Error propagation
Error propagated into a computed area from errors in two sides
A rectangular parcel measured on two sides, and the uncertainty those two measurements put into the acreage — with the fractional form of the same result as a check and the worst case as a bound.
Given
- A rectangular parcel is measured on two adjacent sides.
- Side a = 248.16 ft, with a standard error of ±0.03 ft.
- Side b = 412.55 ft, with a standard error of ±0.04 ft.
- The two measurements were made independently — different setups, different tape settings — so their errors are uncorrelated.
- The corners are square by construction and the squareness is taken as exact for this computation.
- Distances in US survey feet.
Required
- The area and its standard error, in square feet and in acres.
- The relative precision of the area, and how it relates to the relative precisions of the two sides.
- The 95 per cent uncertainty in the area.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Area, and the two partial derivatives
The area is A = a · b. To propagate error into it, take the partial derivative of A with respect to each measured quantity and multiply by that quantity's standard error. The partial with respect to a is b, and the partial with respect to b is a — so the error in one side is amplified by the length of the other.
That amplification is the substance of the problem. An error in the short side gets multiplied by the long side, so the short side's 0.03 ft matters more than its size suggests. It is why measuring the short side casually because it is short is a mistake.
The two contributions are then combined in quadrature, because the two measurements are independent.
A = a · b = 248.16 × 412.55 = 102,378.41 ft²
∂A/∂a = b = 412.55 ft ∂A/∂b = a = 248.16 ft
σ_A = √( (b · σ_a)² + (a · σ_b)² )Evaluate the two contributions
The contribution from the short side is 412.55 × 0.03 = 12.38 ft², and from the long side 248.16 × 0.04 = 9.93 ft². The short side, with the smaller absolute error, contributes more to the area error than the long side does.
Combining in quadrature gives ±15.87 ft². Note that this is less than the sum of the two contributions, 22.30 ft², and more than either one alone — which is the whole behavior of quadrature. It is closer to the sum than it would be if one contribution dominated, because the two are of comparable size.
In acres, ±15.87 ft² is ±0.000364 acre. The parcel is 2.3503 acres, and its uncertainty does not reach the third decimal of an acre — a useful reassurance that reporting an area to 0.001 acre is honest for a parcel measured this well.
| Source | Partial derivative | σ of the measurement | Contribution to σ_A (ft²) | Squared (ft⁴) |
|---|---|---|---|---|
| Side a | b = 412.55 | 0.03 ft | 12.3765 | 153.1778 |
| Side b | a = 248.16 | 0.04 ft | 9.9264 | 98.5334 |
| Σ of squares | 251.7112 | |||
| σ_A = √251.7112 | 15.8654 |
The fractional form, and where the area's precision comes from
Divide the whole propagation by A and it collapses to a statement about relative precisions: the fractional error in a product is the quadrature sum of the fractional errors of its factors. That is the version worth carrying in your head, because it does not depend on the units or the size of the parcel.
Side a is measured to 1:8,272 and side b to 1:10,314. The area is therefore known to 1:6,453 — worse than either side. A product is always known less precisely than its worst factor, which is a fact people routinely get backwards when they reason that errors 'average out'.
The rule of thumb that follows: to know an area to a given relative precision, each side must be measured better than that. Getting the acreage to 1:10,000 on this parcel would need both sides at roughly 1:14,000.
σ_a / a = 0.03 / 248.16 = 1.2089 × 10⁻⁴ (1 : 8,272)
σ_b / b = 0.04 / 412.55 = 9.6959 × 10⁻⁵ (1 : 10,314)
σ_A / A = √( (1.2089×10⁻⁴)² + (9.6959×10⁻⁵)² ) = 1.5497 × 10⁻⁴ (1 : 6,453)
σ_A = 1.5497 × 10⁻⁴ × 102,378.41 = 15.87 ft² ✓ agrees with the direct computationReport the area with its uncertainty
State the area as 102,378 ± 16 ft², or 2.3503 ± 0.0004 acre. Rounding the uncertainty to two significant figures, and the value to match, is the ordinary convention and avoids implying that the uncertainty itself is known to four figures.
The 95 per cent figure is 1.96 σ, or ±31 ft². That is the number to quote when a specification is written in terms of a 95 per cent confidence, which most modern ones are. Quoting a one-sigma figure against a 95 per cent specification silently claims twice the precision.
For a plat, none of this changes the reported acreage — 2.350 acres either way. It changes what can be said if the acreage is later disputed, which is when the computation sheet has to stand on its own.
σ_A = ±15.87 ft² → report ±16 ft²
in acres: 15.87 / 43,560 = 0.000364 ac → ±0.0004 ac
E95 = 1.96 × 15.87 = 31.1 ft² → ±31 ft², or ±0.0007 ac
area = 102,378 ± 16 ft² (1σ) = 2.3503 ± 0.0004 acAnswer
- Area = 248.16 × 412.55 = 102,378.41 ft², or 2.3503 acres.
- σ_A = √((412.55 × 0.03)² + (248.16 × 0.04)²) = √(153.1778 + 98.5334) = √251.7112 = ±15.87 ft².
- Report as 102,378 ± 16 ft², or 2.3503 ± 0.0004 acre at one sigma; ±31 ft² at 95 per cent.
- The area is known to 1 : 6,453, which is worse than either side (1 : 8,272 and 1 : 10,314). A product is always less precise than its worst factor.
- The short side contributes more to the area error (12.38 ft²) than the long side does (9.93 ft²), because each side's error is amplified by the length of the other.
Check
Perturb both sides by their standard errors in the same direction and see what the area does: 248.19 × 412.59 = 102,400.71 ft², which is 22.30 ft² above the nominal area. That is the worst case, where both errors conspire, and it equals the arithmetic sum of the two contributions 12.38 + 9.93 = 22.30 ft².
The 0.0012 ft² by which the perturbation exceeds the sum of the two linear terms is the second-order product σ_a · σ_b = 0.03 × 0.04 = 0.0012 ft². Its being negligible confirms that the linearised propagation is appropriate here — first-order propagation is only valid while the second-order term is small, and it is smaller than the first-order terms by four orders of magnitude.
The quadrature answer of 15.87 ft² is 71 per cent of the worst case of 22.30 ft². For two independent contributions of comparable size the expected ratio is between 1/√2 = 0.707 and 1, and 0.71 sits right at the lower bound as it should when the two contributions are nearly equal.
The fractional route and the direct route give 15.87 ft² independently, using different arithmetic on the same inputs.
More error propagation
All in this category- Standard deviation, standard error and a rejection decision on repeated distancesTen EDM observations of one baseline, reduced to a residuals table, a standard deviation and a standard error of the mean — and one suspect observation tested against a criterion that can actually reject it.
- Error in a taped distance measured in n tape lengths, against a single EDM shotThe same 648 ft line measured two ways. Series propagation over seven tape settings against one EDM observation with a constant-plus-ppm specification, and what the comparison says about when taping is still adequate.
- Error propagated into latitude and departure from errors in an angle and a distanceOne course, one distance uncertainty and one azimuth uncertainty, carried through the general propagation formula with partial derivatives — and a rotation argument that checks the answer without repeating it.