Field

Fluids at work

Turbines, pipes, weirs, balls, sails, arteries and blades. What the conservation laws say about machines, which is more than a designer expects and less than a brochure claims.

The tube widens because the air slows. The streamtube through an actuator disc at an induction factor of 0.333. The three radii are not drawn to taste: each is fixed by requiring the same mass to pass every station, and the slowest station is therefore the widest. The tube widening in front of a wind turbine is why some of the wind goes round it rather than through it, and it is the whole reason a disc cannot take everything.

The most a disc can take

A wind turbine cannot extract more than sixteen twenty-sevenths of the energy passing through the circle its blades sweep. That is not a limit on turbines — it is a limit on anything at all, and it follows from three conservation laws and no engineering.

Efficiency is decided before the engine is chosen. Froude's propulsive efficiency against the ratio of jet speed to flight speed. It is 2/(1+σ) and nothing else — no engine, no fuel, no combustion. A turbojet with a jet at three times flight speed cannot exceed 50% however good its core is, and a propeller moving a great deal of air slowly is above 90% before anybody has designed anything.

A big slow push

The same thrust can be had from a lot of air moved a little or a little air moved a lot, and the two are not equivalent. One number decides which, it contains no engine, and it is why every airliner built since 1970 has a fan far larger than the machine driving it.

Betz's ceiling, and the rotor that cannot reach it. The power coefficient of Glauert's optimum rotor against tip-speed ratio, with Betz's 16/27 drawn as the ceiling it is. The gap is wake rotation: a rotor that extracts power applies a torque, a torque leaves the wake spinning, and that rotational energy never reaches the shaft. It falls as the rotor is geared up and is never zero — which is why large wind turbines turn so slowly and yet have such fast tips.

The wake that has to spin

A rotor that takes power out of the wind must apply a torque to it, and a torque applied to air is angular momentum left behind. The axial theory has nowhere to put that energy, so Betz's ceiling is unreachable at every finite tip-speed ratio — and the gap is computable.

The chart, with one exact line on it. The friction factor of a pipe against Reynolds number, for five relative roughnesses. Every curve here except one is Colebrook's correlation, solved by iteration rather than read off a chart. The exception is the short straight line at the left: f = 64/Re is the laminar solution and it is exact. The curves flatten to the right because once the roughness pokes out of the viscous layer the Reynolds number has nothing left to change.

The roughness a wall cannot feel

A rough pipe and a polished one carry the same flow for the same pressure over three decades of Reynolds number, and then suddenly they do not. What changed is not the pipe. It is the thickness of the film of fluid at the wall, which is the only part of the flow that can see the roughness at all.

The box, and the one thing assumed about it. The control volume across a sudden enlargement. Mass and momentum crossing the two ends are known exactly. The only modelling statement in the whole derivation is written on the annular step: the pressure there is taken to be the upstream pressure, because the fluid in the corner is nearly stationary. Measurement supports it well. Nothing else is assumed, and in particular nothing at all is assumed about the eddy that lives in that corner — which this figure therefore does not draw.

A loss with no viscosity in it

Where a pipe suddenly widens, energy is destroyed. The amount is exact, it has been known since 1766, and the derivation never mentions viscosity, Reynolds number or roughness — because momentum does not care where the energy went, only that it left.

The shock in a river

Shallow water is a gas whose ratio of specific heats is two. The white water below a weir is a shock wave, momentum is conserved across it exactly, energy is not, and one direction is forbidden for the same reason an expansion shock is forbidden — which makes the analogy exact to first order and wrong at the second.

The depth that costs least

For a given flow there are two depths that carry it at any energy above a floor, and exactly one at the floor. That one depth is where the Froude number is one, the least energy is exactly three halves of it, and a bump in the bed that asks for more than the flow has does not thin the water — it backs it up.

The cliff a rough ball reaches sooner. The drag coefficient of a sphere against Reynolds number, on log axes. The smooth curve is Morrison's correlation, which is a fit to measurements and is drawn in the colour this site reserves for a borrowed claim. The other is the same curve shifted along the Reynolds axis by a factor of 6 — a stated model of what roughness does, which is to trip the boundary layer early, and not a measurement of any real ball.

The drag that falls as it speeds up

There is a band of speeds in which a smooth ball experiences less drag the faster it goes. Not a smaller coefficient — a smaller force. Dimples move that band down to where a golf ball actually flies, and they do it by making the friction worse.

Two sides of one ball, at different pressures. The surface pressure coefficient round a ball, measured from the front stagnation point, on the side the seam trips and on the side it does not. Up to separation both follow the exact potential-flow distribution 1 − (9/4)sin²θ. After it both take the same wake pressure, which is what a manometer measures rather than what the ideal theory predicts. The asymmetry is the shaded area between them, and integrating it gives a side force of 0.2949 towards the later-separating side.

A ball that swings without spinning

A cricket ball curves in flight with no spin about any useful axis. The mechanism is not the Magnus effect; it is a seam tripping the boundary layer on one side so that side lets go later. Which way the ball then goes depends on one borrowed number, and this site's own inviscid solver supplies the value that gets it wrong.

Three times the wind, on a reach. The polar diagram: boat speed in every direction, as a multiple of the true wind speed, for drag angles of 14 and 6 degrees. The shaded wedge at the top is the no-go zone, whose half-angle is exactly the sum of the two drag angles. Everywhere outside about twice that angle the boat is faster than the wind, and the maximum is 2.92 times the wind at 110 degrees — which is 1/sin λ at 90° + λ, both checked.

Faster than the wind that drives it

An ice yacht in a fifteen-knot breeze does forty. That is not a trick and it does not need a special sail — it follows from two drag angles and a triangle, and the best speed a boat can reach is one over the sine of their sum.

The fastest way is not the straight one

A boat racing to a mark dead upwind sails seventy per cent further than the distance to it, and arrives first. The best angle is forty-five degrees plus half the apparent wind angle, it comes out of one line of trigonometry, and the same argument says to gybe downwind rather than run.

The radius that costs least

A vessel that carries a flow costs two things to own — the power to push fluid along it and the price of the tissue itself. Minimising the sum gives a best radius, the best radius makes flow proportional to radius cubed, and the rule that follows is a statement about a photograph that came out of a cost function.

The stretch of surface that is boiling. The pressure coefficient along both surfaces of a section at 4 degrees, computed from the same potential-flow solution as the other ideal-flow aerofoil figures. The horizontal line is the vapour pressure at a cavitation number of 1: wherever the suction curve is above it, the liquid there has been pulled below its vapour pressure and is boiling at whatever temperature it happens to be. This section cavitates at any σ below 1.428.

When a body tears the water

A propeller blade moving fast enough pulls the pressure at its own surface below the vapour pressure of the liquid, and the water boils at whatever temperature it happens to be. Where that happens is decided by an inviscid calculation of the pressure along the blade.

$Ninety microseconds, and most of it spent barely moving. The radius of a collapsing cavity against time, both as fractions of their own totals. The bubble spends most of the collapse near its original size and the last few per cent of the radius in the last fraction of a per cent of the time. The total is 91.47 microseconds for a millimetre cavity at one bar, computed by quadrature and agreeing with the closed form in gamma functions to a part in 10⁹.$

The bubble that hammers

A vapour cavity swept into higher pressure does not deflate. It collapses, in ninety microseconds for a millimetre bubble, and the model that describes the collapse predicts a wall speed that reaches the speed of sound in water at three per cent of the original radius — which is to say it predicts its own failure, and locates it.

