The θ–β–M relation
tan θ = 2 cot β (M1² sin²β − 1) / (M1² (γ + cos 2β) + 2)
The wave angle β lies between the Mach angle μ = arcsin(1/M1) and 90°. For a turning angle below the maximum deflection there are two solutions: a weak shock (smaller β, usually still supersonic behind the wave) and a strong shock (β closer to 90°, usually subsonic behind). At M1 = 2, θ = 10°, γ = 1.4, the weak wave is at β ≈ 39.31° and the strong wave at β ≈ 83.70°. Downstream conditions come from a normal shock at Mn1 = M1 sin β, then M2 = Mn2 / sin(β − θ), using the same identities as the normal shock calculator.
Sources
NACA Report 1135 (oblique-shock charts and the θ–β–M equation); NASA Glenn compressible-flow notes; Anderson, Modern Compressible Flow. For a centered expansion, see the Prandtl–Meyer calculator. For isentropic stagnation relations, see isentropic flow.
Frequently asked questions
Which solution, weak or strong, should I use?
The weak shock is the usual attached solution on a wedge or ramp in an unbounded flow. The strong solution appears when the downstream pressure is forced high, for example by a back pressure in a duct. This page reports both.
What happens if I ask for 30° of turn at Mach 2?
That is above θmax (about 23° at M = 2, γ = 1.4), so an attached oblique shock cannot turn the flow that far. The page flags the shock as detached rather than inventing a bow-shock stand-off distance.
How do you get pressure and temperature ratios?
The component of Mach normal to the wave, Mn1 = M1 sin β, is put through the normal-shock relations (the same ones as the normal-shock page). Tangential velocity is unchanged, which is why M2 = Mn2 / sin(β − θ).