Oblique Shock Wave Calculator

Find the deflection angle theta, downstream Mach number, and shock jump ratios from upstream Mach number and shock angle beta.

🔺 Oblique Shock Wave Calculator
Upstream Mach number (M1)2.5
-
1.056
Shock angle (β)30
deg
190
-
Deflection Angle θ
Downstream Mach M2
Pressure Ratio p2/p1
Density Ratio ρ2/ρ1
Temperature Ratio T2/T1
Stagnation Pressure Ratio p02/p01
Mach Angle μ
Step-by-step working

🔺 What is the Oblique Shock Wave Calculator?

This oblique shock wave calculator finds the deflection angle theta, downstream Mach number M2, and the pressure, density, temperature, and stagnation pressure ratios across an oblique shock. Enter the upstream Mach number M1, the shock angle beta, and the ratio of specific heats gamma, and it returns every downstream quantity along with a chart of the theta-beta-M relation for that Mach number.

An oblique shock forms when supersonic flow meets a wedge, a cone, or a compression corner and is turned into itself. Unlike a normal shock, which stands perpendicular to the flow and always makes the downstream Mach number subsonic, an oblique shock is inclined at an angle beta and only decelerates the velocity component normal to the shock, so the downstream flow often stays supersonic. This is the geometry behind supersonic wedge airfoils, engine inlet compression ramps, and the shock cone standing off a supersonic aircraft's nose.

This calculator takes the shock angle beta as an input rather than the deflection angle theta. Solving for theta from a known M1 and beta has a direct closed-form formula, while solving for beta from a known theta requires iterative root-finding, because two different shock angles (a weak and a strong solution) can produce the same theta for the same M1. Taking beta as the input keeps the whole calculation direct and instant.

This calculator is a general compressible-flow reference tool for gas dynamics coursework, supersonic wedge and inlet analysis, and comparing weak-shock behavior against the limiting normal-shock case at beta = 90 degrees.

📐 Formula

tan(θ)  =  2cot(β) × (M1²sin²β − 1) / (M1²(γ+cos2β) + 2)
M1 = upstream (supersonic) Mach number, β = oblique shock angle, θ = flow deflection angle
γ = ratio of specific heats (1.4 for air)
Validity: requires β greater than the Mach angle μ = asin(1/M1), and β ≤ 90°
Mn1  =  M1 sin(β)
Mn1 feeds the standard normal-shock relations to give Mn2, p2/p1, ρ2/ρ1, T2/T1, and p02/p01, exactly as in a normal shock but using the Mach number normal to the shock instead of M1 directly.
M2  =  Mn2 / sin(β − θ)
Example: M1=2.5, β=30°, γ=1.4: θ ≈ 7.99°, Mn1 = 1.25, M2 ≈ 2.1688, p2/p1 ≈ 1.6563, p02/p01 ≈ 0.9871.

📖 How to Use This Calculator

Steps

1
Enter the upstream Mach number. Use the slider or type an upstream Mach number M1 greater than 1.
2
Enter the shock angle. Use the slider or type a shock angle beta greater than the Mach angle and up to 90 degrees.
3
Read the deflection angle and shock ratios. See theta, M2, p2/p1, rho2/rho1, T2/T1, p02/p01, and the Mach angle, along with the theta-beta-M chart.

💡 Example Calculations

Example 1 - Moderate wedge shock (default widget values)

1
M1 = 2.5, β = 30°, γ = 1.4 (air). Mach angle μ = asin(1/2.5) = 23.58°, and 30° exceeds it, so the shock is valid.
2
θ = 7.99°, Mn1 = M1sinβ = 1.2500
3
M2 = 2.1688 (stays supersonic), p2/p1 = 1.6563, ρ2/ρ1 = 1.4286, T2/T1 = 1.1594, p02/p01 = 0.9871
θ = 7.99°, M2 = 2.1688, p2/p1 = 1.6563
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Example 2 - Steeper shock angle, larger deflection

