Aerodynamics in the amusement park: interpreting sensor data for acceleration and rotation

Figure 1 shows a SkyRoller ride in motion. A plane in the outer ring moves in a circle with a radius $r\approx 9.6$ m. This value is obtained from the total diameter, 21.2 m [1] but accounting for the horizontal distance between the center of the plane and the outer wing tip. According to the data sheet [1], the plane moves around the tower 11 times min⁻¹, giving a period of $T=60~\text{s}/11\approx 5.5$ s. We can also introduce an angular velocity $\mathbf{\Omega}=-\Omega {{e}_{Z}}=-\left(2\pi /T\right){{e}_{Z}}$, where the negative sign holds for the clockwise motion and the stationary axis e_Z marks the upwards direction of the tower. We thus find a speed $v=2\pi r/T=r\Omega \approx 11$ m s⁻¹ for the planes in the outer ring. The acceleration due to this circular motion is given by ${{a}_{\text{c}}}={{v}^{2}}/r\approx 12~\text{m}~{{\text{s}}^{-1}}\approx 1.2~\text{g}$. A park can also choose to run the ride with 9.5 or 8.5 turns min⁻¹ [6], which would lead to correspondingly smaller accelerations. Smaller values for velocity and acceleration are also obtained for planes in the inner ring.

To counteract the force, $m\mathbf{g}$, due to gravity, while also providing the force $m\mathbf{a}$ necessary for the acceleration $\mathbf{a}$ of a rider with mass m, the force from the ride must be $m\left(\mathbf{a-g}\right)$. For a centripetal acceleration ${{a}_{\text{c}}}\approx 1.2~\text{g}$, the size of the force from the ride becomes 1.6 mg, directed upwards and towards the centre, at an angle of about ${{51}^{\circ}}$ from the vertical, as in figure 3. A rider will perceive the direction of this force from the ride as 'up', (figure 3), due to the equivalence principle, as discussed, e.g. in [7–9].

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Figure 4 shows data collected with smartphone built-in sensors [3] during the start of the ride in a one of the planes closest to the tower, with an estimated radius of rotation $r\approx 7$ m. It gives a first illustration of the forces and rotations, expressed in the comoving coordinates. The air pressure gives an indication of elevation, and drops from 1016.7 hPa to 1013.2 hPa as the ride moves toward the top of the tower. This change in air pressure can be converted to an elevation gain, giving $\Delta h=\Delta p/\rho g\approx 30$ m.

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As long as the planes (and sensors) do not tilt, the rotation is only around the vertical axis, both in the stationary and rotating system. This rotation in relation to the body is referred to as 'yaw', and the negative sign indicates clockwise motion. From the graphs we can see that the ride starts moving upwards before the rotation starts, at around 93 s in the graph. As the plane starts to move faster around the tower the total force on the rider becomes larger, as seen by the increasing red curve in the top figure. However, the sideways (lateral) force remains close to zero—since the plane is free to rotate, to align with the vector $\mathbf{g-a}$. The tilt of the plane is most evident in the angular velocity curve for the 'pitch', which rises around 18 s, before the effect is clearly noticeable in the force on the rider. The rotation axis, e_Z, for the main rotation around the tower can be expressed in terms of the comoving coordinates as ${{e}_{Z}}=\cos \theta {{e}_{z}}+\sin \theta {{e}_{y}}$, if the plane is tilted by an angle θ. Since the angular velocity around the tower is negative and the angle θ is positive, both the z component (yaw) and the y component (pitch) are negative.

To start the rolling of the plane, the wings must be tilted as in figure 5. The torque from the oncoming air for a tilt by an angle ϕ is proportional to ${{\sin}^{2}}\phi \cos \phi$, which has a maximum [11] for $\phi =\arctan (\sqrt{(}2)-\sqrt{(}3))\approx {{55}^{\circ}}$, which is larger than the maximum tilt angles allowed by the construction ($\pm {{34}^{\circ}}$ for airplanes in the outer circle and $\pm {{47}^{\circ}}$ for the inner circle [6]).

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By keeping the two wings in opposite orientation, a torque τ can be exerted on the plane, causing it to rotate slightly from the equilibrium angle defined by direction of the the vector $\mathbf{a-g}$. As soon as the plane is at an angle θ from this equilibrium orientation, there is an opposing torque, due to gravity and acceleration. This opposing torque causes the rotation to slow down and stop and then, the plane to swing back, past the equilibrium orientation. When the plane changes its direction of rotation, the orientations of the wings must be reversed, and the plane can then rotate to a larger angle next time. With experience and practice, this change can be made more quickly and timely, leading to faster increase in the maximum angle of the swing, and ultimately to a fast rolling.

Figure 6 shows the angular velocity together with the 'vertical' and 'lateral' components of the vector $\mathbf{(a-g)}/g$ for the start of a ride, up to the transition to fast rolling. These data were collected by one of the authors (ML) with a home-built dedicated system built into a small custom-made case (figure 7).

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The total torque on the airplane can be expressed as ${{\tau}_{\text{tot}}}=\tau -M|\mathbf{g-a}|h\sin \theta$, where M is the mass of the plane (including the rider), h is the distance between the center of mass and the axis of rotation, and τ is the torque from the two wings which is independent of the angle $\theta$, but changes sign when the orientations of the wings are interchanged. It should also be noted that the torque from the wings is reduced for fast rotations, since the rotations reduce the relative velocity between air and wings.

