1 Introduction

A beer mat (also known as drink coaster) is a commodity most elegantly used to rest a glass on in order to protect a table surface [1]. However, not only for a physicist there are more exciting applications for a beer mat, one of which is to let it fly. Usually, it is a circular piece of cardboard with a diameter of about \({10}\,\hbox {cm}\), though there are also other shapes in use. For the following we will idealise it as a disc with radius r, mass m and negligible thickness, for a sketch see Fig. 1.

If one tries to throw a beer mat, one quickly realises that one can only achieve reasonable flight distances if the mat rotates around the axis perpendicular to it’s circular area, depicted as \(\varvec{D}\) in Fig. 1. The such generated angular momentum stabilises the orientation of the disk via angular momentum conservation preventing chaotic rotations around one of the two other rotation axes of the disc, for which it is known from classical mechanics that rotations around these are in practice unstable, given the relatively small mass of the beer mat.

Now one could expect the mat to fly similarly to a frisbee, i.e. with angular momentum pointing up or down, called ‘sidespin’ from now on, however, it still turns out to be difficult to predict the path of the flying mat: a seemingly random time \(\tau \) after the flight started the mat begins to either turn left or right, depending on it’s direction of rotation, and—if it does not hit ground beforehand—ends up flying with backspin (i.e. with rotation axis pointing sidewards perpendicular to the direction of flight with the upper side rotating against the direction of flight). With a few more experiments one realises that starting the flight with backspin is stable, while a flight with topspin is not.

Fig. 1
figure 1

Sketch of the disk and the most relevant coordinates

The observation of the aforementioned seemingly random times \(\tau \) might lead one to hypothesise there to be a chaotic effect in the flight of a beer mat. However, it will turn out that this effect comes from the inability of humans to throw the disk reproducibly.

In principle, the theory of a rotating thin disk moving in air is known. Unfortunately, solving the corresponding equations including effects from turbulence is analytically intractable and even numerically highly demanding. An effective treatment is, therefore, in order by treating the disk as a point-like object to a large extend, which allows one to avoid to solve the full fluid dynamics equations. The closest to the situation we are considering here comes the literature about the flight of a frisbee, see for instance [2,3,4,5,6].Footnote 1 However, a frisbee weighs significantly more and has modified edges to stabilise it’s flight characteristics such that the effects that we will study here are sufficiently suppressed to be neglected.

The aim of this paper is threefold: we will give a qualitative explanation of the observed behaviour in the following Sect. 2. From this section the understanding of the phenomenon should be possible without much expert knowledge. Next, we will derive an effective formalism based on a few assumptions describing the flight of a beer mat. This formalism allows us to make predictions, which can be tested experimentally. Thus, in the third part of this paper we present experimental results. For the experiments to be reproducible we have designed and constructed an apparatus which allows us to throw beer mats with variable angular and forward momentum. A high-speed camera is used to record the mats’ flights and the recordings are used to reconstruct the corresponding flight trajectories. These trajectories are then compared to the predicted trajectories from our effective theory.

Eventually, we compare the experimental results to our theory predictions, after a number of parameters are fitted to the experimental data. The predictions are well confirmed, validating the assumptions the effective theory was built on. This allows us to generalise our findings to other types of disks, which could be tested experimentally.

2 Qualitative discussion

Let us start in a situation where a spinning diskFootnote 2 with its angular momentum \(\varvec{L}\) perpendicular to the disk’s surface is moving through a medium (air) without gravitational force. Drag will cause the disk to orientate itself such that it minimises air resistance by minimising its surface exposed to the airflow. This means the axis \(\varvec{D}\) is perpendicular to the flight direction. Thus, any orientation with slightly tilted rotation axis compared to the situation above will automatically adjust back. In the equilibrium state no torque is available to change the direction of angular momentum.

