How thick is thin

The layer next to a surface is where almost everything happens, and the first thing anybody wants to know about it is how thick it is.

There is no honest answer, because the layer has no edge. The velocity climbs from zero at the wall and approaches the free stream asymptotically, getting closer forever without arriving. Asking where it ends is like asking where a hillside stops.

One profile for every station #

Before any thickness can be discussed, something remarkable has to be noticed about that figure.

It is one curve, and it describes the layer at every distance along the plate and at every Reynolds number. The layer at ten millimetres from the leading edge and the layer at a metre have the same shape; only the scale differs.

That is a similarity solution, and it is why the problem is tractable at all. The two-dimensional partial differential equations for the layer collapse into a single ordinary differential equation in one variable, because the only length in the problem is the one the flow itself builds — √(νx/U) — and heights measured in units of that length give the same profile everywhere.

The variable is written η = y √(U/νx), and the whole boundary-layer problem becomes

$$f''' + \tfrac{1}{2} f f'' = 0$$

with f(0) = f′(0) = 0 at the wall and f′ → 1 far away. Two conditions at one end, one at the other, which is a boundary-value problem rather than an initial-value one and is what makes it interesting to solve.

Solving it by aiming #

An ordinary differential equation with a condition at infinity cannot be marched from the wall, because the wall does not supply enough information: f(0) and f′(0) are known and f″(0) is not.

The shooting method guesses it. March from the wall with a trial value, see where the far end lands, and correct. Two guesses nowhere near the answer converge in about six secant steps, and the whole solve takes sixteen milliseconds — which is worth knowing, because it means the profile can be computed afresh for every figure rather than tabulated.

The value that comes out is f″(0) = 0.332057. It is one of the few numbers in fluid mechanics that gets memorised, and it is the slope of the velocity profile at the wall in similarity units. Multiply it by the right combination of density, viscosity and speed and it is the skin friction.

What the number at the wall is for #

Before leaving the wall slope, it is worth converting it into the quantity it exists to produce.

The shear stress on the surface is the viscosity times the velocity gradient there, and putting the similarity variable back into ordinary units gives

$$c_f = \frac{0.664}{\sqrt{Re_x}}$$

where 0.664 is twice 0.332057 and nothing else has been introduced. So a single number computed at η = 0 governs the friction at every station on every flat plate at every speed, once the Reynolds number is accounted for.

That is worth pausing on. The similarity argument has reduced an entire class of flows — every laminar plate that has ever existed — to one curve and one constant, and the constant is a slope at a point. Integrating cf along a plate gives the total friction, which is 1.328/√Re_L, and that number is the whole of laminar skin-friction drag.

Three thicknesses, and only one of them is a height #

Now the original question can be answered, three times.

δ₉₉ is the height at which the speed has reached 99% of the free stream. It is the honest but arbitrary answer — 99 is a choice, and choosing 95 or 99.5 gives a different number. In similarity units it is η = 4.910.

δ* is the displacement thickness, and it is not a height at all. It is the distance the wall would have to be moved outward to pass the same mass flow as an ideal flow with no layer. It measures how much flow the layer has taken away. In similarity units it is 1.7208.

θ is the momentum thickness, the same construction applied to momentum rather than mass. It is 0.6641.

The three sit in a fixed order and always will: θ < δ* < δ₉₉. That ordering is not a coincidence of this profile. Momentum deficit is always smaller than mass deficit, because the momentum integrand carries an extra factor of u/U that is less than one everywhere inside the layer.

Why the second two are the useful ones #

δ₉₉ is what gets quoted and it is the least useful of the three.

Displacement thickness is useful because it says what the outer flow sees. To the inviscid solution outside, a body with a boundary layer on it looks like a slightly fatter body — fatter by δ* — and the standard way of coupling the two theories is to solve the ideal flow round the body, compute the layer, add δ* to the shape, and solve the ideal flow again. Two or three passes and the answers stop moving.

Momentum thickness is useful because it is what the drag is. The momentum missing from the layer had to be given to the wall, so the friction drag on a plate of length L is exactly ρU²θ(L), with no constant of proportionality and nothing approximated.

That last statement deserves emphasis because it is unusual. Most relations in this subject carry an approximation somewhere; this one does not. Drag is momentum thickness, in the same way that mass is mass. It is the reason θ is the variable every integral boundary-layer method tracks, and the reason a wind tunnel can measure drag by traversing a probe across a wake rather than by weighing anything.

