# How far are you able to soar from a swing?

*by*Phil Tadros

August 18, 2023

Dialogue on HackerNews.

Some folks identified some flaws in my modelling (eg: assuming zero distance from swing to ground) which I’ve tried to repair. The unique most distance estimation was round $1m$.

This summer time I’ve spent an absurd period of time studying and studying concerning the physics of swings. Sure, you learn it proper, I’ve been studying concerning the bodily processes that occur when a child is enjoying with a swing within the park. Blame it on my youngsters and the numerous hours spent having fun with these moments with them. Particularly, I learn concerning the physics of pumping a swing and concerning the physics of leaping from a swing. Amidst my deep dive into swing physics, I got here up with a brand new Olympic sport through which you begin seated on a swing with size $L$, your ft comfortably touching the bottom. As a countdown of $T$ seconds commences, you embark on the artwork of swing-pumping. Your problem is to execute a skillful leap earlier than the countdown reaches zero. Along with your soar, you journey a distance $d$ out of your preliminary level, aiming to attain the best attainable $d$.

The query is then, which is the very best methodology to maximise $d$?

Earlier than I current you with the reply to the query I’ll summarize the learnings I bought from studying concerning the physics of a swing. As common, you could find all of the code I used for this submit in my repo.

Discover that ^{ and reply a really comparable query: what’s the optimum time $t$ to leap off, in order to achieve farthest. Nevertheless, these references don’t cope with the pumping of a swing, they only assume that you simply begin swinging at some angle $lambda$ and soar at an angle $phi$, with out pumping the swing at any level. Fixing this drawback is fascinating, however I feel it’s extra thrilling to resolve it when the individual swinging can management the system. This makes it really feel extra like an actual sport you could play in a park or on the Olympic Video games.}

There are a number of papers concerning the pumping of a swing ^{, , and and far more. On this part, I’ll focus particularly on .}

The mannequin for a swing I’ll use is a inflexible dumbbell made up of three lots, suspended by a inflexible rod of size $l_1$ hooked up to the center mass $m_1$. The distances from $m_1$ to the opposite lots $m_2$ and $m_3$ are $l_2$ and $l_3$ respectively. The angle of the rod $l_1$ with the vertical is $phi$ and the angle of the dumbbell with the rod is $theta$. Within the subsequent determine, you’ll be able to see a diagram of the system

The Lagrangian of this technique is

[begin{align}mathcal{L} = & frac{1}{2} I_1 dotphi^2 + frac{1}{2} I_2 left(dotphi + dotthetaright)^2 – l_1 N dotphileft( dot phi + dottheta right) cos theta

&+ M l_1 g cos phi – N g cosleft(phi + thetaright)

end{align}]

the place $M = m_1 + m_2 + m_3$, $N = m_3 l_3 – m_2 l_2$, $I_1 = M l_1^2$, and $I_2 = m_2 l_2^2 + m_3 l_3^2$. Due to this fact, the Lagrange’s equation for $phi$ is

[begin{align}(I_1 + I_2) ddot phi + M l_1 sin phi = & -I_2 ddot theta – l_1 N dottheta^2 sin theta

& + l_1 N dottheta^2 cos theta + N g sin(phi + theta)

& – 2 l_1 N dot theta dot phi sin theta

& + 2l_1 N ddot phi cos theta

end{align}]

The paper proceeds by assuming the swinger pumps the swing by forcing $theta(t) = theta_0 cos(omega t)$, the place $omega$ is the pure angular frequency of the pendulum. Then, they present that there are two regimes, one the place $phi < phi_{textual content{crit}}$ the place the motion is like an harmonic pushed oscillator, and one other one the place $phi > phi_{textual content{crit}}$ the place the motion is an harmonic oscillator with parametric phrases. They then analyze the completely different regimes and clear up their equations. In abstract, they present that for small amplitudes the swing follows the equation

[ddot phi + omega_0^2phi = F cos(omega t)]the place

[begin{align}&omega_0= K_0/I_0

&I_0=I_1 + I_2 – 2l_1 N (1 – theta_0^2/4 )

&K_0=Ml_1g – Ng(1 – theta_0^2/4)

&F=theta_0 left[ (omega^2 I_2 + N(g – omega_0^2l_1)(1 – theta_0^2/8)right]/I_0

finish{align}]

The differential equation has the answer

[phi(t) = left(frac{F}{omega_0^2 – omega^2}right)left(cos omega t – cos omega_0tright)]which appears like for small $t$

This answer is nice sufficient for our strategy since I assume $T$ to be sufficiently small to keep away from the swinger pumping the swing to huge $phi$ values.

Now let’s examine how ought to a swinger soar from a swing to maximise the traveled distance. The evaluation offered right here is predicated on the work of Jason Cole ^{ and Hiroyuki Shima . Discover that the naive answer of leaping at $phi=pi/4$ just isn’t optimum. For example, think about a swing that oscillates within the vary $pm pi/4$, then it’s clear that leaping at $pi/4$ is suboptimal for the reason that swinger will begin its flight with a pace of zero.}

Discover I’m including a brand new variable $h$ that represents the gap from the swing to the bottom. This variable just isn’t current on Jason’s weblog nevertheless it’s on Hiroyuki paper. When you soar from the swing, the equations of movement for the horizontal and the vertical instructions are

[begin{cases}x(t) = l_1 sin phi + v tcosphi

y(t) = h + l_1(1 – cos phi) + vt sin phi – frac{1}{2}gt^2

end{cases}]

Now, we will compute the whole flight time by fixing $y(t_{textual content{flight}})=0$ after which compute the flight distance as $d = x(t_{textual content{flight}})$. The flight time is then

[t_{text{flight}} = frac{vsinphi pmsqrt{v^2sin^2phi+2g(h+l_1(1-cosphi))}}{g}]now discover that solely the optimistic root has bodily that means (we don’t need destructive occasions), so the gap is

[begin{align}d = & l_1 sinphi + frac{v^2 sinphicosphi}{g}

& + sqrt{frac{2v^2cos^2phi(h+l_1(1-cosphi))}{g}+left(frac{v^2sinphicosphi}{g}right)^2}

end{align}]

We now have now all of the items we have to clear up the issue. On one hand, we will compute the swinger angle $phi$ for any given $t$, and we will additionally compute the gap that the swinger will journey when leaving the swing at an angle $phi$.