Work out of a change of swirl

The work a rotor does per unit mass is the blade speed times the change in swirl, and that is all of it — no blade shape, no pressure, no efficiency, no gas properties. It is the same equation for a pump, a compressor, a turbine and a fan, and it follows from angular momentum on a box with nothing assumed about the inside.

The jet is 61.1% of the hole. Flow out of a slot in a plane wall, solved by Kirchhoff's free-streamline method. The outer curve is not a wall and not a guess: it is the streamline on which the pressure is ambient, and where it goes is part of the solution. It leaves the edge of the slot travelling straight down the wall and turns through ninety degrees, settling to a jet whose width is π/(π+2) = 0.6110 of the opening. Every streamline drawn is a level set of the streamfunction the conformal map supplies.

The hole that halves the flow

A jet leaving a sharp-edged hole is narrower than the hole, and by an amount that is not measured but computed. One geometry gives exactly one half from momentum alone; another gives exactly π/(π+2) from a conformal map in which the shape of the free surface is part of the answer.

The jet divides 75% to 25%. A jet striking a plate at 60 degrees. Both sheets leave at the jet's own speed, because their surfaces are at ambient pressure and Bernoulli allows nothing else, and the plate can exert no force along itself because the fluid has no viscosity. Momentum along the plate then fixes the split at (1 + cos β)/2 = 0.7500, and the normal force at ṁV sin β = 0.8660. Nothing about the plate's material, size or roughness enters either.

What a jet cannot push sideways

A jet striking a plate divides in two, and how it divides is fixed by a single sentence — an inviscid fluid exerts no force along a surface. That one statement, plus mass, gives the split exactly — and the same sentence turns a flat plate into a bucket worth twice as much.

Half the jet speed takes everything

A bucket standing still feels the largest force and does no work; a bucket running with the jet does no work either. Between them the power peaks at exactly half the jet speed, for every bucket shape and every flow rate — and at that speed a perfect bucket leaves the water motionless.

The same reading, and only one of them gives it back. A Venturi and an orifice plate at the same diameter ratio, with the pressure along the axis drawn beneath each. Both narrow the flow by the same amount, both read the same difference between the pipe and the narrowest section, and both infer the same flow rate from it. Downstream they part company: the Venturi's diffuser turns the throat's speed back into pressure, and the orifice's jet expands into the pipe and destroys 73% of the reading. The picture is a section rather than a solved field: nothing here computes the jet, and the recirculating corner is not drawn.

The price of knowing the flow rate

Two flowmeters can narrow a pipe by the same amount, read the same pressure difference and infer the same flow rate, and cost pressures that differ by an order of magnitude. What separates them is not viscosity, and not workmanship — it is whether the flow is decelerated or abandoned.

The tunnel makes the stream 7.4% faster. The flow past a cylinder between two walls, solved from the closed form of an infinite image row rather than from a truncated sum. The walls are streamlines exactly — that is what the images are for, and the normal velocity on them is zero to the last bit — and the flow beside the body is squeezed between the body and the wall, which is the whole of the blockage effect. The stream at the model is 7.40% faster than the speed the tunnel's own instruments report far upstream.

The instrument in the answer

A model in a wind tunnel is not a model in the sky. The walls are supplied by an infinite row of reflections, the stream at the model is faster than the tunnel's own instruments report, and the correction is not an empirical fudge — it is a series with a closed form.

12 bar from stopping one metre per second. The head at the valve after it shuts, computed by the method of characteristics on a 600 m pipe. The rise is 122.4 m of water, which is ρaΔV/ρg to 1.4e-14 m — and the scheme was told neither ρaΔV nor anything else about the answer. The wave then runs to the reservoir and back every 2.000 s, and with no friction in the model it never decays: a real pipe damps this out in a few tens of cycles.

Stopping water costs more than moving it

Shut a valve on water running at one metre per second and the pressure that appears is twelve bar — not because the water was pushing hard, but because the only way to stop a column of fluid is to send a message back along it, and the message travels at the speed of sound in the pipe.

$Nothing can deliver more than h/H, and here that is 10.0%. The fraction of the supply a ram can deliver, against the height it is asked to deliver to, from a supply falling 2 m. The upper curve is the exact ceiling h/H, which follows from the energy audit with every loss set to zero and can be reached by no real machine; the lower one is what a ram at 65% efficiency actually sends. Asking for twice the height halves the delivery, exactly, and there is no design that escapes it.$

A pump with no engine

A hydraulic ram lifts water uphill using nothing but the water that is already falling. It has one moving part and no power supply, and everything it can and cannot do follows from an energy audit that fits on one line — including a ceiling nothing about its design can move.

A pump curve out of the momentum theorem. The pressure an ejector delivers, against how much it is entraining, at a fixed nozzle. It has the shape of every pump characteristic ever measured — a shut-off pressure with no flow, falling to no pressure at free delivery — and it was obtained from a momentum balance on a tube with nothing in it. The shut-off value here is 36.0 kPa and the machine at its best power runs at 4.56 times its own motive flow.

Mixing is a pump

Two streams at different speeds mixing in a tube destroy energy — the same Borda–Carnot expression a handbook prints beside a sudden enlargement, with two streams in it instead of one. And while they destroy it the pressure rises, which makes the loss the mechanism of a machine with no moving parts.

1.139 asks for a Francis. The specific-speed axis, with the four machine types on it and one duty marked: 3 m³/s at 60 m, on a shaft turning at 750 rev/min. The number is 1.1387, and the choice of runner follows from it before any blade has been drawn. What the number contains is a ratio of flow to head; what it does not contain is any size at all, which is why one axis serves a garden pump and a gigawatt turbine.

One number picks the machine

A flow rate, a head and a shaft speed contain exactly one dimensionless combination with no size in it. That combination decides whether a duty wants an impulse wheel, a Francis runner or a propeller — before anything has been drawn, sized, or costed.

$Two velocities, 2.50 apart, and only one of them is anybody's. The velocity in Darcy's law is the flow rate divided by the whole cross-section, solid included: a speed no fluid particle ever has, since the fluid occupies only the fraction ε of that area. The speed the fluid actually averages is larger by exactly 1/ε — 2.50 times here — and it is the one that belongs in a residence time, in a pore Reynolds number and in any statement about when a tracer arrives. The grains are drawn to say that the pore-scale flow is not computed here: nothing above resolves it.$

A velocity nobody has

The velocity in Darcy's law is the flow rate divided by the whole cross-section, solid included — a speed no fluid particle in the bed ever has. Averaging buys a linear law and charges for it in exactly this coin, and the constant that comes with it is an area of about a square micron.

Two terms, one with viscosity in it and one without. Ergun's two contributions to the pressure gradient, against the pore Reynolds number, on logarithmic axes. The viscous term rises with the first power of the velocity and the inertial one with the square, so on these axes they are straight lines of slope one and two and there is exactly one crossing. The second term contains no viscosity at all — it is the price of accelerating fluid into every pore and out again — which is why a linear resistance law has to fail eventually whatever the fluid is.

