Question 12.2: Consider hypersonic flow past a sharp cone where -θc ≤ α ≤ θ......

For simplicity, let us first calculate the forces in a body-fixed coordinate system. As shown in Fig. 12.16, the forces in the body-fixed coordinate system are A, the force along the axis of the body, and N, the force normal to the body axis. Once A and N have been calculated, the lift and the drag can be calculated since

L=N \cos \alpha-A \sin \alpha   (12.29)

and

D=N \sin \alpha+A \cos \alpha   (12.30)

Applying equations (12.24) through (12.28)

\cos \eta=\frac{\vec{V}_{\infty} \cdot \hat{n}}{\left|\vec{V}_{\infty}\right||\hat{n}|}    (12.24)

\vec{V}_{\infty}=U_{\infty} \cos \alpha \hat{\imath}-U_{\infty} \sin \alpha \hat{\jmath} (12.25)

\hat{n}=\hat{\imath} \sin \theta-\hat{\jmath} \cos \theta \cos \beta-\hat{k} \cos \theta \sin \beta (12.26)

\cos \eta=\cos \alpha \sin \theta+\sin \alpha \cos \theta \cos \beta (12.27)

C_p=C_{p, 12} \cos ^2 \eta     (12.28)

specifically for the sharp cone flow depicted in Fig. 12.15,

C_{p}=C_{p, t 2}\left(\cos \alpha \sin \theta_{c}+\sin \alpha \cos \theta_{c} \cos \beta\right)^{2}     (12.31)

since the deflection angle is the cone semivertex angle, \theta_{c}, which is a constant.

The axial force coefficient is found by integrating the pressure force over the entire (closed) surface of the cone:

C_{A}=\frac{1}{\left(\frac{1}{2} \rho_{\infty} U_{\infty}^{2}\right)\left(\pi R_{b}^{2}\right)} \oiint_{S}\left(p-p_{\infty}\right)(\hat{n} d S) \cdot \hat{\imath}   (12.32)

Recall that, since we are integrating over a closed surface, the net force in any direction due to a constant pressure acting on that closed surface is zero. Hence, subtracting p_{\infty}, as is done in equation (12.32), does not affect the value of the axial force. A differential element for the surface of the cone (d S) is

d S=r d \beta d s=\left(x \tan \theta_{c}\right) d \beta \frac{d x}{\cos \theta_{c}}    (12.33)

Combining equations (12.31) through (12.33) with equation (12.26) and noting that, since C_{p \text {, base }}=0, the limits of integration are 0 \leq \beta \leq \pi (if we multiply by 2) and 0 \leq x \leq x_{L}, the axial force coefficient is

Integrating first over \beta yields

Integrating with respect to x gives us

C_{A}=\frac{2 C_{p, t 2}}{R_{b}^{2}}\left(\cos ^{2} \alpha \sin ^{2} \theta_{c}+\frac{1}{2} \sin ^{2} \alpha \cos ^{2} \theta_{c}\right) \tan ^{2} \theta_{c} \frac{x_{L}^{2}}{2}

Since R_{b}=x_{L} \tan \theta_{c},

C_{A}=C_{p, 2}\left[\sin ^{2} \theta_{c}+\frac{1}{2} \sin ^{2} \alpha\left(1-3 \sin ^{2} \theta_{c}\right)\right]      (12.34)

Let us calculate the normal force coefficient due to the pressures acting over the closed surface of the cone. Since the normal force is positive in the negative y direction,

C_{N}=\frac{1}{\left(\frac{1}{2} \rho_{\infty} U_{\infty}^{2}\right)\left(\pi R_{b}^{2}\right)} \oiint_{S}\left(p-p_{\infty}\right)(\hat{n} d S) \cdot(-\hat{\jmath})   (12.35)

Combining equations (12.26), (12.31), (12.33), and (12.35) and using the limits of integration, 0 \leq \beta \leq \pi and 0 \leq x \leq x_{L} (as discussed previously), we have

