The cheapest way to stay up

A wing pays two drags: one for having a surface, which is flat in lift and rises with speed, and one for having ends, which falls as the square of the speed. Their sum has a minimum, and that minimum is where the two are equal.

That speed is the answer to a question, and it is not the question most people think they are asking. Drag is a force. Staying in the air for an hour costs power, and power is force times speed, so the two questions have different answers.

Two curves from one budget #

In steady level flight the lift equals the weight, so the lift coefficient goes as $1/V^2$: flying faster means flying at a smaller angle of attack to carry the same load. Substituting that into the drag budget gives a drag of the form

$$D = A V^2 + \frac{B}{V^2}$$

where the first term is friction, growing with speed because the dynamic pressure does, and the second is induced drag, falling with speed because the required lift coefficient does.

Minimising that sum gives $V_{md} = (B/A)^{1/4}$, and at that speed the two terms are equal — which is the result the rung below establishes, and it is worth noticing that it drops out of the algebra rather than being imposed.

Power is drag times speed:

$$P = A V^3 + \frac{B}{V}$$

and the cube changes everything. Setting the derivative to zero gives $3AV^2 = B/V^2$, so $V_{mp} = (B/3A)^{1/4}$, and dividing one by the other leaves

$$\frac{V_{mp}}{V_{md}} = 3^{-1/4} = 0.7598$$

with $A$ and $B$ gone. The aircraft has cancelled. Its weight, its wing area, its aspect ratio, its skin friction, the air density it is flying in — all of it is in $A$ and $B$, and none of it survives into the ratio.

What the solver computed, and how it was checked #

The two coefficients are not invented for the occasion. $A$ comes from the friction drag of a laminar flat plate, which is $1.328/\sqrt{Re}$ per side and is computed here from the Blasius solve — the wall slope of 0.332057 and the momentum thickness of 0.664115, which are the same number twice and are checked against each other wherever they are used. $B$ comes from Prandtl’s lifting line, with a span efficiency of 1.0000 for elliptic loading.

At a chord Reynolds number of $10^6$ and aspect ratio 8, that gives a friction coefficient of 0.002656, a best-glide lift coefficient of 0.2584, and a glide ratio of 48.6 to 1.

The rest is arithmetic on those, and the numbers it produces are worth tabulating because the qualitative point is easy to overstate:

Flying at the minimum-power speed costs 15.5% more drag and saves 12.3% of the power. Neither is a dramatic number, and the existence of the distinction matters more than its size — because the two speeds answer different questions and confusing them means optimising for the wrong one.

Which speed answers which question #

Distance. An aircraft gliding without power converts height into distance at a rate equal to its lift-to-drag ratio, so the furthest glide is at the speed of least drag. A pilot who has lost an engine and wants to reach an airfield flies $V_{md}$.

Time. An aircraft with an engine burning fuel at a rate proportional to power stays airborne longest at the speed of least power. A pilot who wants to loiter, or a glider pilot who wants to sink as slowly as possible while circling in a thermal, flies $V_{mp}$.

Sinking speed. For an unpowered glider, the rate of descent is the power required divided by the weight, so minimum sink is at $V_{mp}$ and best glide is at $V_{md}$ — two speeds every sailplane pilot knows and which appear on the placard as different numbers.

The 24% speed difference between them is large enough to matter in practice and small enough that flying halfway between the two is a defensible compromise. What is not defensible is believing there is one “efficient speed”.

Where the friction number comes from #

The coefficient $A$ deserves more than a citation, because it is the one number in this essay that comes from solving a differential equation rather than from algebra.

The friction coefficient of one side of a flat plate is $1.328/\sqrt{Re_c}$, and the 1.328 is twice the momentum thickness of the Blasius solution. That solution is obtained here by shooting: the equation $f''' + \tfrac{1}{2}f f'' = 0$ is integrated outward from the wall with a guessed wall slope, and the guess is corrected until the profile reaches the free stream.

The check on it is an identity that nothing in the code arranges. Von Kármán’s momentum integral for a plate with no pressure gradient reduces, after the similarity substitution, to $\theta_\eta = 2f''(0)$ — the momentum thickness is exactly twice the wall slope. One of those is where the integration started and the other is a trapezoid rule over its whole output, and they agree to $2 \times 10^{-4}$.

So the number 0.002656 that anchors the whole power curve is traceable: it is $2 \times 2 \times 0.664115 / \sqrt{10^6}$, with the 0.664115 solved rather than tabulated.

