1.1.4. Unit Conversions
Table of Contents
Why Unit Conversions Matter in Physics
Unit conversions appear everywhere in physics. Experimental data comes in many unit systems, textbooks switch between different conventions, and real-world problems often mix units from engineering, everyday life, and science.
This chapter focuses on how to convert units correctly and systematically. We assume you already know what a physical quantity is and what units are, and that you have seen SI base and some derived units. Here we concentrate on the method of converting between units and typical patterns that occur in physics.
Unit conversion is not just an arithmetic trick. It is a way of checking that you understand what a quantity really represents. If you can change its units correctly, you probably understand the quantity.
Important principle of unit conversions:
You can multiply any physical quantity by a fraction equal to 1, written as a ratio of two equal quantities expressed in different units, without changing the physical quantity itself.
For example,
\[
1 = \frac{1000 \ \text{m}}{1 \ \text{km}} = \frac{1 \ \text{inch}}{2.54 \ \text{cm}} = \frac{3600 \ \text{s}}{1 \ \text{h}}.
\]
These expressions are the core of the factor–label method, a systematic technique for unit conversions.
The Factor–Label Method
The factor–label method, also called dimensional analysis in a narrow sense, is a procedure where you treat units as algebraic symbols that can cancel.
You follow three steps:
- Write the quantity with its original units.
- Multiply by conversion factors that equal 1, arranged so that unwanted units cancel.
- Simplify to obtain the quantity in the desired units.
For example, convert 5.0 km to meters:
\[
5.0 \ \text{km} \times \frac{1000 \ \text{m}}{1 \ \text{km}} = 5.0 \times 1000 \ \text{m} = 5.0 \times 10^{3} \ \text{m}.
\]
The symbol “km” appears in the numerator and denominator and cancels, leaving “m” as the final unit.
In this method you use the same algebraic rules for symbols like “m” and “s” as for regular variables. Units that appear both in the numerator and denominator can be canceled, and units can be distributed over products and powers.
Conversion Factors
A conversion factor is built from an equality between two different unit expressions for the same physical quantity. Each equality produces two possible fractions, both equal to 1. You choose the orientation that cancels the unwanted unit.
For example, from
\[
1 \ \text{h} = 3600 \ \text{s}
\]
you can form
\[
\frac{1 \ \text{h}}{3600 \ \text{s}} = 1, \quad \frac{3600 \ \text{s}}{1 \ \text{h}} = 1.
\]
The first one is useful when you want to remove seconds and obtain hours. The second one is useful when you want to remove hours and obtain seconds.
The following table summarizes some frequently used conversion equalities. The detailed meaning of these physical quantities is discussed in other chapters; here we focus only on the unit relationships.
| Quantity type | Equality (definition of factor) |
|---|---|
| Length | $1 \ \text{km} = 1000 \ \text{m}$ |
| Length | $1 \ \text{m} = 100 \ \text{cm}$ |
| Length | $1 \ \text{in} = 2.54 \ \text{cm}$ (exact) |
| Length | $1 \ \text{ft} = 12 \ \text{in}$ |
| Time | $1 \ \text{min} = 60 \ \text{s}$ |
| Time | $1 \ \text{h} = 60 \ \text{min} = 3600 \ \text{s}$ |
| Mass | $1 \ \text{kg} = 1000 \ \text{g}$ |
| Mass | $1 \ \text{tonne} = 1000 \ \text{kg}$ |
| Energy | $1 \ \text{kWh} = 3.6 \times 10^{6} \ \text{J}$ |
When you construct a conversion factor, always start from a correct equality of units.
Using Prefixes in Unit Conversions
Metric prefixes give you a very compact way to express large or small multiples of a base unit. In earlier chapters you will see the definitions of common prefixes. For conversions, you treat a prefixed unit as a plain numerical factor times the base unit.
For example,
\[
1 \ \text{km} = 10^{3} \ \text{m}, \quad 1 \ \text{ms} = 10^{-3} \ \text{s}.
\]
To convert between prefixed units, proceed through the base unit or combine the powers of ten directly.
Example: Convert $3.2 \ \text{ms}$ to seconds.
\[
3.2 \ \text{ms} = 3.2 \times 10^{-3} \ \text{s}.
