–James Clerk Maxwell, “On a particular case of the descent of a heavy body in a resisting medium,” 1853

–James Clerk Maxwell, in a letter to William Thomson (Lord Kelvin), 1857

Dropping paper is a favorite pastime of theoretical physicists. (For another favorite pastime of theoretical physicists, see the pitch of squeaking chalk.) Here, we consider dropping playing cards of length \(l\), width \(w\) (roughly equal to \(l\)), thickness \(d\), and density \(\rho\). We are interested in dropping them in air with density \(\rho_a\); before worrying about fluid dynamics and vortices, let’s take a brief tangent to enjoyably estimate the density of air without looking up any constants.

Since air is mostly nitrogen (which is the seventh element, so \(N_2\) has a mass of \(2\times2\times7\) amu), we know that the ideal gas law gives an estimate for the density of air: $$\begin{aligned} \rho_a \sim \frac{pm}{k_BT} \sim \frac{(1~\mathrm{atm})(30~\mathrm{amu})}{k_BT}. \end{aligned}$$ But what is the value of \(k_BT\) at room temperature? (During the first year of my PhD, my advisor challenged me to an order-of-magnitude physics computation that required the value of \(k_BT\). When I drew a blank, he made me estimate the quantity for half an hour in as many ways as I could think of.)

To figure out \(k_BT\), let’s recall some basic facts from high school chemistry:

- It takes 4 joules/g to heat up water by one degree Celsius. (This is the definition of calorie, so it’s easy to remember.)
- Counting in a cavalier fashion, there are three atoms that can translate and rotate in three dimensions, giving a total of 18 degrees of freedom. Each degree of freedom contributes \(\frac{1}{2}k_BT\) of energy per molecule.
- Water has a molecular mass of 18 amu.
- Avogadro’s number is \(N_A = 6 \times 10^{23}\) (which is somehow much easier to remember than Boltzmann’s constant).

Perhaps the most naive path towards explaining the motion of a falling card is to treat it is a buoyant pendulum. In the “fluttering” mode of falling, a card rocks side to side without turning over. In a normal pendulum of length \(l\), dimensional analysis (observing *à la Galileo* independence from mass) gives a period of \(\sqrt{l/g}\). Here, we must introduce the buoyancy-corrected mass to scale \(g\), where
$$\begin{aligned}
M^{\prime} &= wld(\rho - \rho_a).
\end{aligned}$$
The effective scaling by \(M^{\prime}/M\) gives the buoyant pendulum a time period of
$$\begin{aligned}
\tau_p = \sqrt\frac{Ml}{M^{\prime}g} = \sqrt\frac{l\rho}{g(\rho-\rho_a)}.
\end{aligned}$$
The fluttering motion of the buoyant pendulum represents the motion of a falling card in a fluid *without* vortices. Since \(\tau_p\) is the time a card takes to undergo such oscillatory motion undisturbed by vortices, it provides (roughly) a timescale at which vortices dissipate.

Unlike the simple fluid dynamics behind buoyancy, the falling of the card due to gravity can cause vortex shedding. As air passes rapidly across the surface of the card, the flow lines peel off the surface of the card, producing a wake. Air in that wake experiences a shear force, introducing a vortex. The resulting von Kármán vortex street that emerges from this repeated pattern can cause the card to oscillate in a tumbling motion, where the characteristic timescale of the vortex production is given by the terminal velocity. Performing the explicit computation, the terminal velocity \(v\) is given by balancing the drag (which scales like \(\rho_a wl v^2\) given a high Reynolds number) and gravity adjusted for buoyancy (\(M^{\prime}g\)):
$$\begin{aligned}
v &= \sqrt\frac{M^{\prime}g}{\rho_a wl} = \sqrt\frac{dg(\rho - \rho_a)}{\rho_a}.
\end{aligned}$$
Since the length of the card is the only relevant length scale (the thickness is irrelevant, and the width is roughly the length), the characteristic time for vortex production is
$$\begin{aligned}
\tau_v &= \frac{l}{v} = l\sqrt\frac{\rho_a}{dg(\rho - \rho_a)}.
\end{aligned}$$
If the dissipation time is much larger than the vortex production time, the card will flutter instead of tumble. However, if the vortices are being produced at a similar rate or faster compared to the dissipation time, tumbling will occur. This ratio is known as the Froude number; also, as a sidenote, our analysis only holds in the high-Reynolds regime (i.e., not swimming in corn syrup), so we compute both quantities:
$$\begin{aligned}
\mathrm{Fr} &= \frac{\tau_p}{\tau_v} = \sqrt\frac{d\rho}{l\rho_a}\\
\mathrm{Re} &= \frac{\rho_a v l}{\eta}
\end{aligned}$$
for air viscosity \(\eta\).
For the card, we take \(l = 3.5\) inches, \(d=0.2\) mm, \(\rho = 1~\mathrm{g~cm^{-3}}\). Air has \(\rho_a = 1~\mathrm{g~L^{-1}}\) and \(\eta = 2\times 10^{-5}~\mathrm{kg.m^{-1}~s^{-1}}\). Plugging in numbers, we find that \(\mathrm{Fr} \sim 1\) and \(\mathrm{Re} \sim 10^4\). The Reynolds number is large, as we assumed in our drag force calculation and in our tumbling analysis. Additionally, the Froude number shows that vortices are produced approximately as fast as they dissipate; since this gives the vortex a chance to act on the card, we expect it to tumble [1]. Of course, we already knew that dropping a playing card causes it to tumble, but now we know *why* it tumbles from the principles of vortex shedding!

The numbers above also give a terminal velocity of \(v = 1~\mathrm{m~s^{-1}}\). To “shuffle” the deck, we might ask that the cards perform \(O(10)\) tumbles. How quickly does the tumbling occur? The path of the card simply follows the vortices that are shed, and thus we can estimate the rate of vortex shedding. Naively, we might expect vortex shedding to occur at a frequency \(1/\tau_v\), inversely with the time it takes to produce a new vortex. This is true, but unfortunately only up to a constant — one of the casualties from our cavalier discarding of \(\pi\) and other factors. Regardless of geometry, most objects at \(\mathrm{Re}\sim10^4\) have a Strouhal number of 0.2, where \(\mathrm{St} = f\tau_v\) for vortex shedding frequency \(f\). Consequently, we expect the card to tumble at a rate of \(\sim 5\) Hz. Since reaching terminal velocity is fast compared to tumbling for two seconds, we can roughly estimate the initial height at 2 meters.

[1] More carefully, the tumbling behavior holds for Fr above around 0.7, while the card would instead rock like a pendulum without vortex shedding if \(\mathrm{Fr} < 0.7\) [2]. At \(\mathrm{Fr} \approx 1\), we’re definitely in the tumbling regime. [2] Belmonte, Andrew, Hagai Eisenberg, and Elisha Moses. “From flutter to tumble: inertial drag and Froude similarity in falling paper.”

[3] Pesavento, Umberto.