Mach Angle Calculator — μ = sin⁻¹(1/M)

🔺 Mach Cone & Wavefront Animation

📝 Configuration

🌬️ Gas Selection

Gas Preset:

🔺 Flow Conditions

Mach Number ($M$):

Specific Heat Ratio ($\gamma$):

📐 Upstream State & Geometry

Static Pressure ($p$) [Pa]:

Static Temperature ($T$) [K]:

Gas Constant ($R$) [J/kg·K]:

Reference Distance ($x$) [m]:

Key Equations:

Mach angle:
$\mu = \sin^{-1}\!\left(\dfrac{1}{M}\right) = \sin^{-1}\!\left(\dfrac{a}{u}\right)$

Cone radius at distance $x$:
$r = x \cdot \tan\mu = \dfrac{x}{\sqrt{M^2 - 1}}$

Full cone opening angle:
$2\mu$

Limiting cases:
$M = 1 \Rightarrow \mu = 90°$ (plane wave)
$M \to \infty \Rightarrow \mu \to 0°$ (infinitely narrow cone)

📊 Results & Visualization

Configure inputs and click Calculate to view results.

ℹ️ About the Mach Angle & Mach Cone

When an object (or point disturbance source) travels through a gas at supersonic speed (M > 1), the pressure disturbances it creates cannot propagate upstream. Instead they are confined within a conical region called the Mach cone.

Key concepts:
- Mach angle μ: the half-angle of the cone, given by μ = sin⁻¹(1/M).
- Zone of action: the region inside the cone where disturbances can be felt.
- Zone of silence: the region outside the cone, undisturbed.
- At M = 1, μ = 90° (plane wave — entire downstream region is affected).
- As M → ∞, μ → 0° (infinitely narrow cone).
- A Mach wave is an infinitesimally weak wave inclined at angle μ — it is isentropic.

📘 Calculation Methodology

Mathematical Model & Theory

The Mach angle is the fundamental geometric property of supersonic flow. It represents the angle between the Mach wave (an infinitesimally weak pressure wave) and the local flow direction.

$$\mu = \sin^{-1}\!\left(\frac{1}{M}\right) = \sin^{-1}\!\left(\frac{a}{u}\right)$$ $$\tan\mu = \frac{1}{\sqrt{M^2 - 1}}$$ $$\text{Cone radius at } x: \quad r = x\cdot\tan\mu = \frac{x}{\sqrt{M^2-1}}$$ $$\text{Zone of action area: } A = \pi r^2 = \frac{\pi x^2}{M^2-1}$$

The Mach wave is the weakest possible oblique shock (θ → 0). All stronger waves (oblique shocks) are inclined at angles β > μ with respect to the flow.

Worked Engineering Example

Problem:
A bullet travels at Mach 2.0 through air at T = 300 K, p = 101.325 kPa. Compute the Mach angle and the cone radius 1 m behind the bullet.

Step-by-step:
1. Speed of sound: $a = \sqrt{1.4 \times 287.058 \times 300} = 347.2$ m/s

2. Flow velocity: $u = 2.0 \times 347.2 = 694.4$ m/s

3. Mach angle:
$\mu = \sin^{-1}(1/2) = 30.00°$

4. Full cone angle: $2\mu = 60.00°$

5. Cone radius at x = 1 m:
$r = 1.0 \times \tan(30°) = 0.5774$ m

Result: μ = 30°, cone radius = 0.577 m at 1 m downstream.

Assumptions & References

Assumptions: Inviscid flow. Infinitesimally small disturbances (linear acoustics). Steady supersonic flow (M ≥ 1). Calorically perfect gas. Uniform freestream. The Mach wave itself is isentropic (no entropy change).

References:

Anderson, J. D. Modern Compressible Flow, McGraw-Hill, 4th ed. — Ch. 4 (Mach waves and Mach angle).
Shapiro, A. H. The Dynamics and Thermodynamics of Compressible Fluid Flow, Vol. 1, Wiley, 1953.
Gupta, S. C. Applied Computational Fluid Dynamics — §7.11 (Mach angle definitions).
Liepmann, H. W., & Roshko, A. Elements of Gasdynamics, Dover, 2001.

ThermoFluidCalc Report

🔺 Mach Angle Calculator — μ = sin⁻¹(1/M)