# How to calculate the drag coefficient using terminal velocity?

I was wondering if it were possible to calculate the drag coefficient by allowing an object to reach terminal velocity. Can you rearrange the terminal velocity formula to give the drag coefficient?

Yes you could. Since the force on an object from drag is given by

$$F_D = \frac{1}{2}\rho v^2 A C_D$$

where $C_D$ is the drag coefficient, then all you would need to know are your velocity ($v$), your fluid density $\rho$, your cross sectional area ($A$) and the force of gravity on the body, which would then be equivalent to $F_D$ since the body would have no net force on it. Thus, you could isolate for $C_D$, getting

$$C_D = 2mg/\rho v^2 A$$

• Don't you mean $v^2$? – Bernhard Jul 30 '14 at 7:46
• Same question here, Berhard. This should not be the accepted answer. – David Hammen Jul 30 '14 at 9:25
• I have corrected the formula. Also, one could replace $F_g$ with $mg$, but I leave it up to the author. – gigacyan Jul 30 '14 at 12:05
• Thanks for the edit, definitely meant $v^2$ and I'll change it to mg – Spaderdabomb Aug 1 '14 at 5:07

Following on Spaderdabomb answer, the drag force $F_D$ acts on the body and, as such, it balances with the other forces and the inertia of the body ($\sum\vec{F} = m\tfrac{d\vec{u}}{dt}$). When you consider a situation where the body is at terminal velocity, this means its velocity is constant (at least in that direction) and its acceleration is null.

Consider the case of a parachute free falling at terminal velocity $\vec{u} = \{0,\,0,\,-w\}$, thus at constant $w$. As such $\tfrac{d\vec{u}}{dt} = 0$ and the force balance becomes $\sum\vec{F} = 0 \Leftrightarrow F_D^z - m g = 0$, noting that it is the drag that is sustaining the parachuter. From the perspective of the parachuter, the flow speed is $\vec{u}_\infty = \{0,\,0,\,w_\infty\}$ where $w_\infty = w$. As the drag is given by $F_D = C_D \tfrac{\rho}{2} w_\infty^2 A$, substituting in the force balance returns Spaderdabomb answer, $C_D = \tfrac{2 m g}{\rho w_\infty^2 A}$.