# How does dimensional analysis work?

Dec 26, 2014

Dimensional analysis is simply a way of testing whether the base units of a given equation work out. It operates on a simple principle: the units you have on one side of an equation must match those that you have on the other.

It's analogous to baking a cake. Your cake can only have what ingredients you put in to it.

Consider the following equation:

$F = m a$

Force in base units is $k g \cdot \frac{m}{s} ^ 2$
Mass is just $k g$
Acceleration is $\frac{m}{s} ^ 2$

So now let's evaluate this by plugging in these units into the equation:

$k g \cdot \frac{m}{s} ^ 2 = k g \cdot \frac{m}{s} ^ 2$

This equation works, therefore it can be classified as dimensionally correct.

Let's look at another equation:

$\Delta x = {v}_{o} t + \frac{1}{2} {a}^{2} {t}^{2}$

$\Delta x$ is a distance, measured in meters ($m$)
${v}_{o}$ is a velocity, measured in meters per second ($\frac{m}{s}$)
$a$ is acceleration, measured in meters per second squared ($\frac{m}{s} ^ 2$)
$t$ is time, measured in seconds ($s$)

Now let's just plug everything in:

$m = \frac{m}{s} \cdot s + {\left(\frac{m}{s} ^ 2\right)}^{2} \cdot {s}^{2}$

Notice that I have not included the $\frac{1}{2}$. This is because coefficients do not matter in dimensional analysis because they don't really change the dimension (i.e. half a mass is still a mass).

Now we simplify:

$m = m + \left({m}^{2} / {s}^{4}\right) {s}^{2}$

$m = m + \left({m}^{2} / {s}^{2}\right)$

Since this equation does not work out, this equation is not dimensionally correct.

Hope that helped :)