Combining measurements and their uncertainties

By UK A Level Physics · independent study · myAudience

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Combining measurements and their uncertainties

A derived result inherits uncertainty from its inputs. This lesson assumes fractional uncertainty and rearranging a simple formula. For this cohort, Δ denotes a positive absolute uncertainty and the rules are conservative, first-order school estimates. They are not NIST's statistical standard-uncertainty calculation, which may include correlations and quadrature. Stating this assumption prevents two different uncertainty conventions from being mixed in one answer.

For addition or subtraction of quantities measured separately, add their absolute uncertainties. If a length is obtained as b − a, uncertain endpoints can move in opposite directions, so the uncertainty does not cancel just because the values are subtracted. For b = 82.4 ± 0.5 mm and a = 12.1 ± 0.5 mm, the difference is 70.3 ± 1.0 mm. If both readings share an identical zero offset, that particular component may cancel; identify the shared component rather than pretending every source cancels.

For multiplication or division, add fractional uncertainties, or equivalently percentage uncertainties. With z = xy, Δz/|z| ≈ Δx/|x| + Δy/|y|. The same positive sum is used for z = x/y. A division does not subtract uncertainties. Exact numerical factors such as 2 or π introduce no measurement uncertainty. Compute the nominal result first, combine the fractional contributions, then multiply that total fraction by the nominal result's magnitude.

For a power z = xⁿ, multiply the fractional uncertainty by |n|. A squared diameter therefore contributes twice its fractional uncertainty to a calculated circular area, while a square root contributes half. In an illustrative density model ρ = m/(πr²L), the contributions are Δm/m + 2Δr/r + ΔL/L. This explains why a small uncertain radius can dominate a result even when the mass is measured precisely. It also suggests a targeted improvement: improve the radius measurement rather than every instrument equally.

These rules approximate small uncertainty effects; large relative intervals or complicated dependence may require evaluating endpoint values and reconsidering the model. does not require uncertainty combinations involving trigonometric or logarithmic functions here. Retain intermediate digits and round the final uncertainty sensibly, then align the value's decimal place. A small numerical uncertainty also cannot establish that a model, such as constant speed or a perfect cylinder, is physically suitable.

Read the two rulers separately. The endpoints are 24.8 and 25.2 mm in the first panel and 1.8 and 2.2 mm in the second. In each case the full interval is 0.4 mm wide, whereas the absolute uncertainty is its half-width, 0.2 mm. The percentages compare that half-width with the central value: 100(0.2/25.0) = 0.8% and 100(0.2/2.0) = 10%. The larger percentage does not mean the second interval is physically wider. It means the stipulated bound is large relative to the quantity being reported. Separate scales prevent the drawings from implying that the two objects themselves have equal length.

The area panel requires two length conversions, one for each side. A square with sides 1 cm and 1 cm has area (0.01 m)(0.01 m) = 0.0001 m². The grid is schematic, not a calibrated photograph. A prefix acts on the unit before the unit is squared: cm² means (cm)². Converting a stated uncertainty to another unit requires the same factor as the central value. Thus 25.0 ± 0.2 mm becomes 0.0250 ± 0.0002 m; its percentage uncertainty stays 0.8%. A change of units cannot improve precision.

**Worked subtraction and addition.** Two independently bounded position readings are x₁ = 12.4 ± 0.3 cm and x₂ = 28.6 ± 0.4 cm. Find the displacement and its conservative bound. The centre is 16.2 cm. The lowest difference is 28.2 − 12.7 = 15.5 cm; the highest is 29.0 − 12.1 = 16.9 cm. Therefore report 16.2 ± 0.7 cm. Adding the positions instead gives 41.0 ± 0.7 cm. For sums and differences the absolute contributions add under this worst-case bound convention; subtracting the central values does not subtract their uncertainties.

**Worked multiplication with exact endpoint check.** Let a = 20.0 ± 0.2 mm and b = 10.0 ± 0.1 mm be independently bounded positive lengths. The central area is 200 mm². First-order fractional propagation gives 0.2/20 + 0.1/10 = 0.02, hence approximately 200 ± 4 mm². Exact endpoint products are 19.8×9.9 = 196.02 and 20.2×10.1 = 204.02 mm². The asymmetry exposes the discarded product of small errors; it does not invalidate the useful first-order approximation. An exact enclosing interval is [196.02, 204.02] mm². In SI the first-order report is 0.000200 ± 0.000004 m², not 0.0200 m². This is a bound calculation, not a standard deviation or a 95% interval.

**Transfer check.** A zero offset z is common to both readings, so x₁ = r₁ − z and x₂ = r₂ − z. Does the uncertainty in z contribute twice to x₂ − x₁? Explain algebraically before calculating. Answer: (r₂ − z) − (r₁ − z) = r₂ − r₁, so this exactly shared additive offset cancels. Treating it as two independent contributions would discard known correlation. Scale-factor errors do not generally cancel from a difference. School propagation rules need an explicit measurement model; they are not universal statistical rules.

**All-prefix independent practice.** Convert to SI and normalised standard form: 2.4 THz; 3.6 GHz; 7.2 MJ; 5.0 kV; 8.1 cm; 4.3 mm; 6.2 μA; 9.4 nm; 1.7 pF; 2.8 fs. Answers in order: 2.4×10¹² Hz, 3.6×10⁹ Hz, 7.2×10⁶ J, 5.0×10³ V, 8.1×10⁻² m, 4.3×10⁻³ m, 6.2×10⁻⁶ A, 9.4×10⁻⁹ m, 1.7×10⁻¹² F, 2.8×10⁻¹⁵ s. These are the ten prefixed symbols T, G, M, k, c, m, μ, n, p and f required here. A separate unprefixed case is 0.0065 s = 6.5×10⁻³ s. Convert 3.0 mm³ too: 3.0×10⁻⁹ m³. Confusing milli with mega changes a value by a factor of 10⁹.

Worked example: An illustrative speed is v = d/t with d = 1.20 ± 0.01 m and t = 0.80 ± 0.02 s. Estimate v and its uncertainty.

Nominal v = 1.20/0.80 = 1.50 m s⁻¹.

Fractional uncertainty = 0.01/1.20 + 0.02/0.80 = 0.033333….

Absolute uncertainty ≈ 1.50 × 0.033333… = 0.05 m s⁻¹.

Answer: v = 1.50 ± 0.05 m s⁻¹; the time contribution is larger.

Check your understanding: A radius has 3% uncertainty. What uncertainty does it contribute to πr²?

Answer: 6%, under the stated first-order rule.

Why: Squaring doubles the fractional contribution; the exact π is not another uncertain measurement.