A sliding drop is held harder the faster it goes

The force a contact line holds is a range found how large a ridge of liquid a tilted plate can hold. A contact line can sit at any angle between a receding value and an advancing one, and a ridge on a slope leans downhill until its front edge reaches the advancing angle and its back edge the receding one. At that moment its weight along the plate, per unit length, equals the surface tension times the difference of the two cosines, and it slides. In capillary units the ridge’s cross-section is a Bond number, and the threshold is a Bond number times the sine of the tilt equal to $\cos\theta_r - \cos\theta_a$.

That essay said nothing about what the ridge does next, and it named the question worth asking. A block on a ramp breaks free when static friction is overcome and then slides on a smaller kinetic friction, so it keeps going at tilts that would not have started it. Does a drop? It depends on how the contact angles respond to motion, and that response is set somewhere no photograph of a drop can resolve: in the viscous wedge at each contact line, where a no-slip wall would never let a drop spread at all.

Two edges, two corners, one law #

The ridge is the earlier essay’s: a two-dimensional cross-section of liquid, long across the slope, slow enough that its shape is hydrostatic and its inertia negligible. When it moves at a speed $U$, its downhill edge is an advancing contact line and its uphill edge a receding one, and at each the liquid is sheared in a wedge that narrows to the line. The viscous stress in that wedge bends the free surface, so the angle seen at the scale of the drop — the apparent angle — differs from the one at the solid. Cox’s law gives the difference:

$$g(\theta_{\text{app}}) = g(\theta_{\text{static}}) \pm \mathrm{Ca}\,\ln(L/\lambda),$$

plus at the advancing edge and minus at the receding one, with $\mathrm{Ca} = \mu U/\gamma$ the capillary number, $L$ the ridge’s size and $\lambda$ a molecular slip length that cuts off the viscous stress’s divergence at the line. The function $g$ rises from zero; at small angles it is $\theta^3/9$, and the law becomes Voinov’s cube law, $\theta^3 - \theta_s^3 = 9\,\mathrm{Ca}\ln(L/\lambda)$, which the calculation reproduces within two parts in ten thousand at three small angles. The slip length enters only through the logarithm — for a ridge two millimetres across and a slip length of a nanometre, it is 14.5 — which is why the result is insensitive to what happens at molecular scale.

With the angles given by the speed, the force balance that set the threshold now sets the speed:

$$A\sin\alpha = \cos\theta_r(U) - \cos\theta_a(U).$$

A check comes with it. The balance was derived for any ridge, whatever its shape; the ridge’s actual shape at speed can be shot from Young–Laplace along its surface, starting at the uphill edge at the dynamic receding angle and requiring it to land at the dynamic advancing one. Its area then matches the area the balance used to two parts in $10^8$ at three tilts and sizes.

The front steepens, the back flattens, and the back fails #

The first figure is the response on a plastic-like surface, advancing at 90° and receding at 70° at rest. As the speed rises the advancing angle climbs — 93° at a capillary number of 0.003, 99° at 0.01 — and the receding angle falls faster, to 64° and then 45°. The asymmetry is in $g$: near 70° the function is flatter than near 90°, so the same viscous bending moves the receding angle further.

The receding angle does not level off. It reaches zero at a capillary number of 0.0136, and there the uphill contact line cannot keep up with the ridge: a steeper back is impossible, a flatter one is a film. Beyond that speed no ridge slides with a clean trailing edge. In a real three-dimensional drop this is the speed at which the back of the drop draws out into a corner and then a tail that sheds a trail of smaller drops — the “pearling” of sliding drops — and the calculation places it with the same law that sets every speed below it.

Why the back fails and the front does not #

The two edges are not mirror images, and the asymmetry decides which one gives out. At the advancing edge the viscous bending steepens the angle towards 180°, and with the gas treated as inviscid it approaches that limit only as the speed grows without bound. Only the gas’s own viscosity, small as it is, lets the advancing line fail at a finite speed, by entraining a film of air beneath it: with air’s viscosity included, for water on a 90° surface, that happens at a capillary number of 0.093. The receding line, including the same air, fails at 0.0134 — nearly seven times sooner. The back of a drop always goes first.

