Post-shock properties for an oblique shock. Solve given β (wave angle) or θ (deflection angle) — the θ mode returns both the weak and strong solutions.
Compressible Flow · Oblique ShockInput angle type:
β from μ (Mach angle) to 90° — β = 90° gives normal shock
cp/cv — use gas presets above
Key relations:
P₂/P₁, T₂/T₁, ρ₂/ρ₁, P₀₂/P₀₁ are computed from normal shock Rankine-Hugoniot relations applied to M₁ₙ (not M₁).
Weak and strong solutions:
Detachment angles θmax — air (γ = 1.4):
| M₁ | μ [deg] | θmax [deg] | β at θmax [deg] | M₂ at θmax |
|---|---|---|---|---|
| 1.5 | 41.8 | 12.1 | 66.6 | 0.921 |
| 2.0 | 30.0 | 23.0 | 64.7 | 0.924 |
| 2.5 | 23.6 | 29.8 | 64.8 | 0.940 |
| 3.0 | 19.5 | 34.1 | 65.2 | 0.954 |
| 4.0 | 14.5 | 38.8 | 66.1 | 0.972 |
| 5.0 | 11.5 | 41.1 | 66.6 | 0.981 |
Oblique shocks occur on wedges, compression corners, and inlet ramps in supersonic flow. The shock angle β always lies between the Mach angle μ = arcsin(1/M₁) and 90°. Unlike normal shocks, the downstream flow can remain supersonic (weak solution). Total pressure loss P₀₂/P₀₁ is smaller than the equivalent normal shock — this is why supersonic inlets use ramps to process the flow through weaker oblique shocks rather than one strong normal shock.