Why the layer displaces rather than blocks #

The displacement thickness has a name that invites a wrong picture, and the wrong picture is worth dismantling.

It does not mean the layer blocks a channel of height δ*. The fluid inside the layer is moving; it is simply moving more slowly than it would have. What δ* measures is the shortfall, converted into the height of a strip of free stream that would carry the same shortfall.

The consequence for the outer flow is a shape change rather than an obstruction. A wing with a layer on it presents to the outer flow an effective surface displaced outward by δ*, and since δ* grows along the chord, the effective section is slightly thicker and its trailing edge is slightly blunter than the metal one. On a well-behaved wing that is a fraction of a per cent and it changes the lift by a few per cent; near separation it is much larger, and predicting lift near the stall is largely a matter of getting the displacement right.

What the solver computed, and how it was checked #

There is a difficulty in checking a boundary-layer solve, and it is that the interesting outputs are the ones nobody has an independent value for.

The check used here is an identity that has to hold and that the code does nothing to arrange. Von Kármán’s momentum integral, for a flat plate with no pressure gradient, says

$$\frac{c_f}{2} = \frac{d\theta}{dx}$$

Substituting the similarity forms makes every dimensional quantity cancel and leaves

$$\theta_\eta = 2 f''(0)$$

The left-hand side is a trapezoid rule over the entire computed profile. The right-hand side is where the RK4 march started. They have nothing in common: one is an integral of the output, the other is the input the shooting converged on.

They agree to seven figures. θ = 0.664 114 5 and 2f″(0) = 0.664 114 7.

The displacement thickness gets the same treatment, against its closed form η − f(η) in the limit, and agrees to 10⁻⁷. And the gate has rejection tests beside both: handed a profile whose far-field value is 0.93 instead of 1, the assertion refuses; handed one whose momentum thickness has been inflated by 5%, it refuses.

The shape factor #

The ratio of the first two is a number worth carrying:

$$H = \frac{\delta^*}{\theta} = 2.591$$

It is dimensionless, it is independent of position and Reynolds number for this profile, and it describes the shape rather than the size of the layer.

That makes it a diagnostic. A fuller profile — one pressed against the wall by a favourable pressure gradient — has a lower H. A hollowed-out profile, being pushed backwards by an adverse gradient, has a higher one. H climbs from 2.59 for the flat plate towards about 4 as separation is approached, and reaching 4 is the standard warning that a laminar layer is about to let go.

A turbulent layer has H around 1.4, which is a completely different regime and is why turbulent layers survive adverse gradients that laminar ones cannot.

The practical use is that H is measurable. A traverse across a real layer gives a profile, integrating it gives δ* and θ, and their ratio says how close that layer is to giving up — without any need to know the pressure gradient, the history, or anything else about how it got there. It is a one-number state of health for a boundary layer, and it is the reason experimentalists traverse.

How thick, in millimetres #

The similarity variable is elegant and it does not answer the question a reader actually asked.

Converting: δ₉₉ = 4.910 x / √(Re_x). For air at twenty metres a second, one metre back from a leading edge, the Reynolds number is about 1.4 million and the layer is 4.2 millimetres thick.

That is the number that makes the subject strange. Four millimetres, on a wing a metre and a half across, and inside those four millimetres is every bit of the drag, all of the vorticity, and the entire question of whether the aircraft flies.

The displacement thickness at the same station is 1.5 millimetres and the momentum thickness is 0.56 of a millimetre. That last figure is the one that sets the drag: half a millimetre of missing momentum per metre of plate, and multiplying it by ρU² gives the friction force on that metre directly.

The square root, and what it means #

The growth is as √x, and that shape has a consequence worth reading off the figure.

The layer is thickest, in proportion to distance travelled, right at the leading edge. It grows fast and then reluctantly. Half of its final thickness at the trailing edge was reached in the first quarter of the chord.

The friction follows the same rule inverted: cf = 0.664/√Re_x is largest at the front. Most of the skin friction on a plate is generated in the first fraction of it, which is why leading-edge finish matters far more than trailing-edge finish, and why an insect on a leading edge costs more than the same insect further back.