Fig. 2
figure 2

a Airflow around the disk. b Forces acting on the disk. c Angular momentum and orientation of rotation. d Torque forcing disk into backspin

However, if one now includes the gravitational force pointing into the negative z-direction, the disk will be accelerated downwards. Hence, the airflow approaches it slightly from below (see Fig. 2a). This induces a lifting force \(\varvec{F}^\mathrm {lift}\) upwards [7]. This force now does not attack the disk at its centre, but \(R = \pi r/8\) towards the leading edge [8]. Intuitively, this can be understood by the air stream being broken near the front edge of the disk, so that the lift is acting stronger at the front. Therefore, \(\varvec{F}^\mathrm {lift}\) induces a torque \(\varvec{M} = \varvec{R} \times \varvec{F}^\mathrm {lift}\) which will cause a precession of the disk (Figure 2, C).Footnote 3 The lifting force and, thus, the torque vanishes only in back- or topspin position. But, since the direction of this precession is such that the disk approaches backspin orientation independently of the direction of the angular momentum, the topspin orientation is meta stable and the backspin orientation stable.

Therefore, if the flight of the disk is started not in backspin position, one expects it to perform an oscillation around the backspin position. This oscillation, however, is strongly damped in any real-world experiment.

A secondary but nonetheless important stabilising effect is the Magnus force [10]. Together with drag it is responsible for the aforementioned damping. The Magnus force acts orthogonal to both flight direction and rotation axis \(\varvec{D}\). It does not introduce any torque and, thus, cannot be primarily responsible for the trend towards backspin. However, it can change the flight direction in such a way that backspin is preferred. Imagine the spinning disk simply being dropped without initial spacial velocity and \(\varvec{D}\) pointing in the y-direction, i.e. the leading edge pointing precisely downwards. Once the disk begins to fall, it is accelerated in the x-direction by the Magnus effect.Footnote 4 The sign of this acceleration in x-direction prefers backspin again. Small deviations from backspin can therefore be corrected by the Magnus effect adjusting the flight direction.

3 Effective theory

In this section, we will translate the qualitative arguments developed in the previous section into formulae effectively describing the flight of a beer mat (or any other thin disk). Thereafter, we will test these formulae experimentally in the following section.

Note that this set of equations is not formulated with the goal to explain the reality. Rather it is a means to predict movements with reasonable precision. Terms without an immediate physical intuition have to be read in this light.

3.1 Equations of motions

For a thin disk with radius r and mass m the element of the moment of inertia tensor for rotations around the symmetry axis \(\varvec{D}\) (see Fig. 1) reads \(I_{D} = mr^2/2\). For the other two axes lying in the plane of the disk one finds \(I = I_{D}/2\). We denote the modulus of the velocity of the centre of the disk with \(v = |\varvec{v}|\) and its area with A.

For the angles given in Fig. 1, we will denote \(\omega _0 = {{\dot{\psi }}}\) and assume \(\omega _0\) to be a constant of motion. Our next assumption is that the disk’s area always moves in an orientation with minimised air resistance as explained above, i.e. \(\varvec{v} \perp \varvec{D}\) and starts out in the x-z-plane without loss of generality. The angle \(\theta \in [0,\pi ]\) does not differentiate between top- and backspin, but only together with \(\phi \). In order to describe everything with only a single angle, we introduce \(\vartheta \in [-\pi ,+\pi ]\) such that \(\vartheta = \pm \theta \) with \(\vartheta =+\pi /2\) corresponding to topspin and \(\vartheta \ =-\pi /2\) to backspin. Now the only non-trivial motion is the one described by \(\vartheta \).

As argued in the last section, the acceleration \({\ddot{\vartheta }}\) vanishes for top- and backspin, with backspin the stable state. We expect \({\ddot{\vartheta }}\) to be largest at \(\vartheta = 0\) and symmetric around this point. An easy choice to obtain such a behaviour is a force proportional to \(\cos \vartheta \). Now, there are two contributions to \({\ddot{\vartheta }}\): the first stems from the lifting force \(\varvec{F}^\mathrm {lift}\), which is tilting the disk. We expect the lifting torque to be \(c^\mathrm {pot}_\vartheta \rho r Av^2\) with the density of air \(\rho \) and some dimensionless constant \(c^\mathrm {pot}_\vartheta \). This is the well-known formula for lift combined with a factor r to account for the lever. The second contribution, which reads \(c_\vartheta ^\mathrm {rot} \rho r A^2 \omega _0^2\) with another constant \(c^\mathrm {rot}_\vartheta \), cannot be motivated in such a naive way. It might stem from sub-leading effects, such as the Magnus force, but we introduced it as an a posteriori adjustment to the empirical evidence.