\]
Example: Convert $5.0 \ \text{km}$ to centimeters.
You can do this in one step:
\[
1 \ \text{km} = 10^{3} \ \text{m}, \quad 1 \ \text{m} = 10^{2} \ \text{cm}.
\]
Therefore
\[
1 \ \text{km} = 10^{3} \times 10^{2} \ \text{cm} = 10^{5} \ \text{cm}.
\]
Then
\[
5.0 \ \text{km} = 5.0 \times 10^{5} \ \text{cm}.
\]
You can also show this with explicit conversion factors:
\[
5.0 \ \text{km} \times \frac{10^{3} \ \text{m}}{1 \ \text{km}} \times \frac{10^{2} \ \text{cm}}{1 \ \text{m}} = 5.0 \times 10^{5} \ \text{cm}.
\]
Rule for prefixes:
Replace a prefixed unit with its factor times the base unit. Combine powers of ten carefully. For example, $1 \ \text{mm} = 10^{-3} \ \text{m}$, $1 \ \mu\text{s} = 10^{-6} \ \text{s}$.
Converting Composite Units
In physics most quantities use composite units, such as velocity ($\text{m/s}$), acceleration ($\text{m/s}^2$), or force ($\text{N}$ which is $\text{kg}\,\text{m/s}^2$). To convert composite units, you must convert each part of the unit correctly, including exponents.
You will regularly encounter units that are products or quotients of simpler units:
\[
\text{m/s}, \quad \text{m}^2, \quad \text{m}^3, \quad \text{N/m}^2, \quad \text{J/kg}\,\text{K}, \quad \text{W/m}^2.
\]
The factor–label method still applies, but you must remember that:
- Units in the numerator and denominator both can be converted.
- Exponents on units apply to the numerical conversion factor as well.
Example: Speed Conversion
Suppose a car travels at $90 \ \text{km/h}$. Convert this speed to $\text{m/s}$.
We know that
\[
1 \ \text{km} = 1000 \ \text{m}, \quad 1 \ \text{h} = 3600 \ \text{s}.
\]
Write
\[
90 \ \frac{\text{km}}{\text{h}} \times \frac{1000 \ \text{m}}{1 \ \text{km}} \times \frac{1 \ \text{h}}{3600 \ \text{s}}.
\]
Now cancel “km” and “h”:
\[
90 \times \frac{1000}{3600} \ \frac{\text{m}}{\text{s}} = \frac{90\,000}{3600} \ \frac{\text{m}}{\text{s}} = 25 \ \frac{\text{m}}{\text{s}}.
\]
You can imagine the cancellation visually: the “km” in the numerator cancels with “km” in the denominator of the first factor; the “h” in the denominator cancels with “h” in the numerator of the second factor. Only “m” in the numerator and “s” in the denominator remain.
Example: Area and Volume Conversions
Whenever a unit has an exponent, such as $\text{m}^2$ or $\text{cm}^3$, the conversion factor must also be raised to that power.
Take the equality
\[
1 \ \text{m} = 100 \ \text{cm}.
\]
For area,
\[
1 \ \text{m}^2 = (1 \ \text{m})^2 = (100 \ \text{cm})^2 = 100^2 \ \text{cm}^2 = 10^{4} \ \text{cm}^2.
\]
For volume,
\[
1 \ \text{m}^3 = (1 \ \text{m})^3 = (100 \ \text{cm})^3 = 100^3 \ \text{cm}^3 = 10^{6} \ \text{cm}^3.
\]
This explains why area and volume conversions often differ by large powers of 10 compared with length conversions.
Rule for powers of units:
If $1 \ \text{A} = k \ \text{B}$, then $1 \ \text{A}^n = k^n \ \text{B}^n$. You must raise both the numerical factor and the unit to the same power.
Step-by-Step Strategy for Composite Conversions
When you face more complicated unit conversions, such as $\text{g/cm}^3$ to $\text{kg/m}^3$ for density, use a step-by-step strategy:
- Identify all the different unit symbols in the expression.
- Decide what the final units should be.
- Write conversion factors for each unit you need to change.
