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

The sheath outlasts the body grew a boundary layer along a sphere-cylinder at Mach 15 and followed it eating outward through the hot, thin gas the nose’s nearly normal shock had left beside the body. The answer was that it almost never finishes: a laminar layer takes about a third of a Reynolds number of diameters to swallow the sheath, which at any ordinary Reynolds number is hundreds of thousands of diameters, and a turbulent one hundreds. Any real afterbody behind a blunt nose is heated by the nose’s gas all the way down.

It ended on the shape that matters more. Most hypersonic vehicles are not cylinders behind spheres. They are slender cones whose tips are blunted by a small radius — because a perfectly sharp tip melts — and their bodies are hundreds of nose radii long. The question is whether a cone’s boundary layer, too, spends its whole length inside the nose’s gas, and if not, how blunt a tip has to be before it does.

The sheath on a cone #

Behind a sphere-cone the bow shock starts as the sphere’s, nearly normal at the axis, and bends back towards the sharp cone’s own shock angle. Billig’s correlation describes it as a hyperbola whose vertex is the sphere’s and whose asymptote is the conical shock that Taylor and Maccoll’s solution gives for the cone’s half-angle: 8.51° for a 7° cone at Mach 15. A streamline that started a distance $y$ from the axis upstream crosses the shock where it leans at an angle set by $y$, and carries the entropy that crossing gives it all the way along the body.

On the axis the shock is normal and the entropy rise is $\Delta s/c_p = 2.21$. Far out the shock is the cone’s, nearly a Mach wave at a normal Mach number of 2.2, and the rise is 0.137. Half the excess is gone at 2.36 nose radii: inside that radius is the sheath, gas that passed the nose’s shock and is many times hotter, at the cone’s surface pressure, than the gas that passed the cone’s. The sheath is a small thing in absolute terms, a few nose radii across upstream, and on a cone tipped with a nose of a centimetre it is a stream tube a few centimetres wide.

A boundary layer round a widening circle #

The boundary layer grows by entraining the gas at its edge, so it eats outward through the sheath one streamline at a time, and its edge conditions — temperature, density, viscosity, speed — are those of whichever streamline it has reached. The laminar layer is taken locally similar, as it was on the cylinder, in Lees and Dorodnitsyn’s variables, which carry the edge history in one integral,

$$\xi(x) = \int_0^x \rho_e\mu_e u_e\,r_b^2\,\mathrm{d}x,$$

and the layer’s mass flux is $2\pi f_\delta\sqrt{2\xi/\mathrm{Re}_n}$ in units of the free stream’s through a circle of one nose radius. The edge streamline is the one that carried the same mass upstream, $\pi y_e^2$. The difference from the cylinder is the factor $r_b^2$, Mangler’s transformation for a body of revolution: on a cylinder $r_b$ is fixed, and on a cone it grows as $1 + x\tan\theta_c$.

That one factor is the whole of the result. With uniform edge conditions $\xi$ grows as $x$ on a cylinder and as $x^3$ on a cone, and since $y_e^4 \propto \xi/\mathrm{Re}_n$, the distance to reach a given streamline grows in proportion to the Reynolds number on a cylinder and as its cube root on a cone. The cone’s widening perimeter spreads the boundary layer round a growing circle, and the mass it must collect to reach the sheath’s edge is drawn from a circumference that grows with every nose radius travelled.

The march is checked against that closed form. With edge conditions held uniform, it finds the swallowing station of the exact integral of $(1 + x\tan\theta_c)^2$ to two parts in a million at nose Reynolds numbers of $10^8$ and $10^{10}$, and the far streamlines carry the sharp cone’s entropy to thirteen figures.

Tens to hundreds of nose radii #

At Mach 15 on a 7° cone the laminar layer reaches the sheath’s edge at 28 nose radii for a nose Reynolds number of $10^3$, 70 at $10^4$, 160 at $10^5$ and 353 at $10^6$ — growing by a factor of 2.2 to 2.5 a decade, close to the cube root’s 2.15, the edge’s falling temperature making the layer slightly slower to grow than with uniform conditions.

Set beside the sphere-cylinder with the same nose, the contrast is three orders of magnitude. At a nose Reynolds number of $10^5$ the cylinder’s laminar layer swallows the sheath at $1.4\times10^5$ nose radii; the cone’s at 160. A turbulent layer on the cone, estimated from the one-seventh power law round the same widening perimeter, swallows it at 21 to 37 nose radii over four decades of Reynolds number. The dial shows the Mach number moving the answer only a little above Mach 12: the sheath’s edge moves out slowly as the nose’s shock strengthens, and the cone’s shock angle, which sets the floor of the entropy, hardly changes.