Where Darcy stops

A linear resistance law has to fail eventually, because pushing fluid into a pore and out again costs energy that has nothing to do with viscosity. Where it fails is one dimensionless number, and that number is 150/1.75 — read out of a correlation's own constants rather than measured.

The pressure drop stops rising at 0.213 m/s. The pressure drop across a bed of 500 µm sand, against the velocity through it, in units of the fluidisation velocity. The rising branch is Ergun's resistance and the flat one is the bed's buoyant weight, which the flow cannot exceed however hard it is pushed: past the corner the bed expands rather than resisting more. That flat line is the reason fluidisation is unmistakable in practice — the corner is a crossing of two curves rather than a gradual departure, and it can be read off a gauge.

The bed that weighs itself

Blow hard enough through a pile of sand and the pressure drop stops rising. It cannot rise — a control volume round the bed says the drop can never exceed the buoyant weight of the solid in it, and at the velocity where the two meet the bed stops being a structure and starts being a fluid.

A factor of 1.8 between the extremes. The same numbers as a bar. The spread is exactly 96 over 160/3, which is 1.8 — and it is a spread in a quantity that a single correlation using the hydraulic diameter treats as a constant. A rectangle twenty times as wide as it is deep is 78 per cent more resistant than a circle of the same hydraulic diameter.

A number that is only the shape of the hole

The friction factor times the Reynolds number for laminar flow in a duct is a pure number that depends on the cross-section's shape and on nothing else — not the fluid, not the size, not the flow rate. It is exactly 64 for a circle and exactly 96 for a slot, and the correlation that treats them as the same is wrong by a third.

The section that decides the river

A reach of open channel has a normal depth and a critical depth, and the water has neither. What it has is a profile obeying a first-order equation, which needs exactly one condition — and whether that condition belongs at the upstream end or the downstream end is not a choice, because the equation is stable in one direction and unstable in the other.

Two different moments of one distribution. The permeability and the specific surface of a log-normal bundle, against the width of its pore-size distribution at a fixed median. The permeability rises by five orders because it is a fourth moment and the widest tubes dominate it; the surface falls because it is a first moment and the narrowest tubes dominate that.

A permeability that is only the geometry

Kozeny–Carman says that a porous medium's permeability follows from its porosity and its specific surface. Both are real, both are exactly measurable, and they do not determine the answer: forty-nine tube bundles built with identical values of each span a factor of eighty-two in permeability.

The junction the minimisation is over. A parent vessel entering from the left and two daughters leaving to fixed points. The radii are settled by Murray's law; what is left free is where the branch point sits, and the cost of the junction depends on it. The point drawn is the one the minimisation finds.

The angle a junction chooses

Murray's law fixes the radii at a branching vessel and is where every account of it stops. The same minimisation fixes the angles completely — 74.93 degrees for a symmetric bifurcation, a right angle for a vanishing side branch — and it does so as a triangle of forces, with tensions proportional to the squares of the radii.

Five approach profiles a meter might be looking at. The velocity across the pipe upstream of a contraction, for a uniform flow, fully developed laminar flow, two turbulent power laws and an annular jet of the kind a bend or a partly open valve leaves. All five carry the same volume flow. The meter reads a pressure difference and cannot see any of this.

The profile a meter cannot see

A differential-pressure flowmeter measures a force balance and reports a flow rate. The step between them needs two integrals of a velocity profile the instrument has no access to — and two profiles differing by half the mean velocity across the pipe give identical readings, which is why the standards specify straight pipe rather than a correction.

How much the peak cares about the valve at each instant. A small bump is added to the valve's opening at one moment and the peak is recomputed; the difference, divided by the bump, is plotted against the moment. It is flat and zero for the first 0.45 seconds — that part of the closure is invisible to the peak — rises through the reflection arrivals, and falls to nothing after the peak has happened, which is the one part of the shape that needs no fluid mechanics.

The part of the closure a pipe cannot see

A surge specification says how long the valve takes to shut. The line does not integrate the duration, it integrates the shape — and it is measurably blind to the first half-second of an eight-second closure, while three closures of identical length give peaks 97 per cent apart.

When the tracer leaves, given when it went in. The residence-time distribution of three beds with the same mean residence time and different Peclet numbers. Every one of them has its mean at exactly one, and they are nothing alike: the loosest lets a tenth of the tracer out before a third of the mean time has passed, and the tightest is nearly the spike a plug-flow calculation assumes.

The outlet is the inlet, a while ago

A bed has a mean residence time and everybody quotes it. Six beds with the same mean let their first hundredth through at 0.20 and at 0.85 of it, mix a window of inlet history between 1.53 and 0.18 wide, and convert a first-order reaction by amounts the mean cannot distinguish.

A threshold that is a curve, not a pressure. The tension a rectangular pulse has to reach to make a five-micron bubble run away, against how long the pulse lasts. At a tenth of a microsecond it takes forty bar; at a hundred microseconds it takes 1.05, which is within about one per cent of the static threshold. There is no such thing as the cavitation pressure of this bubble on its own.

A threshold that is also a duration

The cavitation number treats inception as a pressure: below it the liquid tears, above it does not. A five-micron bubble asked to grow in 0.3 microseconds needs 43 bar of tension and the same bubble given a hundred needs 1.05, because it has to make a journey and not merely respond.

$Spin against distance flown, not against time. The spin of a struck golf ball as a fraction of its launch spin, against how far it has flown. The integrated flight sits exactly on an exponential in distance, because the spin-down torque is proportional to speed times spin and the speed cancels. The constant is 1,202 metres and the drive is 200, so the ball arrives with five sixths of the spin it left with.$

The ball that never forgets its spin

Commentary explains a late-swerving ball by saying the spin is dying away. The spin-down torque goes as speed times spin, so spin decays over a distance rather than a time — 1,202 metres for a golf ball against a 200-metre drive — and what dies away is the speed, which makes the swerve stronger.

Two ways to move a duty, along one axis, in opposite directions. A duty at a specific speed of 0.01 and what splitting it does. Dividing the head between stages in series multiplies each stage's specific speed by the number of stages to the three-quarter power, because the group carries (gH)^(−3/4); dividing the flow between units in parallel divides it by the square root of the number of units, because the group carries √Q. Both exponents are read off the group rather than remembered, and they are why the two operations are not interchangeable: three stages buy a factor of 2.28 and three units cost a factor of 0.58. The bands are drawn in the colour this site reserves for a borrowed claim, because where a Francis runner stops is practice rather than a result.

The duty that had no machine

Some duties have a specific speed outside every band, and no runner will do them at any size because the number contains no size. That is true and it is not the end. The same group says how to split the duty until it fits, and the two ways of splitting move it along the same axis in opposite directions, by exponents read straight off it.

The shaft speed is a window, and cavitation closes the top of it. Two groups against shaft speed for the same duty: the specific speed, which must be inside a band for a runner to exist, and the suction specific speed, which must be below about 3 for the impeller not to cavitate. Both rise with the shaft speed, so raising it to reach a band is also raising it towards the cavitation limit. The window here runs from 274.98 rpm to 962.31 rpm and cavitation sets its top. A duty whose window is empty needs something other than a different machine — a booster, a lower installation, or an inducer.

