Every essay — page 12
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Circulation and lift
Where lift actually comes from. The Kutta condition, the Joukowski aerofoil, and lift derived rather than assumed.
Two rings leapfrog only if they start alike
Two coaxial smoke rings passing through each other in turn is the most famous thing vortices do, and it is not what two rings do in general. Set them off from one plane with different radii and below a definite ratio the smaller one draws ahead and never comes back. Which of the two happens is decided before either ring moves, by whether two separate rings could hold the pair's energy.
One coefficient, two errors
A closed tunnel's walls push a measurement one way and an open jet's push it the other, so a wall somewhere between them should push it neither. There is such a wall, its coefficient is a length, and it nulls the blockage at 1.184 tunnel half-heights — and the streamline curvature at 1.590.
The aeroplane rises out of its own gust
A gust pushes a wing up, and the wing, being attached to an aeroplane that is free to move, starts rising — which lowers its angle of attack and takes away part of the load the gust was delivering. How much depends on one number, and the formula every airworthiness code used for forty years is a fit to the calculation this essay does again.
The span takes away the infinity, not the gust
A two-dimensional wing in turbulence reverses its load infinitely often, because the gust spectrum never falls fast enough. A real wing averages the gust across its span, and the average keeps ninety-five per cent of the gust's strength and removes the infinity. The load's crossing rate settles at about two per turbulence scale flown and stops depending on where anyone cut the spectrum off.
A wake ends by bending, not by fading
The two vortices behind an aeroplane do not simply weaken until they are gone. In still air each one bends in the other's strain, the bend grows by a factor of e every thirty-three seconds behind an airliner, and after a couple of minutes the pair has pinched itself into a chain of rings. The wavelength it chooses is eight and a half times the spacing, and the vortex cores hardly enter.
The bird at the point pays for the V
A flock's saving in a V is fixed by where its members sit across the stream, and the angle of the V cannot change it by a single per cent. What the angle changes is who gets the saving. It flows backwards through the formation, so that in a V swept forty-five degrees the leader pays nine-tenths of what it would pay alone while every bird behind it pays under four-tenths.
A fair V is a curved V
In a straight V of birds somebody always pays more than somebody else: at the fairest angle the leader and the two birds at the tips pay half again what the birds between them pay. The V in which every bird pays exactly the same can be solved for, and it is not straight. Its arms leave the leader almost abreast and bend back ever more steeply, to thirty-two degrees at the tips of a flock of nine. The share every bird then pays is the flock's average, fixed by Munk's theorem before any position is chosen — so fairness costs nothing, and the only thing that can make the flock cheaper is flying closer together sideways.
![The number of passes grows in proportion to Reynolds number. Passes the two rings make before their cores are close enough to merge, against the vortex Reynolds number, marched with the merging threshold at 0.24 of the core separation and at 0.30, and the closed form [(0.24 dₘᵢₙ)² − σ₀²] Re ÷ 4Tₚₐₛₛ, with the closest approach dₘᵢₙ and the time between passes Tₚₐₛₛ read from the inviscid pair. Both thresholds give a count in proportion to Re: one or two passes at a few thousand, ten or thirty at thirty thousand.](/f/the-number-of-passes-grows-in-proportion-to-reynolds-number-671303d0.svg)
Leapfrogging rings end by merging, not by parting
Two smoke rings that leapfrog in an ideal fluid do it for ever, because their energy binds them. A real fluid drains the energy, and the obvious guess is that the rings drift out of the bound state and part. They do not: diffusion drains the pair's energy and the energy of every possible pair of free rings together, and the margin that binds them never goes negative. What viscosity does instead is fatten the cores until, as one ring threads the other, the two are close enough to merge. That takes a number of passes in proportion to the Reynolds number — one to three at the few thousand of a laboratory smoke ring.
The follower beats in time with the wake, not the bird
A gliding bird can sit in its neighbour's upwash and stay there. A flapping bird's wake is a wave left in the air — the path its wingtip traced, rising and falling with every beat — and a bird behind gains only if its own wing is where that wave is when it arrives. The best timing is a rule with no aerodynamics in it: lag the bird ahead by the time the wake took to come, so that each wingtip retraces the path of the one before. Directly behind, the rule flips by half a beat, and it buys a smaller loss rather than a gain.
A box wing's fins earn their keep in the spar
Constrain a box wing's root bending moment and its fins stop saving drag: it becomes a biplane. But that constraint counted the lift's moment, not the spars', and the fins join the spars into one frame. Joined at the tips, the two spars hand part of the moment to a couple across the gap — tension in one, compression in the other — which costs far less material than bending. With fins as stiff as the spars that is three-tenths of the moment; with rigid fins it stops at a third, whatever the gap, because a tip joint can only guide a tip. Counted in the spar, the box has less drag and a lighter spar than the elliptic monoplane at once.
A canard pays for its stability in induced drag
The argument for a canard is that both of its surfaces lift upwards, while a tail pushes down and makes the wing carry the difference. Munk's stagger theorem turns the question into arithmetic: two surfaces' least induced drag depends only on how the lift is split between them, not on which is in front. Static margin sets the split. At a margin of a tenth of a chord a tail carries a small upload and costs under one per cent; a canard must carry a fifth of the weight on a third of the span and costs twelve, and the more stable it is made, the more it pays.
A third surface is worth a square
A canard and a tail each leave an aircraft one free share of its lift, and the static margin spends it. Give an aircraft both and one share stays free after trim, so the split can slide along a line to the point of least induced drag. What that slide is worth turns out to be a square: half the bowl's curvature times the square of how far the tail aircraft already trims from twice the bowl's foot. For a cruising wing the saving is under one per cent and less than the canard's own skin friction; it pays only for a wing whose pitching moment is as large as a flapped section's.