Oblique Shock Calculator (θ–β–M: shock angle, M₂ and pressure ratios)

Understanding the oblique shock calculator — concept and principle

This calculator solves the θ–β–M relation for a straight oblique shock: from the upstream Mach number, the flow deflection (wedge half-angle) and γ, it finds the shock angle and the downstream Mach number, pressure, density, temperature and total-pressure ratios. It also gives the maximum deflection for that Mach number. It is used in supersonic aerodynamics and inlet design courses.

Only the velocity component normal to the shock is shocked, so the flow slows but stays supersonic.

How it is calculated

tan θ = 2 cot β (M₁² sin²β − 1) ÷ (M₁² (γ + cos 2β) + 2), solved numerically for β

Normal Mach number M_n1 = M₁ sin β

p₂/p₁ = 1 + 2γ (M_n1² − 1) ÷ (γ + 1)

ρ₂/ρ₁ = (γ + 1) M_n1² ÷ ((γ − 1) M_n1² + 2)

M₂ = M_n2 ÷ sin(β − θ), with M_n2 from the normal-shock relation

Weak or strong solution can be chosen; perfect gas with constant γ.

The principle

When supersonic flow meets a wedge, it cannot "feel" the obstacle ahead, so it turns abruptly through a shock wave inclined at β. Only the velocity component normal to the shock changes; the tangential component carries through unchanged. That lets the oblique shock be treated as a normal shock at Mach M₁ sin β. For each M₁ there is a maximum turning angle; beyond it the shock detaches. Below it, two angles satisfy the equations, and the weak one is what normally forms on a wedge.

Worked example

Default inputs: M₁ = 2, θ = 10°, γ = 1.4, weak shock.

Solving θ–β–M gives β = 39.314°, well above the Mach angle 30°.
M_n1 = 2 × sin 39.314° = 1.2671.
p₂/p₁ = 1.7066; ρ₂/ρ₁ = 1.4584; T₂/T₁ = 1.1702.
M₂ = 1.6405; total-pressure ratio = 0.9846.
Maximum deflection for M₁ = 2 is 22.974°.

Each point is the calculator re-run with only “Flow deflection (wedge half-angle) θ (°)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Flow deflection (wedge half-angle) θ (°)	Shock wave angle β (°)
2.5	32.073
5	34.302
7.5	36.706
10	39.314
12.5	42.169
15	45.344
20	53.423

How to read the result

A weak oblique shock keeps the flow supersonic and loses little total pressure, here only about 1.5%, which is why supersonic inlets use several oblique shocks rather than one normal shock. The strong solution gives subsonic flow behind the shock and much larger losses. If θ exceeds the maximum, a curved detached bow shock forms instead.

Common mistakes

Using the strong solution for an attached wedge shock.
Entering the full wedge angle instead of the half-angle.
Applying perfect-gas relations at very high Mach numbers.

Keep in mind: Results assume a perfect gas, a straight shock and two-dimensional flow. Real inlet and vehicle design needs detailed analysis by qualified engineers.

Frequently asked questions

What is the difference between weak and strong shocks?

For the same M₁ and θ, the weak shock has a smaller angle and usually supersonic flow behind it.

What happens if θ is too large?

The shock detaches from the wedge and becomes a curved bow shock.

Why use the normal component?

Only the velocity perpendicular to the shock is changed by it.

Why is total pressure lost?

A shock is irreversible and raises entropy.

Sources: NASA Glenn Research Center, NPARC/WIND validation program oblshk.f (θ–β–M and normal-shock relations)

Checked 9 Nov 2026. NASA Glenn Research Center, NPARC/WIND validation program oblshk.f (θ–β–M relation; normal-shock relations on M₁ sin β; M₂ = M_n2/sin(β − θ)). Total-pressure ratio: standard normal-shock relation.

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