Transformation from Hodograph to Physical plane

Of course, the hodograph method would be useless if we could not transform the solution back to the physical plane. For completeness, the reverse transformation is given.

$$x_\tau = \left(\frac{\gamma-1}2\right)^{1/2}\frac1{a_0} \left... ...\sin\theta}{\tau^{1/2}(1-\tau)^{1/(\gamma-1)}}\psi_\tau\right)$$ (22)

$$x_\theta = \left(\frac{\gamma-1}2\right)^{1/2}\frac1{a_0} \le... ...rac{\tau^{1/2}}{(1-\tau)^{\gamma/(\gamma-1)}} \psi_\tau\right)$$ (23)

$$y_\tau = \left(\frac{\gamma-1}2\right)^{1/2}\frac1{a_0} \left... ...\cos\theta}{\tau^{1/2}(1-\tau)^{1/(\gamma-1)}}\psi_\tau\right)$$ (24)

$$y_\theta = \left(\frac{\gamma-1}2\right)^{1/2}\frac1{a_0} \le... ...rac{\tau^{1/2}}{(1-\tau)^{\gamma/(\gamma-1)}} \psi_\tau\right)$$ (25)

After Equation 14 or 21 is employed to find $\psi$, Equations 22-25 are integrated to find the x and y coordinates of the streamline.