Units, conversions and compound measures

A unit tells us how a quantity is measured. A compound measure combines two or more quantities, so its unit combines their units too. For example,

$$\text{speed}=\frac{\text{distance}}{\text{time}}$$

may be measured in metres per second, written $\text{m/s}$ or $\text{m s}^{-1}$.

Correct unit work is not decoration. It determines which conversion factor to use, exposes impossible calculations, and gives meaning to a numerical answer.

Prerequisites

You should be able to:

• multiply and divide by powers of $10$;
• calculate with fractions and decimals;
• substitute into and rearrange formulae;
• use ratio and direct proportion.

Review exact arithmetic, ratio, proportion and rates of change or rearranging formulae if necessary.

Quantities, values and units

A measurement has a numerical value and a unit:

$$\text{quantity}=\text{numerical value}\times\text{unit}.$$

Thus $3.5\text{ m}$ and $350\text{ cm}$ describe the same length. The number changes because the size of the unit changes.

Only quantities of the same kind can sensibly be added or compared. Convert them to a common unit first:

$$2.4\text{ m}+75\text{ cm} =2.4\text{ m}+0.75\text{ m} =3.15\text{ m}.$$

The choice of common unit is usually flexible. Choose the one that makes the arithmetic clearest.

Converting linear units

Useful metric relationships include

$$\begin{aligned} 1\text{ km}&=1000\text{ m},\\ 1\text{ m}&=100\text{ cm}=1000\text{ mm},\\ 1\text{ kg}&=1000\text{ g},\\ 1\text{ litre}&=1000\text{ ml},\\ 1\text{ hour}&=60\text{ min}=3600\text{ s}. \end{aligned}$$

A reliable method is to multiply by a conversion fraction equal to $1$. Arrange it so the unwanted unit cancels.

Worked example 1: converting a length

Convert $4.72\text{ km}$ to metres.

Since $1000\text{ m}=1\text{ km}$,

$$4.72\text{ km}\times\frac{1000\text{ m}}{1\text{ km}} =4720\text{ m}.$$

The unit $\text{km}$ cancels. The result is larger numerically because metres are smaller units than kilometres.

Worked example 2: converting a time

Convert $2\text{ h }18\text{ min}$ to hours.

The $18$ minutes are

$$18\times\frac{1\text{ h}}{60\text{ min}}=0.3\text{ h}.$$

Therefore

$$2\text{ h }18\text{ min}=2.3\text{ h}.$$

It is not $2.18$ hours. Decimal notation is based on tenths and hundredths, whereas an hour contains $60$ minutes.

Self-check 1

1. Convert $0.086\text{ m}$ to millimetres.
2. Convert $7350\text{ g}$ to kilograms.
3. Convert $1\text{ h }42\text{ min}$ to hours.
4. Convert $2.35\text{ h}$ to hours and minutes.

Answers

1. $86\text{ mm}$
2. $7.35\text{ kg}$
3. $1+\frac{42}{60}=1.7\text{ h}$
4. $0.35\times60=21$, so $2\text{ h }21\text{ min}$

Area and volume conversions

This is the main conceptual hurdle in unit conversion. A length conversion factor must be squared for area and cubed for volume.

Since

$1\text{ m}=100\text{ cm},$

a square of side $1\text{ m}$ has area

$$1\text{ m}^2=(100\text{ cm})^2=10\,000\text{ cm}^2.$$

Similarly, a cube with side $1\text{ m}$ has volume

$$1\text{ m}^3=(100\text{ cm})^3=1\,000\,000\text{ cm}^3.$$

In general, if $1A=kB$, then

$$1A^2=k^2B^2, \qquad 1A^3=k^3B^3.$$

Worked example 3: converting area

Convert $2.7\text{ m}^2$ to $\text{cm}^2$.

Use the squared conversion:

$$\begin{aligned} 2.7\text{ m}^2 &=2.7(100\text{ cm})^2\\ &=2.7\times10\,000\text{ cm}^2\\ &=27\,000\text{ cm}^2. \end{aligned}$$

Multiplying by only $100$ would treat an area as though it were a length.