The larger \(\omega _0\), the harder it will be to tilt the disk. The corresponding force must depend on the sign of \({{\dot{\vartheta }}}\), but not on the sign of \(\omega _0\). Thus, we expect \(c_\vartheta ^\mathrm {damp} I \omega _0^2{{\dot{\vartheta }}}\), which is in the form of laminar damping. We assume \({{\dot{\vartheta }}}\) to be relatively small such that damping proportional to \({{\dot{\vartheta }}}^2\) can be neglected.

In summary, for \(\vartheta \) we arrive at the following equation of motion with \(K=\rho r A/I\)

$$\begin{aligned} {\ddot{\vartheta }}&= -K\left( c^\text {pot}_\vartheta v^2+c^\text {rot}_\vartheta A \omega _0^2\right) \cos \vartheta - 2c^\text {damp}_\vartheta \omega _0^2 {{\dot{\vartheta }}}\,. \end{aligned}$$

Next we focus on the centre of mass coordinate \(\varvec{x} = (x,y,z)\). The aforementioned condition of minimised air resistance prevents the disk to move in the direction of \(\varvec{D}\). This translates to an effectively reduced gravitational force \(-mg\left( \hat{ \varvec{e}}_z-\varepsilon \varvec{D} \cos \vartheta \right) \). The coefficient \(\cos \vartheta \) takes the dependence of the air resistance in \(\varvec{D}\)-direction on the projected area into account. The case \(\varepsilon =0\) would correspond to a free fall, i.e. the leading edge would be aligned only with the horizontal velocity component and completely ignore the vertical one. Strict alignment would on the other hand imply \(\varepsilon =1\). The truth, as so often, lies somewhere in between.

As before we introduce a damping force with the proportionality constant \(c^\text {damp}_x\) of dimension velocity. Our last assumptions are \(v_y=0\) at \(t=0\) without loss of generality and \(|v_y|\ll |v_x|\) at all times. The latter assumption is not always justified, but usually people throw disks with enough forward momentum that any change of direction is a second-order effect. These assumptions allow to write \(\varvec{D}=\left( 0,\sin \vartheta ,\cos \vartheta \right) ^\top \) significantly simplifying the equations of motion which then read

$$\begin{aligned} m {\ddot{\varvec{x}}}&= -mg\left( \!\begin{array}{c}0\\ -\varepsilon \sin \vartheta \cos \vartheta \\ 1-\varepsilon \cos ^2\vartheta \end{array}\!\right) -c^\text {damp}_x \rho A {\dot{\varvec{x}}}\,. \end{aligned}$$

3.2 Approximate analytic solution

Equations (1) and (2) still form a set of four nonlinear coupled differential equations, but they can be solved numerically much easier and more stably than the full equations of motion described in the appendix. Furthermore they can be approximately solved analytically.

To this end we first assume that \(v\approx v_0=\text {const.}\) over the complete duration of the flight. Then Equation (1) decouples and can be solved independently of the trajectory. We are going to do so in the harmonic approximation about the potential minimum at \(\vartheta =-\pi /2\). It suffices to gain a good understanding of the different stages of the flight. The differential Equation (1) simplifies to

$$\begin{aligned} {\ddot{\vartheta }} + 2c^\text {damp}_\vartheta \omega _0^2 {{\dot{\vartheta }}} + K\left( c^\text {pot}_\vartheta v_0^2+c^\text {rot}_\vartheta \omega _0^2\right) \left( \vartheta +\frac{\pi }{2}\right)&= 0\,. \end{aligned}$$

