Question 5.7: The pilot of an F-16 wants to maintain steady, level (i.e., ......
The pilot of an F-16 wants to maintain steady, level (i.e., in a constant altitude, horizontal plane) unaccelerated flight. Recall from the discussion at the start of this chapter, for flight in a horizontal plane, where the angles are small, the lift must balance the weight and the thrust supplied by the engine must be sufficient to balance the total drag acting on the aircraft. For this exercise, we will assume that the total drag coefficient for this aircraft is given by equation (5.46) :
C_{D 0}+k C_L^2
Consider the following aerodynamic characteristics for the F-16:
table 1
Note that, since we are using equation (5.46) to represent the drag polar, the tabulated values of C_{D 0} include both the profile drag and the compressibility effects. Thus, the values of C_{D 0} presented in the preceding table include the two components of the drag not related to the lift. Had we used equation (5.47),
C_D=C_{D 0}+k C_L^2+\Delta C_{D M} (5.47)
the table would include individual values for the profile drag [the first term in equation (5.47)] and for the compressibility effects [the third term in equation (5.47)]. The parasite drag can be calculated as
\operatorname{Para} D=C_{D 0}\left(0.5 \rho_{\infty} U_{\infty}^2\right) S
The induced drag can be calculated as
\text { Ind } D=C_{D \text { ind }}\left(0.5 \rho_{\infty} U_{\infty}^2\right) S
where the induced-drag coefficient is given by
C_{\text {Dind }}=k C_L^2
In order for the aircraft to maintain steady, level unaccelerated flight, the lift must balance the weight, as represented by equation (5.42a).
W=L=C_L\left(0.5 \rho_{\infty} U_{\infty}^2\right) S (5.42a)
Therefore, one can solve equation (5.42b)
C_L=\frac{2 W}{\rho_{\infty}\left(M_{\infty} a_{\infty}\right)^2 S} (5.42b)
for the lift coefficient as a function of the Mach number.
Referring to equation (5.46), the total drag is the sum of the parasite drag and the induced drag.
Additional aerodynamic characteristics for the F-16 are
\begin{aligned} & \text { Aspect ratio }(A R)=3.0 \\ & \text { Wing (reference) } \operatorname{area}(S)=300 \mathrm{ft}^2 \\ & \text { Airplane efficiency factor }(e)=0.9084 \end{aligned}
Consider an F-16 that weighs 23,750 pounds in steady, level unaccelerated flight at an altitude of 20,000~ \mathrm{ft} (standard atmospheric conditions). Calculate the parasite drag, the induced drag, the total drag, and the lift-to-drag ratio as a function of Mach number in Mach-number increments of 0.1. Use linear fits of the tabulated values to obtain values of C_{D 0} and of k at Mach numbers other than those presented in the table.
table 1
\begin{array}{ccc} \hline \begin{array}{c} M_{\infty} \\ (-) \end{array} & \begin{array}{c} C_{D 0} \\ (-) \end{array} & \begin{array}{c} k \\ (-) \end{array} \\ \hline 0.10 & 0.0208 & 0.1168 \\ 0.84 & 0.0208 & 0.1168 \\ 1.05 & 0.0527 & 0.1667 \\ 1.50 & 0.0479 & 0.3285 \\ 1.80 & 0.0465 & 0.4211 \\ \hline \end{array}The first step is to use straight lines to generate values of C_{D 0} and of k in Mach-number increments of 0.1 . The results are presented in the first three columns of the accompanying table. Note, as mentioned in the problem statement, the values of C_{D 0} for Mach numbers greater than 0.84 include a significant contribution of the wave drag \Delta C_{D M} to the components of drag not related to the lift. The inclusion of the wave drag causes the drag coefficient C_{D 0} to peak at a transonic Mach number of 1.05 . Note also that the value of k for a Mach number of 0.10 is consistent with equation (5.48). That is,
k C_L^2=\frac{C_L^2}{\pi(A R) e} (5.48)
k=\frac{1}{\pi e A R}=\frac{1}{\pi(0.9084)(3)}=0.1168
The other tabulated values of k incorporate the effects of compressibility.
The free-stream density \left(0.001267~ \mathrm{slugs} / \mathrm{ft}^3\right) and the free-stream speed of sound (1036.94 ~\mathrm{ft} / \mathrm{s}) for standard atmospheric conditions at 20,000~ \mathrm{ft} are taken from Table 1.2b.
The computed values of the parasite drag, of the induced drag, and of the total drag are presented in the table and in Fig. 5.31. Note that when the total drag is a minimum (which occurs at a Mach number of approximately 0.52 ), the parasite drag is equal to the induced drag.)
Since the lift is equal to the weight, the lift-to-drag ratio is given by
\frac{L}{D}=\frac{\text { Weight }}{\text { Total Drag }}=\frac{C_L}{C_{D 0}+C_{D \text { ind }}} (5.49)
The computed values for the lift-to-drag ratio are presented in the table and in Fig. 5.32. Since the weight of the aircraft is fixed, the maximum value of the lift-to-drag ratio occurs when the total drag is a minimum. The fact that the induced drag and the parasite drag are equal at this condition is an underlying principle to the solution of Problem 5.3.
