The size a drop is allowed

Surface tension is the odd force in this subject. Everything else scales with a volume or an area — weight with the first, pressure and drag with the second — and tension scales with a length. So its importance against anything else is decided by nothing but size, and the crossover is not a matter of judgement but a computable length:

$$\ell = \sqrt{\frac{\sigma}{\rho g}},$$

which is 2.7 mm for water against air, 1.9 mm for mercury, and about 1.7 mm for most organic liquids. Below it a body of liquid is held by its skin; above it, by its weight. Almost everything counter-intuitive about small-scale flow is that sentence, and the dimensionless form of it is the Bond number $\mathrm{Bo} = \rho g R^2/\sigma$, which is $(R/\ell)^2$ and nothing else.

The number is one where weight and tension are equal. Which raises the question this collection keeps asking: is that where anything happens?

The equation the shape has to satisfy #

A static drop has no flow in it anywhere, which makes it the simplest object this collection draws and one of the least often computed. Its shape is fixed by two statements. Young and Laplace: the pressure jump across an interface is the surface tension times the sum of the two principal curvatures. And hydrostatics: the pressure inside varies with depth as $\rho g z$.

Put them together for an axisymmetric drop, measure $\bar z$ downward from the apex and let $\varphi$ be the angle the surface tangent makes below the horizontal, and in units of the capillary length:

$$\frac{d\varphi}{d\bar s} = \frac{2}{\bar b} + \bar z - \frac{\sin\varphi}{\bar x},\qquad \frac{d\bar x}{d\bar s} = \cos\varphi,\qquad \frac{d\bar z}{d\bar s} = \sin\varphi.$$

Every term earns its place. $2/\bar b$ is the pressure the apex’s own curvature supports, $\bar z$ is the head that depth adds, and $\sin\varphi/\bar x$ is the azimuthal curvature — the term that makes this a drop rather than a two-dimensional ridge. There is exactly one parameter, the apex radius of curvature $\bar b$, and integrating from the apex until $\varphi$ reaches the contact angle gives the whole shape.

The depth that has no size in it #

The largest drop in that figure is the interesting one, and its behaviour has a closed form.

Far from its edge a wide sheet of liquid is flat, so its azimuthal curvature vanishes and the equation collapses to $\sin\varphi\,d\varphi = \bar z\,d\bar z$, which integrates to $\cos\varphi = 1 - \bar z^2/2$ with no drop size in it anywhere. Setting $\varphi$ to the contact angle $\theta$ gives the depth:

$$h = 2\ell\sin\frac{\theta}{2}.$$

For water on a perfectly non-wetting surface that is 5.4 mm, and it cannot be exceeded by any volume of liquid whatever. Pour more on and it spreads.

That is a threshold of a completely different character from the aspect-ratio one, and it is worth naming the difference. It is a maximum rather than an onset: there is no tolerance in it, nothing approaches it gradually, and it is exact. Deep puddles of a non-wetting liquid do not exist, and the depth of the ones that do is set by a property of the liquid rather than by how much was poured.

What the aspect-ratio thresholds mean in millimetres #

The Bond number here is formed on the radius of the sphere of the same volume, which is the only choice that lets drops of different shapes be compared. Converting the thresholds:

A two-millimetre water drop resting on a waxed surface is visibly a bun, and its Bond number is 0.13. The eye is a good instrument here: it notices about ten per cent, which is why the folklore threshold feels approximately right to anybody who has looked at drops — and why it is a hundred times out for anybody who needs a per cent.

That is the general shape of the failure: the rule of thumb is calibrated to the eye, and the eye is calibrated to ten per cent.

Why the departure is quadratic and starts so early #

The aspect ratio falls as $1 - \mathrm{const}\times\mathrm{Bo}$ for small Bond number, and the constant is not small. That is what puts the one-per-cent threshold so far down, and the reason is worth stating because it applies to the whole family.

A sphere’s Laplace pressure is $2\sigma/R$. Gravity’s contribution across the drop is $\rho g R$, which relative to the Laplace pressure is $\rho g R^2/2\sigma = \mathrm{Bo}/2$. So the pressure perturbation is half the Bond number. But the shape responds to a pressure perturbation through the curvature operator, and for the lowest deformable mode of a sphere that operator contributes a factor of four — so the leading distortion is of order $\mathrm{Bo}/8$ rather than $\mathrm{Bo}$.

