Measurement accuracy and dimensional checks

Learning outcome

Interpret rounded measurements, bound calculated results, and use dimensional checks to reject unsuitable formulas.

Concepts and assumptions

A measured value is not automatically exact. Resolution is the smallest change an instrument distinguishes; accuracy concerns closeness to the true value. Extra calculator digits do not improve accuracy.

Assume uncertainty comes only from stated rounding; actual instruments may introduce other errors. For positive values exactly halfway between two choices, round upward.

A reading x rounded to step q represents values from x − q/2 inclusive to x + q/2 exclusive. Its rounding-error magnitude is at most q/2: crossing a midpoint changes the rounded reading.

For positive rectangular dimensions, area increases with either dimension. Multiply lower bounds to obtain the minimum area and upper bounds for the exclusive upper limit. Keep intermediate arithmetic exact.

Dimensions describe quantity types. Perimeter requires length units, area squared units, and volume cubed units. Added terms need compatible dimensions. Wrong dimensions disprove a formula; correct dimensions alone do not prove it.

Worked examples

Example 1 — A rounded reading. A length is recorded as 12.6 cm to the nearest 0.1 cm. Find its range.

Half-step = 0.1/2 = 0.05 cm. Lower bound = 12.6 − 0.05 = 12.55 cm. Upper limit = 12.6 + 0.05 = 12.65 cm. Thus 12.55 cm ≤ length < 12.65 cm; rounding-error magnitude is at most 0.05 cm.

Example 2 — Area uncertainty. A rectangle is recorded as 12 cm by 8 cm, each to the nearest centimetre.

Nominal area = 12 × 8 = 96 cm². Length lies from 11.5 inclusive to 12.5 exclusive; width from 7.5 inclusive to 8.5 exclusive, in cm. Minimum area = 11.5 × 7.5 = 86.25 cm². Upper limit = 12.5 × 8.5 = 106.25 cm². Therefore 86.25 cm² ≤ area < 106.25 cm². These calculated bounds do not imply hundredth-square-centimetre measurement accuracy.

Example 3 — Checking formulas. For a 5 m by 3 m rectangle, test proposed area formulas 2(length + width) and (length + width)².

The first gives 2(5 + 3) = 16 m: length units, so it cannot be area. The second gives (5 + 3)² = 64 m²: suitable dimensions, but incorrect. Actual area = 5 × 3 = 15 m². Dimensional agreement is necessary, not sufficient.

Common mistakes

Calling rounded readings exact; confusing resolution with accuracy; adding incompatible quantities; trusting formulas merely because their units match; rounding intermediate bounds.

Practice questions

1. A length is 7.2 m to the nearest 0.1 m. Give its range and rounding-error bound.
2. A square’s side is 10 cm to the nearest centimetre. Find its nominal area and area range.
3. For lengths l, w and h, check the dimensions of lwh, 2(lw + wh + hl), and lwh + l.
4. A rectangle is recorded as 2.4 m by 1.8 m, each to the nearest 0.1 m. Find its nominal perimeter and range. Is exactly 8.400 m justified?

Worked solutions

1. Half-step = 0.05 m. Thus 7.15 m ≤ length < 7.25 m, with rounding-error magnitude at most 0.05 m.
2. Side range: 9.5 cm ≤ side < 10.5 cm. Nominal area = 100 cm². Squaring gives 90.25 cm² ≤ area < 110.25 cm².
3. lwh has volume dimensions; 2(lw + wh + hl) has area dimensions. lwh + l is invalid: volume and length cannot be added.
4. Nominal perimeter = 2(2.4 + 1.8) = 8.4 m. Bounds are 2(2.35 + 1.75) = 8.2 m and 2(2.45 + 1.85) = 8.6 m. Hence 8.2 m ≤ perimeter < 8.6 m. Exactness is unjustified.