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The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeysThe relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set $s=-\tau >0$. Its circle has center $(-Us,0)$ and radius $cs$. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse $Us$ and opposite side $cs$, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as $M$ increases: its derivative is $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.

Let $U_1,U_2$ denote the total flow speeds, keeping the normal components distinct from them. The original diagram measures $\beta$ from the initial wall direction to the shock wave itself. The outgoing velocity is turned by $\theta$, so it makes angle $\beta -\theta$ with the shock.

Resolve each velocity tangent and normal to the oblique shock. Tangential momentum conservation, with nonzero mass flux and no tangential surface force, makes the tangential velocity continuous:The normal flow obeys the strong normal shock wave result from part (b):Divide the second relation by the first to obtainThe approximation is controlled by the upstream normal Mach number $M\sin \beta \gg 1$, not just by $M\gg 1$. This distinction matters on the small-angle weak branch in part (d). The angle convention is also important: formulas using the angle to the shock normal instead would interchange sine and cosine.

Rewrite the relation from part (c) asThe angle-addition identities reduce this toFor $0<\theta <\arcsin (1/\gamma )$, the right side lies strictly between zero and one. There are two physically admissible strong oblique-shock angle branches:Both lie between $\theta$ and $\pi /2$. They merge when $\gamma \sin \theta =1$, proving the maximum deflection through a strong perfect-gas shock,For small positive $\theta$, the formal strong-shock-limit asymptotics areThe weak branch is normally observed for an unconfined supersonic stream turning through a modest compression corner. It has a smaller normal component of upstream velocity, hence a smaller downstream pressure increase; for small turns the downstream flow remains supersonic. The nearly normal strong branch can be selected by suitable downstream pressure conditions, so it is not mathematically forbidden.

The two limits must be ordered honestly. At finite $M$, the weak oblique shock approaches the Mach angle $\arcsin (1/M)$ as $\theta \to 0$, rather than zero. The displayed weak-branch asymptotic applies when the large-$M$ limit is taken first, or in the overlapping range $1/M\ll \theta \ll 1$, where $M\sin \beta \gg 1$ remains valid.

For $\theta >{\theta }_{\max }$, $\gamma \sin \theta >1$, so there is no real solution for an attached straight oblique shock satisfying the assumed geometry. The shock detaches and becomes curved, with a locally strong, nearly normal portion and a subsonic region allowing the flow to adjust to the wall. The precise detached structure depends on the global geometry and downstream conditions; the attached-shock formula cannot be continued to these turns.

For an oblique shock, boost parallel to the front by the upstream tangential speed. The transformed upstream flow is normal, so part b applies. Inviscid momentum balance has no tangential stress and therefore makes the tangential velocity continuous. Transforming back giveswhile all normal velocity, density, and pressure relations from part b remain unchanged with $M_x$ the normal Mach number.

A shock polar is the locus of possible downstream velocity vectors, or an equivalent pressure/deflection relation, as the orientation of an oblique shock varies for fixed upstream conditions. Entropy admissibility selects physical compressive shocks. The strong-shock perfect-gas velocity polar is a circle, while finite-Mach polars have a different shape.