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Unit Converter
Choose the quantity and the two units, then type a number. The converter shows the conversion factor, the multiplication or division it uses, and the result in all the other units, rounded to 6 significant figures. You can type decimals (2.5), fractions (3/4) and numbers in scientific notation (1.6e-19).
What Is a Conversion Factor?
A conversion factor is the number you multiply a measurement by to express it in another unit. For example, ${1\ \text{in}=2.54\ \text{cm}},$ so the conversion factor from inches to centimeters is $2.54,$ and the factor from centimeters to inches is ${1\div 2.54\approx 0.3937}.$
To change the unit, multiply by a fraction that equals 1: the new unit on top, the old unit at the bottom. The old unit cancels.
$$8\ \text{in}=8\ \cancel{\text{in}}\times\frac{2.54\ \text{cm}}{1\ \cancel{\text{in}}}=20.32\ \text{cm}$$The tables below give the conversion factors for length, area, volume, mass, speed, density, force, energy, power and pressure in metric (SI), US and imperial units, and the formulas for temperature. In the tables, ${=}$ means that the value is exact and ${\approx}$ means that it is rounded to 4 significant figures. After the tables you will find how to convert step by step, a unit converter that shows each step, and practice problems.
On this page: Converter · Prefixes · Length · Area · Volume · Mass · Speed · Density · Force · Energy · Power · Pressure · Temperature · How to convert · Practice
Metric Prefixes
In the metric system, the units of one quantity differ by powers of 10, and the prefix in front of the unit tells you which power. The same prefix works with every unit: kilo means 1000 in the kilometer ($1000$ meters), the kilogram ($1000$ grams) and the kilowatt ($1000$ watts).
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| giga | G | $10^9$ | ${1\ \text{GW}=10^9\ \text{W}}$ |
| mega | M | $10^6$ | ${1\ \text{MJ}=10^6\ \text{J}}$ |
| kilo | k | $1000$ | ${1\ \text{km}=1000\ \text{m}}$ |
| hecto | h | $100$ | ${1\ \text{hL}=100\ \text{L}}$ |
| deca | da | $10$ | ${1\ \text{dam}=10\ \text{m}}$ |
| — | — | $1$ | $\text{m},$ $\text{g},$ $\text{L}$ |
| deci | d | $0.1$ | ${1\ \text{dm}=0.1\ \text{m}}$ |
| centi | c | $0.01$ | ${1\ \text{cm}=0.01\ \text{m}}$ |
| milli | m | $0.001$ | ${1\ \text{mg}=0.001\ \text{g}}$ |
| micro | µ | ${10^{-6}}$ | ${1\ \mu\text{m}=10^{-6}\ \text{m}}$ |
| nano | n | ${10^{-9}}$ | ${1\ \text{nm}=10^{-9}\ \text{m}}$ |
Because the metric units differ by powers of 10, a conversion inside the metric system only moves the decimal point. When you change to a smaller unit, the number gets bigger: ${3.75\ \text{km}=3750\ \text{m}}$ (the point moves 3 places to the right, because ${1\ \text{km}=1000\ \text{m}}$). When you change to a bigger unit, the number gets smaller: ${250\ \text{g}=0.25\ \text{kg}}.$
Length
| Metric units | |