- Multiply by these factors, placing them so that unwanted units cancel.
- Simplify the resulting numerical factor and units.
Example: Density Conversion
Convert a density of $1.2 \ \text{g/cm}^3$ to $\text{kg/m}^3$.
First convert grams to kilograms:
\[
1 \ \text{kg} = 1000 \ \text{g} \quad \Rightarrow \quad 1 \ \text{g} = 10^{-3} \ \text{kg}.
\]
Second, convert $\text{cm}^3$ to $\text{m}^3$:
\[
1 \ \text{cm} = 10^{-2} \ \text{m} \quad \Rightarrow \quad 1 \ \text{cm}^3 = (10^{-2})^3 \ \text{m}^3 = 10^{-6} \ \text{m}^3.
\]
Now start from the original quantity:
\[
1.2 \ \frac{\text{g}}{\text{cm}^3}.
\]
Replace g and $\text{cm}^3$ using the equalities:
\[
1.2 \ \frac{\text{g}}{\text{cm}^3}
= 1.2 \ \frac{10^{-3} \ \text{kg}}{10^{-6} \ \text{m}^3}
= 1.2 \times 10^{-3+6} \ \frac{\text{kg}}{\text{m}^3}
= 1.2 \times 10^{3} \ \frac{\text{kg}}{\text{m}^3}.
\]
Thus the density is $1.2 \times 10^{3} \ \text{kg/m}^3$.
You can also write this with explicit conversion factors:
\[
1.2 \ \frac{\text{g}}{\text{cm}^3}
\times \frac{1 \ \text{kg}}{1000 \ \text{g}}
\times \frac{(100 \ \text{cm})^3}{1 \ \text{m}^3}.
\]
The “g” and “cm” units cancel appropriately, leaving “kg” and “m”.
Converting Between SI and Non-SI Units
Although the SI system is standard in scientific contexts, many everyday and engineering quantities are expressed in non-SI units. In physics problems you may need to convert between them.
Here are some common examples:
| Quantity | Non-SI unit | SI equivalent |
|---|---|---|
| Length | inch (in) | $1 \ \text{in} = 2.54 \ \text{cm}$ |
| Length | foot (ft) | $1 \ \text{ft} = 0.3048 \ \text{m}$ |
| Speed | km/h | $1 \ \text{km/h} \approx 0.278 \ \text{m/s}$ |
| Speed | mph | $1 \ \text{mph} \approx 0.447 \ \text{m/s}$ |
| Pressure | atmosphere (atm) | $1 \ \text{atm} = 1.013 \times 10^{5} \ \text{Pa}$ |
| Energy | calorie (cal) | $1 \ \text{cal} \approx 4.184 \ \text{J}$ |
| Energy | electronvolt (eV) | $1 \ \text{eV} \approx 1.602 \times 10^{-19} \ \text{J}$ |
The same factor–label method works. The only difference is that numerical values sometimes are approximate rather than exact, so your final answer might also be approximate.
Example: Convert a speed of $55 \ \text{mph}$ to $\text{m/s}$.
Use
\[
1 \ \text{mph} \approx 0.447 \ \frac{\text{m}}{\text{s}}.
\]
Then
\[
55 \ \text{mph} \approx 55 \times 0.447 \ \frac{\text{m}}{\text{s}} \approx 24.6 \ \frac{\text{m}}{\text{s}}.
\]
Whenever you use approximate conversion factors, remember that your result is also approximate and should not have more significant figures than the factors you used. The chapter on significant figures explains how to decide the appropriate number of digits.
Inverse and Reciprocal Units
Units can appear in the denominator, such as $\text{s}^{-1}$ or $\text{m}^{-1}$. These are often called inverse or reciprocal units. They occur in rates, frequencies, and many other physical contexts.
To convert such units, you can either:
- Convert the underlying unit first, then take the inverse.
- Use conversion factors placed in the denominator.
For example, convert a rate of $5 \ \text{s}^{-1}$ to $\text{min}^{-1}$.
We know $1 \ \text{min} = 60 \ \text{s}$ so
\[
1 \ \text{s}^{-1} = \frac{1}{1 \ \text{s}} = \frac{1}{\frac{1}{60} \ \text{min}} = 60 \ \text{min}^{-1}.
