Let us calculate the pressure coefficient on each panel of the airfoil in Example 10.1 using Busemann’s theory. We will use equation (10.23).
C_p=\frac{2 \theta}{\sqrt{M_{\infty}^2-1}}+\left[\frac{(\gamma+1) M_{\infty}^4-4 M_{\infty}^2+4}{2\left(M_{\infty}^2-1\right)^2}\right] \theta^2 (10.23a)
or
C_p=C_1 \theta+C_2 \theta^2 (10.23b)
Since panel 1 is parallel to the free stream, C_{p 1}=0 as before. For panel 2 ,
\begin{aligned} C_{p 2} & =\frac{2(-20 \pi / 180)}{\sqrt{2^2-1}}+\frac{(1.4+1)(2)^4-4(2)^2+4}{2\left(2^2-1\right)^2}\left(\frac{20 \pi}{180}\right)^2 \\ & =-0.4031+0.1787 \\ & =-0.2244 \end{aligned}
As noted, the airfoil in this sample problem is relatively thick, and therefore the turning angles are quite large. As a result, the differences between linear theory and higher-order approximations are significant but not unexpected. For panel 3,
\begin{aligned} C_{p 3} & =0.4031+0.1787 \\ & =0.5818 \end{aligned}
For panel 4,
C_{p 4}=0
since the flow along surface 4 is parallel to the free stream.
Having determined the pressures acting on the individual facets of the double-wedge airfoil, let us now determine the section lift coefficient:
C_{l}=\frac{\sum p \cos \theta\left(0.5 c / \cos \delta_{w}\right)}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c} (10.24a)
where \delta_{w} is the half-angle of the double-wedge configuration. We can use the fact that the net force in any direction due to a constant pressure acting on a closed surface is zero to get
C_{l}=\frac{1}{2 \cos \delta_{w}} \Sigma C_{p} \cos \theta (10.24b)
where the signs assigned to the C_{p} terms account for the direction of the force. Thus,
\begin{aligned} C_{l} & =\frac{1}{2 \cos 10^{\circ}}\left(-C_{p 2} \cos 20^{\circ}+C_{p 3} \cos 20^{\circ}\right) \\ & =0.3846 \end{aligned}
Similarly, we can calculate the section wave-drag coefficient:
C_{d}=\frac{\sum p \sin \theta\left(0.5 c / \cos \delta_{w}\right)}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c} (10.25a)
or
C_{d}=\frac{1}{2 \cos \delta_{w}} \Sigma C_{p} \sin \theta (10.25b)
Applying this relation to the airfoil section of Fig 10.5, the section wave-drag coefficient for \alpha=10^{\circ} is
\begin{aligned} C_{d} & =\frac{1}{2 \cos 10^{\circ}}\left(C_{p 3} \sin 20^{\circ}-C_{p 2} \sin 20^{\circ}\right) \\ & =0.1400 \end{aligned}
Let us now calculate the moment coefficient with respect to the midchord of the airfoil section (i.e., relative to x=0.5 c ). As we have seen, the theoretical solutions for linearized flow show that the midchord point is the aerodynamic center for a thin airfoil in a supersonic flow. Since the pressure is constant on each of the facets of the double-wedge airfoil of Fig. 10.5 (i.e., in each numbered region), the force acting on a given facet will be normal to the surface and will act at the midpoint of the panel. Thus,
\begin{aligned} C_{m_{0.5 c}}=\frac{m_{0.5 c}}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}}=\left(-p_{1}\right. & \left.+p_{2}+p_{3}-p_{4}\right) \frac{c^{2} / 8}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}} \\ & +\left(p_{1}-p_{2}-p_{3}+p_{4}\right) \frac{\left(c^{2} / 8\right) \tan ^{2} \delta_{w}}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}} & (10.26) \end{aligned}
Note that, as usual, a nose-up pitching moment is considered positive. Also note that we have accounted for terms proportional to \tan ^{2} \delta_{w}. Since the pitching moment due to a uniform pressure acting on any closed surface is zero, equation (10.26) can be written as
\begin{aligned} C_{m_{0.5 c}}=\left(-C_{p 1}+C_{p 2}\right. & \left.+C_{p 3}-C_{p 4}\right) \frac{1}{8} \\ & +\left(C_{p 1}-C_{p 2}-C_{p 3}+C_{p 4}\right) \frac{\tan ^{2} \delta_{w}}{8} & (10.27) \end{aligned}
The reader is referred to equations (5.11)
M_0=\oiint p x d x d y (5.11)
through (5.15)
C_{M_0}=\frac{M_0}{q_{\infty} S c} (5.15)
for a review of the technique. Thus,
C_{m_{0.5 c}}=0.04329