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Mathematical analysis of the trapeze swing
Introduction
In a the trapeze swing, the flyer moves their body to increase the total energy in the swinging system, and therefore the height of the swing. The most obvious way is by raising or lowering the flyer’s center of mass (CoM), doing work against gravity, as in the analysis at www.flying-trapeze.com. In addition, as the muscles cause body parts to rotate around the axes defined by the joints, they apply torques and change the angular momentum of parts of the flyer (although the total angular momentum is conserved). This is definitely the way flyers flip over, for exampe in a face-off or a miss, but it is unclear if the same effect can significantly affect the swing.
Here I quantitatively analyze the effects of various motions on swing height.
The swing
Geometry of the flyer
Some detailed measurements of human body part masses and weights are listed in Tables I and II. Based on these measurements I calculated the height of the CoM of the flyer when fully extended, and when raising the legs by bending $180^\circ$ at the hips, as in the initial part of a force out. A diagram of these two body positions and the resulting CoM positions is shown in Fig. 1. The position of the extended flyer’s center of mass, measured from the fly bar (ends of arms) in terms of fraction of body height is
h_\mathrm{extended} = 0.155 f_\mathrm{arms} + 0.205 f_\mathrm{head} + 0.44 f_\mathrm{torso} +
0.7 f_\mathrm{ul} + 0.965 f_\mathrm{ll} = 0.495
where $f_\mathrm{part}$ is the mass fraction of each body part from Table I, “ul” stands for upper legs, and “ll” stands for lower legs. For the force-out position the CoM position is
h_\mathrm{extended} = 0.155 f_\mathrm{arms} + 0.205 f_\mathrm{head} + 0.44 f_\mathrm{torso} +
0.44 f_\mathrm{ul} + 0.175 f_\mathrm{ll} = 0.364
Table I: body part volumes, based on https://msis.jsc.nasa.gov/sections/section03.htm Fig. 3.3.6.3.2-1 panel 2 of 12. Note that for limbs, the values below include the sum of right and left. | body part | fraction of total volume | | ———-| ————— | | head + neck | 6.5 % | | torso | 56 % | | upper legs | 15.5 % | | lower legs | 11.5 % | | upper arms | 6 % | | lower arms + hands | 4.5 % |
Table II: Body part lengths, based on https://msis.jsc.nasa.gov/sections/section03.htm Fig. 3.3.1.3-1 with some additional personal observations. From panel 2 of 12: overall height. Directly from height and panel 8 of 12 (or from its differences): hips-floor (upper + lower legs). From cited measurements and estimated offset between shoulder joint and cited figure shoulder height: shoulder joint-hips (torso), shoulder joint-top of head (head and neck). Estimated from personal observation: fraction of overall legs (hip-floor) accounted for by the upper and lower legs, arm length. | body part | fraction of height | | ——— | —————— | | head and neck | 0.21 | | torso | 0.26 | | lower legs | 0.27 | | upper legs | 0.26 | | arms | 0.31 |
Assuming that the flyer is 1.8 m tall (5’11”), the distance from the hands to the center of mass fully extended is $0.89~\mathrm{m}$, and when raising legs it is $0.66~m$. Below I will approximate these as $0.9~\mathrm{m}$ and $0.65~\mathrm{m}$. In addition to these dimensions I assume below that the mass of the flyer is $75~\mathrm{kg} = 165~\mathrm{lbs}$.
The swing of a still, rigid flyer
The geometry of the rig and the flyer is shown in Fig. 2, with (at best) approximate dimensions. The lowest possible center of mass position is with a fully extended flyer at the bottom of the swing (pulse), 4.54 m below the fly-bar crane. The initial position of the flyer is approximated as a “7”, with horizontal arms.
For simplicity, I assume that the system starts with only potential energy, determined by the initial flyer center of mass height, ignoring the fact that the body shape is not extended. The initial height of the CoM above its lowest point is roughly estimated to be $(4.54 - 1.65)~\mathrm{m} = 2.89~\mathrm{m}$. The initial potential energy (relative to the lowest height) is therefore
V(t = 0) = m \, g \, \Delta h = 75~\mathrm{kg} \, g \, (4.54~\mathrm{m} - 1.65~\mathrm{m}) = 2.124~\mathrm{kJ}
where $g = 9.8~\mathrm{m/s^2}$ is the acceleration of gravity, and the mass of the flyer is 75 kg (165 lbs). At the bottom of the swing the potential energy is $V = 0$, so (assuming the flyer remains rigid) at that point the kinetic energy is equal to the initial potential energy. The rotational kinetic energy for a point mass rotating about an axis is
K = \frac{1}{2} I \omega^2 = \frac{1}{2} \omega^2 m r^2
where $I = m r^2$ is the moment of inertia, $\omega = \dot{\theta}$ is the angular velocity, and the centrifugal pseudo-force the flyer feels is
F = m \omega^2 r.