The group with no head in it

The number that picks a machine says nothing about whether the machine can exist. A second group formed from the same variables, with the delivered head replaced by the margin available at the inlet, decides that — and the delivered head has left the expression entirely, so how far a pump lifts is irrelevant to whether it tears the liquid apart at its own entrance.

Where a rotor's pressure rise comes from, as the radius moves. The static pressure rise across a rotor, split into the two terms rothalpy gives it. The diffusion term is held at the de Haller limit throughout — the blade is being asked to slow the relative flow as hard as a boundary layer will allow — so it is a flat 11558.4 Pa at every radius ratio. Everything above that line is the centrifugal term, which costs no diffusion and has no limit of its own. At a radius ratio of 2 it supplies 49.92 per cent of the rise and at 3, 72.66 per cent. An axial machine, at a ratio of exactly one, gets none of it.

What a turning frame keeps

Euler's equation prices the work and says nothing about where the pressure comes from. In the frame turning with the blades — which is accelerating, and carries two fictitious forces — a Bernoulli-like quantity survives both of them, and it splits the pressure rise into a term a boundary layer limits and a term that is free if the radius moves.

A blade root at 412.51 megapascals, and the chord is not in it. The centrifugal stress at a blade root against shaft speed, for one annulus of 0.36 m², in four materials. Integrating the blade's own weight outward gives σ = 2π ρ_b A N² with a taper relief — and the chord has cancelled, the blade count has cancelled, and every property of the gas has cancelled. What is left is an area times the square of a speed, capped by a material. The horizontal lines are each material's allowable stress, and where a curve crosses its own line is the fastest that annulus may be turned in that metal. At 12000 rpm this blade carries 412.51 MPa with a tip speed of 439.82 m/s.

The stress that picks the aerodynamics

The free term in a rotor's pressure rise wants radius and speed, and both are capped by something with no fluid in it. Integrate a blade's own weight outward and the root stress comes out as the annulus area times the square of the shaft speed — with the chord cancelled, the blade count cancelled, and every property of the gas cancelled.

A 204 m hammer traded for a 8.57 m swing over 299 seconds. The water level in a 10 m surge tank at the end of a 2 km tunnel 3 m across, carrying 2 m/s, after the turbine is shut off at once, with the level measured from the reservoir's. Without friction it rises to V₀√(L Aₜ/g Aₛ) = 8.57 m and swings with a period 2π√(L Aₛ/g Aₜ) = 299.1 s, the integration agreeing with both closed forms. With the tunnel's 5 m of friction the level starts 5 m below the reservoir, peaks at 5.61 m after 98 s, falls to −3.70 m, and decays. The same tunnel shut at its end with no tank would take the Joukowsky rise of 204 m. The tank does not remove the column's momentum; it gives it a free surface to push against, slowly.

A tank that turns a hammer into a swing

Shut a turbine at the end of a two-kilometre tunnel in two seconds and the valve takes a rise of 256 metres of head. Put a shaft open to the air beside it and the rise is 51, the tunnel never carries the closure as a wave at all, and its water slows instead against a level that climbs for a minute and a half — to a height that is a closed form with the tank's area under a square root.

Below Thoma's 6.07 m² the governed tank's swing grows; above it, it dies. The tank level after the turbine's power demand drops by two per cent, with a governor holding the power constant, for tanks of 0.7 and 1.3 times Thoma's area of 6.07 m² — a tank 2.78 m across. The smaller tank's swing grows by a factor of 1.47 every 74 s cycle and has reached −12.20 m by 427 s; the larger one's keeps 0.75 of itself every 100 s and is barely visible. Carried on, the smaller tank's run is refused at 794 s, where the head at the turbine has fallen below a quarter of its design value and the governor would be asking for a flow no turbine passes. The instability has nothing to do with the tank's height: it is the governor drawing more water as the level falls, which feeds the swing, against the tunnel's friction, which is the only thing damping it.

The better tunnel needs the bigger tank

A turbine governed to hold its power opens further when the head at it falls, and draws the tank down harder. That makes it a negative resistance, the tunnel's friction is the only thing damping the swing against it, and so the smallest stable tank grows as the friction shrinks — 2.78 metres across for five metres of friction, 6.09 for one.

$A force ceiling ends the similarity: a breeze makes the boat slower as a share of the wind. Speed made good to windward as a fraction of the true wind, on the best course for each wind, for a rig that is flattened once the righting moment binds and for one that is reefed, with the hull's drag angle held at 6°. Below 3.70 m/s the two are the same boat, and the fraction does not depend on the wind — 1.122 at every speed, which is the similarity the two-angle polar rests on. Above it the flattened rig's fraction falls: 0.918 at 6 m/s, 0.572 at 10 m/s, 0.321 at 15 m/s and 0.169 at 20 m/s. The reefed rig's stays at 1.122, because a reefed sail keeps its least drag angle and the hull's angle is held. Once a force is limited, the triangle is no longer the same shape in every wind.$

A breeze the boat cannot use

The two-angle polar makes a boat's speed a fixed fraction of the wind, in any wind. A righting moment ends that at 3.7 metres a second on the beat. Past it the crew must spill force, a flattened sail's drag angle climbs, and the best course to windward moves closer to the wind rather than away from it — which 45° + λ/2 cannot say, because λ now depends on the course.

$A keel pulls the best beat 12 degrees closer to the wind. Speed made good as a fraction of the wind against the course, in 3 m/s: for the boat with its keel solved as a wing, and for a boat whose λ is fixed at 25.69°, the value the keeled boat has on its own best course. The search puts the keeled boat's best beat at 45.83°; the fixed-angle rule puts it at 45° + λ/2 = 57.84°. Pointing higher loads the keel towards its best lift coefficient — 0.130 at 40° against 0.076 at 60° — and lowers its drag angle, so the curve peaks early and falls away faster on the far side. The fixed-angle curve promises 0.653 of the wind; the keeled boat can make 0.553, and only by sailing twelve degrees higher than the rule says.$

A keel flies wherever the course puts it

A keel is a wing whose lift is whatever side force the rig happens to make, so its lift coefficient is chosen by the course and the wind rather than by its designer. Solved that way, the hull's drag angle stops being a property of the boat: it pulls the best beat twelve degrees closer to the wind, halves the reaching speed a fixed angle predicts, and finally charges for reefing.

121 m after the cavity closes, against 69 m from the closure. The head at a valve shut instantly on water flowing at 0.36 m/s through 600 m of 100 mm pipe, a = 1200 m/s, with a steady head of 25 m. The closure raises it to 69.1 m, the Joukowsky head; the reflection returns at one round trip, 1.00 s, and takes the head down to the vapour head, −10.1 m, where a cavity opens (shaded). It closes 2.146 round trips after the closure, and the first pulse after it reaches 121.3 m — 52.3 m above the Joukowsky head — for 146 ms. The step line is the exact solution between events; the thin line is a 240-reach grid solver that was told nothing about it and agrees with its first pulse to better than a millimetre.

