Two connected balloons part over a range of stretch, and even out on either side

Most of the misconceptions about suction come from reading a pressure as a pull. A force forgets the datum and a stress cannot shows what is lost when the ambient is dropped from the account, and where a liquid does pull records the few places where a pull is actually what happens. Elsewhere, a flow goes from where the pressure is high to where it is low, and that rule is correct.

The inference most people draw from it is not. Two balloons of different sizes on the ends of a tube with a clip in it: open the clip, and which way does the air go? The common answer is that the bigger balloon is stretched harder, so its pressure is higher, so air flows out of it until the two match in size. With soap bubbles the answer is famously the other way. A soap film’s tension does not grow as it stretches, so its Laplace pressure goes as one over the radius, and the small bubble empties into the large. That demonstration, and the whole of the capillary physics behind it, is a standard one, treated with Laplace’s law and its consequences elsewhere.

It is usually left there, with the implied lesson that the common answer was simply backwards. It was not backwards. For a rubber balloon, the common answer is right at some sizes and wrong at others, and the reason it can be both is the point of this essay. The question that decides any pair, of any material, is the sign of one slope.

A single slope #

Connect any two volumes by a tube. Air flows until their pressures are equal, and the volumes then rest. Equal pressures can be had at equal sizes, always, so an equal pair is always a stationary state. Whether it is a stable one is a matter of what happens to a small transfer. Move a little air from one to the other. The one that gained is now larger. If that has raised its pressure, the flow returns and the pair recovers. If it has lowered its pressure, more air follows, and the pair parts.

So everything is in the sign of $dp/dR$ on the branch the pair is on: positive, equalising; negative, parting. Soap has one sign everywhere, because its pressure is $4\sigma/R$. A balloon has both.

With a thin tube of radius $a$ and length $\ell$ the flow is Poiseuille’s, so the volumes change in proportion to the pressure difference, with a time scale set by the tube, $\tau\sim\mu\ell R^3/(a^4\,\Delta p)$. In a soft latex balloon a tube of the usual size finishes in seconds. The tube decides the pace. It does not decide the direction, and nothing in this essay depends on it.

A curve with a hill in it #

A real balloon is stretched elastic material, whose tension does grow with stretch, so the elastic-bag intuition is partly correct. For a thin neo-Hookean membrane with a Mooney–Rivlin stiffening $\alpha$, stretched from an unstressed radius $R_0$ to a stretch $\lambda=R/R_0$, the pressure is

$$p = \frac{2Gt_0}{R_0}\left(\frac{1}{\lambda}-\frac{1}{\lambda^7}\right)\left(1+\alpha\lambda^2\right).$$

Three things are competing. The stress in the rubber rises with the stretch. The rubber thins, by volume conservation, as $\lambda^{-2}$. And the radius of curvature grows, which lowers the pressure a given tension can hold. At small stretch the first dominates. Beyond a stretch of about one and a half, the thinning and the curvature win. Far out, the long chains in the rubber run out of slack, and the stiffening term takes over.

That curve is what everybody learns with their lungs. A balloon is hard to start, and suddenly it gives. The hard part is the climb to the peak. The give is the fall beyond it, down which a balloon needs less pressure as it gets bigger. A fairground balloon-blower knows the shape without any of the algebra, and stretches the balloon by hand first, which softens the rubber and lowers the peak.

So a horizontal line drawn across that curve, at any pressure between the trough and the peak, cuts it at three stretches. Three different sizes hold one pressure. That single fact is the whole of the rest.

How much stiffening removes the hill #

The curve’s middle branch isn’t guaranteed. As the stiffening ratio rises, the peak moves out and the trough moves in, and for a ratio above about 0.21 the two have met. The curve then rises all the way. A pair of balloons made of a rubber that stiffens that quickly would never part, at any stretch, and the common answer would be right for the whole of its life.

So the falling branch that makes a party balloon interesting belongs to a soft rubber with a long way to go before its chains run out of slack, and not to rubber as such. A stiff elastic bag, a car’s inner tube at low stretch, and a fire hose are all on a rising branch where they work, and a pair of any of those shares air just as anyone would expect. Where peak and trough come together is a fold in the curve, the same structure as the fold a hanging drop falls at, reached there by changing the load rather than the material. In both cases a state stops being available, and in both the object then moves abruptly a long way to the next one that is.

