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\begin{equation}\label{eqn:angle}  \psi+\alpha = - \theta \ .  \end{equation}  Again, Figure~\ref{fig:sketch} shows how two of the angles in this equation can be calculated, so that $\psi$ can be determined. First,   \begin{equation}   \tan\theta = \frac{\omega}{z}\ ;   \end{equation}   second, the angle $\alpha$ is given by the derivative of the position of the shock front:   \begin{equation}\label{eqn:deriv}   \frac{\rm{d}omega}{\rm{d}z} = \frac{\sin \alpha}{\cos \alpha} = \tan{\alpha}   \end{equation}   Inserting equation~\ref{eqn:rho} to \ref{eqn:deriv} into eqn.~\ref{eqn:4}, we arrive at an ordinary differential equation, that describes the functional form of the shape of the shock front:   \begin{equation}   P(z) = \frac{3}{4}\rho_0\v_0^2 = \frac{3}{4} \frac{\dot{M}}{4\pi\v_{\inf}(z^2+\omega^2)} v_{\inf}^2 \sin^2(\psi)   \end{equation}