Discover that to compute $d$ we have to know $v$. In different sources that examine this drawback, they get $v$ by utilizing vitality conservation, nonetheless, in our case, we all know $phi(t)$ and we will get the preliminary velocity after leaving the swing as $v(t) = l_1 frac{d}{dt} phi(t)$

[v(t) = frac{l_1F}{omega^2_0 – omega^2}(omegacosomega t – omega_0cos omega_0 t)]Now, placing all the pieces collectively we’ve got this set of equations

[begin{cases}d(t) =l_1 sinphi(t) + frac{v^2 sinphi(t)cosphi(t)}{g} + sqrt{frac{2v^2cos^2phi(h+l_1(1-cosphi))}{g}+left(frac{v(t)^2sinphi(t)cosphi(t)}{g}right)^2}

phi(t) = frac{F}{omega_0^2 – omega^2}left(cos omega t – cos omega_0tright)

v(t) = frac{l_1F}{omega^2_0 – omega^2}(omegacosomega t – omega_0cos omega_0 t)

end{cases}]

To compute $d(t)$ we simply must compute $phi(t)$ and $v(t)$ and substitute the values within the first equation. I’ll use the next constants: $M=1$, $m_1=0.4M$, $m_2 = 0.2M$, $m_3 = 0.4M$, $l_1 = 2$, $l_2 = 0.4$, $l_3 = 0.4$, $h=l_3$, $theta_0=1$, $g= 9.8$, $T=2 pi sqrt{l1 / g}$, and $omega= 2pi/T$

With these parameters, we will now plot the traveled distance as a operate of the leaping time.

Let’s keep in mind that we’re within the optimum leaping time $t^*$ for a given most time $T$. To take action we simply want to repair a time $T$ and discover at which $t^* < T$ the gap $d(t)$ is maximized. I did that numerically and plotted the leads to the subsequent picture

After all, the optimum leaping time follows a ladder-like curve. It is because you’re not concerned about leaping backward, and generally it’s higher to leap some seconds earlier than $T$ than to attend for $T$ and end up in a worse place.

Lastly, we will get additionally the utmost traveled distance as a operate of $T$.

For instance, if $T=20s$, which looks as if an inexpensive worth to make the game fascinating, one would anticipate to attain $dapprox 2m$.

Effectively, that’s all for at the moment. On this submit, I’ve offered a brand new Olympic sport that consists of pumping on a swing throughout a given period of time after which leaping and attempting to attain the utmost distance. With the parameters utilized in my simulations, I’d anticipate the world document to be round two meters.

The evaluation offered right here is filled with simplifications. Right here I record a number of the ones I’m conscious of

- The swinger mannequin is oversimplified. For example, authors in
^{ current a mannequin which is extra correct than the one used right here. Nevertheless, I needed to maintain the evaluation “easy”.} - The swinger is assumed to function within the regime of small $phi$, which permits us to make use of an analytical equation for $phi(t)$. Nevertheless, an experimented swinger might obtain huge oscillation angles in a brief period of time, after which our simplification wouldn’t be legitimate anymore.
- As a superb physicist I’ve uncared for any type of friction (swing-rod, swinger-air, and many others.).

Even with all this simplification, I feel the evaluation stills carry some mild to the issue of maximizing the flight distance. Now, the one step nonetheless lacking is the experimental one: go to a park and attempt to beat the theoretical most distance. In line with my numbers, you wouldn’t have the ability to beat the 2-meter mark.

All of this jogs my memory of the anecdote of the mathematician and his spouse transferring a settee. The mathematician spent plenty of time computing if it was attainable to maneuver the couch from one room to the opposite, and at last proved it was inconceivable. Then he went to indicate it to his spouse, which had already moved the couch to the opposite room. So I’m fairly certain that it’s going to be attainable to beat my theoretical most distance. Sadly, I don’t have a swing close to me proper now, so I’ll have to attend till the subsequent go to to the park.

Some days after publishing this submit I began questioning which was the mix of lots $m_1$, $m_2$, and $m_3$ that allowed for the very best outcomes, aka how ought to a Swing Leaping world champion appear to be. To take action I’ve fastened all of the parameters besides the lots, and I’ve compelled $1 = m_1 + m_2 + m_3$ for the reason that habits of the system is unbiased of the whole mass $M$.

Within the subsequent plot we see how the utmost distance depends upon $m_2$ and $m_3$. The crimson star exhibits the place the mix of lots that maximized the gap.

The perfect mixture of lots is $m_1=0$, $m_2 approx 0.625$ and $m_3 approx 0.375$. After all this isn’t an answer that’s possible – possibly . Setting a minimal worth for $m_1 > 0.1$ we get a unique optimum distribution of lots, ie: $m_1approx 0.2$, $m_2 approx 0.5$, and $m_3 approx 0.3$. So we see that the optimum answer is at all times to attenuate $m_1$.

With these new lots the maxiumum distance is round $3m$ which is significantly larger than our first consequence.

We might additionally analyze the very best mixture of lengths $l_*$ and lots more and plenty $m_*$ that maximize the gap, nonetheless I don’t assume it’s going so as to add plenty of worth to the examine so I’ll go away the evaluation as it’s now.

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