The chain that climbs out of the jar

The crack at the end of a whip followed a fold in a flexible line until its tip passed the speed of sound, and the pile that lands heavier than it weighs found the extra force a pile receives from a chain falling into it. It also listed, as a later question, the behaviour in which a chain lifted out of a beaker rises above the rim. The chain fountain is the reverse of a falling chain’s pile. Instead of a chain landing on a surface, it is picked up from a surface, and one might expect the problem of a chain leaving a pile to be the easier of the two.

A long chain of small metal balls on thin rods, the sort used for bathroom-light pulls, is coiled in a jar held some way above the floor. The end is pulled over the rim and let go. Gravity on the falling leg pulls more chain out, and soon the whole chain is pouring out at a steady speed. That is no surprise. Steve Mould put a video of this online in 2013, and the striking thing in it is that the chain does not skim over the rim of the jar. It leaps up from the jar in a rising loop, stands in a stationary arch many centimetres above the rim, and comes down again. The shape holds steady while the chain flows through it. Nothing visible holds it up.

For a physics problem so simple, it was hard to find anyone who was sure of the reason. John Biggins and Mark Warner, in Cambridge, gave the answer in 2014.

What a steady chain has to do #

Take a chain with mass per unit length $\lambda$, moving at steady speed $v$ along a fixed curve. Any curve will do. Each link moves along that curve as a train follows a track, at constant speed but with changing direction. A piece at a place where the track bends needs a centripetal force, and the tension in the chain provides it. A remarkable classical result holds: a chain whose tension is exactly $T = \lambda v^2$, with no gravity, can follow any curve at all, since the sideways component of the tension at each bend exactly provides the required acceleration, whatever the curvature. A loop of chain spun fast keeps its shape when it runs off a spinning pulley and rolls across a table, a well-known demonstration that depends on this. It means the arch shape itself is no great mystery. With tension about equal to $\lambda v^2$, the chain can hold any arch at the top. Gravity’s effect is small if the speed is large enough.

Now what is needed is a tension balance. At the top of the arch, the tension is near $\lambda v^2$. Moving along the chain from there, in a steady state, tension falls with height in exactly the way it does in a hanging chain, since each piece of chain is moving at constant speed along a straight line and has to have its weight supported:

$$T_{\text{jar}} = \lambda v^2 - \lambda g h_1, \qquad T_{\text{floor}} = \lambda v^2 - \lambda g (h_1 + h_2).$$

The arch’s top is at height $h_1$ above the pile and $h_1 + h_2$ above the floor. That gives two statements about the tension at the two ends of the chain, and the physics now resides entirely in those two ends.

What the pick-up must supply #

At the jar, a link at rest in the pile has to be set moving at speed $v$. That needs a momentum per unit time of $\lambda v^2$ supplied to the chain, which is what the classical variable-mass argument of the pile that lands heavier than it weighs gives, and which applies to any such pick-up. Who supplies it?

For a string, or a chain with very small links, the only available force is the tension: $T_{\text{jar}} = \lambda v^2$. Comparing this with the balance above gives $\lambda g h_1 = 0$, so $h_1 = 0$. No arch. So the naive reasoning produces the correct answer for a string, and a real string does simply pour over the rim. If that were the whole story, the classical model would be complete.

Biggins and Warner saw what a stiff link does that a string cannot do. A link in a bead chain is a rod. It lies on the pile, or lies on the links below it. If one end of it is pulled sharply upward, it starts to rotate about its centre, and its far end pushes down into the pile. The pile pushes back. That reaction provides part of the momentum, with the tension supplying the rest.

The calculation for a uniform bar lying on the pile is straightforward, and it uses the same impulse argument as the point that takes the blow without a jolt. The far end is held at the pile, so the bar pivots about it. The lifted end must reach speed $v$, so the angular speed is $v/b$ for a bar of length $b$. Its moment of inertia about the far end is $mb^2/3$, and that takes an impulse $J$ at the lifted end of $mv/3$. Its centre of mass needs $mv/2$ of momentum, so the pile must supply the remaining $mv/6$. The bar has a further swing to complete before all of it reaches speed $v$, but already the essential point has emerged: tension can do less than the full job, since the pile does some of it.

Let the pile’s share of the momentum per unit time be a number $\alpha$. Then the tension at the pick-up is

$$T_{\text{jar}} = (1 - \alpha)\lambda v^2.$$

Combining that with the tension balance gives immediately

$$\alpha v^2 = g h_1.$$

So a chain with no pile reaction has no arch. A chain with such a reaction must have an arch, of a height that goes with the speed squared. That is the core result. The tension at the pick-up has been reduced, and in a steady state something has to account for this reduction. The tension’s difference from $\lambda v^2$ has to be made up by the weight of chain in a rising section, which is an arch. The pile’s upward push is, in effect, what is holding the arch up.