|---|---|
| 1 kilometer (km) | $=1000\ \text{m}$ |
| 1 meter (m) | $=10\ \text{dm}$ $=100\ \text{cm}$ $=1000\ \text{mm}$ |
| 1 centimeter (cm) | $=10\ \text{mm}$ $=0.01\ \text{m}$ |
| 1 millimeter (mm) | $=0.001\ \text{m}$ |
| 1 micrometer (µm), also called a micron (µ) | $=10^{-6}\ \text{m}$ |
| 1 nanometer (nm), formerly called a millimicron (mµ) | $=10^{-9}\ \text{m}$ |
| 1 angstrom (Å) | $=0.1\ \text{nm}$ $=10^{-10}\ \text{m}$ |
| US and imperial units | |
| 1 inch (in) | $=2.54\ \text{cm}$ |
| 1 foot (ft) | $=12\ \text{in}$ $=30.48\ \text{cm}$ |
| 1 yard (yd) | $=3\ \text{ft}$ $=36\ \text{in}$ $=0.9144\ \text{m}$ |
| 1 mile (mi) | $=5280\ \text{ft}$ $=1760\ \text{yd}$ $=1.609344\ \text{km}$ |
| 1 mil (a thousandth of an inch) | $=0.001\ \text{in}$ $=0.0254\ \text{mm}$ |
| 1 nautical mile (nmi) | $=1852\ \text{m}$ $\approx1.151\ \text{mi}$ |
| From metric to US and imperial units | |
| 1 centimeter (cm) | $\approx0.3937\ \text{in}$ |
| 1 meter (m) | $\approx39.37\ \text{in}$ $\approx3.281\ \text{ft}$ $\approx1.094\ \text{yd}$ |
| 1 kilometer (km) | $\approx0.6214\ \text{mi}$ |
Since 1959 the inch has been defined as exactly $2.54\ \text{cm}.$ So every factor between the inch, the foot, the yard or the mile and the metric units of length follows exactly from $2.54;$ the table rounds the ones that are long or never end, such as ${1\ \text{m}}={\frac{100}{2.54}\ \text{in}}\approx{39.37\ \text{in}.}$
Area
An area unit is a square: $1\ \text{cm}^2$ is the area of a square with side $1\ \text{cm}.$ Since ${1\ \text{cm}=10\ \text{mm}},$ this square is made of ${10\times 10=100}$ small squares of $1\ \text{mm}^2.$ So the factor for area is the square of the factor for length: ${1\ \text{m}^2}={(100\ \text{cm})^2}={10{,}000\ \text{cm}^2,}$ not $100\ \text{cm}^2.$
${1\ \text{cm}^2}={10\times 10\ \text{mm}^2}={100\ \text{mm}^2}$
| Metric units | |
|---|---|
| 1 square centimeter (cm²) | $=100\ \text{mm}^2$ |
| 1 square decimeter (dm²) | $=100\ \text{cm}^2$ |
| 1 square meter (m²) | $=100\ \text{dm}^2$ $=10{,}000\ \text{cm}^2$ |
| 1 are (a) | $=100\ \text{m}^2$ |
| 1 hectare (ha) | $=100\ \text{a}$ $=10{,}000\ \text{m}^2$ |
| 1 square kilometer (km²) | $=100\ \text{ha}$ $=10^{6}\ \text{m}^2$ |
| US and imperial units | |
| 1 square inch (in²) | $=6.4516\ \text{cm}^2$ |
| 1 square foot (ft²) | $=144\ \text{in}^2$ $=929.0304\ \text{cm}^2$ |
| 1 square yard (yd²) | $=9\ \text{ft}^2$ $\approx0.8361\ \text{m}^2$ |
| 1 acre | $=43{,}560\ \text{ft}^2$ $=4840\ \text{yd}^2$ $\approx4047\ \text{m}^2$ $\approx0.4047\ \text{ha}$ |
| 1 square mile (mi²) | $=640\ \text{acres}$ $\approx2.590\ \text{km}^2$ |
| From metric to US and imperial units | |
| 1 square meter (m²) | $\approx10.76\ \text{ft}^2$ $\approx1.196\ \text{yd}^2$ |
| 1 hectare (ha) | $\approx2.471\ \text{acres}$ |
| 1 square kilometer (km²) | $\approx0.3861\ \text{mi}^2$ |
