Given Two Angles, δ and θ

It is sometimes useful to obtain a relationship where the two angles are known. The first upstream Mach number, $M_1$ is

$${M_1}^2 = { 2 ( \cot \theta + \tan \delta ) \over \sin 2 \theta - (\tan \delta) ( k + \cos 2 \theta) }$$ (13.59)

The reduced pressure difference is

$${2(P_2 - P_1) \over \rho U^2} = {2 \sin\theta \sin \delta \over \cos(\theta - \delta)}$$ (13.60)

The reduced density is

$${\rho_ 2 -\rho_1 \over \rho_2} = {\sin \delta \over \sin \theta \cos (\theta -\delta)}$$ (13.61)

For a large upstream Mach number $M_1$ and a small shock angle (yet not approaching zero), $\theta$ , the deflection angle, $\delta$ must also be small as well. Equation (13.51) can be simplified into

$$\theta \cong {k +1 \over 2} \delta$$ (13.62)

The results are consistent with the initial assumption which shows that it was an appropriate assumption.