Integrating first with respect to \beta gives us

C_{N}=\frac{2 C_{p, i 2}}{\pi R_{b}^{2}} \int_{0}^{x_{L}}\left(2 \sin \alpha \cos \alpha \sin \theta_{c} \cos \theta_{c} \frac{\pi}{2}\right) \tan \theta_{c} x d x

Integrating with respect to x yields

C_{N}=\frac{2 C_{p, i 2}}{R_{b}^{2}}\left(\sin \alpha \cos \alpha \sin \theta_{c} \cos \theta_{c}\right) \tan \theta_{c} \frac{x_{L}^{2}}{2}

so that

C_{N}=\frac{C_{p, t 2}}{2} \sin 2 \alpha \cos ^{2} \theta_{c}    (12.36)

Referring to equation (12.29), the lift coefficient is

C_{L}=C_{N} \cos \alpha-C_{A} \sin \alpha

Using the coefficients presented in equations (12.34) and (12.36), we have

Correspondingly,

To calculate the pitching moment, note that the moment due to the incremental pressure force acting at a radial distance \vec{r} from the origin is

\overrightarrow{d M}=\vec{r} \times \overrightarrow{d F}=\vec{r} \times p \hat{n} d S

Note that the incremental moment \overrightarrow{d M} is a vector, which can be written in terms of its components:

\overrightarrow{d M}=d L \hat{\imath}+d N \hat{\jmath}+d M \hat{k}    (12.39)

In equation (12.39), L is the rolling moment (which is positive when causing the right wing to move down), N is the yawing moment (which is positive when causing the nose to move to the right), and M is the pitching moment (which is positive when causing a nose-up motion).

If we want to take the moments about the apex of the cone, the moment arm is

\vec{r}=x \hat{\imath}+x \tan \theta_{c} \cos \beta \hat{\jmath}+x \tan \theta_{c} \sin \beta \hat{k}  (12.40)

Let us examine the \vec{r} \times \hat{n} term of the moment expression:

The first term (i.e., the contribution to the rolling moment) is always zero, since, in the absence of viscous forces, the only forces are the pressure forces that act normal to the surface and therefore act through the axis of symmetry for a body of revolution. The negative sign for the \hat{k} term (i.e., the pitching moment) results because a pressure force acting in the first and fourth quadrants (\pi / 2 \geq \beta \geq-\pi / 2) produces a nose-down (negative) pitching moment. For this example we are interested in the pitching moment about the apex, which is considered positive when nose up,

M_{0}=\oiint_{S} \vec{r} \times\left(p-p_{\infty}\right) \hat{n} d S \cdot \hat{k} (12.42)

Taking the resultant cross product of the moment arm from the apex (\vec{r}) and the inward-directed unit normal to the surface area \hat{n} with the dot product with \hat{k}, we obtain

(\vec{r} \times \hat{n}) \cdot \hat{k}=-x \cos \theta_{c} \cos \beta-x \sin \theta_{c} \tan \theta_{c} \cos \beta

Thus, the pitching moment coefficient

We can simplify this expression if we focus on the terms containing the cone-half-angle \left(\theta_{c}\right), so that

\tan \theta_{c} \frac{1}{\cos \theta_{c}}\left(\cos \theta_{c}+\sin \theta_{c} \tan \theta_{c}\right)=\tan \theta_{c}\left(1+\tan ^{2} \theta_{c}\right)=\frac{\tan \theta_{c}}{\cos ^{2} \theta_{c}}

Substituting the modified-Newtonian-flow expression for the pressure coefficient and integrating with respect to \beta yields

Integrating with respect to x and using the fact that R_{b}=x_{L} \tan \theta_{c}, we obtain

C_{M_{0}}=-\frac{C_{p, 2} \sin 2 \alpha}{3 \tan \theta_{c}}   (12.44)

Although the resultant moment is produced by the integration of the distributed aerodynamic forces over the vehicle surface, it can be represented as composed of two components, one effectively due to the normal-force component and the other due to the axial-force component. Thus, we can isolate the two terms of equation (12.43):