Why the exponent is a third and not something else #

The factor of three comes from the cube in the power expression, and tracing where it enters is worth doing because it says how general the result is.

Power has one more factor of $V$ than drag. Differentiating $AV^3$ brings down a 3 where differentiating $AV^2$ brings down a 2, while the induced term goes from $-2B/V^3$ to $-B/V^2$. The ratio of the two coefficients in the stationarity condition therefore changes by a factor of 3, and because the whole thing is a fourth-power relationship the speed changes by $3^{1/4}$.

That reasoning depends on exactly two things: that the parasite term goes as $V^2$ and that the induced term goes as $V^{-2}$. Both are consequences of the drag budget rather than of any particular aircraft, so the result holds for anything obeying it — a glider, an airliner, an albatross.

It fails where the budget fails. A wing with substantial pressure drag has a parasite term that is not simply quadratic; a wing near its critical Mach number has a term that rises steeply and suddenly. Both change the exponent, and both are outside the model here.

What the model is leaving out, and by how much #

The friction here is laminar, on both sides of a flat plate, with no pressure drag at all. That is an optimistic account of a wing and a wildly optimistic account of an aircraft.

Turbulent friction is three to five times the laminar value at these Reynolds numbers. A real wing’s boundary layer is turbulent over most of its chord, so $A$ is several times larger than the value used here.

Pressure drag — the incomplete recovery behind a body with thickness — is absent entirely, because the ideal theory says it is zero and the site’s grid cannot measure it from the wake.

Everything that is not the wing. Fuselage, tail, undercarriage, interference between them: on a light aircraft the wing is perhaps half of the parasite drag, and every one of those contributions is mostly pressure drag from separated flow rather than friction.

The glide ratio of 48.6 to 1 should be read in that light. It is the ratio a wing alone would achieve with a laminar boundary layer and no thickness, and it lands between a good sailplane’s 45 and a very good one’s 60 by coincidence rather than by accuracy. What the model gets right is the shape of the curves and the ratio between the two speeds, both of which are insensitive to the size of $A$ and $B$.

The same argument for a bird, and for a swimmer #

Because the ratio survives the removal of every aircraft-specific quantity, it applies to anything that stays up by making lift and pays for it with friction and induced drag.

Birds. A soaring bird circling in a thermal is minimising sink, so it flies near $V_{mp}$, slow and at high lift coefficient with its wings spread. The same bird crossing between thermals is maximising distance for the height it has, so it flies faster and often reduces its span by sweeping its wings — which raises $A$ and lowers the effective span, moving both speeds up. The observed behaviour is the theory: the sweeping is a variable-geometry solution to a problem whose optimum moves with the objective.

Flapping. Powered flight in a bird costs the induced power of hovering plus the parasite power of moving, and the same two-term structure gives the familiar U-shaped power curve. Its minimum is $V_{mp}$, and migrating birds are observed to fly closer to $V_{md}$ — the range-maximising speed — which is what an animal covering distance rather than loitering should do.

Swimming and rowing. Where the induced term is absent the whole structure collapses: with drag purely quadratic, power goes as $V^3$ and there is no minimum at all, only a monotone rise. That is the useful contrast — the existence of a best speed depends entirely on there being a cost that falls with speed, and for a wing that cost is the price of having ends.

The glide, drawn as a triangle #

The reason minimum drag gives maximum distance is worth one paragraph of geometry, because it makes the whole result visible without any calculus.

An aircraft gliding steadily descends along a path whose angle is set by the balance of forces: weight down, lift perpendicular to the path, drag along it. Resolving gives $\tan\gamma = D/L$, so the glide angle is the drag-to-lift ratio and nothing else. Distance covered per unit height lost is therefore $L/D$, and maximising it means minimising $D$ at the lift the weight demands — which is exactly the minimum-drag point.

The tangent construction is worth knowing because it survives every complication that breaks the algebra. Whatever the polar’s shape — with pressure drag, with compressibility, measured rather than computed — best glide is where a line from the origin touches it, and minimum sink is where a line from the origin touches the power curve. Neither construction cares that the two-term model happened to give a clean answer.

What the picture cannot show #

Both curves are drawn for steady level flight at one weight and one altitude. Fuel burn changes the weight, which moves $B$ and therefore both speeds; altitude changes the density, which moves both by the same factor and leaves their ratio alone.

The power axis is also the power required to fly, and says nothing about the power available from an engine, which varies with speed and altitude in its own way. Where the two curves cross is what sets the maximum and minimum speeds of an aircraft, and neither of those points is on this page.