The metric area units on the way from $\text{m}^2$ to $\text{km}^2$ go up in steps of 100: ${1\ \text{a}=100\ \text{m}^2},$ ${1\ \text{ha}=100\ \text{a}},$ ${1\ \text{km}^2=100\ \text{ha}}.$ A hectare is the area of a square with side $100\ \text{m}.$
Volume
A volume unit is a cube. A cube with edge ${10\ \text{cm}=1\ \text{dm}}$ is made of ${10\times 10\times 10=1000}$ small cubes of $1\ \text{cm}^3,$ and it holds exactly one liter. The factor for volume is the cube of the factor for length: ${1\ \text{m}^3}={(100\ \text{cm})^3}={1{,}000{,}000\ \text{cm}^3.}$
${1\ \text{L}}={1\ \text{dm}^3}={10\times 10\times 10\ \text{cm}^3}={1000\ \text{cm}^3}$
| Metric units | |
|---|---|
| 1 milliliter (mL) | $=1\ \text{cm}^3$ |
| 1 liter (L) | $=1\ \text{dm}^3$ $=1000\ \text{cm}^3$ $=1000\ \text{mL}$ |
| 1 cubic meter (m³) | $=1000\ \text{L}$ $=10^{6}\ \text{cm}^3$ |
| US and imperial units | |
| 1 cubic inch (in³) | $=16.387064\ \text{cm}^3$ |
| 1 cubic foot (ft³) | $=1728\ \text{in}^3$ $\approx28.32\ \text{L}$ $\approx7.481\ \text{gal}$ |
| 1 cubic yard (yd³) | $=27\ \text{ft}^3$ $\approx0.7646\ \text{m}^3$ |
| 1 US fluid ounce (fl oz) | $\approx29.57\ \text{mL}$ |
| 1 US cup | $=8\ \text{fl oz}$ $\approx236.6\ \text{mL}$ |
| 1 US pint (pt) | $=2\ \text{cups}$ $=16\ \text{fl oz}$ $\approx473.2\ \text{mL}$ |
| 1 US quart (qt) | $=2\ \text{pt}$ $\approx0.9464\ \text{L}$ |
| 1 US gallon (gal) | $=4\ \text{qt}$ $=231\ \text{in}^3$ $\approx3.785\ \text{L}$ |
| 1 imperial pint (imp pt) | $\approx568.3\ \text{mL}$ |
| 1 imperial gallon (imp gal) | $=8\ \text{imp pt}$ $=4.54609\ \text{L}$ $\approx1.201\ \text{gal}$ $\approx277.4\ \text{in}^3$ |
| From metric to US and imperial units | |
| 1 liter (L) | $\approx1.057\ \text{qt}$ $\approx0.2642\ \text{gal}$ $\approx61.02\ \text{in}^3$ $\approx0.03531\ \text{ft}^3$ |
| 1 cubic meter (m³) | $\approx35.31\ \text{ft}^3$ $\approx264.2\ \text{gal}$ |
The US gallon and the imperial (British) gallon are different units: the imperial gallon is about 20% larger. The cup in the table is the US customary cup; nutrition labels in the US use a cup of $240\ \text{mL}.$
Mass
| Metric units | |
|---|---|
| 1 metric ton, or tonne (t) | $=1000\ \text{kg}$ |
| 1 kilogram (kg) | $=1000\ \text{g}$ |
| 1 gram (g) | $=1000\ \text{mg}$ |
| US and imperial units | |
| 1 ounce (oz) | $\approx28.35\ \text{g}$ |
| 1 pound (lb) | $=16\ \text{oz}$ $=0.45359237\ \text{kg}$ $\approx453.6\ \text{g}$ $\approx0.03108\ \text{slugs}$ |
| 1 stone (st) | $=14\ \text{lb}$ $\approx6.350\ \text{kg}$ |
| 1 short ton (US ton) | $=2000\ \text{lb}$ $\approx907.2\ \text{kg}$ |
| 1 long ton (UK ton) | $=2240\ \text{lb}$ $\approx1016\ \text{kg}$ |
| 1 slug | $\approx32.17\ \text{lb}$ $\approx14.59\ \text{kg}$ |
| From metric to US and imperial units | |
| 1 kilogram (kg) | $\approx2.205\ \text{lb}$ $\approx35.27\ \text{oz}$ $\approx0.06852\ \text{slugs}$ |
| 1 metric ton, or tonne (t) | $\approx2205\ \text{lb}$ $\approx1.102\ \text{short tons}$ |