Writing the total pitching moment as the sum of the two components, we have

C_{M_{0}}=C_{M_{0, N}}+C_{M_{0, A}}     (12.45)

where

C_{M_{0, N}}=-\frac{2 \tan \theta_{c}}{\pi R_{b}^{2} R_{b}} \int_{0}^{x_{L}}\left(\int_{0}^{\pi} C_{p} \cos \beta d \beta\right) x^{2} d x  (12.46)

and

C_{M_{0, A}}=-\frac{2 \tan ^{3} \theta_{c}}{\pi R_{b}^{2} R_{b}} \int_{0}^{x_{L}}\left(\int_{0}^{\pi} C_{p} \cos \beta d \beta\right) x^{2} d x  (12.47)

Substituting the modified Newtonian-flow expression for the pressure coefficient yields

Integrating gives us

\begin{aligned} & C_{M_{0, N}}=-\frac{C_{p, 22} \sin 2 \alpha}{R_{b}^{2} R_{b}} \tan \theta_{c} \sin \theta_{c} \cos \theta_{c} \frac{x_{L}^{3}}{3} \\ & C_{M_{0, N}}=-\frac{C_{p, 22} \sin 2 \alpha \cos ^{2} \theta_{c}}{3} \frac{x_{L}}{R_{b}} \end{aligned}

Using equation (12.36) for the expression for C_{N}, we can rewrite

C_{M_{0, N}}=C_{N}\left(-\frac{2}{3} \frac{x_{L}}{R_{b}}\right)   (12.48)

Similarly,

Integrating yields

C_{M_{0, A}}=-\frac{C_{p, t 2} \sin 2 \alpha \sin ^{2} \theta_{c}}{3} \frac{x_{L}}{R_{b}}

Using equation (12.34) for the expression for C_{A}, we can rewrite

C_{M_{0, A}}=C_{A} \frac{-\sin 2 \alpha \sin ^{2} \theta_{c}}{3\left[\sin ^{2} \theta_{c}+0.5 \sin ^{2} \alpha\left(1-3 \sin ^{2} \theta_{c}\right)\right]} \frac{x_{L}}{R_{b}}     (12.49)

Following this division of the pitching moment into components, we can represent it as an effective net force acting at a “center of pressure,” as a shown in Fig. 12.17. The pitching moment may be represented by

M_{0}=M_{0, N}+M_{0, A}=-x_{\mathrm{cp}} N-y_{\mathrm{cp}} A    (12.50)

The two terms represent the effects of the normal force and of the axial force, respectively. In terms of the coefficients,

C_{M_{0}}=\frac{M_{0}}{q_{\infty} S R_{b}}=-C_{N} \frac{x_{\mathrm{cp}}}{R_{b}}-C_{A} \frac{y_{\mathrm{cp}}}{R_{b}}   (12.51)

Comparing the formulations presented in equations (12.48), (12.49) and (12.51), we find that

x_{\mathrm{cp}}=\frac{2}{3} x_{L}      (12.52)

and

y_{\mathrm{cp}}=\frac{\sin 2 \alpha \sin ^{2} \theta_{c} x_{L}}{3\left[\sin ^{2} \theta_{c}+0.5 \sin ^{2} \alpha\left(1-3 \sin ^{2} \theta_{c}\right)\right]}       (12.53)

If the vehicle is to be statically stable, the center of pressure should be located such that the aerodynamic forces produce a restoring moment when the configuration is perturbed from its “stable orientation.” Thus, if the vehicle pitches up (i.e., \alpha increases), the net pitching moment should be negative (causing a nose-down moment), which decreases \alpha.

The parameter

\text { S.M. }=\frac{x_{\mathrm{cp}}-x_{\mathrm{cg}}}{x_{L}}  (12.54)

is the static margin. The static margin must be positive for uncontrolled vehicles. For high-performance, hypersonic vehicles, the static margin is usually 3 to 5 \%. Note, however, as illustrated in Fig. 12.17, both the axial force and the normal force contribute to the pitching moment. Thus, it is possible that the vehicle is statically stable when x_{\mathrm{cp}}=x_{\mathrm{cg}}, if y_{\mathrm{cp}} is below y_{\mathrm{cg}} (as is the case in the sketch of Fig. 12.17). In this case the axial force will produce the required restoring moment.