We observe that the system is clearly overdamped in the case of a stable flight (no oscillations are visible). The corresponding stability condition for such a flight reads

$$\begin{aligned} \left( c^\text {damp}_\vartheta \omega _0^2\right) ^2&\overset{!}{>} K\left( c^\text {pot}_\vartheta v_0^2+c^\text {rot}_\vartheta A\omega _0^2\right) \,. \end{aligned}$$

In this case we obtain the well-known solution

$$\begin{aligned} \vartheta (t) \approx -\frac{\pi }{2} + \left( \vartheta (0)+\frac{\pi }{2}\right) \mathrm {e}^{-\lambda t}\,, \end{aligned}$$

with \(\lambda \) given by

$$\begin{aligned} \lambda = c^\text {damp}_\vartheta \omega _0^2 - \sqrt{\left( c^\text {damp}_\vartheta \omega _0^2\right) ^2 - K\left( c^\text {pot}_\vartheta v_0^2+c^\text {rot}_\vartheta A\omega _0^2\right) }\,. \end{aligned}$$

In the overdamped case \(\lambda \) can be simplified to

$$\begin{aligned} \lambda&\approx \frac{\rho r A\left( c^\text {pot}_\vartheta v_0^2+c^\text {rot}_\vartheta A\omega _0^2\right) }{2c^\text {damp}_\vartheta I\omega _0^2} \end{aligned}$$
$$\begin{aligned}&{=}{:}\frac{\rho r}{m}\left( \lambda _1 \frac{v_0^2}{\omega _0^2} + \lambda _0 A\right) \end{aligned}$$

by using \(A = \pi r^2\) and \(I=mr^2/4\). Thus, the convergence depends solely on two universal constants \(\lambda _0,\lambda _1\) and the four variables mass, radius, speed and angular speed. It is very remarkable that only the ratio \(v_0/\omega _0\) is relevant.

With these approximations the remaining differential equations can be solved analytically using hypergeometric functions. However, for the sake of gaining intuitive understanding, we approximate even further. With sidespin initial conditions

$$\begin{aligned} \vartheta (0) = {{\dot{\vartheta }}}(0)=0\,,\quad \varvec{x}(0)=\left( \!\begin{array}{c}0\\ 0\\ h\end{array}\!\right) \,,\quad {\dot{\varvec{x}}}(0)=\left( \!\begin{array}{c}v_0\\ 0\\ 0\end{array}\!\right) \end{aligned}$$

we focus on the y-component. Exploiting our condition of slow movement in this direction allows us to neglect the damping term. This leaves us with

$$\begin{aligned} {\ddot{y}}&= g\,\varepsilon \,\sin \vartheta \cos \vartheta \,. \end{aligned}$$

Using the solution for \(\vartheta \) from Equation (5), we find that there is no acceleration in the y-direction at zero and again at infinite time. In between the acceleration becomes maximal at \(\vartheta =-\pi /4\) with \(|{\ddot{y}}|=g\varepsilon /2\) at the characteristic time scale \(\tau {:}{=}\log 2/\lambda \). In fact, the acceleration is effectively nonzero only for a short time \(t_i\ll \tau \) around \(\tau \) defining two asymptotic regimes: before \(\tau \) we have \({\dot{y}}=0\) and after \({\dot{y}}=\mathrm {const}\ne 0\). The transition between the two regimes takes place exponentially

$$\begin{aligned} y(t)&\approx -c_2g\varepsilon \tau t_i\log \left( 1+\mathrm {e}^{\left( t-\tau \right) /t_i}\right) \end{aligned}$$
$$\begin{aligned}&\approx {\left\{ \begin{array}{ll} 0 &{} t<\tau \\ -c_2g\varepsilon \tau \left( t-\tau \right) &{} t>\tau \end{array}\right. } \end{aligned}$$

with some numeric prefactor \(c_2\) of order one depending on the exact form of \(\vartheta \), which can be obtained by evaluating the integral of Eq. (9).

