xacstrake = yM.A.C.strake tan LEstrake + 0.25 M.A.C.strake (subsonic)
= (0.33 ft) tan 80o + 0.25 ( 6.4 ft) = 3.5 ft
x acstrake = yM.A.C.strake tan LEstrake + 0.50 M.A.C.strake (supersonic)
= (0.33 ft) tan 80o + 0.50 ( 6.4 ft) = 5.1 ft
but these are defined relative to the leading edge of the strake root chord, not the wing root chord. From Table 6.3, the strake root is 8 ft forward of the wing root, so relative to the wing:
x acstrake = -4.5 ft (subsonic)
x acstrake = -2.9 ft (supersonic)
and:
= 6.5 ft (subsonic)
= 9.1 ft (supersonic)
Now, adding the effect of the fuselage, using Cl wing/strake = 0.068/o (predicted in Section 4.7) and the fact that the wing root leading edge is 20 ft aft of the fuselage nose, so that lacwing/strake = 20 ft + xacwing/strake :
= 6.4 (subsonic)
To perform the supersonic calculation, supersonic lift curve slope must be predicted. A specific Mach number must be chosen. For M = 1.5:
= 0.051/o
= 9.0 ft (supersonic)
Next, the aerodynamic center of the F-16A stabilator is located:
stabilator = ctip / croot = 2 ft /10 ft = 0.2
= 6.9 ft
= 3.5 ft
xacstab = yM.A.C. stab tan LE stab + 0.25 M.A.C. stab (subsonic)
= (3.5 ft) tan 40o + 0.25 ( 6.9 ft) = 4.7 ft
xacstab = yM.A.C. stab tan LE stab + 0.4 M.A.C. stab (supersonic)
= (3.5 ft) tan 40o + 0.50 ( 6.9 ft) = 6.4 ft
These are defined relative to the leading edge of the stabilator root chord. From Table 6.3, the stabilator root is 17.5 ft aft of the wing root, so relative to the wing:
x ac stab = 22.2 ft (subsonic)
x ac stab = 23.9 ft (supersonic)
But the distance of interest for the stabilator is lt, the distance from the stabilator’s aerodynamic center to the aircraft center of gravity. Table 6.3 lists the center of gravity as 0.35 M.A.C., so relative to the wing root:
xcg = yM.A.C. tan LE + 0.35 M.A.C.
= (5.875 ft) tan 40o + 0.35 ( 11.4 ft) = 8.9 ft
and:
lt = x ac stab - xcg = 22.2 ft - 8.9 ft = 13.3 ft (subsonic)
= 23.9 ft -8.9 ft = 15 ft (supersonic)
It is now possible to calculate tail volume ratio:
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