The drop falls at a fold

The size a drop is allowed reaches the drops everyone meets — the ones that fall from a tap — through Tate’s law of 1864. A drop hangs from a tube of radius $R$ by the tension acting round the wetted rim, and it falls when its weight reaches that tension:

$$\rho g V = 2\pi R\,\sigma .$$

That essay notes that a correction factor of about 0.6 is needed in practice, for the liquid left behind on the tube, and that the factor had to be tabulated because it depends on the tube. This essay asks the prior question, which the correction factor conceals: is the static part of Tate’s law right at all? Is the largest drop a tube can hold, before anything dynamic happens, the one whose weight equals $2\pi R\sigma$?

It is not, and the reason is not a correction to the balance. It is that no balance decides it.

What a tube can hold #

A hanging drop at rest is a surface of revolution whose curvature is set by the pressure difference across it, and the pressure inside falls with height. Measured in capillary lengths $\ell = \sqrt{\sigma/\rho g}$, with $z$ running up from the apex of the drop and $\phi$ the angle the meridian makes with the horizontal, the shape obeys

$$\frac{dx}{ds} = \cos\phi, \qquad \frac{dz}{ds} = \sin\phi, \qquad \frac{d\phi}{ds} = \frac{2}{b} - z - \frac{\sin\phi}{x},$$

where $b$ is the radius of curvature at the apex. Start at the apex with any $b$, integrate up the meridian, and wherever it reaches $x = R$ the liquid below that plane is a drop the tube can hold.

Measured in capillary lengths the problem has one parameter, the tube’s radius — the Bond number is its square — so every result below is a curve against that one number and holds for any liquid. So a tube does not have a drop. It has a family of them, one for each place a meridian can meet its rim, and feeding liquid into the tube walks along that family. The awkward part is that neither of the obvious labels for a member is monotone along it. The apex radius falls to about $R$ at the hemisphere and then rises again as the drop bulges below the rim; the arc length from apex to rim turns back near the end on a fine tube. The family was followed here as a curve in the plane of those two quantities, stepping along its tangent and correcting back onto it — the standard way to follow a family that folds, and the reason the calculation does not jump between branches.

The family has a largest member #

The volume along the family rises from zero, passes through a maximum, and falls. The maximum is a fold: two equilibria — one on each side of it — approach each other and annihilate, and beyond it there is none. For a drop being fed from a syringe, which adds volume and lets the pressure do what it likes, that is the end. One more increment of liquid and there is no shape the drop can take, so it takes none: it necks and falls.

That is a different kind of statement from Tate’s. A force balance says where two quantities are equal. A fold says where a family of solutions runs out. The first can be right while the second is what happens; here the first is not even right.

The largest drop does not look the same on every tube, and the difference is the first hint of why a single balance cannot describe them all. On a tube a tenth of a capillary length in radius the largest drop is a ball 0.96 capillary lengths across hanging from a thread: its equator is nearly five times the tube’s radius and it narrows all the way up to the rim. On the tube in the first figure the bulb is 1.4 capillary lengths across, the neck closes to just under the tube’s radius a sixth of a capillary length below the rim, and the surface flares out again to meet it. On a tube one capillary length in radius or wider there is no neck at all: the largest drop is widest at the rim itself and hangs like a sagging membrane, and there is nothing for a thread to form from until it has started to fall.

The fold was located by following the family until the volume had fallen three per cent below its largest value, then bisecting along the family between the neighbours of the largest member. Halving the integration step moves the fold volume on a tube of half a capillary length by $2\times10^{-7}$ of itself.

The balance that holds for every drop #

The weight of each drop is carried at the rim, and writing that exactly shows what Tate’s law is. Cut the drop at the tube mouth. The surface meets the rim at an angle $\psi$ to the horizontal, so the tension round the rim pulls up with $2\pi R\sigma\sin\psi$. The liquid inside the mouth is at a pressure $\Delta p$ above the air, and that pressure acts on the mouth’s area, pushing the drop down with $\pi R^2\Delta p$. So, exactly,

$$\rho g V = 2\pi R\,\sigma\sin\psi \;-\; \pi R^2\,\Delta p_{\text{mouth}} .$$

In capillary units $\Delta p_{\text{mouth}}$ is $2/b - z_a$, the apex pressure less the hydrostatic drop to the rim. That identity was checked for every member of three families — 475 drops, from the flattest cap to past the fold — against the volume integrated along the meridian, and the two agree to $3.5\times10^{-7}$ of $2\pi R$.

Tate’s law is this identity with $\sin\psi = 1$ and the pressure term thrown away. Both omissions have a physical reading. The first assumes the surface leaves the rim vertically, as though the drop hung from a cylinder of its own. The second assumes the liquid at the mouth is at the pressure of the air. Neither is true at the fold, and which one matters depends on the tube.