The point where it all goes wrong, and does not matter #

The square-root law has an awkwardness at one end that the growth figure cannot show. At $x = 0$ the thickness is zero and the friction coefficient, which goes as $1/\sqrt{x}$, is infinite. The celebrated solution is singular at the leading edge.

The singularity is not a defect of the solve; it is the boundary-layer approximation reporting its own limit. That approximation needs the layer to be thin compared with the distance travelled, which is the statement that the local Reynolds number is large — and near enough to the leading edge it never is, whatever the free stream is doing. There is always a region where the thin-layer equations do not apply and the full equations are needed.

Two things make it harmless. The region is tiny: it ends where $Re_x$ is of order one, which for air at twenty metres a second is under a micron from the edge, against a plate a metre long. And the singularity is integrable: the total friction is the integral of $x^{-1/2}$, which converges, so the drag on the plate is finite and the micron at the front contributes essentially none of it.

Which is a general property of a similarity solution rather than a quirk of this one. A solution with no length of its own has nothing to set a scale at the origin, so it must be singular there — the singularity is the price of the collapse onto a single curve.

What the picture cannot show #

The first figure has an axis in similarity units, which is what makes it universal and also what makes it hard to read. Nothing on it is a length. A reader wanting to know how thick the layer is on their own wing has to do the conversion, and the figure’s right-hand column does it for one case only.

The profile is also drawn to η = 6.2 and stops, which is not where the layer ends — nothing is where the layer ends. It is where the curve has got close enough to 1 that continuing would draw a straight vertical line.

And the shaded region in the second figure is an area, which is being used to represent two different integrals at once. The displacement thickness is that area; the momentum thickness is a different integral over the same profile and its area is not drawn, because two overlapping shaded regions on one plot would be unreadable. The figure shows one of the two constructions honestly and labels the other.

Where the model stops #

Blasius is laminar, two-dimensional, steady, incompressible, and on a flat plate with no pressure gradient. Every one of those is doing work.

Laminar is the first to go. Real layers become turbulent somewhere between Re_x of half a million and a few million depending on how disturbed the flow is — which is to say, it depends on which Reynolds number is being quoted and what length is in it — and a turbulent layer is thicker, has far more friction, and has a completely different profile shape. Everything in this essay is the low-Reynolds-number case.

No pressure gradient is the assumption that matters most for aerodynamics, because a wing has one everywhere. Falkner and Skan generalised it and the generalisation is where separation comes from.

Flat plate is less restrictive than it sounds. So long as the surface curvature is large compared with the layer thickness — which on a wing it overwhelmingly is — the layer does not know it is on a curve, and the flat-plate solution applies locally with the local free-stream speed. That local speed comes from the ideal solution outside, which is the division of labour the whole approach rests on.

Two-dimensional rules out anything with span-wise flow in the layer, which includes every swept wing. A swept wing’s boundary layer drifts outboard as it travels back, thickening towards the tip and introducing an instability that has no counterpart here.

Incompressible holds to about Mach 0.3 in the layer, after which the temperature rise from friction changes the viscosity and density enough to matter, and the profile shape changes with it.

Who found it, and when #

Prandtl set out the boundary-layer idea in 1904, in a paper of eight pages delivered at a mathematics congress in Heidelberg, and it is the single most consequential paper in the subject’s history.

Heinrich Blasius was his student, and the similarity solution is his doctoral work of 1908. He solved the equation by series expansion from both ends and matched in the middle, which was a substantial undertaking by hand; the shooting method used above is a computer’s approach to the same problem and would have been useless to him.

The reception was slow. Prandtl’s paper was not translated into English for twenty years, and the English-speaking aerodynamics community continued to treat the boundary layer as a curiosity well into the 1920s. G. I. Taylor later described the delay as the most expensive translation failure in engineering.

Blasius himself left fluid mechanics within a few years for teaching, and published almost nothing further. The equation has his name on it and very little else does.

Where the ladder goes next #

Next rungs on this anchor: the momentum-integral method, which gets useful answers for arbitrary pressure gradients without solving anything, by assuming a profile shape and tracking only θ; Falkner–Skan, which is this solution with a pressure gradient added and which locates separation; the turbulent layer, where the profile is logarithmic and the whole similarity argument has to be rebuilt; and transition, which is where one becomes the other and is the least predictable thing in the subject.

Then across to the thin layer, which is why any of this matters, and to the drag budget, where the momentum thickness computed here becomes one half of a wing’s total drag.