Fluid Flow & Aerodynamics
why things fly, and why the usual explanation is wrong
Concept

Hypersonic — where it appears

The regime at Mach numbers of about five and above, where the shock lies close to the body and the gas behind it is hot enough to change chemically. Pressure there depends mostly on the local surface angle, which is the one place Newton's impact theory is nearly right.

Named by 10 essays across 2 fields — each of them below, with the objects they name alongside it.

The lift curve that made flight look impossible. Lift coefficient against incidence, by Newton's impact theory and by thin-aerofoil theory. One is quadratic in the angle and the other linear, so at small incidence — which is where aircraft fly — they differ by more than an order of magnitude. Newton's version says a wing large enough to carry a man would need an engine nobody could build, and for a century that arithmetic was taken as settling the question.

The theory that forbade flight

Newton treated air as a hail of particles that give up their normal momentum on impact, and got a lift coefficient of 2sin²α cos α. At five degrees that is a thirty-sixth of what a wing actually makes, and the quadratic is why powered flight looked arithmetically impossible for two centuries. The same formula is exact at Mach twenty.

misconceptions · Newtonian Four bodies, four drag coefficients, and no flow was solved. Four bodies with their Newtonian drag coefficients, each computed as a quadrature over its own surface with no flow solution anywhere. The cone's answer is exactly 2sin²δ, checked against the closed form to nine decimal places; the flat disc's is exactly 2, since every element of it faces the stream; and the sphere's is 1 against the classical Newtonian value of 1. Every one of those is an integral of one expression over a shape, and none of them required knowing what the air was doing anywhere.

The only theory simple enough to optimise

Whether Newton's sine-squared law is right has two answers — hopeless at the speeds he argued about, nearly exact behind a strong shock. This asks a different question about the same formula. Its pressure depends only on the local surface angle, so a shape's drag is a quadrature rather than a solution, and the best shape can be found by calculus.

misconceptions · Newtonian The temperature behind a shock, which is not the jump condition's. The static temperature along the flow behind a Mach 6 normal shock, with the frozen value the jump conditions give and the equilibrium value they give a long way behind. The gas arrives at 2382 K and settles at 2059, over about four tenths of a millimetre.

A gas that has not finished being shocked

The jump conditions give the state a long way behind a shock. Immediately behind it the molecules have not started vibrating yet, so the temperature is 2,382 K where the equilibrium answer is 2,059 — and the gas takes four tenths of a millimetre to get from one to the other.

compressible · Real gas The shaded face's share of the force is set by K, and it is not small until K is. The fraction of a flat plate's normal force carried by its leeward face against K = M sin α, at Mach 3, 5, 10, 20, with the hypersonic small-disturbance value and the share the leeward face would have at vacuum (dashed). Newtonian theory puts it at zero. The curves collapse on K: the small-disturbance share is 35.7 per cent at K = 0.5, 24.2 at 1, 11.2 at 2 and 3.4 at 4, and the vacuum bound is 28.8, 11.4 and 3.4 per cent at 1, 2 and 4. The zero is a good approximation only where K is large — which is also the only place the Newtonian windward pressure is itself accurate.

The face Newton left in shadow

Newtonian theory gives a surface turned away from the stream a pressure coefficient of exactly zero, and at hypersonic speed the rest of the theory is nearly right. The shaded face is not. Computed exactly on a flat plate, its share of the force depends on the similarity parameter K = M sin α rather than on the Mach number, it is a quarter of the force at K = 1, and it moves a hypersonic plate's best lift-to-drag ratio from 5 to 7 at Mach 10.

misconceptions · Newtonian The worst jet amplifies the stagnation pressure by about the Mach number. The largest amplification of the stagnation pressure, over every incident turn, against the free-stream Mach number, on a logarithmic axis: the type IV jet, the best single turning shock followed by a normal shock, and the lossless ceiling. The jet's peak runs close to the line equal to the Mach number itself, from 3.5 at Mach 4 to 12.4 at Mach 12. The ceiling grows as the Mach number to the power of three and a half and is never approached. The estimate with one turning shock falls further behind the jet as the Mach number rises.

The spot a local theory cannot see

Newtonian theory gives every panel of a hypersonic vehicle a pressure set by its own angle to the stream, and no panel more than the stagnation pressure behind a normal shock. Let a shock from one part cross the bow shock of another and a supersonic jet forms that reaches the surface through weaker shocks. At Mach 8 it stagnates at 8.6 times the ceiling — and the worst amplification at every Mach number is close to the Mach number itself.

misconceptions · Newtonian A bow shock that grows as a line explosion does. The radius of the bow shock around a hemisphere-nosed cylinder, in body diameters, against distance behind the nose: the blast-wave analogy, R/d = 0.795 C^¼ (x/d)^½ with C, the drag coefficient, 0.919, which contains no Mach number, and Billig's correlation of measured bow shocks on spheres at Mach 5, 10 and 20, anchored at the nose and continued as a hyperbola to the Mach cone. Both grow as the square root of the distance; at Mach 20 the analogy's shock is a steady 0.71 of Billig's.