Twice the margin, on top of the hammer

Shut a valve on a line whose pressure is low and the returning wave boils the water beside it. When that cavity closes, the head at the valve can pass the Joukowsky rise — by up to twice the margin that let the water boil, in a sawtooth that jumps each time one more round trip fits into the cavity's life, and for a time that is shortest exactly when the pulse is tallest.

Matched at one speed and at no other

Every stage of a compressor passes the same mass flow, and the annulus behind each one is cut for the density the air will have reached there at design speed. Slow the shaft and the air is less dense than the metal expects, so the rear stages carry more volume than they were drawn for while the front ones starve — and below a definite speed no throttle setting keeps all of them working at once.

Between hover and twice the hover inflow

A rotor's momentum balance has an answer for every climb and for every fast descent, and none for descending at anything between zero and twice its own hover inflow. There one root sends air out of both ends of its streamtube and the other root is not a real number, and at each edge of the band one end of the tube stops moving — which is where the vortex ring state lives, and where a wind turbine's thrust coefficient of one sits.

$A pump slowed into a system with a static lift leaves its own specific speed. The specific speed of the operating point, as a share of its value at the best point, against the fraction of the design flow delivered, for a pump drawn for 0.1 m³/s against 40 m at 1450 rpm, specific speed 0.545 at its best point. Under speed control into a system whose static lift is 0 per cent, 30 per cent, 60 per cent, 90 per cent of the design head (lines), and under a throttle at design speed into the 60 per cent system (dashed). With no static lift the speed-controlled pump stays exactly at its best point and its specific speed never moves. At half the design flow it has fallen to 1.000, 0.800, 0.708, 0.652 of the design value as the static share rises, and to 0.598 under the throttle.$

The specific speed a pump spends its life at

A pump is chosen by its specific speed at its best point and then run somewhere else. Written in the pump's own coefficients the number is √φ/ψ^¾, a position along its characteristic, and a variable-speed drive keeps it there only when the system it pumps into has no static lift. Every metre of lift moves a slowed pump along its own curve, towards shut-off, and a throttle moves it further.

Six nozzles, six pump curves, and where each one is best. The head ratio a water jet pump delivers against the flow ratio it entrains, for six area ratios from a narrow nozzle to one filling four-fifths of the throat. A wide nozzle makes a tall, steep curve that is finished at a small flow; a narrow one makes a low, long curve. The dots are each curve's best-efficiency point. Nozzle, suction, throat-friction and diffuser losses are included at borrowed representative values, and the mixing loss is computed.

The nozzle that is best at one thing

Put the four losses back into a jet pump and three questions get three answers. The most head comes from a nozzle four-fifths of its throat, in closed form; the best efficiency from one a quarter of it; and the most flow from whichever nozzle is smallest, because flow has no optimum at all.

The characteristic gets a wall, and the wall does not care about the discharge. One jet pump's characteristic, at an area ratio of 0.275, with the flow ratio at which its throat entry reaches vapour pressure drawn for three values of the cavitation parameter σ = (Pₛ − pᵥ)/(Pₘ − Pₛ). Left of a wall the machine runs on its curve. At the wall no lower discharge pressure raises the flow: the head ratio can fall to zero along the vertical and the flow ratio stays where it is. The wall's position contains the nozzle and suction losses and nothing downstream of the throat entry.

The wall the suction puts in the curve

A liquid jet pump's lowest pressure is where the entrained stream enters the throat, and when that reaches vapour pressure the pump curve stops being a curve. The flow ratio freezes at a value no lower discharge pressure can move — and raising the motive pressure, the obvious cure, brings the wall closer.

Flat until the back pressure reaches a value, then gone. Entrainment ratio against back pressure, as a multiple of the suction pressure, for three mixing-section sizes of one steam ejector driven from a motive supply a hundred times its suction pressure. Each is exactly flat while the entrained stream is choked beside the jet, up to its critical back pressure. The dashed lines join that point to the back pressure at which the entrainment has fallen to nothing, 1.4 per cent higher for the middle machine. The model fixes those two ends and not the path between them. A larger mixing section entrains more and breaks at a lower back pressure.

A choke that belongs to two streams

A steam ejector entrains a fixed amount of gas whatever its back pressure, up to a pressure where it stops. The flat part is a choke, and the entrained gas is not at Mach one when it happens: it is at 0.886, because the supersonic jet beside it is part of the same throat. One area ratio then trades that entrainment for compression, and the trade decides how a vacuum train is built.

The current that runs across the river

The standard channel calculation is one-dimensional, and a river is not. Put a bend in it and the surface tilts by a few centimetres, which is invisible — and a helix forms, a few per cent of the main flow, which decides where the river will be in a hundred years.

$The depth at the edge, against how much pressure is left there. The brink depth as a fraction of the critical depth, against the share of the hydrostatic pressure the brink is taken to retain. A free jet — no pressure at all — gives exactly two thirds, to 1e-16. The measured ratio of 0.715 is met at 40 per cent of hydrostatic, which is more residual pressure than a falling nappe looks as though it should have.$

The depth at the edge is not the critical one

Every open-channel calculation assumes the pressure is hydrostatic, and usually says so nowhere. At a free overfall it is not, the assumption's failure can be measured, and the measurement is one number — a brink depth that is 0.715 of the critical depth where a hydrostatic reading would make it one.

$The vapour fraction that would stop a cavity, against temperature. How much vapour a cavity would have to make, as a multiple of the liquid volume it cools, before the cooling had used up the whole driving tension. In cold water it is six hundred and the effect is unreachable; at the boiling point it is below one and the effect is the dominant thing in the problem. Four decades across the range a pump might see, out of a vapour density that rises by a factor of two hundred.$

A cavity that cools the water it came from

The usual account takes the vapour pressure as a property of the liquid, looked up once. It is a property of the liquid at whatever temperature the cavity has left it — and making vapour costs latent heat, so a cavity suppresses itself. In cold water that is unmeasurable and in liquid hydrogen it is the dominant term.

The rudder pays for the keel's wake

A rudder angled to carry side force feels drag out of all proportion to its share, and that is the sensation behind the rule that weather helm is slow. The pair's drag does not care who is billed. It is least with the rudder carrying an eighth of the load, at under a degree of helm, and that share does not change with how hard the boat is pressed.

An open rotor's slipstream shrinks; a duct holds it open. The radius of the slipstream behind a hovering rotor of radius R, against the distance behind the disc: for an open rotor, from the vortex-cylinder model, contracting towards R/√2 so that the wake ends with half the disc's area; and for a rotor in a straight duct, whose slipstream leaves at the full area, and in a duct that widens to 1.3 times it. The same thrust from a wider jet needs a slower one, and a slower jet wastes less energy.

A duct is worth the square root of two

An open rotor squeezes its slipstream to half its own area and pays for the fast jet that results. Put the same rotor in a straight duct and the slipstream leaves at the rotor's full area, the jet is slower, and hovering costs 29 per cent less power. The duct is not a passive guard: it carries half the thrust itself, on the suction round its inlet lip. In cruise almost all of the advantage disappears.

