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In this post: https://math.stackexchange.com/a/438618/66675 , Ross Millikan explains how one can use dimensional analysis to check for errors even if the equation itself has no units.

Can someone provide an example of an error that dimensional anaylsis would be able to catch?

He explains how to do so here:

I find unit checks very useful. Even if there are no obvious units, you may be able to find some. Solving a quadratic ax2+bx+c=0 you might say there are no units, but it has to be true even if x is a length. Then a is length−2 and so on. Make sure all your terms match appropriately. This will catch some errors.

But I can't think of an example on how this will be useful if we just make up units. I make a ton of errors and I would like to improve my error-checking skills.

Community
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user1251385
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3 Answers3

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A funny and useful little book from Barry Cipra may help :
"Misteaks... and how to find them before the teacher does..."

(google preview)

Raymond Manzoni
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Let's say you're trying to compute the constant acceleration of an object, and you know its initial velocity $v_0$, it's initial position, $x_0$, the total distance it went $d$, and the time over which you've observed its travel, $t$.

You might know that $d = \frac12 at^2+v_0t+x_0$, and you need to solve for $a$.

So then you gather terms together:

$$d-x_0 -v_0t = \frac12 at^2,$$ $$2\left(d-x_0-v_0t\right) = at^2$$ $$\frac{2\left(d-x_0-v_0t\right)}{t^2} = a.$$

Let's say $d = 10 \textrm{m}, x_0 = 0 \textrm{m}, v_0 = 4 \textrm{m/s}, and t=20 \textrm{s}$.

So you do all this, and you simplify to get

$$a = \frac{2d}{t^2} - \frac{2x_0}{t^2} -2v_0 \implies a = \frac{2\cdot 10}{400}-0-8.$$

So you get $a = -7.95 \textrm{m/s/s}$.

This, however, is the wrong answer. Can you spot the mistake?

The mistake was that when I simplified $\frac{2v_0t}{t^2}$, I ignored the square in the denominator.

If we kept units in the whole time, we'd know that there was an error right away.

$$a\ \frac{\textrm{m}}{\textrm{s}^2} = \frac{2d}{t^2} \frac{\textrm{m}}{\textrm{s}^2} - \frac{2x_0}{t^2} \frac{\textrm{m}}{\textrm{s}^2} - \frac{2v_0t}{t^2} \frac{\textrm{m/s}\cdot \textrm{s}}{\textrm{s}^2}.$$

Notice in this term, all the units are meters per second squared. This helps us avoid making mistakes like canceling a $t$ with a $t^2$. Otherwise, the last term would have units of $\frac{\textrm{m}}{\textrm{s}}$, which would not add to units of $\frac{\textrm{m}}{\textrm{s}^2}$.

Emily
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Another way to do this is to leave out a lot of the constants - BUT, you must leave in the exponents. The following is very close to what Arkamis has, but for more complicated problems I find it helps to not use, in that case, the constant mutiliplier 2 and the 1/2.

Since your answer should be meters (m), just do the right side as follows:

$\frac{1}{2}a{t^2} + {v_0}t + {x_0}$

$\frac{m}{{{t^2}}}{t^2} + \frac{m}{t}t + m$

$m + m + m$

Which is meters and is the same unit as the left side. You could solve for a, and basically do the same thing. This just shows you what it looks like without some of the extra constants. Remember though, you have to keep the exponents!

Mitchell Kaplan
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