0.1) \times 10^{3}\,\text{mm}^{2}$ .

Worked Example 4 — division for density

A metal block:

$m = (50.00 \pm 0.05)\,\text{g}$ .
$V = (18.5 \pm 0.3)\,\text{cm}^{3}$ .

Calculate the density and its absolute uncertainty in $\text{kg m}^{-3}$ .

Solution.

$\rho = \frac{m}{V} = \frac{50.00}{18.5} = 2.703\,\text{g cm}^{-3} = 2703\,\text{kg m}^{-3}.$

Percentage uncertainties:

$\%m = \frac{0.05}{50.00}\times 100 = 0.10\%, \qquad \%V = \frac{0.3}{18.5}\times 100 = 1.62\%.$

$\%\rho = 0.10 + 1.62 = 1.72\%.$

Absolute uncertainty:

$\Delta\rho = \frac{1.72}{100}\times 2703 = 47\,\text{kg m}^{-3} \approx 50\,\text{kg m}^{-3}\,(\text{1 s.f.}).$

Quote $\rho = (2700 \pm 50)\,\text{kg m}^{-3}$ .

Worked Example 5 — using powers (kinetic energy)

A student measures the mass and speed of a trolley:

$m = (1.20 \pm 0.01)\,\text{kg}$ .
$v = (2.50 \pm 0.05)\,\text{m s}^{-1}$ .

Calculate the kinetic energy and its uncertainty.

Solution.

$E_k = \tfrac{1}{2}mv^{2} = 0.5 \times 1.20 \times (2.50)^{2} = 3.75\,\text{J}.$

Percentage uncertainties:

$\%m = \frac{0.01}{1.20}\times 100 = 0.83\%, \qquad \%v = \frac{0.05}{2.50}\times 100 = 2.00\%.$

$v$ is squared, so its contribution doubles:

$\%(v^{2}) = 2 \times 2.00 = 4.00\%.$

The factor $\tfrac{1}{2}$ is a dimensionless constant — no uncertainty contribution.

$\%E_k = \%m + \%(v^{2}) = 0.83 + 4.00 = 4.83\%.$

$\Delta E_k = \frac{4.83}{100}\times 3.75 = 0.18\,\text{J} \approx 0.2\,\text{J}\,(\text{1 s.f.}).$

Quote $E_k = (3.8 \pm 0.2)\,\text{J}$ — value rounded to 1 dp to match the uncertainty.

Notice how the square on $v$ doubles its contribution. In many physics formulas (area, volume, kinetic energy, gravitational PE, electrical power $P = I^{2}R$ ), one quantity appears squared or cubed and dominates even when its raw percentage uncertainty is modest.

Worked Example 6 — a cube root

The radius of a sphere is calculated from its volume via

$r = \sqrt[3]{\frac{3V}{4\pi}}.$

If $V = (4.5 \pm 0.3)\,\text{cm}^{3}$ , find $r$ and its percentage uncertainty.

Solution.

$r = \sqrt[3]{\frac{3 \times 4.5}{4\pi}} = \sqrt[3]{\frac{13.5}{12.566}} = \sqrt[3]{1.074} = 1.024\,\text{cm}.$

Percentage uncertainty in $V$ :

$\%V = \frac{0.3}{4.5}\times 100 = 6.67\%.$

The cube root corresponds to $n = 1/3$ , so

$\%r = \frac{1}{3}\times 6.67\% = 2.22\%.$

$\Delta r = \frac{2.22}{100}\times 1.024 \approx 0.02\,\text{cm}.$

Quote $r = (1.02 \pm 0.02)\,\text{cm}$ .

Fractional powers reduce the percentage uncertainty. This is why square-root and cube-root dependencies (pendulum period, escape velocity, simple-harmonic frequency) are relatively forgiving experimentally.

Worked Example 7 — full Young modulus calculation

The Young modulus of a wire is determined via

$E = \frac{F L}{A x}, \qquad A = \pi \left(\frac{d}{2}\right)^{2} = \frac{\pi d^{2}}{4}.$

Measurements:

$F = (20.0 \pm 0.1)\,\text{N}$ .
$L = (1.500 \pm 0.005)\,\text{m}$ .
$d = (0.34 \pm 0.01)\,\text{mm}$ .
$x = (1.8 \pm 0.1)\,\text{mm}$ .

Calculate $E$ and its percentage uncertainty.

