Measurement, Units, Scale, and Error

Measurements connect mathematics to physical quantities: a length, time, mass, temperature, or electrical value. A number without its unit is incomplete, and a measurement without an uncertainty can imply more certainty than the instrument supports. This lesson develops reliable conversion, estimation, and reporting habits. Familiarity with ratios, powers of ten, and percentages is helpful.

Quantities, Units, and SI Prefixes

A quantity is what is measured; a unit is the agreed reference used to state it. The International System of Units (SI) has base units such as metre (m), kilogram (kg), second (s), ampere (A), and kelvin (K). Other units are derived from them: speed has unit m/s, area m², and volume m³.

A prefix multiplies a unit by a power of ten. Prefixes change the size of the unit, not the quantity being described.

Prefix	Symbol	Factor	Example
giga	G	109	1 GW = 109 W
mega	M	106	1 Mm = 106 m
kilo	k	103	1 km = 103 m
centi	c	10−2	1 cm = 10−2 m
milli	m	10−3	1 mL = 10−3 L
micro	µ	10−6	1 µs = 10−6 s
nano	n	10−9	1 nm = 10−9 m

Symbols are case-sensitive: m means milli when used as a prefix, while M means mega. Also, 1 mg is 10⁻⁶ kg, because 1 g = 10⁻³ kg and milli means another factor of 10⁻³.

Worked example: prefix conversion

Convert 4.72 km to centimetres.

4.72 km × (1000 m / 1 km) × (100 cm / 1 m)
= 472 000 cm
= 4.72 × 10⁵ cm

The units km and m cancel, leaving cm. This cancellation is a useful check that the conversion has been set up correctly.

Unit Conversion by Dimensional Analysis

Dimensional analysis treats units like algebraic factors. Multiply by a conversion factor equal to 1, written so unwanted units cancel. For example, 60 s/1 min equals 1 because 60 s and 1 min name the same duration. Choose its orientation to cancel the starting unit.

Worked example: compound units

A cyclist travels at 18.0 km/h. Convert the speed to m/s.

18.0 km/h × (1000 m / 1 km) × (1 h / 3600 s)
= 5.00 m/s

The result is smaller numerically because one metre is much shorter than one kilometre, while one second is much shorter than one hour. Keeping factors as fractions prevents reversing a conversion by guesswork.

Worked example: squared units

Convert an area of 2.50 m² to cm².

2.50 m² × (100 cm / 1 m)²
= 2.50 m² × 10 000 cm² / 1 m²
= 25 000 cm²
= 2.50 × 10⁴ cm²

Do not merely multiply by 100: both dimensions are converted, so the factor is 100². For volume, cube the linear factor.

Dimensions as a Reasoning Tool

A dimension describes the type of quantity independently of the chosen unit. Length has dimension L, time T, and mass M. A speed has dimension L/T; an acceleration has L/T². An equation can only add or equate terms with the same dimensions.

Worked example: test a formula

Could d = v + t describe a distance d in terms of speed v and time t?

[d] = L
[v + t] would require L/T + T

The terms L/T and T are different dimensions, so they cannot be added.
The formula is dimensionally impossible.

In contrast, d = vt is plausible because (L/T)(T) = L. Dimensional agreement does not prove a formula is physically correct, but disagreement proves that it is wrong.

Significant Figures and Rounding

Significant figures communicate the resolution of a measured value. All nonzero digits are significant; zeros between significant digits are significant; leading zeros are not; and trailing zeros after a decimal point are significant. Scientific notation removes ambiguity: 3.00 × 10⁴ has three significant figures, whereas 30000 may be unclear.

• For multiplication and division, round the final result to the fewest significant figures in the measured inputs.
• For addition and subtraction, round the final result to the least precise decimal place in the inputs.
• Keep extra guard digits during calculation and round once at the end.

Worked example: two rounding rules

Multiplication: 3.42 cm × 1.6 cm = 5.472 cm²
Fewest significant figures is 2, so result = 5.5 cm².

Addition: 12.11 g + 0.3 g + 4.006 g = 16.416 g
The least precise place is tenths, so result = 16.4 g.

Exact counted quantities and defined conversion factors, such as 12 items in a dozen or 100 cm in 1 m, do not limit significant figures.

Accuracy, Precision, and Tolerance

Accuracy is closeness to an accepted or true value. Precision is the closeness of repeated measurements to one another. A set can be precise but inaccurate if a scale is consistently miscalibrated. Repeated readings help reveal random variation; calibration helps address systematic error.

A tolerance gives an acceptable interval. A rod labelled 50.0 cm ± 0.2 cm is accepted if its actual length is from 49.8 cm to 50.2 cm, inclusive. The ± value is an absolute tolerance in the same units as the measurement.

