How this instrument works
An oblique shock is the angled wave a supersonic flow forms against a wedge, ramp, or swept leading edge, rather than the flat shock a blunt nose takes head-on. The upstream flow, moving at Mach number M1, meets a wave standing at angle β to its own direction; crossing it, the flow slows, compresses, and bends toward the surface by the deflection angle θ. The θ-β-M relation is the algebra that locks these three numbers together for a fixed ratio of specific heats γ, held here at 1.4, the value for air.
The formula's shape comes from a standard trick: split the upstream velocity into a piece perpendicular to the shock, M1 sinβ, and a piece parallel to it. Only the perpendicular piece behaves like a plain normal shock — it slows and compresses; the parallel piece passes through unchanged, because nothing at a shock front pushes sideways along its own surface. Recombining the two afterward is what produces the cotβ and sin²β terms in the numerator, and why the deflection depends on both the shock angle and the approach Mach number rather than either alone.
The relation only holds between two boundaries. Below the Mach angle, μ = asin(1/M1), no attached shock can stand at all — a disturbance simply cannot outrun the flow far enough upstream to form one, so β must exceed μ. And for any given M1 there is a maximum deflection, θmax, that no wedge angle can beat; push the ramp past it and the attached oblique shock cannot survive, so it detaches into a curved bow shock standing off the nose — the situation a blunt reentry capsule lives in permanently, and the reason supersonic inlet ramps are never cut steeper than the flow allows.
- Enter the Upstream Mach number M1 for the approaching flow; it must exceed 1, since an oblique shock is a purely supersonic phenomenon.
- Enter the Shock wave angle β, the angle the wave itself makes with the oncoming flow direction, somewhere between the Mach angle and 90°.
- Read the Flow deflection angle θ, the angle the flow bends through — equivalently, the wedge or ramp half-angle that produces exactly that shock.
- Raise β toward 90° to see θ climb, peak at θmax, then fall back toward zero at the normal-shock limit, where the wave stands square across the flow.
Worked example — a 30° shock in Mach 2.5 flow
Take a supersonic inlet ramp meeting an approach flow at Mach 2.5, with the shock standing at β = 30° (0.5236 rad) to the oncoming air. First cot β = cot 30° = 1.7321, and M1² sin²β − 1 = 6.25 × 0.25 − 1 = 0.5625, so the numerator is 2 × 1.7321 × 0.5625 = 1.9486. The denominator is M1²(γ + cos 2β) + 2 = 6.25 × (1.4 + 0.5) + 2 = 13.875, which gives tan θ = 1.9486 ⁄ 13.875 = 0.1404.
Taking the arctangent gives θ = 0.1395 rad, which is 7.99° — call it 8° of flow turning for a shock angled at 30°. That is the figure a ramp designer reads straight off this instrument: cut the first ramp surface 8° into the flow and the shock that forms sits at 30°, exactly where the intake was sized for. It also sits well inside the weak-shock branch — at Mach 2.5 the maximum possible deflection is close to 29.8°, reached near β = 65°, so an 8° ramp is nowhere near the detachment limit that would push the shock off the inlet lip.
Questions
What's the difference between the shock angle and the deflection angle?
The shock angle β is measured from the approaching flow direction to the wave itself; the deflection angle θ is how far the flow actually turns after crossing it. β is always the larger of the two — they only converge in the limit of a vanishingly weak shock, where the wave angle collapses toward the Mach angle and θ shrinks toward zero.
Why is gamma fixed instead of being a field I can set?
Gamma (γ) is the ratio of specific heats of the gas crossing the shock, and this instrument fixes it at 1.4, the value for air and other diatomic gases at ordinary flight temperatures. Combustion products, helium, or hot dissociating air need a different γ and a separate calculation; the formula's shape is unchanged, only that one constant moves.
What happens if I enter a shock angle below the Mach angle?
No physical shock can exist there. Oblique shocks only stand at angles between the Mach angle, μ = asin(1/M1), and 90° for a normal shock; below μ a disturbance cannot outrun the flow to reach that far upstream. At M1 = 2.5 the Mach angle is about 23.6°, so the 30° shock angle in the worked example sits safely inside the valid range.
Is there a limit to how far a shock can deflect the flow?
Yes — for every upstream Mach number there is a maximum deflection, θmax, reached at a shock angle roughly two-thirds of the way from the Mach angle to 90°. At Mach 2.5 that peak is close to 29.8°, near β = 65°, well above the 8° in the worked example. Ask for more turning than θmax allows and no attached oblique shock can supply it; the flow forms a detached bow shock instead.
Does the wedge's own angle equal the shock angle or the deflection angle?
The deflection angle. A wedge with a 10° half-angle turns the flow by exactly 10°, so θ = 10° describes the wedge; the shock it produces stands at whatever steeper angle β the θ-β-M relation demands for that Mach number. Mixing the two up is the single most common error in reading an oblique-shock chart.
Why does entering β directly avoid the weak-versus-strong shock ambiguity?
Because the formula runs forward: for any β you choose, tan θ has exactly one value, so nothing needs choosing between. The weak-and-strong split only shows up when working backward from a target θ to find β — that equation has two roots, and picking between them needs an extra criterion this angle-in, angle-out instrument never has to apply.