A fountain in proportion to the drop #

The other end supplies the second equation. Links landing on a floor have their momentum removed by the floor and the chain’s tension together. If the chain lands on a floor that simply absorbs it, as a loose chain falling onto a hard surface largely does, then the tension at the bottom is zero:

$$T_{\text{floor}} = 0 \implies v^2 = g(h_1 + h_2).$$

Combining these,

$$h_1 = \frac{\alpha}{1 - \alpha}h_2,$$

and that is a straight line through the origin.

The arch is in proportion to the drop. This is a fine prediction, since it can be tested simply by moving the jar, and it contains no chain-specific quantities apart from $\alpha$. With a share of about 0.12, which the experiments indicate for the common bead chain, the arch stands at about a seventh of the drop. Biggins’s measurements show this straight line over a range of heights up to several metres, with an increasingly unsteady arch as the height rises. High-speed video has also shown the expected motion in the pile: individual links flick up as they are picked up, and a small dent appears in the pile. Links that have been made longer or stiffer, or longer in relation to their spacing, do make a higher fountain. A rope, or a flexible thread with beads too close to pivot, makes none.

The simple bar gives a share of about 1/6, or 0.17, and experiments suggest 0.12. That is a reasonable agreement for a calculation which treats a chain as independent bars on a hard floor. A real pile gives way when pushed, and neighbouring links get in the way. A pile can yield to the link’s push, which reduces the reaction it can apply. A pile in a jar is more rigid than a loose pile on a table, and pushing a jar’s chain across a bed of soft foam would be expected to reduce the arch. This is what happens.

The same chain, the other way up #

At the floor, a similar reaction may work in the opposite sense. As the pile that lands heavier than it weighs noted, experiments show a chain falling on to a table can be pulled down faster than free fall by a similar effect: a link hits the table at one end and rotates, and its other end pulls the chain downward. Grewal, Johnson and Ruina found this in 2011 by comparing a chain falling on a table with a free-falling ball. In the notation above, this adds a further fractional share at the floor, making $T_{\text{floor}}$ actually negative in the sense of a pull on the chain, called a downward kick. Biggins and Warner showed this effect would add to the fountain’s height, and that floor kick can be either present or absent, depending on the surface. That is a neat experimental handle: placing the falling chain on a soft surface removes the kick and lowers the arch.

The same physics applies at the two ends of the chain, then, but with opposite signs and on different surfaces, and it is one stiffness argument that links them. The rod that friction cannot decide about deals with some of the peculiar things that can happen in a rigid-body model with contact, and the assumption of an impulsive contact on a bar that may or may not slip has similar limitations here.

The result also has an interesting energy budget. At the pick-up the pile has done no work: it pushed against a point that does not move. So energy has been saved compared with the case of a string, in which the tension provides all the momentum. With a string, half of the work done by the pull is lost at the pick-up, in what amounts to the kind of inelastic collision collisions are easier than forces analyses, between each new piece of string and the moving chain. Each new link’s kinetic energy is bought at a lower price with a pivot, by a factor of $1 - \alpha$ or so.

Why it never reaches a falling stone’s speed #

The speed itself is given by $v^2 = gh_2/(1 - \alpha)$. For a falling stone dropped through the same distance, it would be $v^2 = 2gh_2$. With a chain, then, the speed falls well short of what free fall gives, by a factor of about $\sqrt{2(1 - \alpha)}$. The difference has to do with what the falling leg of chain has to accomplish. Its weight is the whole power source, and it must accelerate new links from rest at the top, and then each link hits the floor and stops, losing its energy.

There are two equal losses in the string case. Half of the available energy is spent on the kinetic energy of the moving chain, and a chain’s kinetic energy is lost on impact. The same work is also dissipated at the pick-up, by inelastic jerks. So the chain has half of the energy per unit mass that a stone would have, and it runs at $1/\sqrt{2}$ of a stone’s speed. With the pile’s help the pick-up is less wasteful, so a chain with a higher $\alpha$ runs a little faster, as the middle curve shows. This argument shows that the fountain’s existence and the extra speed are connected, because both come from the same reaction. With the conventional variable-mass rocket equation, every mass increment has to be accelerated, and the rocket that leaves its fuel at home is the reverse of this story.

A chain running at 4.7 m/s through a drop of 2 m means a link moves through the system in less than a second, and a total length of chain of perhaps fifty metres runs out in ten seconds or so. That is about how long the videos last, and the sound is memorable. Beads strike one another and the jar’s rim about a thousand times a second.

