In §1.7 we saw that any unit can be written as a fraction equal to one, and chain-multiplying those fractions cancels unwanted units. That was just the introduction. Here we go through the full recipe, the non-linear traps (offset scales like Fahrenheit), compound units with many names (energy), and the competing unit systems (SI, CGS, imperial, natural). Goal: after this section, you can do any conversion on paper correctly without trusting your memory.
The principle (recap)
Every unit-defining equation can be turned into two fractions, both exactly equal to one:
$$
1\ \mathrm{km} = 1000\ \mathrm{m}
\quad\Rightarrow\quad
\underbrace{\frac{1000\,\mathrm{m}}{1\,\mathrm{km}}}{=\ 1}
\qquad\text{or}\qquad
\underbrace{\frac{1\,\mathrm{km}}{1000\,\mathrm{m}}}
$$
Because each equals 1, multiplying by it does not change the value — it only changes the form. The trick of factor-label is picking the direction that cancels the unwanted unit.
The recipe
For any conversion, follow these four steps:
- Write the starting value with its units. e.g.
72 km/h. - Multiply each unwanted unit by a fraction that has that unit in the denominator and the target unit in the numerator.
- Cancel units algebraically (top and bottom).
- Compute the number last. If the leftover unit is what you expected, the value is right.
Chained example:
$$
72\ \frac{\mathrm{km}}{\mathrm{h}}
\ \times\ \frac{1000\,\mathrm{m}}{1\,\mathrm{km}}
\ \times\ \frac{1\,\mathrm{h}}{3600\,\mathrm{s}}
\ =\ \frac{72 \times 1000}{3600}\ \frac{\mathrm{m}}{\mathrm{s}}
\ =\ 20\ \frac{\mathrm{m}}{\mathrm{s}}
$$
Notice km and h cancel top-and-bottom. If they didn’t, your multiplication was wrong. This is a built-in debugger: if the final unit isn’t what you expected, you either flipped a fraction or dropped a step.
Some non-trivial conversions
Pressure: atm → Pa
$$
1\ \mathrm{atm}\ \times\ \frac{101{,}325\,\mathrm{Pa}}{1\,\mathrm{atm}}\ =\ 101{,}325\ \mathrm{Pa}
$$
Energy (multiple names): kWh → J
$$
1\ \mathrm{kWh}
\ \times\ \frac{1000\,\mathrm{W}}{1\,\mathrm{kW}}
\ \times\ \frac{3600\,\mathrm{s}}{1\,\mathrm{h}}
\ \times\ \frac{1\,\mathrm{J/s}}{1\,\mathrm{W}}
\ =\ 3{,}600{,}000\ \mathrm{J}\ =\ 3.6\ \mathrm{MJ}
$$
Density: g/cm^3 → kg/m^3
$$
\rho = 1\ \frac{\mathrm{g}}{\mathrm{cm}^{3}}
\ \times\ \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}
\ \times\ \left(\frac{100\,\mathrm{cm}}{1\,\mathrm{m}}\right)^{3}
\ =\ \frac{100^{3}}{1000}\ \frac{\mathrm{kg}}{\mathrm{m}^{3}}
\ =\ 1000\ \frac{\mathrm{kg}}{\mathrm{m}^{3}}
$$
Note: the exponent 3 applies to the whole fraction — top and bottom. If you only raise the top, you get the wrong answer. This is the same rule as in §1.6: “the exponent applies to the whole ‘prefix + unit’.”
Non-linear scales: Fahrenheit, Celsius, Kelvin
Factor-label works for units that have a linear-through-zero relationship with the reference unit — that is, $y = k\,x$ (not $y = k\,x + b$).
m↔ft: linear through zero ($1\ \mathrm{ft} = 0.3048\ \mathrm{m}$) — factor-label just workskg↔lb: linear through zero — factor-label OKK↔°C: linear but with an offset: $T_C = T_K – 273.15$ — factor-label alone doesn’t work°C↔°F: has both scale and offset: $T_F = \tfrac{9}{5} T_C + 32$ — factor-label alone doesn’t work
Why? A temperature unit like °F can mean two very different things:
- Absolute temperature (
212°F= boiling point of water) - Temperature difference (
36°Fbetween body temperature and room temperature)
For absolute value conversion, use the full formula, not just fraction multiplication:
$$
100\ ^{\circ}\mathrm{C}\ \Rightarrow\ T_F = \tfrac{9}{5}(100) + 32 = 212\ ^{\circ}\mathrm{F}
$$
For temperature-difference conversion, the offset drops out (since it’s subtraction) and only the ratio remains:
$$
\Delta T = 10\ ^{\circ}\mathrm{C}\ \Rightarrow\ \Delta T_F = \tfrac{9}{5}(10) = 18\ ^{\circ}\mathrm{F}
$$
So for a temperature difference factor-label with ratio 9/5 works; for an absolute value it doesn’t. Keep this distinction in mind for any thermodynamic calculation.
Kelvin is safest: because it starts from absolute zero, both absolute values and differences relate to °C by a constant offset only ($1\ \mathrm{K} = 1\ ^{\circ}\mathrm{C}$ for differences). In physics, use Kelvin almost always — least room for error.
