Fluids

The vessel whose level falls evenly

Water leaves a hole in the floor of a vessel faster when the vessel is full, because the pressure at the hole is the depth of water above it. In a cylinder that makes the level fall quickly at first and slowly at the end, so hour marks on its wall crowd towards the bottom. A level that falls at one rate needs a vessel that widens as the outflow quickens: its radius must grow as the fourth root of the height. The Egyptians who made the oldest surviving water clock, about 1400 BC, gave it sloping sides, which halves the error of a cylinder. They also gave it a scale for every month, because the hours they wanted to count were not all the same length.

Assumes: The pressure that only knows depth · The bottle that pours at one rate however full it is

The bottle that pours at one rate however full it is made water leave a vessel at a steady rate by moving the point at which the pressure is fixed: a sealed bottle with a tube through its stopper holds its outlet at the head of the tube’s tip whatever the level, so the flow does not slow as the bottle empties. That is a way to make the flow uniform. A water clock wants something slightly different, and older: a level that falls uniformly, so that a scale on the vessel’s wall reads the time. The two are not the same, because a level falling uniformly is a statement about the vessel’s shape as well as about its outflow, and the oldest solution to the problem is a statement about shape alone.

Three vessels emptying through the same hole. The water level against time in three vessels that each hold 25 litres to a depth of 35 cm and drain through the same 0.8 mm hole in the floor, the outflow following Torricelli's law. The cylinder empties in 16.0 hours, but its level falls fast at first and slowly at the end — a parabola in time — and it is half empty in depth after only 32% of the run. The bowl whose radius grows as the fourth root of the height lowers its level at one rate from top to bottom, a straight line, and empties in 12.9 hours. A straight-sided cone sloping outwards, the shape a potter can actually throw, comes between: half its depth goes in 49% of its run.
Fig. 1 The water level against time in three vessels each holding 25 litres to a depth of 35 cm and draining through the same 0.8 mm hole. The cylinder empties in 16.0 hours, its level falling as a parabola, half its depth gone in 32% of the run. The quartic bowl, its radius growing as the fourth root of the height, lowers its level in a straight line and empties in 12.9 hours. The sloping-sided cone comes between, half its depth gone in 49% of its run.

Why a cylinder will not do

The pressure that only knows depth set out the rule: the pressure at the floor of a vessel of still water is the weight of the column above it, per unit area, whatever the vessel’s shape. A hole in the floor converts that pressure into speed. Water leaves at about 2gh\sqrt{2gh}, the speed it would have reached falling freely through the depth hh — Evangelista Torricelli’s result of 1643 — multiplied by a coefficient of about 0.61 that accounts for the jet narrowing just after it leaves a sharp-edged hole. The outflow is therefore proportional to the square root of the depth. Halve the depth and the flow falls by about thirty per cent.

In a cylinder the surface area is the same at every height, so the level falls at a rate proportional to the outflow, which is to say to the square root of the depth. That rate falls as the vessel empties, and the level traces a parabola in time: fast at the start, slow at the end. The first figure has the numbers. A cylinder holding twenty-five litres to a depth of thirty-five centimetres, draining through a hole eight-tenths of a millimetre across, takes sixteen hours to empty, and its level has fallen halfway in under a third of that. A cylinder makes a clock whose hours are long at the bottom of the vessel and short at the top.

That clock can still be read, if its marks are placed where the level actually is at each hour rather than spaced evenly. The second figure draws them. On the cylinder they crowd towards the floor, the first hour’s fall six times the second-last’s, and the last few hours can hardly be told apart. A clock-maker who wanted the marks evenly spaced, so that one ruler served, had to change the vessel.

The quartic bowl

Where the hour marks fall. The three vessels in section, each holding 25 litres to 35 cm, with a mark at the level the water reaches after each hour of draining through a 0.8 mm hole (every third mark heavier). On the cylinder the marks crowd towards the bottom: the first hour lowers the level 6.0 times as far as the second-last. On the cone they crowd less (1.2 times). On the quartic bowl they are evenly spaced, so a single straight ruler reads the time. The price is a vessel that is wide at the top and narrow at the bottom in a particular way: its radius grows as the fourth root of the height, so its side is nearly vertical near the rim and flares hard near the floor.
Fig. 2 The three vessels in section, each holding 25 litres to 35 cm, with a mark at the level the water reaches after each hour (every third heavier). On the cylinder the first hour’s fall is 6.0 times the second-last’s; on the cone, 1.2 times; on the quartic bowl the marks are evenly spaced. The bowl’s side is nearly vertical near the rim and flares hard near the floor.

