Formula and Method for the Oblique Shock Calculator
When a supersonic flow (M1 > 1) is turned through an angle by a wedge, ramp, or corner, it passes through an oblique shock wave inclined at a shock angle beta to the oncoming flow. The upstream Mach number M1, the flow deflection angle theta, the shock angle beta, and the specific heat ratio gamma are linked by the theta-beta-M relation: tan(theta) = 2cot(beta) x (M1^2 sin^2(beta) - 1) / [M1^2 (gamma + cos2beta) + 2]. This calculator solves that equation numerically for beta, then applies the normal-shock jump conditions to the velocity component normal to the wave to find the downstream Mach number and the pressure ratios.
How the calculation works
For a chosen M1 and gamma, the theta-beta-M relation has, at most, two valid shock angles for a given deflection angle theta: a smaller weak-shock angle (the solution almost always seen in real supersonic flows over wedges and inlet ramps) and a larger strong-shock angle (only realized when something downstream, such as a blunt duct, forces it). The calculator first finds the maximum deflection angle theta_max the flow can sustain at that M1 by locating the peak of theta(beta), then solves for beta on the requested branch. Once beta is known, only the Mach-number component normal to the shock, M1n = M1 sin(beta), is treated as passing through a normal shock: M2n^2 = [1 + (gamma-1)/2 x M1n^2] / [gamma x M1n^2 - (gamma-1)/2], p2/p1 = 1 + 2gamma/(gamma+1) x (M1n^2 - 1), and p02/p01 follows from the normal-shock stagnation-pressure-loss formula. Because the tangential velocity component is unchanged across the wave, the downstream Mach number is recovered as M2 = M2n / sin(beta - theta).
Common mistakes
- Ignoring the detachment limit: if theta exceeds theta_max for the entered M1, no attached oblique shock solution exists — the shock detaches and curves into a bow shock, which this straight-shock formula does not describe.
- Mixing up M1 and M1n: normal-shock tables and formulas use the normal component M1 sin(beta), not the freestream Mach number M1 itself — plugging M1 directly into normal-shock relations gives the wrong answer.
- Assuming the strong solution: nearly all unconfined supersonic flows (wedges, ramps, fins) settle on the weak-shock branch; the strong-shock branch requires a specific downstream pressure condition to be physically realized.
- Working in the wrong angle units: beta and theta must be in radians inside the trigonometric terms if you evaluate the theta-beta-M relation by hand — this calculator converts degrees internally.
Real-world applications
- Supersonic aircraft and missile nose/inlet design, where oblique shocks compress incoming air with far less total-pressure loss than a single normal shock.
- Multi-ramp supersonic and scramjet inlets, which use a series of oblique shocks to slow and compress flow efficiently before a final near-normal shock.
- Supersonic wind-tunnel and schlieren-photography experiments, where the measured shock angle off a wedge is used to back out the freestream Mach number.
- Wave-drag and lift estimation for supersonic wedge airfoils and control surfaces.
Frequently Asked Questions
What is the difference between a weak and a strong oblique shock?
For a given upstream Mach number and deflection angle, the theta-beta-M relation has two mathematically valid shock angles. The weak-shock solution has the smaller beta, keeps the downstream Mach number M2 above (or only slightly below) 1 in most cases, and is the solution observed in almost all unconfined supersonic flows over wedges and ramps. The strong-shock solution has a much larger beta (closer to 90 degrees), always leaves M2 well below 1, and only occurs when a downstream constraint, such as a duct or high back pressure, forces the flow onto that branch.
What happens if the deflection angle exceeds theta_max?
Every upstream Mach number has a maximum flow deflection angle theta_max it can turn through with an attached, straight oblique shock. If your theta input is greater than theta_max for the given M1, no attached-shock solution exists — the shock detaches from the corner and forms a curved bow shock standing off the body, which is not described by the straight-shock theta-beta-M equation. Reduce theta or increase M1 to bring the flow back within the attached-shock range.
How is the shock angle beta related to the Mach angle?
The Mach angle mu = arcsin(1/M1) is the limiting case of an oblique shock as the deflection angle theta approaches zero — an infinitesimally weak Mach wave. The shock angle beta always satisfies mu ≤ beta ≤ 90 degrees, with beta = 90 degrees corresponding to a normal shock (where theta is again zero, but the flow experiences the maximum possible pressure rise and total-pressure loss for that M1).
Why does the specific heat ratio gamma matter, and what value should I use?
Gamma (the ratio of specific heats, cp/cv) sets how much a gas compresses and heats up for a given Mach number and shock strength. Air behaves as a diatomic gas with gamma of about 1.4 at normal flight temperatures, which is the right choice for most aerospace problems. Use a lower value, around 1.2 to 1.3, for hot combustion gases or CO2-rich mixtures, and a higher value, about 1.667, for monatomic gases such as helium or argon.
Common Tip Percentages
| Service | Typical Tip Range | Notes |
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| Category | % of Net Income | Examples |
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Common Measurement Conversions for Everyday Use
| Quantity | US / Imperial | Metric Equivalent |
|---|
| 1 cup | 8 fl oz | 237 mL |
| 1 pint | 16 fl oz | 473 mL |
| 1 quart | 32 fl oz | 946 mL |
| 1 gallon | 128 fl oz | 3.785 L |
| 1 lb | 16 oz | 453.6 g |
| 1 ft | 12 in | 30.48 cm |