Worked example 4: converting volume in the opposite direction

Convert $4500\text{ cm}^3$ to $\text{m}^3$.

Because $1\text{ m}^3=1\,000\,000\text{ cm}^3$,

$$4500\text{ cm}^3 \times\frac{1\text{ m}^3}{1\,000\,000\text{ cm}^3} =0.0045\text{ m}^3.$$

Litres and cubic units

The key exact relationships are

$$1\text{ litre}=1000\text{ cm}^3=1\text{ dm}^3$$

and

$$1\text{ m}^3=1000\text{ litres}.$$

Worked example 5: capacity and volume

A tank has volume $0.84\text{ m}^3$. Find its capacity in litres.

$$0.84\text{ m}^3\times\frac{1000\text{ litres}}{1\text{ m}^3} =840\text{ litres}.$$

Self-check 2

1. Convert $36\,000\text{ mm}^2$ to $\text{cm}^2$.
2. Convert $0.052\text{ km}^2$ to $\text{m}^2$.
3. Convert $75\text{ cm}^3$ to $\text{mm}^3$.
4. Convert $3.2$ litres to $\text{cm}^3$ and to $\text{m}^3$.

Answers

1. Since $1\text{ cm}^2=100\text{ mm}^2$, the answer is $360\text{ cm}^2$.
2. $0.052\times1000^2=52\,000\text{ m}^2$.
3. $75\times10^3=75\,000\text{ mm}^3$.
4. $3200\text{ cm}^3$ and $0.0032\text{ m}^3$.

Compound units and unit algebra

Units can be multiplied, divided and cancelled like algebraic factors. The notation

$$\text{m/s}=\text{m s}^{-1}$$

means metres divided by seconds. Negative indices are especially useful for several factors:

$$\frac{\text{kg}}{\text{m}^3}=\text{kg m}^{-3}.$$

The word per means divide. Miles per gallon, pounds per kilogram and people per square kilometre are all rates.

Common compound measures are

$$\begin{aligned} \text{speed}&=\frac{\text{distance}}{\text{time}},\\ \text{density}&=\frac{\text{mass}}{\text{volume}},\\ \text{pressure}&=\frac{\text{force}}{\text{area}}. \end{aligned}$$

The same structure appears in unit price, flow rate, population density, frequency density and many other contexts.

Speed, distance and time

If $v$ is average speed, $d$ is distance and $t$ is elapsed time, then

$$v=\frac dt, \qquad d=vt, \qquad t=\frac dv.$$

Average speed uses total distance divided by total time. It is not generally the arithmetic mean of two speeds.

Worked example 6: mixed time units

A runner covers $15\text{ km}$ in $1\text{ h }12\text{ min}$. Find the average speed in $\text{km/h}$.

First convert the time:

$$1\text{ h }12\text{ min} =1+\frac{12}{60} =1.2\text{ h}.$$

Then

$$v=\frac{15}{1.2}=12.5\text{ km/h}.$$

Converting between $\text{m/s}$ and $\text{km/h}$

Starting with $1\text{ m/s}$,

$$1\frac{\text{m}}{\text{s}} \times\frac{1\text{ km}}{1000\text{ m}} \times\frac{3600\text{ s}}{1\text{ h}} =3.6\frac{\text{km}}{\text{h}}.$$

Therefore

$$\boxed{\text{m/s}\times3.6=\text{km/h}}$$

and

$$\boxed{\text{km/h}\div3.6=\text{m/s}}.$$

Worked example 7: distance in consistent units

A train travels at $108\text{ km/h}$ for $35$ seconds. Find the distance travelled in metres.

Convert the speed:

$$108\div3.6=30\text{ m/s}.$$

Now the speed and time use seconds, so

$$d=vt=30\times35=1050\text{ m}.$$

Worked example 8: why averaging speeds is subtle

A cyclist rides $12\text{ km}$ outward at $8\text{ km/h}$ and returns along the same route at $24\text{ km/h}$. Find the average speed for the whole journey.

