= Solution
Let $U=|u_1|$ and $q=(\gamma-1)/(\gamma+1)$, the strong-shock limit of the normal <velocity> ratio. In the plane containing the upstream direction and the shock normal, take $u_1=Ue_X$ and $n=\cos\beta\,e_X+\sin\beta\,e_Y$. The normal <velocity> is multiplied by $q$ while the tangential <velocity> is unchanged, hence
$$
u_2=u_1-(1-q)(u_1\cdot n)n,
\quad u_{X2}=U[1-(1-q)\cos^2\beta],\quad
u_{Y2}=-U(1-q)\cos\beta\sin\beta.
$$
Eliminating $\beta$ gives the <strong-shock polar for a perfect gas>:
$$
\boxed{u_{Y2}^2=(U-u_{X2})(u_{X2}-qU).}
$$
Equivalently this is the circle
$$
\left(u_{X2}-\frac{\gamma U}{\gamma+1}\right)^2+u_{Y2}^2
=\left(\frac U{\gamma+1}\right)^2.
$$
Both signs of the perpendicular component are possible by changing shock orientation. The endpoints $u_{X2}=qU,U$ lie on the upstream axis. An original sketch for $\gamma=5/3$ illustrates the locus and a tangent ray:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-70-shock-polar.png]
{title=Strong-shock velocity polar for gamma five thirds and the tangent giving maximum deflection}
The deflection is the angle of the ray from the origin to the downstream <velocity>, rather than the shock-normal angle $\beta$. Its maximum occurs where this ray is tangent to the circle. The circle's center is at distance $\gamma U/(\gamma+1)$ and its radius is $U/(\gamma+1)$, so the tangent right triangle proves
$$
\boxed{\delta_{\max}=\arcsin(1/\gamma).}
$$
This is the <maximum deflection through a strong perfect-gas shock>; for $\gamma=5/3$ it is about $36.87$ degrees.
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