The pound is defined as exactly $0.45359237\ \text{kg}.$ As a unit of mass, the word "ton" has three meanings: the metric ton (tonne) of $1000\ \text{kg},$ the US short ton of $2000\ \text{lb}$ and the British long ton of $2240\ \text{lb}.$ The slug is the mass that a force of $1\ \text{lbf}$ accelerates at $1\ \text{ft/s}^2.$
Speed
| 1 meter per second (m/s) | $=3.6\ \text{km/h}$ $\approx2.237\ \text{mi/h}$ $\approx3.281\ \text{ft/s}$ |
| 1 kilometer per hour (km/h) | $=\tfrac{5}{18}\ \text{m/s}$ $\approx0.2778\ \text{m/s}$ $\approx0.6214\ \text{mi/h}$ $\approx0.9113\ \text{ft/s}$ |
| 1 mile per hour (mi/h, or mph) | $=1.609344\ \text{km/h}$ $=0.44704\ \text{m/s}$ $\approx1.467\ \text{ft/s}$ |
| 1 foot per second (ft/s) | $=0.3048\ \text{m/s}$ $=1.09728\ \text{km/h}$ $\approx0.6818\ \text{mi/h}$ |
| 1 knot (kn) | $=1.852\ \text{km/h}$ $\approx0.5144\ \text{m/s}$ $\approx1.151\ \text{mi/h}$ |
From km/h to m/s, divide by $3.6.$ From m/s to km/h, multiply by $3.6.$
Where does $3.6$ come from? An hour has $3600$ seconds, so $$1\ \frac{\text{km}}{\text{h}}=\frac{1000\ \text{m}}{3600\ \text{s}}=\frac{1}{3.6}\ \frac{\text{m}}{\text{s}}.$$ A knot is one nautical mile per hour. It is used at sea and in aviation.
Density
| 1 gram per cubic centimeter (g/cm³), the same as 1 g/mL or 1 kg/L | $=1000\ \text{kg/m}^3$ $\approx62.43\ \text{lb/ft}^3$ $\approx1.940\ \text{slug/ft}^3$ |
| 1 kilogram per cubic meter (kg/m³) | $=1\ \text{g/L}$ $=0.001\ \text{g/cm}^3$ |
| 1 pound per cubic foot (lb/ft³) | $\approx16.02\ \text{kg/m}^3$ $\approx0.01602\ \text{g/cm}^3$ |
| 1 pound per cubic inch (lb/in³) | $=1728\ \text{lb/ft}^3$ $\approx27.68\ \text{g/cm}^3$ |
| 1 slug per cubic foot (slug/ft³) | $\approx515.4\ \text{kg/m}^3$ $\approx0.5154\ \text{g/cm}^3$ |
Water has a density of about ${1\ \text{g/cm}^3=1000\ \text{kg/m}^3}.$ This is why one liter of water has a mass of about one kilogram, and one cubic meter of water a mass of about one metric ton.
Force
| 1 newton (N) | $=1\ \text{kg}\cdot\text{m/s}^2$ $=10^{5}\ \text{dyn}$ $\approx0.1020\ \text{kgf}$ $\approx0.2248\ \text{lbf}$ $\approx7.233\ \text{pdl}$ |
| 1 kilonewton (kN) | $=1000\ \text{N}$ $\approx102.0\ \text{kgf}$ $\approx224.8\ \text{lbf}$ |
| 1 dyne (dyn) | $=1\ \text{g}\cdot\text{cm/s}^2$ $=10^{-5}\ \text{N}$ |
| 1 kilogram-force (kgf) | $=9.80665\ \text{N}$ $\approx2.205\ \text{lbf}$ |
| 1 pound-force (lbf) | $\approx4.448\ \text{N}$ $\approx0.4536\ \text{kgf}$ $\approx32.17\ \text{pdl}$ |
| 1 poundal (pdl) | $=1\ \text{lb}\cdot\text{ft/s}^2$ $\approx0.1383\ \text{N}$ |
Under standard gravity, ${g=9.80665\ \text{m/s}^2}$ (close to its value on the Earth's surface), a mass of $1\ \text{kg}$ weighs ${1\ \text{kgf}\approx 9.8\ \text{N}},$ and a mass of $1\ \text{lb}$ weighs $1\ \text{lbf}.$ Older books write nt for the newton, kgwt and lbwt for the kilogram-force and the pound-force, and gm for the gram.