The level falls at a rate equal to the outflow divided by the area of the water’s surface. To keep that ratio constant while the outflow grows as h\sqrt h, the area must grow as h\sqrt h too, and since the area of a circle grows as the square of its radius, the radius must grow as the fourth root of the height:

dhdt=−Cd a2ghπr(h)2=constant⟹r∝h1/4.\frac{dh}{dt} = -\frac{C_d\,a\sqrt{2gh}}{\pi r(h)^2} = \text{constant} \quad\Longrightarrow\quad r \propto h^{1/4}.

The vessel this describes is a bowl whose wall is a quartic curve, h∝r4h \propto r^4. Near the floor it is very narrow and flares quickly; near the rim its wall is almost vertical, because the fourth root changes slowly at large heights. In the second figure its hour marks are evenly spaced from top to bottom, and in the first figure its level falls in a straight line. The calculation behind the figures integrates the level from the outflow law step by step, never from the formula; the bowl’s level falls at one rate to within a per cent until the last centimetre, where the hole’s finite size begins to matter.

The same problem in another form is one that the curve that does not ask where it started solved for a pendulum: find a shape that makes a time independent of something it would otherwise depend on. Christiaan Huygens found the cycloid, along which a sliding bead reaches the bottom in the same time from any starting height. The quartic bowl is the water clock’s cycloid. It makes the level’s rate independent of the depth, by the same strategy of letting the shape do the compensating that a mechanism would otherwise have to do.

One rule for every outflow law

The quartic bowl is one case of a rule that covers every outlet. Suppose the outflow grows as some power of the depth, Q∝hpQ \propto h^p. A level that falls evenly needs a surface area proportional to the outflow at every depth, so the area must grow as hph^p and the radius as hp/2h^{p/2}. Torricelli’s hole has p=12p = \tfrac12 and wants a radius growing as the fourth root of the height. A long, fine tube, in which viscosity sets the flow, has p=1p = 1 and wants a radius growing as the square root. And an outlet whose flow does not depend on the depth at all, p=0p = 0, wants a radius that does not change: a plain cylinder.

That last case is not hypothetical. The hourglass that keeps time found that sand leaves an opening at a rate independent of how much sand is above it, because the walls of the glass carry the weight and the grains at the opening feel almost none of it. For sand the outflow exponent is zero, and so an hourglass needs no shaping at all: any bulb, cylindrical or pear-shaped, empties its sand at a steady rate, and the glass-blower can give it whatever form is pleasing. A water clock of the same shape would be useless. The difference between the two oldest kinds of timer is the exponent in their outflow laws, and the shape a vessel must have is that exponent written in glass or stone.

The rule also says why the even-falling bowl empties sooner than a cylinder of the same volume and height. The bowl keeps most of its water near the top, where the head is greatest and the flow fastest, and has little near the floor, where a cylinder spends its slow last hours. In the first figure the bowl is empty in 12.9 hours against the cylinder’s 16.0. An even scale and a short run come together here, from the same choice of shape.

What the Egyptian clock-makers did

The oldest surviving water clock was found at Karnak and carries the name of Amenhotep III, about 1400 BC. It is an outflow clock of alabaster: water ran out through a small hole near its base, and the time was read from the level against marks on its inner wall. Its sides are not vertical. The vessel is a truncated cone, wider at the top, with walls sloping outward at an angle of the order of seventy degrees from the horizontal, and the slope is there to make the level fall more evenly than it would in a cylinder.

How fast the level falls, at each depth. The rate at which the water level falls, in millimetres an hour, against the depth of water above the hole, for the three vessels. Torricelli's outflow grows as the square root of the depth, so in a cylinder the level falls 40 mm/h when full and 12 mm/h with 3 cm left. The quartic bowl offsets the faster outflow with a wider surface in exact proportion — area growing as the square root of the depth — and its level falls at 27.0 mm/h at every depth. The cone, matched to the bowl at the rim and near the floor, falls at 24.0 mm/h at the top, fastest, 30.3, at about a third of the depth, and slower again near the floor.
Fig. 3 How fast the level falls at each depth. In the cylinder, 40 mm/h when full and 12 with 3 cm left. In the quartic bowl, 27.0 mm/h at every depth. A cone matched to the bowl at the rim and near the floor falls at 24.0 mm/h at the top and 30.3 at about a third of the depth, fastest there and slower again near the floor.