The surprising connection is that it is the afterbody, not the nose, that decides. The nose sets the sheath’s size and its heat, the same for a cylinder and a cone; the body’s perimeter sets how fast the boundary layer gathers mass, and a perimeter that grows in proportion to distance turns a linear law into a cube root. It is the same Mangler factor that makes a cone’s laminar heating $\sqrt3$ times a flat plate’s at the same edge conditions, here acting on mass rather than heat.

Where the line explosion stops #

The cylinder’s sheath lasted because nothing replaced the nose’s shock. Its bow shock was a line explosion, a cylindrical blast wave fed by the drag of the nose and spreading sideways for as long as the body went on, and every streamline near the body had passed the explosion’s front. A cone does not leave a line explosion. Its own surface keeps pushing the air aside at a fixed angle, and its shock, which leans steeply near the nose, settles onto the conical shock’s angle: within ten per cent of it seventeen nose radii out from the axis, about a hundred nose radii behind the tip; after that the shock that new streamlines cross is the cone’s weak one, and the nose’s strong shock is a fixed, finite stream tube of the past. The sheath on a cone is a bounded amount of hot gas, and the boundary layer only has to eat that amount. On a cylinder the explosion keeps the bounded amount from ever being reached, because the body’s perimeter never grows to collect it faster.

That is also why the cone’s entropy profile above has a floor. Far out, the streamlines cross a conical shock at a normal Mach number of 2.2 and are heated by only a seventh of the nose’s entropy rise. A cylinder’s far streamlines cross a shock that weakens towards a Mach wave and carry almost no entropy; a cone’s carry the cone’s, which is why the sharp cone’s gas at the surface is nearly twice the free stream’s temperature even with no nose at all.

Slenderer cones keep their sheath longer #

The perimeter’s growth goes as the tangent of the half-angle, so a slenderer cone gathers mass more slowly and keeps its sheath further back. At a nose Reynolds number of $10^5$ a 3° cone swallows at 408 nose radii, a 7° cone at 160 and a 15° cone at 49. A slender re-entry body and a blunter one with the same nose therefore differ by a factor of eight in how much of their surface the nose’s gas covers — and in the cylinder limit, a half-angle of zero, the distance runs back up to the hundred thousands of the previous essay.

What the sheath does to the edge #

The reason the distance matters is what sits at the boundary layer’s edge until then. The gas beside a hypersonic body remembers its nose found the sheath’s gas ten to twenty times the free stream’s temperature at the afterbody’s pressure; on the cone, ten nose radii back the edge is at 14 times the free stream’s temperature, and the sharp cone’s own gas at the same pressure would be at 1.9. By the swallowing distance the edge has fallen to 5.4, and beyond it the layer is fed by gas that never passed the nose. A boundary layer whose edge is seven times hotter has a lower density and a higher viscosity at its edge, a thicker and slower layer, and a different heat-transfer coefficient; it also has a lower edge Reynolds number, which delays its transition to turbulence. That last effect is the reason blunting a cone’s tip is a known way to move its transition aft, and the swallowing distance is how far aft it can move it.

How blunt a two-metre cone has to be #

Put lengths on it. A cone two metres long with a 7° half-angle has a base radius of 246 millimetres. The swallowing distance in nose radii grows as the cube root of the nose’s Reynolds number, which itself grows with the nose radius, so for each flight condition there is one nose radius at which the sheath is swallowed exactly at the base: blunter, and the whole cone sits in the nose’s gas; sharper, and the rear of the cone is in the cone’s own gas. At a free-stream Reynolds number of $10^6$ per metre — Mach 15 near forty kilometres — that radius is 22 millimetres for a laminar layer and 74 for a turbulent one. At $10^5$ per metre, higher up, it is 40 and 96; at $10^7$, lower down, 12 and 57.

These are nose radii of a few per cent of the base radius, which is where re-entry cone designers actually choose them, and the coincidence is not one. A nose blunt enough to survive the heating is blunt enough to put a measurable part of the cone inside its sheath, and the choice of nose radius is in part a choice of where along the body the boundary layer’s edge stops being the nose’s gas.