Solution. Step 1 — area:

$A = \frac{\pi (0.34 \times 10^{-3})^{2}}{4} = \frac{\pi \times 1.156 \times 10^{-7}}{4} = 9.08 \times 10^{-8}\,\text{m}^{2}.$

Step 2 — Young modulus:

$E = \frac{20.0 \times 1.500}{9.08 \times 10^{-8} \times 1.8 \times 10^{-3}} = \frac{30.0}{1.634 \times 10^{-10}} = 1.836 \times 10^{11}\,\text{Pa} \approx 1.8 \times 10^{11}\,\text{Pa}.$

Step 3 — percentage uncertainties:

Quantity	$\Delta/x \times 100\%$	Power	Contribution
$F$	$0.50\%$	$1$	$0.50\%$
$L$	$0.33\%$	$1$	$0.33\%$
$d$	$2.94\%$	$2$	$5.88\%$
$x$	$5.56\%$	$1$	$5.56\%$

Sum:

$\%E = 0.50 + 0.33 + 5.88 + 5.56 = 12.3\%.$

$\Delta E = \frac{12.3}{100}\times 1.836 \times 10^{11} = 0.226 \times 10^{11} \approx 0.2 \times 10^{11}\,\text{Pa}.$

Quote $E = (1.8 \pm 0.2) \times 10^{11}\,\text{Pa}$ .

Observations the markers want to see in a Top-band answer:

The biggest contributors are $d$ (via $d^{2}$ , $5.88\%$ ) and $x$ ( $5.56\%$ ). Reducing either has the largest leverage.
$L$ is the smallest contributor; investing in a more precise ruler would not help. The diameter and extension measurements are the bottleneck.
The overall $\sim 12\%$ is typical for school Young modulus practicals, explaining the customary $10$ – $20\%$ disagreement with the textbook $2.0 \times 10^{11}\,\text{Pa}$ for steel.

Worked Example 8 — mixed combination

$R = \frac{V_1 - V_2}{I},$

with $V_1 = (6.00 \pm 0.05)\,\text{V}$ , $V_2 = (2.40 \pm 0.05)\,\text{V}$ , $I = (0.240 \pm 0.005)\,\text{A}$ .

Solution. Stage 1 — subtract (absolute first):

$V_1 - V_2 = 6.00 - 2.40 = 3.60\,\text{V}, \qquad \Delta(V_1 - V_2) = 0.05 + 0.05 = 0.10\,\text{V}.$

$\%(V_1 - V_2) = \frac{0.10}{3.60}\times 100 = 2.78\%.$

Stage 2 — convert to percentage; stage 3 — divide:

$R = \frac{3.60}{0.240} = 15.0\,\Omega, \qquad \%I = \frac{0.005}{0.240}\times 100 = 2.08\%.$

$\%R = 2.78 + 2.08 = 4.86\%.$

$\Delta R = \frac{4.86}{100}\times 15.0 \approx 0.7\,\Omega.$

Quote $R = (15.0 \pm 0.7)\,\Omega$ .

Worked Example 9 — resistivity (sum + product + power in one)

Resistivity from a wire experiment combines all three rules in one expression:

$\rho = \frac{R A}{l} = \frac{R \pi (d/2)^{2}}{l} = \frac{\pi R d^{2}}{4 l}.$

A student records:

$R = (4.20 \pm 0.05)\,\Omega$ (computed earlier from $V_1 - V_2$ divided by $I$ , with the difference already incorporated).
$d = (0.36 \pm 0.01)\,\text{mm}$ .
$l = (1.200 \pm 0.005)\,\text{m}$ .

Calculate $\rho$ and its percentage uncertainty.

Solution. Step 1 — value:

$\rho = \frac{\pi \times 4.20 \times (0.36 \times 10^{-3})^{2}}{4 \times 1.200} = \frac{\pi \times 4.20 \times 1.296 \times 10^{-7}}{4.800} = 3.56 \times 10^{-7}\,\Omega\,\text{m}.$

Step 2 — percentage contributions in tabular form:

Quantity	$\Delta/x \times 100\%$	Power	Contribution
$R$	$1.19\%$	$1$	$1.19\%$
$d$	$2.78\%$	$2$	$5.56\%$
$l$	$0.42\%$	$1$	$0.42\%$

Step 3 — sum:

$\%\rho = 1.19 + 5.56 + 0.42 = 7.17\%.$

Step 4 — absolute uncertainty:

$\Delta\rho = \frac{7.17}{100}\times 3.56 \times 10^{-7} = 0.26 \times 10^{-7} \approx 0.3 \times 10^{-7}\,\Omega\,\text{m}.$

Quote $\rho = (3.6 \pm 0.3) \times 10^{-7}\,\Omega\,\text{m}$ .