Where the model stops #

The most important omission is compressibility. Everything here assumes drag is quadratic in speed, and above about Mach 0.7 the wave drag of a transonic wing rises steeply enough to make the parasite term effectively higher order. That moves the optimum speeds and eventually breaks the analysis.

The second is that fuel-burn optimisation is not the same problem as power optimisation. Range for a jet is maximised near the speed for minimum drag over speed, which is a third stationary point again — faster than $V_{md}$, not slower — because a jet’s fuel flow follows thrust rather than power. The result stated here is a propeller and glider result, and the airline version is arrived at by the same method with a different objective.

A third speed, which is the one airlines care about #

Two stationary points have been found so far, and there is a third that matters more commercially than either.

An aircraft with a propeller burns fuel at a rate proportional to power, so its endurance is maximised at $V_{mp}$ and its range at $V_{md}$ — the classic pair. A jet burns fuel at a rate proportional to thrust, and in level flight thrust equals drag, so a jet’s endurance is maximised at $V_{md}$ and its range at the speed that minimises drag per unit distance, which means maximising $V/D$ rather than $1/D$.

Repeating the derivation with that objective gives $V_{mr} = 3^{1/4} V_{md} = 1.316\, V_{md}$ — faster than minimum drag by exactly the reciprocal of the factor that made minimum power slower. The three speeds sit at 0.760, 1.000 and 1.316 in units of the minimum-drag speed, and which of them is “efficient” depends entirely on what is being counted and what kind of engine is doing the counting.

That symmetry is not a coincidence. Each objective differs from the next by one factor of $V$, and each extra factor multiplies the stationarity condition’s coefficient ratio by 3, so each speed is $3^{1/4}$ above the last. A ladder of objectives produces a geometric ladder of speeds, with ratio 1.316, and the whole family falls out of the same two-term budget.

Real airline cruise speeds are higher again, because time has a cost as well as fuel, and the operator is minimising a weighted sum rather than fuel alone. The industry’s name for that weighting is the cost index, and setting it is the same optimisation with a commercial term added.

Loading the aircraft moves the speeds and not the ratio #

The two coefficients hide the weight, and pulling it back out gives the result every glider pilot uses and most explanations of ballast get backwards.

Friction lives in $A$, which carries the density and the wing area. Induced drag lives in $B$, which carries the square of the weight. So the minimum-drag speed goes as the square root of the wing loading divided by the square root of the density, and so does every other speed on the ladder above — all three move together, and their ratios of $3^{1/4}$ are untouched.

The minimum drag itself is proportional to the weight, exactly. Which means the glide ratio does not change at all: it depends on the aspect ratio and the friction coefficient and on nothing else, and a loaded aircraft glides at precisely the same angle as an empty one.

That is why competition sailplanes fill their wings with water. Ballast buys no glide angle whatever — the best $L/D$ is a property of the airframe — but it moves the whole polar to higher speed, so the same angle is flown faster, and a cross-country flight is a sequence of glides at that angle. The cost is that the sink rate at any given point on the polar rises in the same proportion as the speed, so climbing in a weak thermal becomes worse, and the water is dumped when the day turns soft or before landing.

The density half of it is why placards quote indicated airspeed rather than true. Indicated airspeed is measured from dynamic pressure, so it already carries the density, and an aircraft flown at its placarded best-glide indication is at the right point of the polar at any altitude while its true speed rises steadily with height.

Who found it, and when #

The two-term drag polar is Prandtl’s, from the lifting-line work of 1918–1921, and the speeds follow immediately from it — they are in Glauert’s textbook of 1926 and in every performance manual since. The $3^{-1/4}$ ratio is one of those results that is rediscovered by every generation of students because it is a one-line derivation with a memorable answer.

Its practical home is gliding, where the difference between minimum sink and best glide is the whole of cross-country technique: circle at minimum sink, cruise at whatever speed the next thermal’s expected strength justifies. That last refinement — MacCready’s speed-to-fly theory, from 1954 — turns the choice into an optimisation over the whole flight rather than over one leg, and it starts from exactly the two curves on this page.

Where the ladder goes next #

Below this rung, the two drags a wing pays sets up the budget and finds the lift coefficient at which the two halves are equal. This rung takes the same budget, multiplies by speed, and finds that the answer moves.

Beside it, the span is the whole story is about lowering $B$, which lowers both speeds and raises the glide ratio. What neither can address is the term that is missing from the budget altogether — pressure drag — which is a wake measurement this site’s own machinery refuses to make.