The outward time is

$$\frac{12}{8}=1.5\text{ h},$$

and the return time is

$$\frac{12}{24}=0.5\text{ h}.$$

Hence

$$\text{average speed} =\frac{\text{total distance}}{\text{total time}} =\frac{24}{2} =12\text{ km/h}.$$

The answer is not $(8+24)/2=16\text{ km/h}$ because the cyclist spends different amounts of time at the two speeds.

Density, mass and volume

Density measures mass per unit volume:

$$\rho=\frac mV.$$

Rearranging gives

$$m=\rho V, \qquad V=\frac m\rho.$$

Typical units are $\text{g/cm}^3$ and $\text{kg/m}^3$.

Worked example 9: finding mass

A metal block has volume $240\text{ cm}^3$ and density $7.8\text{ g/cm}^3$. Find its mass in kilograms.

The units are already compatible:

$$\begin{aligned} m&=\rho V\\ &=7.8\frac{\text{g}}{\text{cm}^3}\times240\text{ cm}^3\\ &=1872\text{ g}\\ &=1.872\text{ kg}. \end{aligned}$$

The $\text{cm}^3$ factors cancel, leaving mass units.

Worked example 10: converting density

Convert $2.7\text{ g/cm}^3$ to $\text{kg/m}^3$.

Convert both the numerator and denominator:

$$\begin{aligned} 2.7\frac{\text{g}}{\text{cm}^3} &=2.7\frac{10^{-3}\text{ kg}}{10^{-6}\text{ m}^3}\\ &=2.7\times10^3\frac{\text{kg}}{\text{m}^3}\\ &=2700\text{ kg/m}^3. \end{aligned}$$

Thus

$$1\text{ g/cm}^3=1000\text{ kg/m}^3.$$

It is dangerous to convert only grams to kilograms. The cubic unit in the denominator must also be converted.

Pressure, force and area

Pressure is force per unit area:

$$p=\frac FA.$$

In SI units, force is measured in newtons, area in square metres, and pressure in pascals:

$$1\text{ Pa}=1\text{ N/m}^2.$$

Also,

$$F=pA, \qquad A=\frac Fp.$$

Worked example 11: area conversion before substitution

A force of $540\text{ N}$ acts uniformly on an area of $30\text{ cm}^2$. Find the pressure in pascals.

Convert the area:

$$30\text{ cm}^2 =30\times10^{-4}\text{ m}^2 =0.003\text{ m}^2.$$

Therefore

$$p=\frac{540}{0.003}=180\,000\text{ Pa}.$$

A small contact area can produce a large pressure. Using $30$ as though it meant $30\text{ m}^2$ would make the answer wrong by a factor of $10\,000$.

Multi-step compound measure problems

A robust method is:

1. write the defining formula;
2. choose the required output units;
3. convert inputs to a compatible set of units;
4. substitute, keeping units visible where useful;
5. calculate without premature rounding;
6. check the size and unit of the answer.

Worked example 12: flow rate and filling time

Water flows into a cuboid tank measuring $1.2\text{ m}$ by $80\text{ cm}$ by $50\text{ cm}$ at $24$ litres per minute. The tank is initially $15\%$ full. How long will it take to fill?

Convert all lengths to metres:

$80\text{ cm}=0.8\text{ m},\qquad50\text{ cm}=0.5\text{ m}.$

The tank volume is

$$1.2\times0.8\times0.5=0.48\text{ m}^3=480\text{ litres}.$$

Since $15\%$ is already filled, $85\%$ remains:

$$0.85\times480=408\text{ litres}.$$

Hence

$$t=\frac{408\text{ litres}}{24\text{ litres/min}} =17\text{ min}.$$

Worked example 13: density with geometric volume

A solid cylindrical rod has radius $6\text{ mm}$, length $1.5\text{ m}$ and density $8.4\text{ g/cm}^3$. Find its mass in kilograms.