1
M1 = 2.0, β = 40°, γ = 1.4 (air). Mach angle μ = asin(1/2.0) = 30.00°, and 40° exceeds it.
2
θ = 10.62°, Mn1 = 1.2856
3
M2 = 1.6173, p2/p1 = 1.7615, ρ2/ρ1 = 1.4906, T2/T1 = 1.1818, p02/p01 = 0.9818
θ = 10.62°, M2 = 1.6173, p2/p1 = 1.7615
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Example 3 - Higher Mach number, shallow shock angle

1
M1 = 3.0, β = 25°, γ = 1.4 (air). Mach angle μ = asin(1/3.0) = 19.47°, and 25° exceeds it.
2
θ = 7.28°, Mn1 = 1.2679
3
M2 = 2.6381, p2/p1 = 1.7087, ρ2/ρ1 = 1.4597, T2/T1 = 1.1706, p02/p01 = 0.9845
θ = 7.28°, M2 = 2.6381, p2/p1 = 1.7087
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Example 4 - Invalid input: shock angle below the Mach angle

1
M1 = 2.5, β = 20°, γ = 1.4. Mach angle μ = asin(1/2.5) = 23.58°, which is greater than 20°.
2
No oblique shock can exist at a shock angle below the Mach angle for this M1. The calculator shows an error instead of a result.
Error: shock angle must exceed the Mach angle (23.58°)
Try this example →

❓ Frequently Asked Questions

What is an oblique shock wave?+
An oblique shock is a shock wave inclined at an angle beta to the upstream flow direction, formed when supersonic flow is turned into itself by a wedge, cone, or compression corner. Unlike a normal shock, it deflects the flow by an angle theta while also decelerating the component of velocity normal to the shock.
Why does this calculator ask for the shock angle beta instead of the deflection angle theta?+
Given M1 and beta, the deflection angle theta has a direct closed-form formula. Given M1 and theta instead, finding beta requires solving a cubic-like equation numerically since two valid shock angles (weak and strong) can produce the same theta. Taking beta as the input avoids that iterative root-finding step entirely.
What is the Mach angle and why does beta have to exceed it?+
The Mach angle mu = asin(1/M1) is the angle of the weakest possible disturbance (a Mach wave) that supersonic flow can generate. An oblique shock only exists for beta strictly greater than mu; at beta = mu the shock strength drops to zero and it becomes a Mach wave, and beta below mu is not physically possible for this M1.
What happens when beta equals 90 degrees?+
At beta = 90 degrees the oblique shock relations reduce exactly to the normal shock relations: the deflection angle theta becomes zero and Mn1 equals M1. A normal shock is simply the limiting case of an oblique shock at the maximum possible shock angle.
What is the difference between the weak and strong shock solutions?+
For a given M1 and deflection angle theta less than theta_max, two shock angles beta satisfy the theta-beta-M relation: a smaller beta (weak shock, usually leaves M2 supersonic) and a larger beta (strong shock, always leaves M2 subsonic). This calculator takes beta directly as an input, so it always returns whichever solution corresponds to the beta you enter, and the weak branch is what naturally forms on an unconfined wedge.
Can the downstream Mach number M2 stay supersonic after an oblique shock?+
Yes. Unlike a normal shock, where M2 is always subsonic, an oblique shock only decelerates the velocity component normal to the shock. For most weak-shock solutions, M2 remains supersonic, which is why oblique shocks are used in supersonic inlets to slow the flow gradually while avoiding the large total-pressure loss of a single normal shock.
What is theta_max and why does it matter?+
theta_max is the largest deflection angle a given M1 can achieve with an attached oblique shock. If a wedge half-angle exceeds theta_max for the approaching Mach number, no attached oblique shock solution exists, and the shock detaches into a curved bow shock standing off the wedge.
Why does total pressure loss depend on the shock angle?+
The stagnation pressure ratio p02/p01 depends only on the Mach number normal to the shock, Mn1 = M1 sin(beta). A smaller beta gives a smaller Mn1 and therefore a weaker shock with less total-pressure loss, which is exactly why multi-shock supersonic inlets use a series of shallow oblique shocks rather than one steep shock.
Does the ratio of specific heats gamma change the result much?+
Yes, though the effect is secondary to M1 and beta. Air uses gamma = 1.4, but combustion products, monatomic gases, and high-temperature real-gas effects have different gamma values, and this calculator lets you override the default to model any calorically perfect gas.
How is this calculator different from the Normal Shock Wave Calculator?+
A normal shock is a special case of an oblique shock at beta = 90 degrees, always facing the flow straight on and always leaving M2 subsonic. This calculator handles the general inclined case for any beta above the Mach angle, which is the relevant geometry for wedges, cones, and supersonic inlet ramps rather than a shock standing perpendicular to the flow.
Is the Mach angle the same thing as the shock angle?+
No. The Mach angle mu depends only on M1 and marks the angle of an infinitesimally weak disturbance, while the shock angle beta is always steeper than mu and marks the angle of an actual finite-strength shock. Every real oblique shock has beta strictly greater than mu, never equal to it.