That ordering is general in coating, and it runs the other way round from the teapot’s problem, where the danger is a liquid sheet that will not let go of an edge. When the receding line fails, liquid is left behind as a film, and the film’s thickness is then set by the same balance that fixes the film a plate drags out of a bath — a capillary number to the two-thirds power, times whatever length the meniscus has. The sliding drop’s trail and a dip-coated plate’s film are the same deposit, reached from opposite directions.

The resistance only rises #

The second figure is the answer to the question a block on a ramp poses. The resistance is $\cos\theta_r - \cos\theta_a$ at the dynamic angles, and every change the speed makes to the angles increases it: the advancing angle grows, its cosine falls; the receding angle shrinks, its cosine rises. So the resistance starts at exactly the static value — 0.342 of the surface tension on the plastic — and rises monotonically, to 1.217 at the speed at which the receding line fails. It rises at every one of two hundred speeds checked between rest and failure.

A contact line has no kinetic friction lower than its static one. A drop that has broken free is held at least as hard as it was held at rest, and harder the faster it goes. If the tilt is reduced after it starts sliding, it decelerates and stops at the tilt at which it started; there is no range of tilts in which a moving drop keeps moving and a resting one stays put. The block-on-a-ramp picture is wrong in the one feature that makes it a block.

What this law leaves out is what produces stick and slip on real surfaces: chemical and topographic defects that pin the line locally and let go suddenly, which is where measured drop motion becomes jerky. The smooth law describes the line between defects, and it says the jerks come from the surface, not from any velocity weakening of the line itself.

The speed past the threshold #

The third figure puts the law to work on a ridge of one square capillary length — for water, a cross-section of about 7.4 square millimetres. Below each surface’s threshold it is stuck. Past it the speed rises from zero, linearly at first: the slope of the resistance at zero speed has a closed form from Cox’s law, $\ln(L/\lambda)\,[\sin\theta_a/g'(\theta_a) + \sin\theta_r/g'(\theta_r)]$, which the computed resistance matches to four parts in a million, so the speed just past the threshold is the excess weight divided by that slope.

On the plastic the ridge starts at 20° and reaches a capillary number of 0.0072 at 45° and 0.0117 held vertical. On the repellent coating, whose cosines differ by about the same amount but whose angles are larger, it starts at 19.6°, just below the plastic’s 20°, and runs faster at every tilt — nearly twice as fast at 30°. On clean glass the curve barely exists. Its angles are small — 30° advancing, 10° receding — its receding line fails at a capillary number of only $4\times10^{-5}$, and between the threshold at 6.8° and the tilt at which the back fails there is a single degree. A ridge on clean glass either stays put or runs with its back drawn out into a film: which is exactly how a freshly cleaned window sheds rain, in thin fast-running films rather than beads.

The glass figure also says something about the static threshold the earlier essay measured on glass: the sliver between sticking and failing is so narrow that a drop on clean glass essentially never slides as a drop. What the tilting-plate method measures there is the tilt at which the ridge starts to leave a film, and the receding angle it infers is the one at which the line gives way — which is why receding angles on well-wetting surfaces are the hardest contact angles to measure reproducibly.

In physical units the capillary number is the speed over $\gamma/\mu$. For a silicone oil of a hundred centistokes that is 0.22 metres a second per unit, so the ridge on the plastic at 45° slides at 1.6 millimetres a second, where the model’s neglect of inertia is safe. For water $\gamma/\mu$ is 73 metres a second and the same capillary number is half a metre a second — fast enough that the drop’s inertia, not in this model, matters. The law of the angles still holds at the lines; the force balance needs an inertial term the ridge does not have.

The shape at speed #

The fourth figure draws the ridge. At rest on the point of sliding it stands at 90° at its front and 70° at its back, the shape the size a drop is allowed gives to a drop of one square capillary length: flattened by its weight but still more cap than puddle. Sliding at 30° it stands at 93° and 64°; at 60°, at 99° and 45°. The front steepens little, the back flattens a lot, and the ridge’s mass shifts forward as its tail lengthens. That is the shape a sliding drop is photographed with: blunt at the front and drawn out behind, and it is drawn out behind because the receding angle is the one the speed moves most.

Stuck, sliding, or leaving a trail #

The fifth figure maps the three outcomes for every ridge and tilt. The lower curve is the earlier essay’s threshold, ridge size equal to 0.342 over the sine of the tilt. The upper curve is the new one: ridge size equal to 1.217 over the sine of the tilt, where even the largest resistance the contact lines can offer — with the back’s angle at zero — is too little. Between them the ridge slides steadily. A ridge of one square capillary length on this surface slides at every tilt from 20° to vertical; one of two square capillary lengths starts at 10° and fails above 37°; one of five fails at every tilt past 14°.