Worked example: interpreting an instrument

A ruler marked every millimetre gives a length reading of 12.4 cm. If the reading uncertainty is half the smallest division, report it as 12.4 cm ± 0.05 cm. This means a reasonable interval is 12.35 cm to 12.45 cm, not that the object has exactly 12.400... cm length.

Absolute, Relative, and Percent Error

When an accepted value is known, let the measured value be x and the accepted value be a. Then:

Absolute error

|x − a|, stated in the original unit.

Relative error

|x − a| / |a|, a unitless ratio.

Percent error

(|x − a| / |a|) × 100%.

Use the accepted value in the denominator. If it is zero, relative and percent error are undefined, so report absolute error instead.

Worked example: compare two measurements fairly

A voltage is accepted as 9.00 V and measured as 8.82 V.

Absolute error = |8.82 − 9.00| V = 0.18 V
Relative error = 0.18 / 9.00 = 0.020
Percent error  = 0.020 × 100% = 2.0%

An absolute error of 0.18 V may sound small or large depending on the quantity; the 2.0% error supplies that context.

Scale Drawings and Models

A scale compares a drawing or model length with the corresponding actual length in the same units. A scale of 1:50 means 1 unit on the drawing represents 50 of the same units in reality. Convert units before applying the ratio.

For a linear scale factor k from drawing to actual object, lengths multiply by k, areas by k², and volumes by k³. This is why a small-looking plan can represent a very large area.

Worked example: floor plan

A rectangular room measures 7.6 cm by 4.2 cm on a plan at scale 1:50. Find its actual dimensions and area.

Actual length = 7.6 cm × 50 = 380 cm = 3.8 m
Actual width  = 4.2 cm × 50 = 210 cm = 2.1 m

Actual area = (3.8 m)(2.1 m) = 7.98 m²
            = 8.0 m² to 2 significant figures

The 2 significant figures are appropriate because 7.6 cm and 4.2 cm each have 2 significant figures. Multiplying the drawn area by 50, rather than 50², would be a scale error.

Common Mistakes

• Moving a decimal without tracking units: write conversion factors so units visibly cancel.
• Forgetting powers in area or volume: convert m² with a squared factor and m³ with a cubed factor.
• Rounding every intermediate line: premature rounding can noticeably change a result.
• Calling close repeated readings accurate: they demonstrate precision; compare with an accepted value to assess accuracy.
• Using the measured value in percent-error denominator: use the accepted value.
• Applying a linear scale factor to area: square the factor for area and cube it for volume.

Practice Set

1. Write 0.000 047 0 m in scientific notation and state its number of significant figures.
2. Convert 6.35 µA to amperes and 0.0840 L to millilitres.
3. Convert 72.0 km/h to m/s using conversion factors.
4. Convert 0.360 m³ to litres. Use 1 m³ = 1000 L.
5. Use dimensions to decide whether E = mv² could represent an energy when [m] = M and [v] = L/T. Energy has dimension ML²/T².
6. Calculate 14.28 cm − 2.1 cm and report it with the correct decimal-place rule.
7. A part is specified as 25.00 mm ± 0.05 mm. Is a measured part of 25.06 mm within tolerance? State the allowed interval.
8. An experiment gives 3.76 g for a sample whose accepted mass is 3.80 g. Find the absolute error and percent error.
9. A map scale is 1:25 000. Two towns are 6.8 cm apart on the map. Find their actual separation in kilometres.
10. A model car is made at scale 1:18. The real car has a frontal area of 2.16 m². Find the model's frontal area in cm².

Answer Checks

1. 4.70 × 10⁻⁵ m; there are 3 significant figures. The final zero is significant because it is written after a decimal point.
2. 6.35 µA = 6.35 × 10⁻⁶ A. Also, 0.0840 L × 1000 mL/L = 84.0 mL.
3. 72.0 km/h × (1000 m/1 km) × (1 h/3600 s) = 20.0 m/s.
4. 0.360 m³ × 1000 L/m³ = 360 L. Written as 3.60 × 10² L, the three significant figures are explicit.
5. [mv²] = M(L/T)² = ML²/T², which matches energy. The expression is dimensionally plausible.
6. 14.28 cm − 2.1 cm = 12.18 cm, rounded to tenths: 12.2 cm.
7. The interval is 24.95 mm through 25.05 mm. Since 25.06 mm is above the upper limit, it is not within tolerance.
8. Absolute error = |3.76 − 3.80| g = 0.04 g. Percent error = (0.04/3.80) × 100% = 1.05...%, or 1.1%.
9. 6.8 cm × 25 000 = 170 000 cm = 1.7 km.
10. The model area is 2.16 m²/18² = 0.006 666... m². Since 1 m² = 10 000 cm², this is 66.7 cm² to 3 significant figures.