Energy
| 1 joule (J) | $=1\ \text{N}\cdot\text{m}$ $=1\ \text{W}\cdot\text{s}$ $=10^{7}\ \text{erg}$ $\approx0.7376\ \text{ft}\cdot\text{lbf}$ $\approx0.2390\ \text{cal}$ $\approx9.478\times10^{-4}\ \text{Btu}$ |
| 1 kilojoule (kJ) | $=1000\ \text{J}$ $\approx0.2390\ \text{kcal}$ |
| 1 calorie (cal) | $=4.184\ \text{J}$ $\approx3.086\ \text{ft}\cdot\text{lbf}$ $\approx3.966\times10^{-3}\ \text{Btu}$ |
| 1 kilocalorie (kcal), the food Calorie | $=1000\ \text{cal}$ $=4.184\ \text{kJ}$ |
| 1 watt-hour (Wh) | $=3600\ \text{J}$ $=3.6\ \text{kJ}$ |
| 1 kilowatt-hour (kWh) | $=3.6\times10^{6}\ \text{J}$ $=3.6\ \text{MJ}$ $\approx860.4\ \text{kcal}$ $\approx3412\ \text{Btu}$ |
| 1 foot-pound (ft·lbf) | $\approx1.356\ \text{J}$ $\approx0.3240\ \text{cal}$ $\approx1.285\times10^{-3}\ \text{Btu}$ |
| 1 British thermal unit (Btu) | $\approx1055\ \text{J}$ $\approx778.2\ \text{ft}\cdot\text{lbf}$ $\approx252.2\ \text{cal}$ $\approx0.2931\ \text{Wh}$ |
| 1 electronvolt (eV) | $\approx1.602\times10^{-19}\ \text{J}$ |
| 1 erg | $=10^{-7}\ \text{J}$ |
There are several calories. This table uses the thermochemical calorie, ${1\ \text{cal}=4.184\ \text{J}}.$ The International Table calorie is slightly larger, $4.1868\ \text{J}.$ The Calorie with a capital C on food labels is the kilocalorie. The Btu in the table is the International Table Btu.
Power
| 1 watt (W) | $=1\ \text{J/s}$ $=10^{7}\ \text{erg/s}$ $\approx0.2390\ \text{cal/s}$ $\approx0.7376\ \text{ft}\cdot\text{lbf/s}$ $\approx3.412\ \text{Btu/h}$ |
| 1 kilowatt (kW) | $=1000\ \text{W}$ $\approx1.341\ \text{hp}$ $\approx1.360\ \text{PS}$ $\approx737.6\ \text{ft}\cdot\text{lbf/s}$ $\approx3412\ \text{Btu/h}$ |
| 1 horsepower (hp), mechanical | $=550\ \text{ft}\cdot\text{lbf/s}$ $={33{,}000\ \text{ft}}\cdot {\text{lbf/min}}$ $\approx745.7\ \text{W}$ |
| 1 metric horsepower (PS) | $\approx735.5\ \text{W}$ $\approx0.9863\ \text{hp}$ |
| 1 foot-pound per second (ft·lbf/s) | $\approx1.356\ \text{W}$ |
| 1 Btu per hour (Btu/h) | $\approx0.2931\ \text{W}$ |
| 1 calorie per second (cal/s) | $=4.184\ \text{W}$ |
Power is energy per unit of time, so $1\ \text{kWh}$ is the energy used in one hour at a power of $1\ \text{kW}.$ The horsepower (hp) used in the US and the UK is about $745.7\ \text{W}.$ Car engines in Europe are often rated in metric horsepower (PS or CV), which is about $735.5\ \text{W},$ or 1.4% less.