Further details, in particular the discussions of x- and z-components, can be found in “Appendix C”.

4 Experimental method

Fig. 3
figure 3

The beer mat shooting apparatus

The beer mat shooting apparatus is picturised in Fig. 3. It consists of two electric motor powered treadmills which can be programmed to move at a given speed up to \({16}\,{\hbox {m}/\hbox {s}}\) forwards or backwards independently of each other. Each of the two treadmills runs around two gears with radius \({10}\,\hbox {mm}\), one of which (the black ones in Fig. 3) is connected to the electro motor. The speed of the treadmills can be inferred by measuring the rotations per minute of the driving gear using a digital laser non-contact photo tachometer. We denote the rotations per minute of the left and the right driving gear by \(u_l\) and \(u_r\), respectively.

A beer mat put between these treadmills is accelerated until the edges assume the speed of the respective treadmill. We have checked with high speed camera videos that an undamaged beer mat does indeed not slip along the treadmills. During acceleration the beer mat is confined vertically between two plastic surfaces.

During operation the apparatus is placed on a table so that the beer mats are shot from a height of \({0.98}\,\hbox {m}\) with the plastic surfaces horizontally aligned. Then, to the level of accuracy required here the apparatus produces reproducible initial flight conditions. All beer mats shot with the same treadmill configuration followed the same trajectory and hit the same point on the floor, up to minor deviations in the order of \({0.1}\,\hbox {m}\).

Flights were recorded using a high speed cameraFootnote 5 with 500 frames per second. Example videos of flights are available in the supplemental material. We used the program TrackerFootnote 6 to extract the coordinates of the beer mat at any given time from the videos. The measurements cover a broad range of different initial velocities and angular momenta. A summary of the recorded and analysed experimental setups is provided in Table 1 of the appendix.

Fig. 4
figure 4

Horizontal position y orthogonal to the initial flight direction in meters against time in seconds. The flight started with initial conditions as in Eq. (8) with \(v_0={2.6}{\hbox {m}/\hbox {s}}\) and \(\omega _0={49}{\hbox {s}}^{-1}\) (corresponding to 7.9 rotations per second). The effective theory fit follows Eq. (10). For details on the numerical simulation see Appendix 6.7. The lengths are not exactly correct due to perspective distortion

5 Results and discussion

For each of the recorded experiments we have analysed the trajectory of the corresponding flight. First, we observe that the angular velocity \(\omega _0\) changes by less than \(10\%\) during the flights.

Next we reconstruct the horizontal position y(t) from the recorded video. We fit the functional form Equation (10) to this data for y(t) for each experimental setup separately and determine the parameters \(c_2\epsilon , t_i\) and \(\tau \). One representative example of such a y-trajectory and the corresponding fit are shown in Fig. 4. The so determined values for the characteristic time \(\tau \) and \(t_i\) are shown in Fig. 5 as functions of \(v_0/\omega _0\). From this Figure one can already read off that stable flights of beer mats longer than \({0.45}\,{\hbox {s}}\) are hardly possible. The detailed fit results are provided in Table 2 of the appendix.

Note that we also solved the full set of differential equations including lift, drag and sub-leading effects numerically, for details see Appendix 6.7. The corresponding y-trajectory is also depicted in Fig. 4 for means of comparison as the purple dash-dotted line.

In Fig. 6 we show \(\lambda \) determined via \(\lambda = \log 2/\tau \) as a function of \((v_0/\omega _0)^2\). In the same Figure we show a fit of Eq. (7) to the data for \(\lambda \) (with \(\rho ={1.25}{\hbox {kg}/\hbox {m}^{3}}\), \(m={5.9}\,{\hbox {g}}\) and \(A=\pi r^2\), \(r={5.3}\,{\hbox {cm}}\)) yielding \(\lambda _0 = {12.4\pm 1.5}\,{\hbox {s}}^{-1}\) and \(\lambda _1 = {10.7\pm 0.7}\,{\hbox {s}}^{-1}\).

