# How to find the equation with dimensional analysis in this case?

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?

• 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 $mg$ ? Oct 7, 2022 at 20:30
• @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. Oct 7, 2022 at 20:51
• Your problem has one Pi group, which can be taken as $S^3(\rho/m)^2$.
– J.G.
Oct 7, 2022 at 22:10

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: $$[v]=L_{||}T^{-1}$$ $$[g]=L_{||}T^{-2}$$ $$[S]=L_{\perp}^2$$ $$[\rho]=ML_{||}^{-1}L_{\perp}^{-2}$$ $$[m]=M$$