# ACDp Estimation

The data for ACDp given in Table 9.21 were extracted from  and are approximate.

9.19.1 Induced Drag

The formula for induced drag used is:

CDi = Cl2/(3.14 x 3.73) = CL/11.71 (Table 9.22)

9.19.2 Supersonic Drag Estimation

Supersonic flight would have a bow shock wave that is a form of compressibility drag, which is evaluated at zero CL. Drag increases with a change of the angle of attack. The difficulty arises in understanding the physics involved with an increase in the CL. Clearly, the increase – although lift-dependent – has little to do with viscosity unless the shock interacts with the boundary layer to increase pressure drag. Because the very purpose of design is to avoid such interaction up to a certain CL, this book addresses the compressibility drag at a supersonic speed composed of compressibility drag at a zero CL (i. e., CDshock) plus compressibility drag at a higher Cl (i. e., ACdw).

To compute compressibility drag at a zero CL, the following empirical proce­dure is adopted from . The compressibility drag of an object depends on its thick­ness parameter; for the fuselage, it is the fineness ratio and for the wing it is the t/c ratio. The fuselage (including the empennage) and wing compressibility drags are computed separately and then added in with the interference effects. Graphs are used extensively for the empirical methodology (Figures 9.19 through 9.26). Com­pressibility drag at both Mach 0.9 and Mach 2.0 is estimated.

Drag estimation at Mach 0.9 follows the same method as worked out in the civil aircraft example and is tabulated in Section 9.19.8. For the fuselage compressibility drag (including the empennage contribution) at Mach 2.0, the thickness parameter is the fuselage fineness ratio.

Table 9.22. Vigilante induced drag

 Cl 0.2 0.3 0.4 0.5 0.6 0.7 CDi 0.00342 0.00768 0.0137 0.0214 0.0307 0.0418

Stepl: Plot the fuselage cross-section along the fuselage length as shown in Figure 9.17 and obtain the maximum cross-section Sn = 45.25 ft2 and the fuselage base Sb = 12 ft2. Find the ratios (1 + Sb/Sn) = 1 + 12/45.25 = 1.27 and S„/Sw = 45.25/700 = 0.065.

Step 2: Obtain the fuselage fineness ratio l/d = 73.3/7.788 = 9.66 (d is minus the intake width). Obtain (l/d)2 = (9.66)2 = 93.3.

Step3: Use Figure 9.21 to obtain Cm (l/d)2 = 18.25 at Mto = 2.0for(1 + SD/Sn) = 1.27. This gives Cm = 18.25/93.3 = 0.1956. Convert it to the fuselage contribution of compressibility drag expressed in terms of the wing reference area: CDwf = CDn x (Sn/Sw) = 0.1956 x 0.065 = 0.01271.

For the wing compressibility drag at Mach 2.0, use the following steps:

Step 1: Obtain the design CLDES from Figure 9.22 for the supersonic aerofoil for the AR x (t/c)1/3 = 3.73 x (0.05)1/3 = 1.374. This gives CLDES = 0.352. Test data of CLDES from  gives 0.365, which is close enough and used here.

Step 2: Obtain from Figure 9.23 the two-dimensional design Mach number, MDES^D = 0.784. Using Figure 9.24, obtain AMar = 0.038 for 1/AR = 0.268. Using Figure 9.24, obtain AMda/ = 0.067 for А/ = 37.5 deg.

Figure 9.17. Vigilante RA-C5 fuselage cross-section area distribution

 (a) Local skin friction
 Figure 9.19. Flat-plate skin friction coefficient CF variation

Step 3: Make the correction to obtain in the design Mach as MDES = Mdes_2d + AMar + AMda1/4 or Mdes = 0.784 + 0.0.038 + 0.067 = 0.889. Then, AM = MTO – MDES = 2.0 – 0.889 = 1.111.

Step 4: Compute (t/c)5/3 x [1 + (h/c)/10)] = (0.05)5/3 x [1 + (0)/10)] = 0.00679.

Step 5: Compute AR tan ALE = 3.73 x tan 43 = 3.73 x 0.9325 = 3.48.

Step 6: Use Figure 9.25 and the values in Step 5 to obtain [ACDc_wing/ {(t/c)5/3 x [1 + (h/c)/10]}] = 0.675. Compute ACdc_wing = 0.675 x 0.00679 = 0.00458.

Finally, the interference drag at supersonic speed must be added to the fuselage and wing compressibility drag. Following is the procedure for estimating the wing – fuselage interference drag:

Step1: Compute (fuselage diameter at maximum area/wing span) = 7.785/53.14 = 0.1465.

Figure 9.20. Corrections for laminari – zation

Step 2: With the taper ratio, X = 0.19, compute (1 – X) cos Ai/4 = (1 – 0.19) cos 37.5 = 0.643.

Step3: Using Figure 9.26, obtain ACd_[NT x [(1 – X)cos Ai/4] = 0.00048. Compute ACd_[NT = 0.00048/0.643 = 0.00075.

The compressibility drag of the Vigilante aircraft at zero lift is summarized in Table 9.23.

The compressibility drag at Mach 0.9 is computed as for the civil aircraft example and is given in Table 9.24 along with the drag at both Mach 0.6 and Mach 2.0.