An eighth is not far from one, and yet the one-per-cent threshold sits at 0.0079 rather than 0.08. The remaining order of magnitude is the tolerance: a per cent of aspect ratio needs a per cent of distortion, and $\varepsilon/8$ at $\varepsilon = 0.01$ is 0.00125 — which is the right order and under the computed 0.0079 by a factor of six, because the aspect ratio is a ratio of two radii that move in opposite directions and the sensitivity is not simply the distortion.

None of that arithmetic is in the Bond number. It is in the solution, and it has to be computed.

The other number that flattens a drop #

Nothing above has any air moving in it. A drop can also be flattened by aerodynamic pressure, and the group that decides is the Weber number rather than the Bond number.

The two are usually confused, and for falling drops the confusion matters: a drop at terminal speed has its weight carried by the air rather than by the fluid below it, so its internal hydrostatic gradient is nearly absent and its shape is set almost entirely by the pressure distribution of the flow going round it. A falling drop is a bun and not a tear computes that case, and the crossover between the two mechanisms is where $\mathrm{We}$ and $\mathrm{Bo}$ are comparable — which for a drop at terminal speed happens at about two millimetres.

Below that size a drop’s shape is a Bond-number question and its answer is nearly a ball. Above it, the Weber number takes over and the answer stops being nearly anything.

What the shape is worth measuring for #

The reason drop shapes were tabulated by hand for eighty years is that the argument runs backwards as well as forwards. The shape is determined by the Bond number; so the Bond number can be read off the shape; so the surface tension can be measured by photographing a drop, if the density is known.

That is still the standard method. A pendant drop hanging from a needle is imaged, its meridian is fitted to the same equation integrated here, and $\sigma$ falls out. It works because the shape is sensitive to the Bond number over exactly the range where the drop is neither a ball nor a puddle — which is to say between the one-per-cent threshold and the point where it detaches, roughly two decades of Bond number.

Below that range the drop is a sphere to within the measurement error and the fit has nothing to grip; above it the drop falls off. The instrument’s working range is the region between the threshold this essay computes and the physical limit of the configuration, which is a nice example of a measurement whose precision is set by how far a dimensionless group is from its own onset.

The size a drop leaves at, which is the one anybody meets #

There is a third threshold in this problem and it is the one that decides how big the drops in ordinary life are. A pendant drop hanging from a tube grows until it cannot be held, and what holds it is the surface tension acting round the wetted perimeter. Equate that to the weight and

$$W_{\max} = \pi d\,\sigma,$$

which is Tate’s law, from 1864. It contains the tube’s diameter, the liquid’s tension and nothing else — no viscosity, no flow rate, and no reference to how the drop was formed.

Divide by $\rho g$ to get the volume, and the capillary length reappears immediately:

$$V = \psi\,\pi d\,\ell^2 ,$$

with $\psi$ a correction of about 0.6 for the liquid that stays behind on the tube when the neck pinches. The volume of a falling drop is the tube’s diameter times the square of a length that belongs to the liquid. For water from a five-millimetre nozzle that is 69 cubic millimetres, an equivalent sphere five millimetres across — which is what a dripping tap actually produces, and why every dripping tap in the world produces very nearly the same size of drop.

That is worth stating as the general answer to a question this essay has been circling. Drops in nature are a few millimetres across, in every liquid and from every source, because every mechanism that makes a drop is a competition between tension and weight and therefore sets a size of order $\ell$. Rain, dew running off a leaf, condensate falling from a pipe, mercury off a spatula — all of them a few capillary lengths, and the variation between liquids is the square root of $\sigma/\rho g$ rather than anything about how the drop was made.

Run the law backwards and it is an instrument, and a much older one than the photographed meridian. Count the drops that fall from a calibrated tube as a known volume drains through it; the volume per drop gives the tension directly. That is the stalagmometer, it needs no optics and no computer, and it was the standard bench method for a century. Its weakness is exactly the $\psi$ above: the fraction left behind depends on the ratio of the tube’s diameter to the capillary length, so the correction is a tabulated function rather than a constant, and Harkins and Brown spent a good deal of 1919 measuring it.