Solving for $\omega$ in terms of $K$ gives
\omega = \sqrt{\frac{2 K}{m r^2}}
and substituting into the expression for $F$ gives
F = \frac{2 K}{m r^2} m r = \frac{2 K}{r} = \frac{2 \times 2.124~kJ}{4.54~m} = 935~N
and the corresponding acceleration
a = F / m = 12.5~m/s^2 = 1.3~g
in addition to the $1~g$ exerted by gravity.
For completeness, the period of this swing, which is far from the commonly analyzed infinitesimal amplitude limit, is given in wikipedia as
T = \frac{2 T_0}{\pi} K(k)
where $T_0 = 2 \pi \sqrt{\frac{r}{g}}$ is the small amplitude approximate period, $k = \sin \frac{\theta_0}{2}$ and $K$ is the complete elliptic integral of the first kind. For the dimensions in Fig. 2 $r = 4.54~\mathrm{m}$ and $\theta_0 = 70^\circ$. Therefore, $T_0 = 4.3~\mathrm{s}$, $k = 0.57$, $K(k) = 1.918$ Wolfram Alpha, so $T = 5.25~\mathrm{s}$.
The effect of raising and lowering the flyer’s CoM
If the flyer raises or lowers their CoM they can change their potential or kinetic energy, but cannot affect their angular momentum. I therefore compute the change in the swing if the flyer raises their legs at the bottom of the swing, doing work against gravity and the centrifugal pseudo-force, while keeping angular momentum constant.
The increase in potential energy due to the $\Delta~r = 0.25~\mathrm{m}$ raising of the center of mass by lifting the legs is
{\Delta}V = m \, g \, {\Delta}r = 184~\mathrm{J}.
The angular momentum fully extended is
L = I \omega = m r^2 \omega_i
and after raising the legs it is
L = m (r-{\Delta}r)^2 \omega_f
Solving for the final angular velocity gives
\omega_f = \omega_i \frac{r^2}{(r-{\Delta}r)^2}
The corresponding change in kinetic energy is
\begin{eqnarray}
K_f - K_i & = &
\frac{1}{2} \left( m (r - {\Delta}r)^2 \omega_f^2 - m r^2 \omega_i^2 \right) \\
& = & \frac{1}{2} m \omega_i^2 \left( (r - {\Delta}r)^2 \frac{r^4}{(r - {\Delta}r)^4} \omega_i^2 - r^2 \right) \\
& = & \frac{1}{2} m \omega_i^2 \left( \frac{r^4}{(r - {\Delta}r)^2} - r^2 \right) \\
& = & \frac{1}{2} m \omega_i^2 r^2 \left( \frac{r^2}{(r - {\Delta}r)^2} - 1 \right)
\end{eqnarray}
With $\Delta~r = 0.25$ m the fractional change of the kinetic energy of 12%, or 262 J. The increase of the total energy of the system is $184 + 262$ J, to 2.63 kJ. This corresponds to a final height of 3.58 m above the lowest point.
When the flyer reaches peak, $3.58~\mathrm{m} - 2.89~\mathrm{m} = 0.7~\mathrm{m}$ higher than the beginning of the swing, their potential energy is maximized and kinetic energy is zero. If they could extend their legs to increase $r$ back to its initial value without otherwise changing their energy, they could go back to this new, larger height at the back of the the swing, swing forward, and repeat the process. The gain in energy (and therefore height) from the next swing will be even larger: the change in potential energy will be the same, but because the initial height is larger, the value of $\omega$ at the bottom will be larger, so the kinetic energy gain will be larger as well.
Note, however, that this simplistic analysis neglects two factors. The first is that changes in body position at the back end of the swing are required for the flyer to avoid hitting the back of their legs on the board. The other is that if the flyer is not exactly horizontal at peak, extending the legs will lower the CoM somewhat, and lose some potential energy. The change in height of the CoM is exactly $\Delta r \cos \theta$. The peak angle can be determined from $\cos \theta = (4.54 - 3.58) / 4.54 = 0.211$, so the loss of height will be 0.05 m. The flyer will therefore gain only 0.84 m, rather than 0.89 m.
Another complication is the fact that the lines are flexible, not rigid, and moving CoM away from the rotation axis formed by the fly-bar crane requires pushing away from the fly bar. Since the mass of the flyer is much larger than the bar, without a force pulling the flyer away from the axis most of the motion would actually be pushing the bar toward the axis, creating slack in the lines. As a result, the extension has to happen more gradually, using gravity and the centrifugral pseudo-force to pull the flyer awy from the axis and keep the lines taut. A quantitative description of this effect is beyond the scope of this work.