A cone is the shape that can be turned on a wheel or cut with straight tools, and it is a fair compromise. The third figure compares the rate the level falls at each depth. In a cylinder it changes by a factor of three or more over the vessel’s height. In a cone matched to the quartic bowl at its rim and near its floor it varies by about a quarter, fastest about a third of the way down, where the cone is narrower than the bowl would be. The fourth figure turns those rates into clock errors.

The error of an evenly marked water clock. How far each vessel's clock is ahead of the true time, in minutes, if its wall is marked evenly from full to empty — as a simple ruler would mark it — and read by the level. The cylinder runs fast from the start, its level falling quickly, and is 202 minutes ahead at its worst, a little before the middle of the run, before the slow end brings it back. The sloping-sided cone is at most 41 minutes out; the quartic bowl, which an even scale suits, is within 4.7 minutes throughout.
Fig. 4 How far each clock is ahead of the true time if its wall is marked evenly from full to empty. The cylinder is 202 minutes fast at its worst, a little before the middle of the run; the sloping-sided cone at most 41 minutes; the quartic bowl within 4.7 minutes throughout.

With evenly spaced marks the cylinder runs more than three hours fast in the middle of a sixteen-hour run. The cone is never more than about forty minutes out, and the quartic bowl, which the even scale suits, keeps within five minutes, the residue coming from the last centimetre, where the outflow law is no longer exact. The sloping sides of the Karnak vessel buy most of the improvement a perfect shape would.

The Karnak clock was not marked with one even scale. Its inner wall carries twelve columns of marks, one for each month, because the Egyptian hour was a twelfth of the night, and nights are longer in winter than in summer. A clock that measured unequal hours needed a different scale every month, and the marks in each column are spaced to suit the vessel’s own draining as well as the season. The ancient clock-maker’s problem was therefore harder than the one the figures solve: not a uniform hour, but a scale that turned a falling level into hours that changed length through the year.

A clock that keeps the outflow steady instead

The other way to an even scale changes the outflow rather than the vessel. If water is fed into a receiving vessel from a supply held at a constant head, it arrives at a constant rate, and the level in a cylindrical receiver rises evenly. Ctesibius of Alexandria built clocks on this principle in the third century BC: a supply tank was kept brim-full by an overflow, so its head never changed, and the water it delivered raised a float in the receiver, which drove a pointer or a figure along a scale. The principle is the same one the bottle that pours at one rate used, arrived at two thousand years earlier: pin the head and the flow follows.

The two strategies divide the work differently. The outflow clock with a shaped vessel needs no moving parts and no supply, only a vessel cut to the right curve. The inflow clock needs a steady supply and an overflow, but its receiver can be any shape that is easy to make, and its float can turn gears. Medieval and Islamic clock-makers developed the second into astronomical clocks with automata; the shaped outflow vessel never needed more than a potter.

When viscosity drains the vessel

The shape depends on what drains it. The radius an evenly falling vessel needs at each height, as a fraction of its radius at the top, when the flow through the outlet pays both Torricelli's loss (growing as the square of the flow) and the viscous loss of the bore it passes through (growing in proportion to it). Through a 0.8 mm, 2 mm long hole, viscosity takes 5% of the head and the shape is the quartic bowl, r ∝ h^(1/4) (dotted). Through a tube 0.5 mm, 10 mm long it takes 44%, and through one 0.3 mm, 30 mm long, 96%: the flow approaches proportionality to the depth, and the even-falling shape approaches r ∝ h^(1/2) (dashed), a vessel that flares far more gently near the floor. The fine tube pays for it in time: it would take of the order of 13 days to pass 25 litres, against 9 hours for the open hole. A water clock of a useful size works in the first regime.
Fig. 5 The radius an evenly falling vessel needs at each height, as a fraction of its radius at the top, when the outlet pays both Torricelli’s loss and the viscous loss of its bore. Through a 0.8 mm hole viscosity takes 5% of the head and the shape is the quartic bowl (dotted); through a tube 0.5 mm wide and 10 mm long, 44%; through one 0.3 mm wide and 30 mm long, 96%, and the shape approaches r ∝ h^(1/2) (dashed). That tube would take about 13 days to pass 25 litres.

The quartic bowl rests on Torricelli’s law, which describes water accelerating through a hole whose walls play no part. A long, narrow outlet is different. If the water must pass along a fine tube, the friction of its walls dominates, the flow follows the law the fourth power in a pipe derived for viscous flow, and the flow is proportional to the pressure — to the depth itself — rather than to its square root. The even-falling vessel then needs its area proportional to the depth, so its radius grows as the square root of the height: a much gentler flare near the floor than the quartic bowl’s.