The failure mode of that instrument is instructive too. Push the liquid through faster and the drop detaches early, because the momentum of the incoming stream helps to break the neck; the measured tension comes out low, and the error grows with the flow rate. A stalagmometer therefore has to be run slowly enough that the drop is quasi-static at every instant — which is the same condition, in a different currency, as everything else in this essay: the result belongs to a static balance, and what invalidates it is anything that gives the liquid somewhere to put its inertia.

The one thing the number does say correctly #

It is worth being clear about what the Bond number gets right, because it is not nothing.

Scaling. Every threshold in this essay is a Bond number, and a Bond number is a statement about $\rho g R^2/\sigma$ as a whole. Change the liquid and the numbers do not move; the sizes move, exactly as the group requires. Mercury’s capillary length is 1.9 mm against water’s 2.7, so mercury’s one-per-cent drop is 0.34 mm rather than 0.49, and the ratio is $\sqrt{\sigma/\rho g}$ for each.

That is the whole content of a dimensionless group and it is a great deal. What it is not is a threshold, and conflating the two is the mistake: the group says how the threshold moves; only a solution says where it is.

The same distinction rescues the folklore from being simply wrong. “Drops smaller than the capillary length are spherical” is a correct statement about scaling with a badly chosen constant in it. The constant is $1/\sqrt{126}$ for a per cent, $1/\sqrt{18}$ for five per cent and $1/\sqrt{7.5}$ for ten — and the sentence is usually said with the constant equal to one.

The same length, doing other jobs #

The capillary length is not only about drops. It is the length at which surface tension stops being the dominant restoring force for waves, which is why the shortest gravity waves and the longest capillary waves meet at a wavelength of about 1.7 cm on water, and why that wavelength is the slowest wave the surface can carry.

It is the length that decides whether a jet breaks into drops or falls apart in some other way, and it is the length below which a liquid can be held in an inverted container by nothing but the tension across the opening.

What the picture cannot show #

Every drop here is static. No flow, no evaporation, no internal circulation, and no contact-angle hysteresis — a real drop on a real surface has an advancing angle and a receding one that differ by tens of degrees, so “the contact angle” is not a number a surface has.

The integrator has a noise floor. The meridian is walked with a fourth-order scheme and a drop of apex radius 0.005ℓ comes back with an aspect ratio of 1.002 when it is a sphere to eight figures. Any threshold below about half a per cent in aspect ratio is being read out of the arithmetic rather than out of the physics, and the solver refuses to return one.

The Bond number is formed on the equivalent sphere’s radius and not on the drop’s own height or width, both of which change as it flattens. Every threshold in this essay would move by tens of per cent under a different choice — which is the same ambiguity that attends every dimensionless group in the collection, and is why the figures print which length they used.

And the puddle result is two-dimensional. The closed form is for a ridge, where there is one curvature; an axisymmetric drop approaches the same depth from above and does so slowly, being still five per cent deep at six capillary lengths across. Real puddles are neither, and the number to quote for one is the ridge’s.

Who found it, and when #

Young and Laplace both arrived at the pressure–curvature relation in 1805, independently and with some acrimony. Bashforth and Adams tabulated the drop shapes in 1883, by hand, over several years — the tables ran to hundreds of pages and were the standard way of measuring surface tension for the next eighty years, since the shape of a pendant drop determines $\sigma$ if the density is known.

The name “Bond number” is twentieth-century and belongs to Wilfrid Bond, whose 1928 paper was about bubbles rising in viscous liquids and used the group in passing. That is a common pattern: the group gets named for the person who happened to use it in a paper somebody cited, and the threshold it carries gets attached to the name rather than to any calculation.

The surprising connection is with the puddle. That a maximum depth exists at all is not obvious, and that it contains no reference to the amount of liquid is stranger still. It is the same structural result as the maximum a shock can turn a flow or the wind an anticyclone cannot have: a governing equation that simply has no solution past a certain value, so the question stops being how much and becomes whether at all.

Where the ladder goes next #

Above this rung sit the shapes with flow in them — a drop being sheared, a drop on an inclined plane about to run, a drop being pulled off a surface — none of which is static and all of which need the contact line to be modelled rather than prescribed. The contact line is where continuum fluid mechanics is least comfortable, since a moving contact line with a no-slip wall has infinite dissipation and needs a slip length to be finite at all.

Beside it sits the falling drop, whose flattening has no gravity in it, and the jet, where the same tension is fighting inertia. Below it is the theorem that counts the groups, which produces the Bond number as an output rather than introducing it as a name.