Pressure
| 1 pascal (Pa) | $=1\ \text{N/m}^2$ $=10\ \text{dyn/cm}^2$ $\approx9.869\times10^{-6}\ \text{atm}$ $\approx1.450\times10^{-4}\ \text{psi}$ $\approx0.02089\ \text{lbf/ft}^2$ |
| 1 kilopascal (kPa) | $=1000\ \text{Pa}$ |
| 1 megapascal (MPa) | $=10^{6}\ \text{Pa}$ |
| 1 bar | $=10^{5}\ \text{Pa}$ $=100\ \text{kPa}$ $\approx0.9869\ \text{atm}$ $\approx14.50\ \text{psi}$ |
| 1 standard atmosphere (atm) | $=101{,}325\ \text{Pa}$ $=101.325\ \text{kPa}$ $=1.01325\ \text{bar}$ $=760\ \text{mmHg}$ $\approx14.70\ \text{psi}$ $\approx406.8\ \text{inH}_2\text{O}$ |
| 1 millimeter of mercury (mmHg), practically the same as 1 torr | $\approx133.3\ \text{Pa}$ |
| 1 pound-force per square inch (psi) | $\approx6895\ \text{Pa}$ $\approx6.895\ \text{kPa}$ $\approx51.71\ \text{mmHg}$ $\approx27.68\ \text{inH}_2\text{O}$ |
| 1 pound-force per square foot (lbf/ft²) | $\approx47.88\ \text{Pa}$ |
| 1 inch of water (inH₂O) | $\approx249.1\ \text{Pa}$ |
Pressure is force per unit of area: ${1\ \text{Pa}=1\ \text{N/m}^2}.$ The standard atmosphere is a fixed value close to the average air pressure at sea level; it holds up a column of mercury $760\ \text{mm}$ high in a barometer. "psi" means pounds (pound-force) per square inch. The inch of water has several slightly different definitions (they depend on the temperature of the water), so its factors are always shown rounded.
Temperature
Temperature is the only quantity on this page that cannot be converted with a factor alone: the Celsius, Fahrenheit and Kelvin scales have different zero points. Use these formulas:
Example. $25\ {}^\circ\text{C}$ in degrees Fahrenheit: ${\frac{9}{5}\times 25+32}={45+32}={77,}$ so ${25\ {}^\circ\text{C}=77\ {}^\circ\text{F}}.$ And $98.6\ {}^\circ\text{F}$ in degrees Celsius: ${\frac{5}{9}\,(98.6-32)}={\frac{5}{9}\times 66.6}={37,}$ so ${98.6\ {}^\circ\text{F}=37\ {}^\circ\text{C}}.$
| ${}^\circ\text{C}$ | ${}^\circ\text{F}$ | $\text{K}$ | |
|---|---|---|---|
| absolute zero | ${-273.15}$ | ${-459.67}$ | $0$ |
| the same in °C and °F | ${-40}$ | ${-40}$ | $233.15$ |
| water freezes at 1 atm | $0$ | $32$ | $273.15$ |
| room temperature | $20$ | $68$ | $293.15$ |
| body temperature | $37$ | $98.6$ | $310.15$ |
| water boils at 1 atm | $100$ | $212$ | $373.15$ |
A temperature difference (a change of temperature) is converted with the factor alone, without the ${+32}$ or the ${+273.15}\text{:}$ a change of $1\ {}^\circ\text{C}$ is a change of $1\ \text{K}$ and of $1.8\ {}^\circ\text{F}.$
How to Convert Units
Every conversion except temperature uses the same idea. An equation between two units, such as ${1\ \text{in}=2.54\ \text{cm}},$ gives two fractions that are equal to 1:
${\frac{2.54\ \text{cm}}{1\ \text{in}}=1}$ and ${\frac{1\ \text{in}}{2.54\ \text{cm}}=1}$
Multiplying by 1 does not change a quantity, only the unit in which it is written. You choose the fraction that has the old unit at the bottom, so that it cancels.
- Write the quantity with its unit.
- Find a row of the tables that connects the old unit with the new one.
- Multiply by the fraction with the new unit on top and the old unit at the bottom, and cancel the units.
- Check the answer: a smaller unit needs a bigger number. Round the answer to about as many significant figures as the given number has.