Most outlets lie between, paying a little of each loss, and the last figure computes the ideal shape for three of them. The shape depends on which loss dominates, and that is decided by the outlet’s Reynolds number, the ratio of the water’s inertia to its viscosity in the jet. A hole eight-tenths of a millimetre across in a thin floor runs at a Reynolds number of more than a thousand, viscosity takes five per cent of the head, and the quartic bowl is right. A tube a third of a millimetre wide and three centimetres long gives most of the head to viscosity, and the ideal vessel is close to the square-root shape. But such a tube passes water so slowly that the twenty-five-litre vessel would take nearly two weeks to drain. A clock that marks hours in a vessel of a useful size works in the inertial regime, where the quartic bowl is the shape, and that is what the Karnak potter’s cone approximates.

Reading the level

A clock is only as good as the reading of its scale, and a water level is harder to read than it looks. Water climbs a few millimetres up a clean wall, by the capillary rise that how high water will climb computed, so the level against a mark is a curved meniscus rather than a line, and in a narrow part of a vessel the curve can be a sizeable fraction of an hour’s fall. The quartic bowl is narrowest exactly where its marks are closest to the outlet, which is one practical reason a real clock is never run to the last centimetre. Ancient outflow clocks typically marked the hours by small drilled holes or notches on the inner wall, read from above as the falling water uncovered them, which turns the problem into telling whether a hole is wet — easier by lamplight than judging a line.

How far the argument reaches

The figures treat the discharge coefficient as fixed at 0.61, the value for a sharp-edged hole at moderate speeds. It changes as the head falls and the jet through the hole slows, by tens of per cent at the lowest heads, which bends the last part of every curve. They treat the water as pure and the hole as clean; a real clock’s hole silts and furs with lime, which slows it over months, and the ancient clocks are thought to have been recalibrated against the stars. Temperature changes water’s viscosity by about two per cent per degree, which matters only in the viscous regime; in the inertial regime a clock is nearly indifferent to it. And evaporation from the open surface, a few millimetres a day in a warm room, is negligible over one night’s run but not over a week’s.

What the figures cannot show

The figures show vessels as perfect surfaces of revolution, and a real clock’s interior is not. Its hole is not exactly round, its floor not exactly flat, and the small departures from the ideal shape near the outlet, where the vessel is narrowest, matter most. They cannot show the scales the ancient clocks actually carried, or how they were made. Whether the marks on the Karnak vessel were computed, copied from an earlier clock, or set by filling and timing against the stars is not known with certainty. Nor can the figures show how the clock was read: at night, by lamplight, by a priest looking down into a dark vessel at a dark waterline against carved notches, which is a source of error larger than any the figures compute.

Still open: how the ancient scales were made

Surviving outflow clocks from Egypt and from the Greek and Roman worlds differ in their slopes, and their scales differ in how well they match their vessels. Some appear to have been calibrated empirically, by filling them and marking the level against an independent time, and others seem to follow a rule. How much the clock-makers understood about why sloping sides helped, whether any of them knew a curved wall would do better, and how the twelve monthly scales of the Karnak clock were laid out, are questions argued from measurements of the few vessels that survive, of which most are fragments. Experimental reconstructions, filled with water and timed, have checked how well particular vessels keep time. They have not settled how the knowledge was arrived at.

The habit worth carrying away is to ask which of two quantities a design must hold fixed. Water leaves a hole in a vessel’s floor at a rate proportional to the square root of the depth, so a cylinder’s level falls fast at first — sixteen hours to empty, half its depth gone in a third of the time, an evenly marked scale 202 minutes fast — while a bowl whose radius grows as the fourth root of the height lowers its level at one rate, 27.0 mm/h at every depth; a straight-sided cone, the shape of the oldest surviving water clock, cuts the cylinder’s error to 41 minutes, and if viscosity rather than inertia governs the outlet the right shape becomes r ∝ h^(1/2). A clock can keep time by holding its flow steady or by shaping its vessel to the flow it has, and the oldest one did the second.

Part 11 of 12

This essay is one argument about Hydrostatics. The others:

The objects named here

The third axis, after the field and the reading path: the things themselves, and every essay that touches each one.

Discharge coefficientHydrostatic pressureIsochronismPoiseuilleReynolds numberTautochroneTorricellis lawWater clock