The band between the curves is the whole working range of a sliding drop, and it is set by two numbers of the surface: the static hysteresis, which places the lower curve, and the receding angle alone, which places the upper one through the speed at which it can be driven to zero. A surface that sheds water cleanly needs a narrow hysteresis and a large receding angle; the two are different properties and are separately engineered.

What was checked #

The ledger holds four checks: Cox’s law against Voinov’s cube law at small angles; the force balance against the ridge’s own shape shot at its dynamic angles; the onset slope against its closed form; and the resistance’s rise at every step from rest to failure.

What the smooth ridge leaves out #

Three dimensions. A real drop’s contact line is curved, its angles vary round it, and its back fails first at its rear corner, forming the corner and then the pearls; the ridge is the case in which the line is straight and the failure is a film.

Defects. The static angles are properties of an ideal surface, the range a contact line holds at rest; real ones pin the line at defects and release it, which produces stick–slip motion the smooth law does not.

Inertia and the interior. The ridge’s own internal shear is folded into the logarithm’s outer length, and its inertia is neglected; for water above a few centimetres a second, the second fails before the first does.

The slip length. The logarithm makes the answer insensitive to the molecular cut-off — a factor of ten in $\lambda$ moves the speeds by about a seventh — but not independent of it.

Where the law comes from, and why it is logarithmic #

The logarithm in Cox’s law is the moving contact line’s whole story in one factor. A wedge of liquid of angle $\theta$ sheared by a wall moving under it carries a stress that falls as one over the distance from the line — the Huh–Scriven result — and the free surface bends under the pressure that goes with it, turning through an angle proportional to the integral of that stress over distance. Integrated from the molecular scale $\lambda$, where slip cuts it off, to the drop’s own size $L$, one over distance gives $\ln(L/\lambda)$. So the contact angle a drop shows is its molecular angle bent by a logarithm’s worth of viscous stress, and every speed in this essay is a capillary number divided by that logarithm. It is the same logarithm that makes slip a memory of one mean free path enough to rescue the wall’s condition: a cut-off anywhere near molecular size gives nearly the same answer, because it enters only as the log of a very large ratio — and the slip length itself, a length that can only be measured by squeezing a film thin enough to feel it, need not be known well.

That is also why the resistance must rise. The viscous bending always acts to oppose the motion that causes it — steepening the front the drop is pushing into, flattening the back it is pulling away from — and both changes increase the difference of cosines. A velocity-weakening friction would need the bending to reverse sign with speed, and nothing in a viscous wedge does that.

The convention: capillary number and square capillary lengths #

Speeds are capillary numbers, $\mathrm{Ca} = \mu U/\gamma$. Ridge sizes are cross-sectional areas in square capillary lengths, $\sqrt{\gamma/\rho g}$ squared — a Bond number. Angles are measured through the liquid. The resistance is per unit length of ridge, in units of the surface tension. The ridge’s size $L$ in the logarithm is two millimetres and the slip length one nanometre.

Who worked it out #

Huh and Scriven showed in 1971 that the moving contact line’s viscous stress diverges; Voinov in 1976 and Cox in 1986 derived the dynamic-angle law from the matched viscous wedge. Furmidge in 1962 gave the retention-force balance for a drop on a tilted plate. Podgorski, Flesselles and Limat in 2001 photographed sliding drops developing corners and shedding pearls at a critical speed, and Le Grand, Daerr and Limat in 2005 tied that speed to the receding angle’s approach to zero.

Still open: the corner a real drop grows #

A three-dimensional drop does not fail at the back all at once. Its rear edge sharpens into a corner whose opening angle closes as the drop speeds up, and the receding line in the corner moves obliquely, at a normal speed smaller than the drop’s. The next calculation gives the ridge’s back a corner of half-angle $\phi$, applies Cox’s law to the normal speed $U\sin\phi$, and asks at what speed the corner’s opening angle reaches the value at which a cone of liquid is drawn from its tip — whether that speed, for the plastic surface here, is the ridge’s 0.0136 multiplied by a factor that depends only on the angles, and so whether pearling can be predicted from a tilting-plate measurement of two contact angles.