Example 1: Metric Units
Convert $3.2\ \text{km}$ to meters and $450\ \text{mm}$ to meters.
${1\ \text{km}=1000\ \text{m}},$ so ${3.2\ \text{km}}={3.2\times 1000\ \text{m}}={3200\ \text{m}.}$
${1\ \text{mm}=0.001\ \text{m}},$ so ${450\ \text{mm}}={450\times 0.001\ \text{m}}={0.45\ \text{m}.}$
Example 2: Miles to Kilometers
A marathon is $26.2$ miles long. How many kilometers is it?
${26.2\ \text{mi}}={26.2\ \cancel{\text{mi}}}\times {\frac{1.609344\ \text{km}}{1\ \cancel{\text{mi}}}}\approx{42.2\ \text{km}}$
The product is $42.164\ldots,$ but $26.2$ has only 3 significant figures, so the answer is about $42.2\ \text{km}.$
Example 3: Dividing by the Factor
Convert $100\ \text{cm}$ to inches.
Now the old unit is the centimeter, so the centimeters go to the bottom of the fraction:
${100\ \text{cm}}={100\ \cancel{\text{cm}}\times\frac{1\ \text{in}}{2.54\ \cancel{\text{cm}}}}={\frac{100}{2.54}\ \text{in}}\approx{39.37\ \text{in}}$
Multiplying by $\frac{1}{2.54}$ is the same as dividing by $2.54.$
Example 4: A Chain of Factors
How many inches are there in a mile?
No row of the table goes from miles straight to inches, but there are rows from miles to feet and from feet to inches. Multiply by both fractions:
${1\ \text{mi}}\times {\frac{5280\ \text{ft}}{1\ \text{mi}}\times\frac{12\ \text{in}}{1\ \text{ft}}}={63{,}360\ \text{in}}$
Example 5: Area
A room is $15\ \text{ft}$ long and $12\ \text{ft}$ wide. Find its area in square meters.
The area is ${15\times 12=180\ \text{ft}^2}.$ Since ${1\ \text{ft}=0.3048\ \text{m}},$ one square foot is ${0.3048^2\ \text{m}^2}={0.09290304\ \text{m}^2,}$ and
${180\ \text{ft}^2}={180\times 0.09290304\ \text{m}^2}\approx{16.7\ \text{m}^2.}$
You can also convert the sides first: ${15\ \text{ft}=4.572\ \text{m}},$ ${12\ \text{ft}=3.6576\ \text{m}},$ and ${4.572\times 3.6576\approx 16.7}.$ The answer is the same.
Example 6: Volume
A fish tank is $50\ \text{cm}$ long, $30\ \text{cm}$ wide and $40\ \text{cm}$ high. How many liters of water does it hold?
${V}={50\times 30\times 40}={60{,}000\ \text{cm}^3.}$ Since ${1\ \text{L}=1000\ \text{cm}^3},$ the tank holds ${60{,}000\div 1000=60\ \text{L}}.$
Example 7: Speed
Convert $90\ \text{km/h}$ to meters per second.
Both units change, so multiply by two fractions: one cancels the kilometers, the other cancels the hours.
${90\ \frac{\cancel{\text{km}}}{\cancel{\text{h}}}}\times {\frac{1000\ \text{m}}{1\ \cancel{\text{km}}}\times\frac{1\ \cancel{\text{h}}}{3600\ \text{s}}}={25\ \frac{\text{m}}{\text{s}}}$
The numbers give ${90\times 1000\div 3600=25}.$ This is the rule from the Speed section: ${90\div 3.6=25}.$
Example 8: Density
Iron has a density of $7.87\ \text{g/cm}^3.$ Express it in $\text{kg/m}^3.$
Replace $1\ \text{g}$ by $0.001\ \text{kg}$ and $1\ \text{cm}^3$ by ${10^{-6}\ \text{m}^3}$ (because ${1\ \text{m}^3=10^6\ \text{cm}^3}$):
${7.87\ \frac{\text{g}}{\text{cm}^3}}={7.87\times\frac{0.001\ \text{kg}}{10^{-6}\ \text{m}^3}}={7.87\times 1000\ \frac{\text{kg}}{\text{m}^3}}={7870\ \frac{\text{kg}}{\text{m}^3}}$
Example 9: Energy
A $2\ \text{kW}$ heater is on for $3$ hours. How much energy does it use, in kilowatt-hours and in joules?