A hypersonic body leaves a line explosion behind it

A blunt body at hypersonic speed does work on the air at a rate equal to its drag, and each slice of air it passes through is struck once and left to expand. Seen from the ground, that is a line explosion, and Sedov's cylindrical blast wave gives the bow shock's width and the pressure on the afterbody without a Mach number in either. The analogy's classical constants come straight out of the blast solution. So do its limits: it holds only while its own shock stays strong, over a length that grows as the square of the flight Mach number, and it puts the body inside a core hotter than anything the flow can reach.

compressible · Blast wave Next to the body the gas remembers the nose. The gas temperature, over the free stream's, against distance from the axis of a hemisphere-nosed cylinder at Mach 15, at 2, 20 and 200 diameters behind the nose; the body's surface is at a half. The gas next to the body crossed the nearly normal part of the bow shock and carries its entropy: 16.7 times the free stream's temperature at 2 diameters, 9.29 at 200, falling outwards to the gas that crossed the weaker, oblique shock. The blast-wave analogy's core at 2 diameters, dashed, runs off the top of the frame on its way to infinity.

The gas beside a hypersonic body remembers its nose

The blast-wave analogy gets a blunt body's bow shock and afterbody pressure right and its temperature absurdly wrong: it puts the body in the empty core of an explosion, where the temperature has no bound. The real gas beside the body crossed the nearly normal shock at the nose and carries that crossing's entropy all the way down. Expanded to the afterbody's pressure, it is ten to twenty times the free stream's temperature at Mach 15, it never falls below a floor set by the nose alone, and the sheath that carries it is wider than the body.

compressible · Blast wave A laminar boundary layer eats through the sheath a Reynolds number of diameters back. How far out the boundary layer has eaten into the entropy layer, as the upstream radius of the streamline at its edge in nose diameters, against distance behind the nose, at Mach 15 for Reynolds numbers on the diameter of 10⁴ to 10⁷, laminar, and 10⁶ turbulent. The sheath's edge is the streamline whose entropy is half the axis's, 1.22 diameters out. The laminar layer reaches it at 3.3·10³ diameters for the smallest Reynolds number and 3.4·10⁶ for the largest; turbulent at 10⁶, at 536.

The sheath outlasts the body

A blunt hypersonic body wraps itself in a sheath of hot, thin gas from its nose's shock, and its boundary layer grows by eating that gas from the inside. The usual picture has the boundary layer through the sheath within a few nose diameters. It is not: a laminar boundary layer takes about a third of a Reynolds number of diameters to swallow the sheath at Mach 15, which is thousands to millions of diameters, and a turbulent one hundreds. On any body of ordinary length the boundary layer never sees the cooler gas outside, and it is heated by the sheath's gas all the way down.

compressible · Blast wave A cone swallows its sheath as the cube root of the Reynolds number; a cylinder in proportion to it. The swallowing distance in nose radii against the Reynolds number on the nose radius at Mach 15: a 7° sphere-cone, laminar and turbulent, and the sphere-cylinder of the same nose, laminar. At Reₙ = 10⁵ the cone's laminar layer swallows the sheath at 160 nose radii and its turbulent one at 28.1; the cylinder's laminar layer at 1.36·10⁵.

A blunted cone eats its nose's sheath in a few hundred radii

A sphere-cylinder's boundary layer takes a third of a Reynolds number of diameters to get through the hot sheath its nose leaves, which on any real body is never. Put the same nose on a slender cone and the answer collapses: the cone's widening perimeter feeds the boundary layer so fast that it swallows the sheath in tens to hundreds of nose radii, growing only as the cube root of the Reynolds number. On a two-metre cone at Mach 15 that makes a nose of about two centimetres the line between a base still bathed in the nose's gas and one that has forgotten it.

compressible · Blast wave Bluntness delays a cone's transition by a factor that peaks, and the correlation's Mach exponent sets how high. The distance at which a blunted 8° cone's layer transitions, over the sharp cone's 0.3 m, against the nose Reynolds number, at Mach 6 and 2.6·10⁷ per metre, for three correlations of the form Reθ/Mₑⁿ = C. Reθ alone: a largest delay of 4.96 at a nose of 43.9 mm, nose Reynolds number 1.14·10⁶; Reθ / √Mₑ: a largest delay of 2.57 at a nose of 20.1 mm, nose Reynolds number 5.23·10⁵; Reθ / Mₑ: a largest delay of 1.41 at a nose of 7.29 mm, nose Reynolds number 1.9·10⁵. Past a nose Reynolds number of 8.81·10⁵ the Mach-ratio criterion transitions within ten nose radii of the tip, where the model's edge is least trustworthy.

A blunted nose delays its cone's transition only so far, and the correlation decides how far

A blunted tip leaves a sheath of hot, slow gas along a hypersonic cone, and the boundary layer growing inside it sees a lower Reynolds number and a lower Mach number than a sharp cone's. That is why bluntness is said to push transition back. Read the layer's edge through the sheath and the push has a limit: the delay grows with the nose, peaks, and falls again once the transition point lies inside the sheath. How large the peak is depends on whether the transition correlation divides by the edge Mach number — five times the sharp distance at Mach 6 for the momentum-thickness Reynolds number alone, 1.4 for its ratio to the Mach number — and every peak sits forty to sixty-five nose radii from the tip.

compressible · Blast wave

Named alongside it

The objects these essays reach for when they reach for this one.

Model limitEntropy layerBow shockMisconceptionPressure coefficientBoundary layerDragEntropyMach numberNewtonian impactOblique shockReynolds number

All concepts