What this example teaches you that the earlier ones did not is that the squared diameter dominates the budget, just as in the Young modulus practical. The micrometre measurement of $d$ is the bottleneck — a digital callipers with $\pm 0.01\,\text{mm}$ resolution on a $\sim 0.3\,\text{mm}$ wire is the single source of most of the final uncertainty. Examiners look for this experimental-design observation in Top-band Paper-3 commentary.

When to Round Percentage Uncertainties

A common feature of Mid-band answers is quoting a percentage uncertainty to four or five significant figures: " $\%\rho = 7.1683\%$ ". This is a marker of misunderstanding, not precision. The convention OCR follows is:

Quote percentage uncertainties to 1–2 significant figures. Two s.f. is the absolute upper limit; one s.f. is safer for the final headline figure.
Quote absolute uncertainties to 1 significant figure, with the value rounded to match the decimal place of the uncertainty.

The pedagogical reason is that the propagation rules are themselves only conservative estimates — they overstate the true random-error combination by some tens of per cent compared to the rigorous quadrature formula. Quoting " $7.1683\%$ " implies a precision in the uncertainty estimate itself that the A-Level rules do not deliver. The honest figure is " $\sim 7\%$ ".

A useful mid-calculation convention is to carry intermediate percentage uncertainties to one extra figure (so " $7.17\%$ " inside the working, " $7\%$ " or " $0.3 \times 10^{-7}$ " in the final answer). This prevents premature rounding errors propagating through the multiplication step, while still giving an honest final figure.

Example — over-precise versus appropriate

Suppose the calculation gives $\%E = 12.347\%$ on a Young modulus value $E = 1.836 \times 10^{11}\,\text{Pa}$ .

Over-precise (Mid-band): $\Delta E = 0.2266 \times 10^{11}\,\text{Pa}$ — implies the uncertainty itself is known to four s.f.
Appropriate (Top-band): $\Delta E = 0.2 \times 10^{11}\,\text{Pa}$ , value $E = (1.8 \pm 0.2) \times 10^{11}\,\text{Pa}$ — the uncertainty has one s.f., the value matches its decimal place.

Mark schemes routinely deduct one mark for over-precise uncertainty quoting. It is cheap and avoidable.

Dominant Uncertainty

Most experimental error budgets are not democratic. Typically one quantity contributes the majority of the final percentage uncertainty, and the others contribute small corrections. Recognising this lets you:

Focus experimental effort on improving the dominant measurement.
Approximate the propagation by the dominant term alone, useful as a sanity check.
Justify which apparatus upgrade has the largest leverage on the final result.

The first-approximation rule

If the largest percentage contribution is more than $\sim 5\times$ each of the others, the smaller terms can be ignored to a first approximation. The error in doing so is bounded by their sum and is typically only a fraction of the dominant term.

Worked example — dominant uncertainty in practice

Three quantities contribute to a derived value:

$\%A = 10\%$ .
$\%B = 0.5\%$ .
$\%C = 0.4\%$ .

The full A-Level rule gives $\%Z = 10 + 0.5 + 0.4 = 10.9\%$ . The dominant-term approximation gives $\%Z \approx 10\%$ . The discrepancy is $0.9$ percentage points, less than $10\%$ of the dominant figure — well below the precision to which we round.

Combining Uncertainties

Combining Uncertainties

The Three Rules

Rule 1 — Addition and subtraction: add absolute uncertainties

Rule 2 — Multiplication and division: add percentage uncertainties

Rule 3 — Powers: multiply percentage uncertainty by the power

Combining the rules

Which Rule to Use? — A Decision Flowchart

Worked Example 1 — Adding two lengths

Worked Example 2 — subtraction can magnify percentage uncertainty

Worked Example 3 — multiplication for area

Worked Example 4 — division for density

Worked Example 5 — using powers (kinetic energy)

Worked Example 6 — a cube root

Worked Example 7 — full Young modulus calculation

Worked Example 8 — mixed combination

Worked Example 9 — resistivity (sum + product + power in one)

When to Round Percentage Uncertainties

Example — over-precise versus appropriate

Dominant Uncertainty

The first-approximation rule

Worked example — dominant uncertainty in practice

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