Oblique shock

Remarks

The flow is turned towards the shock by a deflection angle $\theta$, given by the $\theta$-$\beta$-$M$ relation $$\tan\theta = 2 \cot\beta \, \frac{M_1^2 \sin^2\beta - 1}{M_1^2 (\gamma + \cos 2\beta) + 2}.$$ For given $M_1$, the deflection $\theta$ has a maximum $\theta_{\max}$. A wedge of half-angle less than $\theta_{\max}$ carries an attached oblique shock, and there are two possible shock angles, the weak and the strong solution. For a larger wedge angle the shock detaches from the tip and is curved. The case $\beta = \pi/2$ is the normal shock, with $\theta = 0$.

Example: for $\gamma = 1.4$, $M_1 = 2$ and $\beta = \pi/4$, one has $\mu_1 = \pi/6 < \beta$ and normal Mach number $\sqrt 2 > 1$. The relation gives $\tan\theta = 2 \cdot 1 \cdot \dfrac{2 - 1}{4 \cdot 1.4 + 2} = \dfrac{2}{7.6}$, so $\theta \approx 14.7^\circ$.

Non-example: if $\beta < \mu_1$, then $M_1 \sin\beta < 1$, the normal component is subsonic and no shock is possible at that angle. The limit $\beta = \mu_1$ gives $\theta = 0$ and a jump of zero strength, a Mach wave.