On fine tubes the surface does leave the rim nearly vertically — $\psi$ is within two degrees of ninety for any tube below a fifth of a capillary length — so Tate’s tension term is right. What it misses is the pressure. The largest drop is several capillary lengths tall compared with its own apex radius, the pressure at its apex is high, and the liquid in the mouth is still above atmospheric: it pushes down on the drop with a force that is a fifth to a quarter of the rim’s tension. That is why fine tubes fall short of Tate.

On wide tubes the picture inverts. On any tube wider than about half a capillary length the largest drop is two to two and a half capillary lengths long, so on a wide tube the hydrostatic drop from apex to rim is larger than the apex pressure and the liquid in the mouth is below atmospheric — the tube is holding the drop partly by suction, a pressure below the air’s that nothing is pulling on, only a weight hanging under a surface. At the same time the surface leaves the rim at a shallow angle, so the tension term falls. The suction helps and the angle hurts, and above 1.83 capillary lengths the suction wins.

How far off Tate is #

The deepest point is not an exotic tube. For water, whose capillary length is 2.73 mm, a ratio of 0.758 at 0.4 capillary lengths is a tube 1.1 mm in radius — a dropper, a burette tip, a pipette. On it the largest static drop is a quarter smaller than the law that is supposed to describe it predicts, before any liquid has been left behind.

Fine tubes approach Tate, but slowly. At a fiftieth of a capillary length, a needle 55 micrometres in radius for water, the ratio is 0.939; the missing six per cent is still the pressure in the mouth, falling off roughly as the square root of the radius. Tate’s law is an exact limit of a real calculation, and it is a limit that is never quite reached by any tube anyone uses.

And the correction factor absorbs both errors without distinguishing them. Harkins and Brown’s measured factor — the volume that actually falls, divided by Tate’s $2\pi R\ell^2$ — runs from about 0.6 to a little over 0.7 across the tubes normally used, and it is usually explained, as the account of drop sizes explains it, by the liquid left behind on the tube when the neck pinches. On the millimetre tube the static part of it alone is 0.76. So the factor is a product of two quite different things, and the static half is not small; the next section finds that on a standard dropper it is most of the whole.

A drop that is a unit of measure #

A drop is a unit of dose in medicine, and pharmacopoeias standardise the dropper that makes it. The European Pharmacopoeia’s normal dropper has a tip three millimetres across and must deliver twenty drops of water to the gram at 20 °C — fifty cubic millimetres a drop. That delivered volume is a borrowed number, fixed by the standard, not computed here. Everything else about the same tip is computed.

Its radius, 1.5 mm, is 0.55 capillary lengths of water. The largest static drop the tip can hold is 53.5 mm³; Tate’s law says 70.1. The drop that falls, fifty, is 93.5 per cent of the largest static drop and 71 per cent of Tate’s.

So of the 29 per cent by which the delivered drop falls short of Tate’s law, about 24 points are statics — the pressure in the mouth, which no amount of care in the dripping can remove — and about 5 are the neck, the liquid genuinely left behind. The usual account has those proportions the wrong way round. It is worth saying what that does and does not show. It does not show that the neck is unimportant: its share depends on the tube, and on a wide tube where the static factor rises towards one the neck’s share grows. It does show that a correction factor explained entirely by what the neck leaves behind is mostly explained wrongly, and that the part it attributes to dynamics is largely a static term Tate’s balance dropped.

That also explains a practical oddity of the standard. A dropper is specified by its outer diameter, not its bore, because the liquid wets out to the outer rim and $R$ is the rim’s radius. Two droppers of the same bore and different wall thickness deliver different drops, by exactly the amount the fold volume changes with $R$, and the pharmacopoeia fixes the outside for that reason.

A syringe and a reservoir lose the drop at different sizes #

The fold is the limit for a drop fed at a controlled volume. A drop fed at a controlled pressure — from a reservoir with a fixed head of liquid behind it — has a different limit, and it arrives much sooner.

The logic is the same as for a soap bubble on a straw. The pressure a small cap needs rises as it bulges, because its curvature rises, until the cap is a hemisphere of the tube’s own radius. Past that the drop widens, its curvature falls, and so does the pressure it needs. A reservoir at fixed pressure that has pushed the drop to the pressure maximum is now supplying more than any larger static drop can resist, so the drop grows without further encouragement. Under volume control each increment is chosen by the syringe, and the drop can follow the falling branch all the way to the fold.

For the half-capillary-length tube the pressure maximum sits at a meridian angle of 82 degrees, near the hemisphere, and at 9.5 per cent of the fold volume. The same tube, the same liquid and the same static family give two limits a factor of ten apart, and which one applies is a property of the plumbing behind the tube rather than of the drop. A drop supplied through a long, fine capillary behaves like a volume-controlled one, because the resistance of the feed sets the flow; one supplied from a wide reservoir behaves like a pressure-controlled one, and grows past the hemisphere under its own momentum long before its tube’s fold.