Time per litre rises in a straight line with the litres already filtered. The elapsed time divided by the filtrate collected, against the filtrate, for a constant-pressure filtration of a silica slurry through a cloth: integrated step by step, and Ruth's closed form. The line's slope is the cake's resistance, which grows with every litre, and its intercept is the cloth's, which does not. After a quarter of a cubic metre the cake is already the larger.

A filter is slowed by what it has caught

Push a slurry through a cloth and within minutes the cloth no longer matters: the particles it has caught form a cake, the cake is a packed bed, and every litre filtered makes the bed thicker for the next. The filtrate grows only as the square root of time. And if the cake is soft, it packs tighter against the cloth the harder it is pushed, until more pressure buys almost nothing.

A washed filter's best cycle is shorter, and still has one. The filtrate a batch filter produces per hour, averaged over filtering, washing and twenty minutes of emptying and refilling, against the filtrate collected in each cycle, for a slurry of 400 kilograms of silica per cubic metre and wash ratios of 0, 1, 2 and 5 pore volumes. Each curve has a best cycle, marked, and the wash both lowers it and moves it to a thinner cake: from 0.307 to 0.217 m³ per m² of cloth.

A washed filter works as long as it rests

A batch filter that has to be emptied between runs has a best cycle, and the rule for it is exact: the time the cake is responsible for equals the time the filter stands idle. Washing the cake does not break the rule, because a wash through a finished cake costs time in proportion to the square of its thickness, just as filtering it did. What a wash changes is the price, and the price is set by the slurry's concentration and by how far the wash front smears, not by the cake's softness.

One fast pump or two slow ones

A pumping station with several identical pumps in parallel has a control that a single pump does not: how many of them are running. With no static lift the answer is all of them, always, slowed together. With lift it is a number that falls with the flow and is almost never whole — the flow times the number of pumps, over the square root of the head the system asks. Staging rounds it, and the right moment to round up comes before the running pumps have run out of speed.

Surge is a loop round the characteristic's peak. The two runs in the plane of flow coefficient and plenum pressure rise, over the characteristic continued to reversed flow. With B = 0.5 the state slides off the peak and settles on the stalled characteristic. With B = 2 it traces a large loop: the flow collapses at nearly constant pressure, reverses while the plenum empties, recovers to the right-hand branch at low pressure, and climbs back up it while the plenum fills, round and round.

The plenum decides whether a compressor surges

Throttle a compressor past the peak of its characteristic and it does one of two things. It settles into rotating stall, a steady state with a cell of dead flow running round the annulus, or it surges, the whole flow through the machine collapsing, reversing and recovering over and over. Which one is not decided by the blades. It is decided by the volume the compressor discharges into, against the inertia of the air in its duct — one number, Greitzer's B, which grows with the blade speed, so the same machine stalls at part speed and surges at full speed.

A duct beats Betz only on the area it chooses

Put a wind turbine inside a flaring duct and it can take more than 16/27 of the wind's power through its rotor, which is the claim. Measure the same power against the duct's exit, the area the device actually fills in the wind, and a plain diffuser takes less than a bare rotor of that size would. It crosses the limit only when the duct holds a suction behind its exit, and the suction it needs is exactly the pressure on the back face of Betz's own disc.

Air at a valve softens the hammer only in quantity

A vapour cavity at a closed valve holds the head at one value, and that is why the pressure after it collapses climbs a staircase of equal steps. Put a pocket of free air there instead and the valve becomes a spring. A millionth of the pipe's volume trims only the tallest, briefest pulses; a hundred-thousandth takes a third off them and moves them; a ten-thousandth can make the pulse taller than it was, even on a line that never cavitates. The air that reliably removes the hammer is a thousandth of the pipe, which is a deliberate air vessel rather than a little dissolved gas.

The exit area is a condition, not a second choice. Critical entrainment and critical compression ratio for a steam ejector with a mixing tube of 60 nozzle throats, against the nozzle's exit area as a multiple of the exit that delivers the jet at exactly the pressure it shares with the entrained gas. Both peak at the matched exit, at every suction drawn, so no exit area buys entrainment at the cost of compression or the other way. Halving the exit costs 2.4 per cent of the entrainment at a hundredth of the motive pressure, doubling it 3.2 per cent.

An ejector's nozzle exit is a condition, not a choice

A steam ejector has two areas a designer can pick, the mixing tube's and the nozzle exit's, and one of them looked like a second way round the trade between entrainment and compression. It is not. Both are largest with the exit matched to the pressure the jet meets, as a rocket's thrust is, and every other exit draws a curve inside the matched one. What moves the machine is loss, and the losses sort themselves: only the nozzle's reaches the entrainment, and the sharp edge of the characteristic belongs to the ideal machine alone.

A phantom biplane partner doubles the induced drag. Induced drag at a given lift, as a multiple of its deep-water value, against the foil's depth in spans. The surface's image of the foil is an identical wing, equally loaded, twice the depth above it, so the real foil pays its own induced drag plus the mutual drag of a biplane with a gap of twice its depth: 1 + σ, with σ Prandtl's biplane factor. The elliptic biplane's 1 + σ and the aspect-ratio-9 foil's lattice both run to two as the depth closes.

A foil under the surface flies with a phantom

At foiling speed the water's surface cannot hold a pressure, and the image that condition requires is not the reversed one a wall makes. For a horizontal foil it is an identical wing, equally loaded, above the surface — a biplane partner that takes lift and gives nothing back, doubling the induced drag as the foil rises. For the board that pierces the surface it is a reversed copy, which makes the board pay two and a half times what a keel under a hull pays. It is the board, not the foil, that decides how deep a foiling boat rides.

Fly level through short waves, follow long ones. The mean drag of the two strategies against wavelength, for a wave 0.4 m high at 8 m/s: flying level at the best mean depth for that sea, and following the surface at the calm boat's best depth, with the flat-water drag as the faint line. Level flight costs about the same whatever the wavelength; following costs the square of the heave it demands and falls away as the waves lengthen. Into the waves the two cost the same at 15.6 m; running with them, where the boat meets each wave slowly, at 5.9 m.

A foil flies level through a short sea and follows a long one

A foiling boat in waves has two ways to fly. It can hold its height and let the surface rise and fall over its foil, or it can follow the surface and heave with every wave. Flying level costs a deeper ride and a few per cent of drag whatever the wavelength; following costs the square of the heave, which grows with the frequency the boat meets the waves at. Into a head sea the two cost the same at a wavelength of about sixteen metres, running with the waves at about six, and the wave's height hardly moves either.

The air a cavity releases is capped by the cavity

When a shut valve pulls the pressure behind it down, the air dissolved in the water starts to come out, and a little gas was already known to soften the hammer. But air leaves the water only while its pressure in the cavity is below the pressure the water was saturated at, so a cavity can fill with at most its own volume of air at that pressure — a seven-thousandth of the pipe here, however much water gives up its air and however fast. That is less than a third of what removes the hammer. Released quickly, the air does not cushion the collapse at all: it turns the vapour cavity into an air cavity, and the pulse becomes the column-separation pulse with the margin measured to the atmosphere — taller, on this main.