The equation for the time dependence of the angle becomes

$$\begin{eqnarray}&&I\ddot{{\theta}}=\tau -M|\mathbf{g-a}|h\sin \theta.\end{eqnarray}$$

The dots denote time derivatives, but we can also use also the notation $\omega =\dot{{\theta}}$ for the angular velocity and $\alpha =\dot{{\omega}}=\ddot{{\theta}}$ for the angular acceleration. This formula can be used to obtain an expression for the rotational energy $I{{\omega}^{2}}/2$, using the relation

$$\begin{eqnarray}&&{\int}_{{{\omega}_{0}}}^{\omega}\omega ~\text{d}\omega ={\int}_{{{\theta}_{0}}}^{\theta}\alpha ~\text{d}\theta \end{eqnarray} \tag{ 1 }$$

analogous to the expression ${\int}_{{{v}_{0}}}^{v}v~\text{d}v={\int}_{{{s}_{0}}}^{s}a~\text{d}s$ for linear motion. We show in the appendix how this equation can be used for a more detailed investigation of the initial rolls.

Figure 8 shows the lateral component of the accelerometer data for the full ride, together with data from the first ride of one of the authors (AMP), collected with a wireless dynamic sensor system (WDSS) carried in a data vest [10]. The lateral component was chosen for the graphs in figure 8, since it is more sensitive to small changes in angle. Smartphone sensors (see e.g. [2–4]), which typically measure up to $\pm 2g$ along each axis, are also well suited for this ride, with its relatively small forces.

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Figure 8 provides a comparison between a ride where the planes never makes it all the way round and one ride with more than 40 full rolls. Fast rolling requires practice in order to achieve the right timing and the fast reversal of the orientation of the wings.

A first observation is that the maximum lateral force is about 1.6 mg, both for the limited turns and for the full turns, where the angular velocity is much larger. Also the vertical components reach essentially the same value (positive, as well as negative). The centripetal acceleration connected with the rolling of the plane must thus be negligibly small. This reflects the position of the sensors close to the axis of rotation. The value for the lateral component thus depends only on the angle relative to the force from the ride.

During the fast rotations, the wings are kept in the same position, and we might have expected a continued increase of angular velocity, if the torque had remained constant. However, for fast rotations, the speed of the outer parts of the wings becomes comparable to the speed of the plane relative to the air, leading to a reduction of the torque. (The wing tip is more than r  =  1.3 m away from the x axis. The speed of the wing tip due to the rotation becomes, $r\omega \approx 10$ m s⁻¹ for the fastest rotation.) As the force from the air is proportional to the square of the speed relative to the surface, this can have a large effect on the rolling, leading in effect to a limitation of the angular velocity. The approximations used in the attempt to model the motion are thus valid only for small angular velocities.

The centripetal acceleration due to the rotation around the x axis does not have a significant influence on the forces measured with a sensor close to the rotation axis. However, the fast spinning causes a sufficiently large acceleration of the centre of mass to keep the body pushed into the seat also in the upside down position (or, rather, the seat pushes the body to provide the additional force required for the centre-of-mass acceleration). In fact, the rotation around an axis close to the heartline can be seen as a body centrifuge. The acceleration of the head, can be estimated using the measured angular velocities:

$$\begin{eqnarray}&&{{a}_{H}}=H{{\omega}^{2}}\end{eqnarray}$$

towards the center of rotation at a distance H. For most of the turns, the angular velocity at the 'bottom', when the plane and rider are 'upright' is around $\omega =6$ s⁻¹, which corresponds to an acceleration ${{a}_{H}}\approx 1.8$ g for a distance $H\approx 0.5$ m. The force on the head required for this acceleration needs to be added—as a vector—to the force required for the accelerated motion of the heartline, which is about 1.6 mg pointing up and towards the centre as shown in figure 3. The body then needs to pull the head (with mass m_H) towards the body with a force ${{m}_{H}}\left(1.8-1.6\right)~\text{g}=0.2{{m}_{H}}g$. This corresponds to a 'negative g force', −0.2g. For the faster rotations, the angular velocity reaches $\omega \approx 8$ s⁻¹ in the 'upright' position, and $\omega \approx 6$ s⁻¹ in the upside-down position. The top of the head then experiences a negative g force around  −3.4g.

Safety standards for amusement rides typically limit negative vertical g forces to  −2g. However, in a rotation around the heartline the lower part of the body accelerates in the opposite direction. We are not aware of standards that apply for rotations around the heartline. However, we may consider the rotation-induced pressure in a liquid with density ρ at a distance r from the axis, rotating with an angular velocity ω can be written as $p(r)={\int}_{0}^{r}\rho r{{\omega}^{2}}\text{d}r=\rho \left({{r}^{2}}/2\right){{\omega}^{2}}=(r/2)\rho \left(r{{\omega}^{2}}\right)$. This be compared with the pressure $\rho gh$ of a stationary liquid at a level h below the surface. Even for the largest angular velocities shown in figure 9, the rotation-induced pressure at 0.5 m from the axis is equivalent to the pressure at 0.8 m below the surface. This is thus significantly less than if the whole body is upside down or exposed to  −1g in other ways. Still, some riders report that they experience a light headache after having achieved a large number of spins. The effect from the rotation around the heartline is added to the forces on a point on the rotation axis.