I'm doing some exercises in dimensional analysis and don't know how to proceed.

The problem is: use the dimensional analysis to find the terminal velocity $v$ of a free-faller. This speed depends on his mass $m$, his cross section $S$, air density $\rho$ and acceleration $g$.

If I try to set $ v = m^{\alpha} S^{\beta} \rho^{\gamma} g^{\delta} $, what I get by comparing the units is a system of three equations with four unknowns. Namely:

$1 = 2 \beta - 3 \gamma + \delta$

$-1 = -2 \delta$

$ 0 =\alpha + \gamma $

I can only solve for $\delta$. How to proceed? What can I do with the other unknowns?

  • 1
    $\begingroup$ This is not unusual. You must make additional assumptions. For example, it is likely, is it not, that the expression for $v$ contains the combination $mg$ ? $\endgroup$ Oct 7, 2022 at 20:30
  • $\begingroup$ @Philip Wood : this is as good as an answer, which you might expand it to, if the OP doesn't take the hint to write his own answer. $\endgroup$ Oct 7, 2022 at 20:51
  • $\begingroup$ Your problem has one Pi group, which can be taken as $S^3(\rho/m)^2$. $\endgroup$
    – J.G.
    Oct 7, 2022 at 22:10

1 Answer 1


I can suggest two different ways to find the answer.

First, as was mentioned in the comments, is to use the further assumption that according to the equations of motion, $g$ and $m$ always appear in the combination $mg$.

Second, which is somehow more general and is used in fluid mechanics. We introduce two different dimensions for length: One in the direction of motion which we cal $L_{||}$ and the other one in the perpendicular direction which we call $L_{\perp}$. So, the lengths in cross section are different from those that appear in $g$ and $v$. While for $\rho$ as it is proportional to inverse volume, and for volume: $$[V]=L_{||}L_{\perp}^2$$ Then, intuitively, we may write: $$[v]=L_{||}T^{-1}$$ $$[g]=L_{||}T^{-2}$$ $$[S]=L_{\perp}^2$$ $$[\rho]=ML_{||}^{-1}L_{\perp}^{-2}$$ $$[m]=M$$

And we will have an extra equation to fix the exponents.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.