Theory — Dimensional analysis
Dimensional analysis is the method of tracking the physical type of a quantity rather than only its numerical value.
Every measured quantity can be reduced to powers of a small set of base dimensions. In SI, the most common base dimensions are
length, mass, time, electric current, temperature, amount of substance, and luminous intensity. A velocity, for example,
is not just a number. It has the dimension of length divided by time, while a force has the dimension of mass times length
divided by time squared. This is why dimensional analysis is useful both for unit conversion and for checking whether an
equation can possibly be correct.
Base-dimension vector.
\[
\begin{aligned}
[L,M,T,I,K,N,J]
\end{aligned}
\]
The rule set is simple. When quantities are multiplied, the exponents of their base dimensions add. When quantities are
divided, those exponents subtract. When a quantity is raised to a power, each exponent is multiplied by that power.
Addition and subtraction are much stricter: only quantities with exactly the same dimensions can be added or subtracted.
That last rule is what makes dimensional analysis such a strong error detector. If one term is a length and another term is a time,
then the expression is already invalid before any number is substituted.
Core algebra of dimensions.
\[
\begin{aligned}
[ab] &= [a][b] \\
\left[\frac{a}{b}\right] &= \frac{[a]}{[b]} \\
[a^n] &= [a]^n
\end{aligned}
\]
Unit conversion
A unit conversion works only when the source and target have the same base-dimension vector. Once that is true, the problem
becomes a scaling problem: compute how many SI units are represented by the source expression, compute how many SI units are
represented by the target expression, and then divide one factor by the other. That ratio is the conversion factor.
This is why dimensional analysis is safer than memorizing isolated conversion formulas. Instead of remembering a special rule
for each case, you reduce both unit expressions to their structure and let the dimensions decide whether the conversion is legal.
General conversion rule.
\[
\begin{aligned}
x_{\text{out}} &= x_{\text{in}} \cdot
\frac{\text{source SI factor}}{\text{target SI factor}}
\end{aligned}
\]
For example, one mile is \(1609.344\ \mathrm{m}\) and one hour is \(3600\ \mathrm{s}\). So a speed written in miles per hour
can be converted to meters per second by replacing the mile with meters and the hour with seconds.
Example: \(60\ \mathrm{mph}\) to \(\mathrm{m/s}\).
\[
\begin{aligned}
60\ \frac{\mathrm{mi}}{\mathrm{h}}
&= 60\cdot \frac{1609.344\ \mathrm{m}}{3600\ \mathrm{s}} \\
&= 26.8224\ \mathrm{m/s} \\
&\approx 26.82\ \mathrm{m/s}
\end{aligned}
\]
This is the classic cancellation picture in dimensional analysis: the mile cancels against the mile in the conversion factor,
and the hour cancels against the hour in the time conversion. Only the desired target unit remains.
Equation checking
Dimensional analysis can also test whether an equation is dimensionally homogeneous. That does not prove the equation is
physically correct, but it is a necessary condition for correctness. Suppose you want to check
\(v = d/t\). Assign dimensions to the variables: velocity is \(L\,T^{-1}\), distance is \(L\), and time is \(T\).
Then the right-hand side reduces to \(L/T\), which matches the left-hand side. The equation passes the dimensional test.
Example: \(v = d/t\).
\[
\begin{aligned}
[v] &= L\,T^{-1} \\
\left[\frac{d}{t}\right] &= \frac{L}{T} = L\,T^{-1}
\end{aligned}
\]
The same method checks more complicated formulas such as \(F = ma\). Mass has dimension \(M\), acceleration has dimension
\(L\,T^{-2}\), so the right-hand side becomes \(M\,L\,T^{-2}\), which is exactly the dimension of force.
Example: \(F = ma\).
\[
\begin{aligned}
[m a] &= M \cdot L\,T^{-2} \\
&= M\,L\,T^{-2} \\
&= [F]
\end{aligned}
\]
Derived units
Many named units are simply shorthand for combinations of base dimensions. A joule is a newton-meter, and a newton is a
kilogram-meter per second squared. Therefore both \(J\) and \(N\cdot m\) reduce to the same base-dimension vector
\(M\,L^{2}\,T^{-2}\). This is why dimensional analysis can recognize that two expressions look different but still represent
the same physical quantity.
Example: \(N\cdot m\) and \(J\).
\[
\begin{aligned}
[N\cdot m]
&= (M\,L\,T^{-2})\cdot L \\
&= M\,L^2\,T^{-2} \\
&= [J]
\end{aligned}
\]
Custom units and practical use
Custom units are useful when a problem uses a specialized scale that is not already in the built-in list. If you define
something like furlong = 201.168 m or fortnight = 14 day, the unit can be inserted into the same
algebra as the built-in units. Because the calculator converts every custom definition back to SI base dimensions, the same
consistency checks continue to work. That means a custom unit is not just a text label. It becomes a new conversion factor
with a proper physical dimension.
In practice, dimensional analysis is a safety tool. It helps catch missing time factors, inverted conversion ratios,
wrong derived units, and algebraic mistakes inside formulas. It does not replace full derivations, but it is one of the
fastest ways to avoid unit mistakes before they propagate through a longer calculation.