Energy — one quantity with many names
Energy probably has the most variety of unit names. The important ones:
| Unit | SI equivalent | Where you see it |
|---|---|---|
J (joule) |
1 J |
SI base; convert everything to this |
cal (calorie) |
4.184 J |
Chemistry, nutrition (nutrition Calorie = kcal) |
kcal (dietary “Cal”) |
4184 J |
Food labels — a chocolate bar 250 kcal ≈ 10^6 J |
eV (electron volt) |
1.602 \times 10^{-19}\ \mathrm{J} |
Atomic and particle physics |
Wh |
3600\ \mathrm{J} |
Battery capacity |
kWh |
3.6 \times 10^{6}\ \mathrm{J} |
Household electricity |
BTU |
1055\ \mathrm{J} |
HVAC, US heating engineering |
erg |
10^{-7}\ \mathrm{J} |
CGS, in older E&M literature |
Rules of thumb:
– Chemistry / atomic physics: eV, kcal
– Electronics / power: kWh, Wh
– Biology / nutrition: kcal (= Cal)
– Thermodynamics: J or cal
– In any project’s calculations, convert everything to J first, do the math, then convert to the target unit at the end. This is the industry-standard workflow.
Competing unit systems
Although SI is the global standard now, you’ll encounter these too:
CGS (centimeter–gram–second)
- Length:
cm, mass:g, time:s - Force:
dyne=10^{-5}\ \mathrm{N} - Energy:
erg=10^{-7}\ \mathrm{J} - In electromagnetism it has multiple variants (Gaussian, Heaviside–Lorentz) that write Maxwell’s equations differently — common in older theoretical literature. A frequent source of confusion.
Imperial / US customary
- Length:
ft,mi,in - Mass:
lb,slug - Absolute temperature:
°R(Rankine — Kelvin built on Fahrenheit’s scale) - Standard in US engineering. For academic physics, convert to SI in the first step and make your life easy.
Natural units — where fundamental constants become 1
These are powerful tools that eliminate fundamental constants from equations, but each pushes you into a different system:
- Planck units: $c = G = \hbar = k_B = 1$. Consequences: length =
1.6 × 10^{-35} m, time =5.4 × 10^{-44} s. Cosmology and quantum gravity. - Atomic units (Hartree): $m_e = e = \hbar = 1/(4\pi\varepsilon_0) = 1$. Length = Bohr radius =
5.3 × 10^{-11} m. Atomic and molecular physics. - Geometrized units: $c = G = 1$. In general relativity, to avoid dragging around factors of $c^2$.
In each, the numbers in the equations get simpler, but you have to know which system you’re in. For any final report, convert back to SI.
Practical strategy
1. Convert to SI early. Before you start any calculation, convert all inputs to m, kg, s, A, K. It takes a moment but eliminates the multiplicative-error class entirely.
2. Keep units on your variables. Even in symbolic algebra, write v_0 = 20 m/s, not just v_0 = 20. Check the final answer’s unit is what you expected.
3. For numeric calculations use factor-label; for symbolic reasoning use dimensional analysis. Factor-label answers “how many m/s in 72 km/h?”. Dimensional analysis (§1.7) answers “how does the pendulum period depend on…?”
4. Memorize a small conversion set — look up the rest. These are worth memorizing:
– $1\ \mathrm{mile} = 1.609\ \mathrm{km}$
– $1\ \mathrm{ft} = 0.3048\ \mathrm{m}$
– $1\ \mathrm{lb} = 0.4536\ \mathrm{kg}$
– $1\ \mathrm{atm} = 101{,}325\ \mathrm{Pa}$
– $1\ \mathrm{cal} = 4.184\ \mathrm{J}$
– $1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
– $T_K = T_C + 273.15$
– $g \approx 9.81\ \mathrm{m/s^2}$ (Earth-surface gravitational acceleration)
Common mistakes
1. Flipping a fraction. If the final unit isn’t what you expected, flip the last fraction you multiplied by.
2. Exponent on a compound unit. cm^3 → m^3 uses (1/100)^3 = 10^{-6}, not 1/100. So $1\ \mathrm{cm}^3 = 10^{-6}\ \mathrm{m}^3$, not $10^{-2}\ \mathrm{m}^3$.
3. Treating a temperature conversion as simple factor-label. 20°C = 68°F (via the formula), not 20 × 9/5 = 36°F. Fahrenheit has an offset.
4. Swapping °C and K in formulas that require absolute temperature. The ideal gas law ($pV = nRT$) needs Kelvin, not Celsius. If you plug in Celsius, the equation becomes meaningless because T = 0 would be absolute zero, not the freezing point of water.
5. Nutrition Calorie vs physics calorie. Cal (capitalized) in nutrition = 1 kcal = 1000 cal. If you compute human energy expenditure and take the label’s Cal as cal, you’ll be off by 1000×.
What you should be able to do
After this section you should be able to:
- Convert any chained expression (length, mass, time, speed, density, energy, pressure) to SI using factor-label
- Recognize the difference between converting an absolute temperature and a temperature difference and apply the correct approach
- Convert
°C ↔ K ↔ °Faccurately in both directions - Move between the ~8 common energy units and know which ones you expect to see in which field
- Know when to use SI vs when to reach for CGS/imperial/natural units, and why converting to SI at the start of a calculation is the safest workflow
📚 See also: Halliday Vol 1, Ch 1 — sections on Changing Units and end-of-chapter unit-conversion problems.
📖 Open reference: NIST — Guide for the Use of the International System of Units (Chapter 6, conversion between non-SI and SI units with official tables).
📖 BIPM — SI Brochure, conversion appendices.
📖 CODATA — Recommended Values of the Fundamental Physical Constants (precise constant values, including eV, calorie, and more).
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