Oblique shocks and the theta-beta-Mach relation

Why it matters

Whenever a supersonic stream is turned into itself — by a wedge, a ramp in an intake, the nose of a missile or a wing leading edge — an oblique shock forms. Oblique shocks compress the flow with much less stagnation-pressure loss than a single normal shock, which is why supersonic intakes use a series of ramps, and why sharp, thin leading edges are used on supersonic wings.

Key ideas

Geometry. A uniform stream at M₁ meets a wedge of half-angle θ (the flow deflection angle). An attached straight shock forms at angle β (the wave angle) to the upstream flow. Downstream the flow is parallel to the wedge surface.

Reduction to a normal shock. Split the velocity into components normal and tangential to the shock. Momentum conservation along the shock shows the tangential component is unchanged: w₁ = w₂. The normal component behaves exactly like flow through a normal shock with upstream Mach number Mₙ₁ = M₁·sin β. So all the normal-shock ratios apply with Mₙ₁ in place of M₁, and Mₙ₁ > 1 is required. This also gives β ≥ μ = sin⁻¹(1/M₁): the weakest oblique shock is a Mach wave.

θ–β–M relation. Geometry plus the normal-component density ratio gives the link between θ, β and M₁ (see Formulas). Features of the θ–β–M chart:

• For a given M₁ and θ below a maximum θmax, there are two solutions: a weak shock (smaller β, usually supersonic downstream) and a strong shock (larger β, always subsonic downstream). In external flow over wedges and ramps the weak solution is what normally occurs; the strong one appears only if a high downstream pressure forces it.
• θ = 0 at β = μ (Mach wave) and at β = 90° (normal shock).
• If θ > θmax (about 23° at M = 2, 29.8° at M = 2.5, 34.1° at M = 3 for air) no attached straight shock exists. The shock detaches and stands ahead of the body as a curved bow shock, normal on the axis and weakening to a Mach wave far away.
• For fixed θ, increasing M₁ reduces β (the shock leans back closer to the surface) and raises θmax.
• The downstream Mach number is M₂ = Mₙ₂ / sin(β − θ).

Reflection and intersection. An oblique shock hitting a flat wall reflects as a second shock that turns the flow back parallel to the wall. If the deflection needed exceeds θmax for the reduced Mach number, a Mach reflection (with a short normal stem) forms instead.

Cone vs wedge. For the same half-angle, a cone produces a weaker shock (smaller β) than a wedge, because the flow can spread around it in three dimensions (Taylor–Maccoll solution; use cone charts).

Formulas

Mₙ₁ = M₁·sin β

tan θ = 2·cot β · (M₁²·sin²β − 1) / (M₁²·(γ + cos 2β) + 2) — θ–β–M relation.

Mₙ₂² = (1 + (γ−1)/2 · Mₙ₁²) / (γ·Mₙ₁² − (γ−1)/2); then M₂ = Mₙ₂ / sin(β − θ).

p₂/p₁ = 1 + 2γ/(γ+1) · (Mₙ₁² − 1)

ρ₂/ρ₁ = (γ+1)·Mₙ₁² / (2 + (γ−1)·Mₙ₁²) = tan β / tan(β − θ)

T₂/T₁ = (p₂/p₁)·(ρ₁/ρ₂); p₀₂/p₀₁ from the normal-shock relation with Mₙ₁.

μ = sin⁻¹(1/M₁) — lower limit of β.

Symbols: θ flow deflection (wedge half-angle), β shock wave angle measured from the upstream flow direction (degrees or radians); M₁, M₂ upstream and downstream Mach numbers; Mₙ Mach number of the normal component; γ ratio of specific heats; p, ρ, T in Pa, kg/m³, K. Valid for a straight attached shock in a calorically perfect gas.

Worked examples

Example 1 (standard). Air at M₁ = 2.5 flows over a wedge with θ = 10°. Find β (weak solution), Mₙ₁ and p₂/p₁. γ = 1.4.

1. Lower bound: μ = sin⁻¹(1/2.5) = 23.6°; θmax ≈ 29.8°, so the shock is attached.
2. Solve the θ–β–M relation by trial for tan θ = 0.1763. At β = 31.85°: M₁²sin²β − 1 = 6.25 × 0.2785 − 1 = 0.7405, M₁²(γ + cos 2β) + 2 = 6.25 × 1.8430 + 2 = 13.519, 2·cot β = 3.2194; product 3.2194 × 0.7405/13.519 = 0.1763. So β ≈ 31.9°.
3. Mₙ₁ = 2.5 × sin 31.85° = 1.319.
4. p₂/p₁ = 1 + 1.1667 × (1.7405 − 1) = 1.864.

Answer: β ≈ 31.9°, Mₙ₁ ≈ 1.32, p₂/p₁ ≈ 1.86 (and M₂ ≈ 2.09).

Example 2 (GATE level). A ramp turns an M₁ = 3.0 air stream by 15°. The weak shock angle is β = 32.24°. Find M₂, p₂/p₁ and p₀₂/p₀₁, and compare with a normal shock at M = 3.

1. Mₙ₁ = 3.0 × sin 32.24° = 1.600.
2. Mₙ₂² = (1 + 0.2 × 2.561)/(1.4 × 2.561 − 0.2) = 1.5122/3.3854 = 0.4467, Mₙ₂ = 0.668.
3. M₂ = 0.668 / sin(32.24° − 15°) = 0.668 / 0.2964 = 2.25.
4. p₂/p₁ = 1 + 1.1667 × (2.561 − 1) = 2.82.
5. Normal-shock table at Mₙ₁ = 1.60: p₀₂/p₀₁ = 0.895.
6. A normal shock at M = 3 gives p₀₂/p₀₁ = 0.328.

Answer: M₂ ≈ 2.25, p₂/p₁ ≈ 2.82, p₀₂/p₀₁ ≈ 0.895, versus 0.328 for a normal shock — the reason intakes compress through oblique shocks first.

Common mistakes

• Using M₁ instead of Mₙ₁ = M₁ sin β in the normal-shock formulas.
• Forgetting the sin(β − θ) step when finding M₂: M₂ ≠ Mₙ₂/sin β.
• Picking the strong-shock root for an attached shock on a wedge in open flow.
• Measuring β from the wedge surface instead of from the upstream flow direction.
• Assuming an attached shock exists for any θ — check θ < θmax first.
• Treating cone and wedge shocks as identical.

For GATE AE

Questions give two of M₁, θ, β and ask for the third (with a θ–β–M chart, a table or direct substitution when β is given), then ask for pressure ratio, M₂ or stagnation-pressure loss via Mₙ₁. Concept questions cover weak and strong solutions, detachment, the Mach-wave limit and reflected shocks. Practise substituting a given β directly into the θ–β–M relation — that is quick by calculator — and remember θmax values roughly.

Quick check

1. What is the smallest possible shock angle at M₁ = 2?
2. If M₁ = 3 and β = 30°, what is Mₙ₁?
3. Is the flow behind a strong oblique shock subsonic or supersonic?
4. What happens if the wedge angle exceeds θmax?
5. Which velocity component is unchanged across an oblique shock?

Answers: 1. 30°; 2. 1.5; 3. subsonic; 4. the shock detaches into a curved bow shock; 5. the tangential component.