The nose’s own heating, and the trade #

The nose is blunted in the first place because a sharp tip cannot survive. The convective heating at a stagnation point falls as the inverse square root of the nose radius, so a tip ten times blunter is heated three times less per square metre; that is the reason for every rounded nose cap since the first re-entry vehicles. Blunting also lengthens the sheath, in proportion to the nose radius times the cube root of its Reynolds number, so the trade has two sides along the body as well as at the tip. The cone behind a blunter nose spends more of its length with a hot, low-density gas at its boundary layer’s edge, which heats it differently and delays its transition; behind a sharper nose the layer reaches the cone’s own gas early, transitions sooner and heats the rear of the cone more. The critical nose radius computed here is the point at which that second effect changes sides at the base, and the numbers say it sits within the range of nose radii that the first effect already asks for.

The heating along the way is driven from a recovery temperature set by the edge conditions, as a wall that heats itself found for any compressible layer, and inside the sheath those conditions are the nose’s; and the heating ratio between sheath and cone gas, about two-thirds on the cylinder, applies here only over the first few hundred nose radii. On the cone, unlike the cylinder, the ratio returns to one before the body ends, and the rear of the cone is heated as a sharp cone would be.

The convention behind the numbers #

Lengths are in nose radii from the nose; the layer starts at the tangency, one nose radius back. The nose Reynolds number is on the free stream’s density, speed and viscosity and the nose radius. The sheath’s edge is the streamline whose entropy excess over the sharp cone’s is half the axis’s, the measure used on the cylinder; a different fraction moves the distances, and they move together. Swallowing means the boundary layer’s edge has reached that streamline, not that the entropy layer has vanished. The gas is perfect with $\gamma = 1.4$ and viscosity proportional to temperature to the 0.7 power.

What the model leaves out #

The nose region’s own pressure. The edge is held at the sharp cone’s surface pressure from the tangency on; in reality the flow overexpands round the shoulder and recompresses within a few nose radii, which changes the first few nose radii of the layer’s growth and very little after. Real-gas effects. At Mach 15 the gas behind the nose’s shock is dissociating and its ratio of specific heats is no longer a number, which changes the sheath’s temperature and its thickness. Wall temperature. A cold wall thins the layer and slows its growth. Transition. The laminar and turbulent answers are bounds on a layer that changes between them, and where it does depends on the sheath. The turbulent estimate is the one-seventh power law with local edge properties, uncorrected for compressibility, which changes it by a factor of order two. A thin shock layer. At higher Mach numbers and larger half-angles the cone’s own shock lies close to its surface, a shock that lies on the body, and the sheath and the boundary layer then share a layer too thin to treat as separate; the model’s picture of a boundary layer eating into an inviscid sheath needs the shock layer to be thick against the boundary layer. And angle of attack, which puts the sheath on one side of the body and the cone’s gas on the other. Ablation. A re-entry nose made of an ablator recedes and rounds as it heats, so its radius is not fixed through a trajectory. Every distance here is in nose radii, so a nose that grows blunter carries the end of its sheath back along the body as it does, and by the argument of the closing section carries the delayed transition with it; the vapour an ablator blows into the layer also thickens it and changes its composition, an effect of the same order as the wall temperature’s and not computed here.

Who mapped the swallowing #

The entropy layer of a blunted body was described in the hypersonic boundary-layer theory of the late 1950s, and the swallowing distance became a working quantity in the re-entry programmes of the 1960s, where Rotta and others estimated it by mass balances of this kind. Stetson’s wind-tunnel experiments on blunted cones in the 1970s and 1980s measured transition moving aft as the nose was blunted and then, past a critical bluntness, moving forward again — the reversal being one of the long-standing puzzles of the subject. Billig’s shock-shape correlations date from 1967; Lees’s similarity form from 1956; Mangler’s transformation from 1948. The calculation here assembles them for one question, and its result, that the cone’s own growth rather than the nose’s sheath sets the answer, is the transformation’s factor read on mass.

Still open: the transition the sheath delays #

The boundary layer inside the sheath has a lower edge Reynolds number than the cone’s gas would give it, which delays its transition; past the swallowing distance the layer meets the cone’s own gas and its Reynolds number jumps. That is the mechanism usually given for the aft movement of transition with blunting, and for its reversal when the nose is blunted too far — the layer then spends so long in the sheath that disturbances it carries from the nose have time to grow. The next calculation follows the layer’s edge Reynolds number and momentum thickness along the cone through the swallowing distance, applies a stated transition criterion, and asks at what nose radius the transition location stops moving aft and starts moving forward — whether the reversal is where the swallowing distance reaches the transition point, as the simplest reading of the experiments suggests.