The same shape in other places #

A family of equilibria that runs out at a fold is the most common way a static state ends, and it turns up in places that look nothing like a drop. A film of oil that heats itself has a family of steady temperatures that folds when the heating outruns the cooling, and past the fold the film has no steady state — which is thermal runaway. The liquid that holds itself up in a tube by tension has a largest negative pressure it can sustain for the same structural reason. In each case the question “which force wins?” has an answer, and it is the wrong question: the right one is where the solutions stop existing.

The same tension-round-a-perimeter balance, cut on a plane where the pressure is atmospheric, has no missing term at all, and the rise of a liquid in a tube is exact for that reason. What makes the pendant drop a clean example is that the family can be computed exactly and the fold read off it, and that the naive force balance — which is a correct identity — can be seen failing term by term. The identity is not wrong; it is being evaluated at a guessed state rather than at the fold.

What the picture cannot show #

Every drop in these figures is static, and the most important event — the fall — is not. The red drop in the first figure is the last static shape, not the shape that falls. In the second or two after the fold the drop accelerates downwards, its neck thins above the equator, and a thread forms and breaks, often leaving a smaller satellite drop behind. None of that is in a static family, and a picture of the fold-point drop is a picture of the moment before anything interesting happens.

The drops are also drawn as though the liquid wets exactly to the outer edge of the rim. A real tube may wet only its inner edge, or creep along the outside, and the effective $R$ in every formula above is then different. Tate knew this; so did Harkins and Brown, who specified tubes ground flat to a sharp edge for exactly that reason.

Where the model stops #

Pinned at the rim. The contact line is fixed at $x = R$ for every member of the family. On a clean, sharp rim that is what happens, because a contact line cannot climb round a sharp edge until its angle has changed by the edge’s own angle — the same pinning that decides whether a pour dribbles down a spout. On a rounded rim or a surface the liquid does not wet, the line can slide, and the family is a different one.

Axisymmetric. Every drop computed is a surface of revolution. On wide tubes a drop can lose its symmetry and drip from one side before the axisymmetric fold is reached, which is one reason the curve is not drawn beyond 2.2 capillary lengths.

Quasi-static. The filling is assumed slow enough that every intermediate drop is at rest, and the air round it still — a drop falling through air is flattened by a different number altogether, the one that sets a raindrop’s shape. A tap run fast enough to make a jet is not described at all — that is a different regime with its own number.

The convention the numbers depend on #

Volumes are in cubic capillary lengths and radii in capillary lengths, with $\ell = \sqrt{\sigma/\rho g}$; for water at 20 °C that is 2.73 mm, so a volume of one is 20.3 microlitres. Tate’s law in these units is $V = 2\pi R$. The angle $\psi$ is measured from the horizontal, so a surface leaving the rim straight down has $\psi = 90°$. The angle is the one at the rim only: a drop with a neck below the rim has a meridian whose angle passes ninety degrees at the equator, keeps rising through the neck, and comes back below ninety as the surface flares out to the rim — which is why the largest drop on a tube of half a capillary length meets its rim at 81 degrees while plainly having a neck.

Who found it, and when #

Thomas Tate stated the law in 1864. Lord Rayleigh noticed in 1899 that it could not be exact and proposed a correction; W. D. Harkins and F. E. Brown measured the correction factor systematically in 1919, and their table is still used for drop-weight measurements of surface tension. The shapes of pendant drops as solutions of Young–Laplace were tabulated by Bashforth and Adams in 1883 — before computers, by hand — and the stability of the family was worked out by Padday and Pitt in 1973, whose paper is the standard account of the fold. The distinction between volume-controlled and pressure-controlled stability is older than any of these for bubbles and was carried over to drops in the same work.

Still open: how much of the largest drop leaves #

The static calculation ends at the fold with a drop of known shape and volume. What leaves is decided by the next fraction of a second: the drop accelerates, the neck forms above the equator, and a thread pinches at a place and time set by the dynamics of the thinning neck. The fraction that goes — about 0.93 of the fold volume, on the evidence of one standard dropper — is the number this calculation cannot reach, and it is known here only at one tube.

The calculation that would follow starts the neck from the fold shape computed here and follows it with a one-dimensional thinning-thread model, the same slender-jet equations that describe where a jet stops being a jet, until it pinches. It would put a computed number where Harkins and Brown’s table has a measured one, and say how much of their factor is statics and how much is the neck. Beside it is the question the reservoir figure raises: what a drop fed at constant pressure actually does once it has passed the pressure maximum, and whether the size it falls at is set by the supply or by the tube.