The arch grows as the chain gets up to speed #

The steady state does not appear at once. Start with a slow chain, with the falling leg in place and no arch. Newton’s law for the chain as a whole, with mass $\lambda(h_2 + 2h_1)$ moving at a speed $v$, is

$$(h_2 + 2h_1)\dot{v} = gh_2 - (1 - \alpha)v^2,$$

and if the arch’s height follows the speed through $h_1 = \alpha v^2/g$, this is a single equation for $v$ that can be integrated. The integration settles at the steady speed, with a time constant about a second, and it was checked numerically against the expression above. The arch’s height grows with the square of the speed, and so it emerges more slowly at first and then quickly. A real chain’s start-up shows the same pattern, with an arch that grows from nothing in about a second.

Real runs are much less tidy than this smooth curve. Videos show the arch wobbling sideways and an uneven pile causing a jump as a tangled section passes through. Those irregularities become more pronounced at a larger height. In the record setups of several metres, the arch often breaks up altogether. This is unsurprising, given that a smooth steady-state solution requires a smooth input. A pile is not a neat source, and its dents and collapses change the local value of $\alpha$ over time.

Why a chain can follow any curve #

The tension’s special value deserves a further look, because it explains why the arch’s shape is not itself a concern. A chain that moves at speed $v$ along a curved path with a local radius of curvature $r$ needs a sideways force per unit length $\lambda v^2/r$. Tension $T$ in a curved chain provides exactly $T/r$ per unit length, a standard result. So if $T = \lambda v^2$ the two are equal at every point for any radius of curvature. Curvature has dropped out of the equation. Any shape whatsoever can be traced by the chain, and the shape is determined by whatever disturbances created it.

This is also the condition under which a transverse wave on the chain moves at the speed $\sqrt{T/\lambda} = v$. A disturbance travelling against the flow of the chain therefore stands still in the laboratory frame. That is why a kink in a fast-moving chain can seem frozen in space, as the wave that is required to stand still shows for a similar case in water. A chain’s stationary arch can be seen as a wave pattern, standing still against the flow, in which the chain moves through a shape that stays where it is.

In a siphon, a liquid rises above its reservoir and pours down; the height a siphon cannot pass discusses how far this can go. The two systems look alike, since both have fluid rising over a high point from a reservoir, driven by a falling leg. A chain, however, has a real tension while a liquid has a pressure, and a chain’s arch requires a force from below where a siphon’s rise relies on atmospheric pressure. Neither needs the other’s mechanism, and the similarity of the name for the chain’s siphoning, sometimes used, is only superficial.

A single number for a whole pile #

The whole account is a steady-state balance for a chain with an idealised pick-up and a floor with no kick. It treats $\alpha$ as a single number. Real links do not start from a neatly horizontal attitude in the pile, and $\alpha$ depends on how a link lies, on what neighbouring links do, on the rigidity of the pile, and on how quickly the picking up happens compared with the rod’s own natural motion. The calculation for the bar assumes a uniform rod, a single impulse and a fixed far end. It ignores a real bead chain’s structure, in which balls and short rods rotate through a limited angle and the chain’s bending stiffness differs in different directions.

Air resistance is ignored, and it becomes significant at high speeds, particularly for the long arches of several metres. The simple tension balance ignores the small contribution of curvature at the top. The treatment also leaves out the sideways motions of the chain and the question of why the arch’s plane is stable. The prediction of a linear height law is what has been tested. Numerical simulation of individual links has been used to confirm the picture, but an actual measured $\alpha$ that follows from first principles for a real pile of bead chain remains out of reach.

Still open: what a real pile does #

The missing piece in the theory is a prediction for $\alpha$ from the geometry of links and the structure of a pile. Biggins measured a range of values and showed they depended on chain type, and Biggins and Warner showed how a hanging chain rather than a loose pile changes things, but no theory exists from which to calculate the number for a given chain. Numerical simulations with many thousands of rigid links have reproduced fountains, and they have shown that the reaction comes from many links collectively rather than from one bar pivoting on a rigid floor.

The theory also touches on a broader issue about the pick-up of any flexible material from a stock, which is an industrial problem as well: threads and fibres are unwound from bobbins, and chains and cables are deployed from ships. In undersea cable laying and in textile processing, a sudden pick-up from a stock causes shocks that can break a line. Whether the reaction effect found in bead chains shows up at an engineering scale, at a usable level, or whether the same mechanism applied to a designed stack can reduce a fibre’s tension at pick-up, has received little study. A demonstration that began as a viral video has thus pointed at a question nobody had considered.

A chain running at steady speed through a fixed curve must have a tension near $\lambda v^2$ at the top, falling with height down each leg. A string picked up from a pile has that same tension at the bottom of the rising leg, so no arch fits. A stiff link pivots on the pile, which pushes back with a share $\alpha$ of the momentum, and the balance then requires $h_1 = h_2\alpha/(1 - \alpha)$. The fountain is held up from underneath, through the very link being lifted.