With a small plenum, stall feedback closes the hysteresis loop. The stall amplitude against the throttle as it is closed slowly past the peak and opened again, at B = 0.2, with no stall feedback and with feedback at twice the critical gain; closing solid, opening dashed. Without feedback the cell appears suddenly near γ₀ = 0.6094 and vanishes only at 0.6428, and the loop between the two legs has an area of 0.078. With feedback the cell grows gradually as the throttle closes and shrinks along the same curve as it opens: the area is 0.0014.

A throttle can hold the peak only in a small plenum

A compressor makes its most pressure at the peak of its characteristic, and at the peak it either surges or stalls. A throttle moved by feedback can fight both. Driven by the plenum's pressure it damps surge, with a gain that grows as the square of Greitzer's B until it meets a ceiling B does not move; driven by the flow, it cannot. Driven by the stall cell's amplitude it turns the sudden jump into deep stall into a gradual one and closes the hysteresis loop — but only while the plenum is small. Above B of about a third the same feedback turns the jump into a cycle.

Heading into the sea, follow the long waves and fly level through the short. The dinghy's mean drag against the crossover encounter frequency below which it follows the surface, in units of the sea's peak frequency, heading into seas of significant height 0.2, 0.4 and 0.6 m. Following everything — the right-hand end — costs the heave's lift, and flying level through everything — the left — costs the foil's swinging depth and, in the roughest sea, breaches. Between them each sea has a best crossover: for 0.4 m at 2.92 times the peak frequency, 67.6 N against 67.9 flying level and 199 following.

A foil in a random sea follows it up to its peak

In a regular wave a foiling boat chooses between holding its height and following the surface. A real sea is a spectrum, and a wand that filters its signal can do both at once: follow the long waves and fly level through the short. The best place to divide them is close to the frequency of the sea's own peak, the divided control beats either pure strategy, and in a rough sea it is the only one of the three that keeps the foil a safe distance under the surface and the drag near its calm-water value.

Every ball's best angle is on one curve. The launch angle that carries a ball furthest over level ground, against β, the square of its launch speed measured in terminal speeds. At small β the air does nothing and the answer is 45°. It falls slowly as β grows: 41° at β = 1, 32.6° at 10 and 20.3° at 1000. A shot put sits at β = 0.009 and 44.9°, a golf ball struck without spin at β = 2.39 and 38.1°, a shuttlecock hit at 30 m/s at β = 19.5 and 30.2°.

The terminal speed sets the best launch angle

In a vacuum every ball goes furthest at forty-five degrees. In air the best angle is lower, and it is lower by an amount that depends on one number only: how many times its own terminal speed the ball is launched at. A shot put, a baseball and a shuttlecock are three points on one curve, which falls from 45° to 41° at once and then needs a thousandfold more speed to reach 20°.

$Lift takes the launch angle down far faster than drag does. The best launch angle against β for a ball whose lift is a fixed fraction ℓ of its drag, from ℓ = 0 — the drag-only curve — to ℓ = 1. At β = 10 drag alone puts the best angle at 32.6°; a lift-to-drag ratio of 0.4 takes it to 15.9° and 0.8 makes a level launch best. Once the curve reaches zero it stays there: the lift lifts the ball, and any angle given at launch is height the lift would have bought for less drag.$

A spinning ball should leave low

Drag alone takes a ball's best launch angle from 45° to the high thirties at the speeds sport is played at. Backspin takes it into the teens, because a ball with lift is a glider and a glider buys its height with lift rather than with angle. Every tenth of lift-to-drag ratio takes a fixed slice off the angle, and above a line that depends on the launch speed the best flight leaves the ground level.

$Past transition the cube law gives way to two and a half. The exponent k in Q ∝ r^k obeyed by the cheapest vessel, against that vessel's Reynolds number, for a smooth wall and for a wall with a millimetre of roughness. While the flow is laminar k is 3, Murray's law. Past transition it drops to 2.45 at Re = 2·10⁴ and 2.4 at 1.01·10⁷, near Blasius's 27/11 = 2.455. A fixed absolute roughness barely moves it, because the cheapest pipes for large flows are large and a millimetre is a small fraction of them.$

The cheapest turbulent pipe follows two and a half, not three

Murray's cube law comes from Poiseuille's resistance, and it fails the moment the flow in a vessel turns turbulent. The same cost minimised with turbulent friction gives a flow that grows as the radius to the power 27/11 in a smooth pipe and 7/3 in a rough one. Murray's invariant, a wall shear the same in every vessel, goes with it, and so does the local rule that made the law credible as biology: a vessel that holds its shear at a set point is twice too wide.

Two small pumps carry two big discs

Split a pumping duty between two pumps of the same head and speed and each one keeps the big pump's impeller diameter, because the head fixes the tip speed. So the pair spins two full-size discs in their casings for the flow one disc used to carry, and the efficiency penalty that a station comparison assumed to be two points is four. That moves the lift at which splitting saves energy from 44 per cent to 65, and below a specific speed of about a half it never does.

At one specific speed a bigger pump is a better one, and by less as it grows. Best efficiency against impeller radius for geometrically similar pumps of 40 m head at three specific speeds; the shaft speed falls as the radius grows, so the tip speed is the same at every size. At Ωₛ 0.55 a 25 mm impeller reaches 0.797, a 0.2 m one 0.869 and a 1.6 m one 0.887: 7.17 points for the first factor of eight, 1.84 for the second.

Moody's fifth root is true at one size

A model pump is less efficient than the full-size machine, and since 1925 the gap has been bridged with a fifth root of the size ratio. Built from its losses, a pump's inefficiency scales three ways at once — leakage as the inverse size where a clearance cannot shrink, friction as a fifth root until the passages go rough, and a fixed share not at all — so the exponent runs from a third down to a twentieth, and the fifth root is right only near one size. The same arithmetic says a duty should be split by speed, not by width.

With lift in proportion to spin, a drive gains from spin until its launch is level; the measured curves give it a best spin where drives are hit. The longest carry of a 70 m/s drive at each spin rate, its launch angle chosen for that spin, for four pairs of coefficients: lift in proportion to the spin ratio with fixed drag; drag rising with spin; lift growing as its 0.4 power; both measured forms together. The fixed pair peaks near 4,800 rpm, where its best launch reaches level, and drag rising alone near 4,400 rpm; lift saturating alone is still rising at 6,000; with both measured forms the carry peaks at 234.6 m at 2273 rpm and 16.8°.

A drive's best spin falls where players hit it only because lift tires and drag grows

With lift in proportion to spin and drag fixed, a golf drive keeps gaining from backspin until its best launch has fallen to level, near 4,800 rpm, and past a few thousand more the model flies a ball the length of eight football pitches. Give lift and drag the curves wind tunnels measure — lift growing ever more slowly with the spin ratio, drag rising with it — and the best spin moves to near 2,300 rpm at 70 m/s, with a launch well above level, inside the range launch monitors recommend. Either curve alone leaves it at twice that or more. And along a ridge of best launches the carry hardly moves while the angle falls by ten degrees, which is why the launch angle is the softer of the two numbers.

A wand that lags its sea makes flying level the better bet

A foiling dinghy whose height control divides the sea sharply follows the long waves, flies level through the short and beats both pure strategies. Give the division the lag a real wand or filter has and the gain disappears in a moderate sea; in a rough one no crossover both keeps the foil under and spares it from pulling down. A second-order filter, which cuts more sharply, does worse rather than better, because it flies through more of the sea near its crossover than the sea contains.