\]
Thus
\[
5 \ \text{s}^{-1} = 5 \times 60 \ \text{min}^{-1} = 300 \ \text{min}^{-1}.
\]
Alternatively, write
\[
5 \ \text{s}^{-1} \times \frac{60 \ \text{s}}{1 \ \text{min}} = 300 \ \text{min}^{-1}.
\]
The key idea is that the conversion factor still equals 1, and the units cancel just as in other examples.
Common Patterns and Shortcuts
Although the systematic factor–label method is always safe, some conversions become so common that physicists learn shortcuts. It is important, however, to understand that these shortcuts come from the same underlying method.
For example, the relation between km/h and m/s is used very frequently. From
\[
1 \ \text{km/h} = \frac{1000 \ \text{m}}{3600 \ \text{s}} = \frac{10}{36} \ \frac{\text{m}}{\text{s}} \approx 0.278 \ \frac{\text{m}}{\text{s}},
\]
you might memorize that
\[
v \ (\text{m/s}) \approx \frac{v \ (\text{km/h})}{3.6},
\]
and
\[
v \ (\text{km/h}) \approx 3.6 \times v \ (\text{m/s}).
\]
These relations are simply the factor–label method applied once and then remembered.
Shortcut for km/h to m/s:
$v \ (\text{m/s}) = \dfrac{v \ (\text{km/h})}{3.6}$, and
$v \ (\text{km/h}) = 3.6 \, v \ (\text{m/s})$.
Shortcuts can save time, but if you ever doubt a shortcut, reconstruct it from basic conversion factors. This reduces the chance of using a wrong factor.
Typical Mistakes and How to Avoid Them
Many errors in physics calculations are not conceptual mistakes, but unit conversion mistakes. Here are some common ones and ways to avoid them.
One frequent mistake is forgetting to convert all parts of a composite unit. For example, when converting $\text{km/h}$ to $\text{m/s}$, some students convert kilometers to meters but forget to convert hours to seconds.
Another common error is forgetting to raise the numerical factor when converting squared or cubed units. For instance, thinking that $1 \ \text{m}^2 = 100 \ \text{cm}^2$ instead of $10^{4} \ \text{cm}^2$.
A third mistake is mixing up the orientation of the conversion factor. If you multiply by $\frac{1 \ \text{km}}{1000 \ \text{m}}$ when you intend to convert kilometers to meters, you will obtain a number that is 1000 times too small.
To avoid these mistakes:
- Write the units explicitly at every step and treat them as algebraic symbols.
- Check that unwanted units cancel, and desired units remain.
- For powers of units, write the conversion factor inside parentheses and raise it to the relevant power, such as $(100 \ \text{cm})^2$.
- Keep track of reasonable magnitudes. If you convert 1 km to meters and obtain 0.001 m, something is clearly wrong.
Consistency of Units in Calculations
Unit conversions are not only about changing the form of a single quantity. They are also essential to make entire calculations consistent. In physics, formulas are derived assuming that quantities are expressed in specific units, typically SI units.
For example, in a later chapter you will see the equation for kinetic energy,
\[
K = \frac{1}{2} m v^2.
\]
If you insert $m$ in kilograms and $v$ in meters per second, the result $K$ comes out in joules automatically, since the joule is defined as
\[
1 \ \text{J} = 1 \ \text{kg} \cdot \text{m}^2/\text{s}^2.
\]
However, if you were to insert $m$ in grams and $v$ in km/h without converting to SI units, the numerical result would not be a value in joules. You would first need to convert all input quantities to consistent units, usually SI, or carefully track the units and convert at the end.
Thus, before using any formula, you should:
- Decide on a consistent set of units for all quantities, most conveniently SI.
- Convert each quantity to those units.
- Perform the calculation.
- Convert the result to any other unit you prefer for interpretation.
This procedure prevents hidden unit errors and makes it easier to compare results.
Summary
In this chapter you have seen how to convert units using the factor–label method, how to handle metric prefixes, how to treat powers and composite units, and how to recognize and avoid common mistakes. Unit conversions are a basic but essential skill, and they underpin almost every quantitative step you will take in physics.
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