What is an oblique shock wave?

An oblique shock is a shock wave inclined at an angle beta to the upstream flow direction, formed when supersonic flow is turned into itself by a wedge, cone, or compression corner. Unlike a normal shock, it deflects the flow by an angle theta while also decelerating the component of velocity normal to the shock.

Why does this calculator ask for the shock angle beta instead of the deflection angle theta?

Given M1 and beta, the deflection angle theta has a direct closed-form formula. Given M1 and theta instead, finding beta requires solving a cubic-like equation numerically since two valid shock angles (weak and strong) can produce the same theta. Taking beta as the input avoids that iterative root-finding step entirely.

What is the Mach angle and why does beta have to exceed it?

The Mach angle mu = asin(1/M1) is the angle of the weakest possible disturbance (a Mach wave) that supersonic flow can generate. An oblique shock only exists for beta strictly greater than mu; at beta = mu the shock strength drops to zero and it becomes a Mach wave, and beta below mu is not physically possible for this M1.

What happens when beta equals 90 degrees?

At beta = 90 degrees the oblique shock relations reduce exactly to the normal shock relations: the deflection angle theta becomes zero and Mn1 equals M1. A normal shock is simply the limiting case of an oblique shock at the maximum possible shock angle.

What is the difference between the weak and strong shock solutions?

For a given M1 and deflection angle theta less than theta_max, two shock angles beta satisfy the theta-beta-M relation: a smaller beta (weak shock, usually leaves M2 supersonic) and a larger beta (strong shock, always leaves M2 subsonic). This calculator takes beta directly as an input, so it always returns whichever solution corresponds to the beta you enter, and the weak branch is what naturally forms on an unconfined wedge.

Can the downstream Mach number M2 stay supersonic after an oblique shock?

Yes. Unlike a normal shock, where M2 is always subsonic, an oblique shock only decelerates the velocity component normal to the shock. For most weak-shock solutions, M2 remains supersonic, which is why oblique shocks are used in supersonic inlets to slow the flow gradually while avoiding the large total-pressure loss of a single normal shock.

What is theta_max and why does it matter?

theta_max is the largest deflection angle a given M1 can achieve with an attached oblique shock. If a wedge half-angle exceeds theta_max for the approaching Mach number, no attached oblique shock solution exists, and the shock detaches into a curved bow shock standing off the wedge.

Why does total pressure loss depend on the shock angle?

The stagnation pressure ratio p02/p01 depends only on the Mach number normal to the shock, Mn1 = M1 sin(beta). A smaller beta gives a smaller Mn1 and therefore a weaker shock with less total-pressure loss, which is exactly why multi-shock supersonic inlets use a series of shallow oblique shocks rather than one steep shock.

Does the ratio of specific heats gamma change the result much?

Yes, though the effect is secondary to M1 and beta. Air uses gamma = 1.4, but combustion products, monatomic gases, and high-temperature real-gas effects have different gamma values, and this calculator lets you override the default to model any calorically perfect gas.

How is this calculator different from the Normal Shock Wave Calculator?

A normal shock is a special case of an oblique shock at beta = 90 degrees, always facing the flow straight on and always leaving M2 subsonic. This calculator handles the general inclined case for any beta above the Mach angle, which is the relevant geometry for wedges, cones, and supersonic inlet ramps rather than a shock standing perpendicular to the flow.