Centimetres are convenient because the density uses $\text{cm}^3$:

$$r=0.6\text{ cm}, \qquad h=150\text{ cm}.$$

The volume is

$$V=\pi r^2h =\pi(0.6)^2(150) =54\pi\text{ cm}^3.$$

Therefore

$$\begin{aligned} m&=\rho V\\ &=8.4(54\pi)\text{ g}\\ &=453.6\pi\text{ g}\\ &\approx1.425\text{ kg}. \end{aligned}$$

Keeping $\pi$ until the final line avoids unnecessary rounding error.

Units as an error check

Units cannot prove that an answer is correct, but inconsistent units prove that something is wrong.

For instance, from $d=vt$,

$$\left(\frac{\text{m}}{\text{s}}\right)(\text{s})=\text{m},$$

so the right side has the required distance unit. By contrast, $d=v/t$ would have units

$$\frac{\text{m/s}}{\text{s}}=\text{m/s}^2,$$

which describes acceleration, not distance.

Quantities joined by addition or subtraction must have compatible units. In a model such as

$$s=ut+\frac12at^2,$$

both terms have length units:

$$(\text{m/s})(\text{s})=\text{m},$$

and

$$(\text{m/s}^2)(\text{s}^2)=\text{m}.$$

This principle becomes increasingly valuable in functions in modelling and mechanics.

Common misconceptions

• Converting after mixing incompatible units. Convert before substituting, unless unit cancellation is being handled explicitly.
• Using a linear conversion for area or volume. Square or cube the entire conversion factor.
• Reading decimal hours as minutes. $2.4$ hours is $2$ hours $24$ minutes, not $2$ hours $40$ minutes.
• Inverting a rate. If speed is distance per time, time per distance is its reciprocal and is a different measure.
• Averaging rates without considering weights. Use total quantity divided by total time, mass, distance or other relevant total.
• Dropping the unit. An answer of $12$ does not distinguish $12\text{ m/s}$ from $12\text{ km/h}$.
• Assuming unit cancellation guarantees correctness. A dimensionally consistent formula may still use the wrong constant or model.

Mixed self-check

1. A car travels $162\text{ km}$ in $2\text{ h }15\text{ min}$. Find its average speed in $\text{km/h}$ and $\text{m/s}$.
2. A liquid has mass $630\text{ g}$ and density $0.84\text{ g/cm}^3$. Find its volume in litres.
3. A $1.8\text{ kN}$ force acts on a rectangle measuring $4\text{ cm}$ by $6\text{ cm}$. Find the pressure in pascals.
4. A tap delivers $7.5$ litres per minute. How many seconds does it take to deliver $2.25$ litres?
5. Explain why the equation $E=mc$ cannot represent kinetic energy if $m$ is mass and $c$ is speed.
6. A journey consists of $30$ minutes at $40\text{ km/h}$ and $90$ minutes at $80\text{ km/h}$. Find the average speed.

Answers

1. $2\text{ h }15\text{ min}=2.25\text{ h}$, so $v=162/2.25=72\text{ km/h}=20\text{ m/s}$.
2. $V=m/\rho=630/0.84=750\text{ cm}^3=0.75$ litres.
3. $1.8\text{ kN}=1800\text{ N}$ and $24\text{ cm}^2=0.0024\text{ m}^2$, so $p=1800/0.0024=750\,000\text{ Pa}$.
4. The rate is $7.5/60=0.125$ litres per second, so $t=2.25/0.125=18\text{ s}$.
5. Its units would be $(\text{kg})(\text{m/s})=\text{kg m s}^{-1}$, the units of momentum. Energy has units $\text{kg m}^2\text{ s}^{-2}$.
6. The distances are $40(0.5)=20\text{ km}$ and $80(1.5)=120\text{ km}$. Therefore the average speed is $140/2=70\text{ km/h}$.

Final checklist

Before accepting an answer, ask:

1. What quantity am I finding?
2. Which formula defines it?
3. Are all input units compatible?
4. Have I squared or cubed a length conversion where required?
5. Do the units of my result match the required quantity?
6. Is the numerical size plausible?

Next steps

Strengthen the algebra behind these formulae in rearranging formulae and the structure of rates in ratio, proportion and rates of change. Then apply unit reasoning to functions in modelling, kinematics language and connected rates of change.