Along the carry's ridge the run hardly moves the best spin, because every drive on it lands alike. The total distance of a 70 m/s drive, carry and run, at each spin rate launched at the angle that carries furthest for that spin, on a firm, a medium and a soft fairway, with the carry alone. The carry peaks at 234.6 m at 2273 rpm; the total along the ridge peaks at 1728 rpm on the firm turf, 274.3 m, 2052 rpm on the medium turf, 249.3 m, 2375 rpm on the soft turf, 239.9 m. Every drive on the ridge lands between 45.2° and 49.2°.

The run moves the best drive below the ridge that carries furthest

A drive's carry is greatest along a ridge of launches on which more spin trades for a lower angle at almost no cost. Every drive on that ridge lands at about the same steep angle, so the run after landing hardly chooses between them. What the run rewards is leaving the ridge: less spin and a launch several degrees lower, a lower flight that lands flatter, gives up thirteen metres of carry and gets back twenty in the run. On an ordinary fairway the best drive in all lands near thirty-five degrees, at every ball speed; on a firm one the model's ground stops being a turf anybody has measured before the run stops paying.

$At the cheapest fork each daughter runs at the parent's velocity along it, so the split loses only the sideways part. The power lost at a symmetric split, as a multiple of the flow times the parent's dynamic pressure, against the daughters' radius as a fraction of the parent's, at each of three cheapest angles. For laminar, k = 3 the angle is 37.5°, the curve's minimum is at a ratio of 0.7937 — the cheapest ratio — and its value is sin²θ = 0.37; for smooth turbulent, 27/11 the angle is 28.4°, the curve's minimum is at a ratio of 0.754 — the cheapest ratio — and its value is sin²θ = 0.226; for rough, 7/3 the angle is 25.1°, the curve's minimum is at a ratio of 0.743 — the cheapest ratio — and its value is sin²θ = 0.18.$

The cheapest fork loses only its sideways speed, and pricing it closes the fork

A turbulent network built at least cost forks at 28.4°, with daughters three quarters of the parent's radius, and those numbers priced the pipes and never the split. Price the split as Borda priced a sudden widening and something exact appears: at the cheapest fork each daughter already runs at the parent's velocity component along it, so the junction loses nothing but the sideways part of the parent's speed, and its loss cannot be reduced by resizing the daughters. It can be reduced by narrowing the fork. A split costing under two per cent of the bill moves the cheapest angle by seven degrees, because the pipes' bill is flat at its best angle and the split's loss is not.

Every airway down to the eleventh generation dissipates several times Poiseuille's loss, because its layer restarts at the fork. Pedley's factor, the developing layer's dissipation over the Poiseuille dissipation of the same airway, generation by generation along Weibel's symmetric tree, at total flows of 0.25, 1 and 5 litres a second. 0.25 L/s: 4.88 in generation 1, 2.57 in generation 5, above one down to generation 8; 1 L/s: 9.77 in generation 1, 5.15 in generation 5, above one down to generation 11; 5 L/s: 21.8 in generation 1, 11.5 in generation 5, above one down to generation 14.

The layer every airway restarts bends Murray's cube towards the turbulent pipe's

A laminar fork does not lose energy the way a turbulent one does, in a burst of mixing at the split. It loses it all the way down each daughter, because the boundary layer starts again at every flow divider and an airway a few diameters long ends before the layer has grown across it. Pedley measured that extra dissipation in 1970, and in the form he wrote it the radius cancels: it depends on the flow and the length alone. Price it into the cheapest tree and Murray's cube law gives way to a flow growing as the radius to the power twelve-fifths or thirteen-fifths, the band the turbulent pipe's exponent sits in, across the first eleven generations of a resting lung.

A strike cannot choose its spin and launch apart: each angle of attack is one line across the plane, and only an upward one reaches low spin with height. Backspin and launch angle of the ball as the dynamic loft is changed from 1° to 24° above the angle of attack, for a 45 m/s head striking at angles of attack of −4°, 0°, +4° and +8°; on each line, the loft that gives the longest total. At −4° the best is 2413 rpm at 4.43°; at 0° the best is 2129 rpm at 7.43°; at +4° the best is 1636 rpm at 9.71°; at +8° the best is 1221 rpm at 12.3°. The free plane's best for a 66.6 m/s ball is 1349 rpm at 12.5°.

A driver reaches the best drive only by striking upward

The longest drive asks for a launch near twelve degrees with little more than a thousand revolutions a minute of backspin, and a club cannot ask for those separately. Its strike sets the spin by the angle between the head's path and its face, and the launch by the face's loft, so a player who swings level can trade spin for launch only along one line — and that line misses the best drive. Striking upward moves the line. At a 45 m/s head speed the longest total comes at an attack of nine degrees upward with fourteen of loft, 4.6 metres past the best level strike, and it still lands near thirty-five degrees.

A travelling bubble's B-factor grows as the square root of its age, and the bubble turns heat-limited when it reaches B*. The vapour a growing bubble has made over the liquid it has cooled, against its age, in water at 80 °C under a tension 10 kPa below the vapour pressure, while inertia limits its growth; the dashed line is B, at which the cooling would use up the whole tension. B is 4.3 at a microsecond and 136 at a millisecond, and meets B = 31.6 at 53.8 µs, when its radius is 0.141 mm.

A bubble is throttled by its age, and a sheet by its length

A cavity cools the liquid it boils, and the cooling lowers the vapour pressure it needed. Whether that matters was a question about one number, the B-factor — vapour made over liquid cooled — and no real cavity's was known. Computed, the two kinds a pump has are nothing alike. A travelling bubble cools a thin shell that thins further as it stretches, so its B grows as the square root of its age and reaches the critical value at Brennen's thermal time: a millisecond in water at 60 °C, four microseconds at the boiling point. An attached sheet keeps renewing the liquid it cools, so its B is a few, fixed by its length and speed. In warm water the bubbles are suppressed forty degrees before the sheet.

$Kicked at ninety per cent of its peak flow, the compressor is held only when both modes have an actuator. The flow through the compressor, as a fraction of the peak flow, after a stall cell of amplitude 0.05 is started at 90 per cent of the peak flow, at B = 2, under four controllers: a tip injector of gain 0.7 and capacity 2 per cent of the peak flow with the throttle fixed; the injector with a pressure-driven throttle; the pressure-and-flow throttle with no injector; and both. injector, throttle fixed: surges or runs away; injector, pressure throttle: surges or runs away; throttle pair, no injector: stalls; injector and throttle pair: recovers.$

A compressor needs one actuator for each way it can fail

A throttle alone cannot hold a compressor left of its peak in a large plenum: it has one degree of freedom and the machine has two ways to fail, and the throttle's own surge gain runs into a ceiling that the plenum's size does not move. Give the stall cell its own actuator — air injected at the rotor tips against the cell — and give the throttle a second signal, the flow, and a machine at B = 2 is held at ninety per cent of its peak flow, recovering from a kicked cell nearly three times the injector's capacity. Let the injection lag the cell by a radian and the line still holds, but almost no disturbance is survived.

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