It also means that a designer of an arterial stent or a pneumatic actuator, who wants a thin membrane that inflates evenly, is asking for the rubber’s stiffening ratio to lie above that value. And an aneurysm in a thin wall is the same falling branch: the bulge that forms in a weakened artery holds a lower pressure as it grows, by much the same geometric argument, which is part of why it does not stop.

Where a pair ends #

The three regimes come straight from the slope. Started on the first rise, below the peak at 1.42, a pair evens out, and the common answer is right. Started on the fall, it parts, as soap does. Started beyond the trough at 4.47, the stiffened rubber behaves like the bag everyone imagined, and the pair evens out again.

What is not soap-like is how the parting ends. A soap bubble has nothing to stop it, and the small one vanishes. A small balloon collapses down the falling branch and on to the first rise, where its pressure can match the larger one’s at a stretch a little above one. There it stops. A pair started at a stretch of two settles at 1.15 and 2.44, at equal pressure and very unequal size. The air has stopped flowing, since the pressures are equal, as the rule said they must be. The size was only ever an assumption.

Such an end state needs a stability of its own, which is not the same condition as before. Transfer a little air, and the change of pressure difference is $p'(a)+p'(b)$, with each slope taken with respect to its own volume. The large balloon, on the falling branch, has a negative slope; the small one, on the rising branch, has a large positive one, because it is so little inflated that its volume responds slowly to pressure. The sum is positive, and the state holds. It is a case of one balloon’s steep branch stabilising the other’s unstable one. That is what lets it sit on a part of the curve no balloon could occupy alone in a pair of equals. Merritt and Weinhaus worked all this out in 1978, and the plain Poiseuille flow in the tube needs no change for any of it.

The pair remembers how it was filled #

The three-crossing curve has a further consequence, and it is the one that is hardest to believe until it has been seen.

Fill a pair slowly from one common supply. It climbs the first rise together, and at the peak, one balloon, whichever is very slightly ahead, goes over the top. From then on, every addition goes into that one. The other sits on the first rise, hardly inflated, while its partner grows to more than five times its unstressed radius. At a mean stretch of 4.5 they stand at 5.65 and 1.12. This is everyone’s experience of blowing up two balloons joined by a tube.

But blow them up separately, each to a stretch of 5, and only then open the tube. Now both are past the trough, on the third branch. They stay equal. Let air out slowly and they remain nearly equal, at a mean stretch of 4.5, until the mean has fallen below the trough and one of them tips. So at the same total amount of air, in the same two balloons, connected by the same tube, the pair can be equal or wildly unequal, and only its history says which. There is no dissipation or friction in this calculation that could be blamed for it, either. Both states are stable, the air is at rest in both, and nothing in either of them has any information about how it got there. Such a thing is what a lost steady state looks like from the other side.

Real balloons have their own rate-dependent hysteresis as well, in the rubber, which only adds to this. It is not needed for any of it.

How a falling slope spreads through a network #

It is worth seeing where else this argument applies, because it is not about rubber.

Any parallel set of elements sharing a common supply, in which an element’s driving pressure falls as it carries more, won’t share evenly. Two parallel channels with boiling in them are the engineer’s classic form of it: a channel with less flow boils harder, resists more and loses more flow, and the pair parts until one is nearly dry. That is Ledinegg’s instability, and it has melted a good many heated tubes. The same sign in the characteristic of a surge-prone compressor gives it its own version, in time rather than in space. An extra pipe is a different failure of the same assumption, that a network will distribute itself in the way that looks reasonable from outside. What is minimised by a network is not what anybody is hoping for.

The pair also marks a boundary with the opposite kind of stability. A surface whose tension falls as it shrinks reverses the soap film’s sign. If $\sigma/\sigma_0=1+\beta(R^2/R_0^2-1)$, near a working radius, then for $p=2\sigma/R$ the slope has the sign of $2\beta-1$, and past $\beta=\frac12$ a surface behaves like a balloon on its first rise.