${E}={2\ \text{kW}\times 3\ \text{h}}={6\ \text{kWh}.}$ Since ${1\ \text{kWh}=3.6\times 10^6\ \text{J}},$ ${E}={6\times 3.6\times 10^6\ \text{J}}={2.16\times 10^7\ \text{J}.}$
Practice Problems
Click a problem to see the answer.
1. Convert $4.2\ \text{km}$ to meters.
${4.2\times 1000=4200\ \text{m}}.$
2. Convert $350\ \text{mm}$ to meters.
${350\times 0.001=0.35\ \text{m}}.$
3. A man is $6\ \text{ft}$ tall. What is his height in centimeters?
${6\times 30.48=182.88\ \text{cm}},$ about $183\ \text{cm}.$
4. How many miles is a $10\ \text{km}$ race?
${10\div 1.609344}\approx{6.21\ \text{mi}}$ (or ${10\times 0.6214\approx 6.21}$).
5. Convert $3\ \text{m}^2$ to square centimeters.
${1\ \text{m}^2=10{,}000\ \text{cm}^2},$ so ${3\ \text{m}^2=30{,}000\ \text{cm}^2}.$
6. A field has an area of $5000\ \text{m}^2.$ How many hectares is that? How many acres?
${5000\div 10{,}000}={0.5\ \text{ha};}$ ${\ 0.5\times 2.471\approx 1.24}$ acres.
7. How many $250\ \text{mL}$ glasses can you fill from a $2.5\ \text{L}$ bottle?
${2.5\ \text{L}=2500\ \text{mL}},$ and ${2500\div 250=10}$ glasses.
8. Convert $72\ \text{km/h}$ to meters per second.
${72\div 3.6=20\ \text{m/s}}.$
9. A sprinter runs at $10\ \text{m/s}.$ What is this speed in km/h and in mi/h?
${10\times 3.6=36\ \text{km/h}};$ ${\ 36\div 1.609344}\approx{22.4\ \text{mi/h}.}$
10. A fuel tank holds $15$ US gallons. How many liters is that?
${15\times 3.785\approx 56.8\ \text{L}}.$
11. A car has a mass of $1200\ \text{kg}.$ What is its mass in pounds?
${1200\div 0.45359237}\approx{2646\ \text{lb}}$ (or ${1200\times 2.205=2646}$).
12. A $60\ \text{W}$ lamp is on for $5$ hours. How much energy does it use, in kWh and in joules?
${60\times 5}={300\ \text{Wh}}={0.3\ \text{kWh};}$ ${\ 0.3\times 3.6\times 10^6}={1.08\times 10^6\ \text{J}.}$
13. An engine has a power of $150\ \text{hp}.$ What is its power in kilowatts?
${150\times 745.7\ \text{W}}={111{,}855\ \text{W}}\approx{112\ \text{kW}.}$
14. A tire is inflated to $32\ \text{psi}.$ What is that in kilopascals and in bars?
${32\times 6.895}\approx{221\ \text{kPa}}\approx{2.21\ \text{bar}.}$
15. Convert $30\ {}^\circ\text{C}$ to degrees Fahrenheit and ${-10\ {}^\circ\text{F}}$ to degrees Celsius.
${\frac{9}{5}\times 30+32=86\ {}^\circ\text{F}};$ ${\ \frac{5}{9}\,(-10-32)}={-\frac{70}{3}}\approx{-23.3\ {}^\circ\text{C}.}$
16. How many cubic centimeters are there in one cubic foot?
${1\ \text{ft}=30.48\ \text{cm}},$ so ${1\ \text{ft}^3}={30.48^3\ \text{cm}^3}\approx{28{,}317\ \text{cm}^3}$ (about $28.3\ \text{L}$).
See also: Exponents · Ratio and Proportion · Percents · Measuring Angles