Fig. 5
figure 5

Characteristic times \(t_i\) and \(\tau \) derived via a fit of Eq. (10) for a variety of different initial conditions in speed \(v_0\) and angular speed \(\omega _0\)

Note that the statistical error from the fit depicted with errorbars has but a small contribution to the overall error. The total error comes mostly from the fact that the flight of a beer mat is not perfectly reproducible with our apparatus and in principal identical initial conditions lead to slightly different results when repeated several times. This discrepancy between statistical and total error is reflected in the large spread of points totally incompatible with the size of their error bars.

Fig. 6
figure 6

Damping factor \(\lambda \) derived via a fit of Eq. (10) for a variety of different initial conditions in speed \(v_0\) and angular speed \(\omega _0\). The fit follows Eq. (7)

As we mentioned above, \(\omega _0\) does change measurably during the flight, however, not significantly in particular in the phase of the flight relevant for our analysis. Therefore, we think the main assumption for our effective theory of constant \(\omega _0\) is fulfilled well enough, even more so as the effective theory, namely Eqs. (10) (and (C2), (C5) from the appendix), describes the flight trajectories extremely well with fitted parameters. We find our earlier expectations confirmed as indeed \(t_i\ll \tau \) for all flights we investigated. Most importantly, the values of \(\lambda \) are in good agreement with a linear dependence on \(v_0^2/\omega _0^2\) plus a constant.

These values for \(\lambda _0\) and \(\lambda _1\) are universal constants for every system describable by the effective theory. Thus, they should allow us to predict the duration of a stable flight of any rotating and thin enough disk. Just using the constant term \(\propto \lambda _0\) solely depending on the disk’s properties allows one to predict the longest possible stable flight duration. We arrive at some very interesting and astonishingly realistic predictions: recall that a beer mat maintains stability for up to about \({0.45}{\hbox {s}}\). Now a standard playing cardFootnote 7 is expected to reach slightly more than half that time, namely about \({0.24}\,\hbox {s}\), a CD holds out for twice a beer mat’s time or \(\sim {0.8}\,\hbox {s}\), a frisbee for ca. \({0.7}\,\hbox {s}\) (without any aerodynamics due to curvature taken into account) and a discus could in principle fly undisturbed for up to ca. \({16}\,\hbox {s}\). These predictions could be tested experimentally.

We find especially the time estimate for the frisbee very interesting because it is not unrealistic for a mediocre thrower to have the frisbee flip towards backspin in the timespan of about one second though this estimate lies rather on the short side and of course the wing form of a frisbee allows it to remain stable for a much longer time when thrown professionally. The reason is that frisbees have their aerodynamic center very near to their center of mass [2] and, thus, experience much less torque.

6 Summary

In this paper, we have investigated the peculiar flight trajectories of beer mats: independently of their initial conditions, beer mats always tilt into backspin position shortly after being thrown. When thrown by hand, the moment this tilting starts is seemingly random.

We have presented an explanation for this effect and developed an effective theory describing the flight of beer mats, or any thin disk, alongside an experimental investigation of such flights. We used the experimental results to estimate the parameters of the aforementioned effective theory, which describes the data very well. The effective theory then makes universal predictions: for instance, the damping factor \(\lambda \) of the tilting motion must depend linearly on \((v_0/\omega _0)^2\) (the centre-of-mass velocity and the angular speed, respectively) and otherwise only on the disk’s radius, mass and the density of air. This is nicely confirmed by our empirical results.

\(\lambda \) is directly related to the time \(\tau \) via \(\lambda =\log (2)/\tau \). \(\tau \) is the time after take off at which the tilting into backspin orientation becomes visible, i.e. \(\tau \) is the time of stable flight. Its apparent randomness solely stems from the inability to reproduce initial conditions when throwing by hand.

Since the effective theory holds for any thin disk we are also able to predict \(\tau \) for other types of disks like playing cards or CDs, which could be tested experimentally. Frisbees, however, do have different aerodynamic properties than beer mats due to their rounded down edges and, thus, enjoy a significantly extended stable flight time.