That is the mechanical content of a surfactant film in a set of connected air sacs, and its physiological story belongs elsewhere. In this context it makes one point: a stable exchange does not need a stiffer material, only one whose restoring force has the right slope at the working point. Rubber has it twice, and the middle of its curve doesn’t.

All of this at a thousandth of a kilopascal #

The sizes of the pressures are worth setting beside the ambient, since this subject has so badly misread absolute pressure. A party balloon’s excess pressure, at its peak, is a few kilopascals; on the falling branch it is a little over half that. The atmosphere round it is a hundred. The difference in pressure between the pair, while it is parting, is a few tens of pascals, a few parts in ten thousand of the ambient.

No gauge in a school laboratory would register that without some care. Nor would the absolute pressure inside either balloon suggest anything. That’s the opposite warning from the one the earlier essays on this subject make. They said that a suction measured from the ambient must not be read as a force. This one says that a tiny pressure difference, negligible as a fraction of the absolute pressure, is the only thing the air is responding to. The air’s motion is decided by the gradient, and it doesn’t take a large difference to move air through a tube of low enough resistance. That’s why a pump with no engine can be driven by so little.

What the picture cannot show #

Rubber’s memory. Natural rubber has hysteresis, and a balloon that has been inflated once is softer the second time round, the Mullins effect. Its peak is lower after the first inflation. So a real balloon’s curve shifts over its first few cycles. The sequence of three branches is what survives. Two balloons from one packet differ in thickness by more than the hundredth of a stretch used to start the pair here, so the choice of which one wins is not an accident in practice: it is the thinner one, nearly always.

The shape of a balloon. The membrane here is a sphere that stays a sphere. A real balloon past the peak does not necessarily inflate evenly; a long one has a visible bulge travelling along it, with a hard unstretched part next to a soft stretched one, in what is itself a phase separation within a single balloon, governed by the same falling branch. A pair of connected spheres is therefore also a model of what happens inside one tube of rubber.

The tube. The time scale is in units of the tube’s, and the flow’s inertia, which would matter in a wide tube during a rapid transfer, is left out. It has no effect on where the pair ends.

A temperature. Air’s own compression and expansion are ignored, on the ground that the pressures involved are a percentage or two of the ambient. Rubber heats as it stretches, and cools on release, which is a measurable effect that a bare hand can feel on a stretched band, and which changes the stiffness a little.

The convention #

Rubber’s pressure is in units of $2Gt_0/R_0$; the stretch is $\lambda=R/R_0$. The Mooney–Rivlin ratio $\alpha$ of 0.05 is chosen as typical of a natural latex balloon. The first figure’s two bodies have pressures of one over the radius, in units of the tube’s relaxation. End states of the pair are those after a long time at a rate six times the first figure’s; the filling cycle keeps an imperfection of a part in a thousand alive at each step, standing in for the differences between real balloons. “Pressure” is the excess over ambient throughout. Air is treated as incompressible, since the excess is a few kilopascals against a hundred.

How it was checked #

The closed form for rubber’s peak comes from setting the derivative to zero, so the numerical search agreeing with it is a test of the arithmetic, and the march’s closure on to the cube-root final radius is a test of its conservation. The band of the pair’s parting was confirmed against the rubber’s own peak and trough: the pair evens out below 1.42 and above 4.47, which are the stretches at which the curve’s slope changes sign. The model refuses a radius of nothing, a tube of no conductance, and a stretch of less than one.

Who found it, and when #

Mooney’s law for rubber is from 1940, and Rivlin’s from 1948. Merritt and Weinhaus’s two-balloon paper is from 1978, and Müller and Strehlow’s account of rubber and balloons, with its treatment of the bulge and of many balloons in a row, from 2004. Ledinegg’s flow instability is from 1938.

Still open: how many balloons share evenly #

Two balloons split at a stretch of two. The next calculation would connect many, say a hundred, each with a slightly different unstressed radius, to one common pipe, and ask how they fill as air is slowly added. It would be expected that one inflates first, runs down its falling branch and takes most of the flow, then another. It would ask whether a filling sequence with this ordering has a universal form, and whether the sequence of pressure drops it shows, each one a balloon going over the peak, has a distribution of the kind found in a material failing one unit at a time.