Question 9.15: Plot the flight envelope (combined V-n diagram) for the foll...

FAR does specify regulations/requirements for gust loads as well as a positive maximum load factor and a negative maximum load factor for acrobatic aircraft. We first need to calculate the aspect ratio and mean geometric chord:

AR=\frac{b^2}{S}\Rightarrow b=\sqrt{AR \cdot S}=\sqrt{7\times 19.33}\Rightarrow b=11.63\space m \quad \quad \quad \quad (3.9) \\ \space \\ \overline{C}=\frac{S}{b}=\frac{19.33}{11.63}\Rightarrow \overline{C}=1.66 \space m \quad \quad \quad \quad (3.9)

The combined V-n diagram is plotted in three steps: (1) basic (maneuver) V-n diagram, (2) gust V-n diagram, and (3) combined V-n diagram.

1. Basic V-n diagram

As discussed in Section 9.9.3, the general shape of the basic V-n diagram resembles Figure 9.30. We just need to determine the coordinates of points K, J, G, F, B, and A. The unit of velocity in the graph is selected to be in knot (KEAS).
Since the aircraft type is acrobatic, the maximum limit load factor, based on FAR 23, is as follows:
(Positive) n_{\max} = +6,
(Negative) n_{\max} = −(0.5 × 6) = −3.
The dive speed of this aircraft from Equation 9.200 is
V_D=1.55V_C=1.55\times 310=480.5\space knot.
Hence, the coordinates of points F and G are (6, 480.5) and (−3, 480.5). To determine coordinates of points A, B, J, K, we need to derive two equations regarding C_{L\max}.

V_s=\sqrt{\frac{2mg}{\rho SC_{L_{\max}}}}=\sqrt{\frac{2\times 2300\times 9.81}{1.225\times 19.33\times 2}}=30.87\space m/s=60 \space KEAS \quad \quad \quad \quad (2.49)

Thus, the coordinate of point A is (1, 60). The top stall curve or the load factor as a function of airspeed (in m/s) is

n=\frac{\rho V^2SC_{L_{\max}}}{2W}=\frac{1.225\times V^2\times 19.33\times 2}{2\times 2300\times 9.81}=0.100105V^2\quad \quad \quad \quad (9.195)

For point B, the load factor is given as 6, so the corresponding speed is determined by

6=0.100105\space V^2 \Rightarrow V=75.6\space m/s=147\space KEAS

Thus, the coordinates of point B is (6, 147). With the same technique, we can derive the equation for the lower stall curve

V_{si}=\sqrt{\frac{-2mg}{\rho S(-C_{L_{\max}})}}=\sqrt{\frac{-2\times 2300\times 9.81}{1.225\times 19.33\times (-1.2)}}=39.85\space m/s=77.5 \space KEAS \quad \quad \quad \quad (2.49)

So, the coordinate of point K is (−1, 77.5). The lower stall curve or the load factor as a function of airspeed (in m/s) is

-n =\frac{\rho V^2 S(-C_{L_{\max}})}{2W}=\frac{1.225\times V^2\times 19.33\times (-1.2)}{2\times 2300\times 9.81}=-0.00063V^2 \quad \quad \quad \quad (9.195)

For point J, the load factor is given as −3; hence, the corresponding speed is determined as

-3 = -0.00063 V² ⇒ V = 69 m/s = 134.2 KEAS

Thus far, we have collected the following coordinates:

O → (0, 0),
A → (1, 60),
B → (6, 147),
F → (6, 480.5),
G → (-3, 480.5),
J → (-3, 134.2),
K → (-1, 77.5).

By using these data, we can plot the basic V−n diagram as illustrated in Figure 9.30.

2. Gust V-n diagram:

The variations of load factor as functions of airspeed (V) and gust speed (V_g) are given by

n=1+\frac{k_gV_{gE}V_Ea\rho S}{2W} \quad \quad \quad \quad (9.206)

Since the cruising speed is given for 10,000 ft, two flight conditions are considered for the maximum load factor. Then, we calculate n for both V_C  and  V_D.

a. Sea level
The aircraft mass ratio is

\mu_g=\frac{2m}{\rho \overline{C}as}=\frac{2\times 2300}{1.225\times 1.66\times 6.3\times 19.33}=18.75 \quad \quad \quad \quad (9.208)

Then, the gust alleviation factor is obtained as

k_g=\frac{0.88\mu_g}{5.3+\mu_g}=\frac{0.88\times 18.75}{5.3+18.75}=0.684\quad \quad \quad \quad (9.207)

From Table 9.9, at sea level with the cruising speed, the gust velocity of ±50 ft/s (i.e., ±15.25 m/s) should be considered; thus, the load factor will be

n=1+\frac{0.684\times (\pm15.25)\times V\times 6.3\times 1.225\times 19.33}{2\times 2300\times 9.81}\Rightarrow n=1\pm 0.03436V\quad \quad \quad \quad (9.206)

Since the cruising speed (V_C) is 310 KEAS,

n = 1 + 0.03436V = 1 + 0.03436 × 310 × 0.5144 = 6.48 (positive value)

n = 1 – 0.03436V = 1 – 0.03436 × 310 × 0.5144 = -4.48 (negative value)

We perform similarly for dive speed. When the aircraft is flying with dive speed (V_D), the gust speed should be ±25 ft/s (i.e. ±7.5 m/s). Hence, the load factor is

n=1+\frac{0.684\times (\pm 7.5)\times V\times 6.3\times 1.225\times 19.33}{2\times 2300\times 9.81}\Rightarrow n=1\pm 0.01688V\quad \quad \quad \quad (9.206)

Since the dive speed (V_D) is 480.5 KEAS,

n = 1 + 0.01688V = 1 + 0.01688 × 480.5 × 0.5144 = 1 + 4.173 = 5.173 (positive value)

n = 1 – 0.01688V = 1 – 0.01688 × 480.5 × 0.5144 = 1 – 4.173 = -3.173 (negative value)

b. 10,000 ft altitude
At 10,000 ft altitude, the air density is 0.9 kg/m³. Parameters μ_g  and  k_g are obtained:

\mu_g=\frac{2m}{\rho\overline{C}as}=\frac{2\times 2300}{0.9\times 1.66\times 6.3\times 19.33}=26.54 \quad \quad \quad \quad (9.208) \\ \space \\ k_g=\frac{0.88\mu_g}{5.3+\mu_g}=\frac{0.88\times 26.54}{5.3+26.54}=0.733 \quad \quad \quad \quad (9.207)

When the gust velocity is ±50 ft/s (i.e., ±15.25 m/s), the load factor will be

n=1+\frac{0.733\times (\pm 15.25)\times V\times 6.3\times 0.9\times 19.33}{2\times 2300\times 9.81}\Rightarrow n=1\pm 0.02715V\quad \quad \quad \quad (9.206)

The cruising speed (V_C) is given as 310 KEAS; therefore,

n = 1 + 0.02715V = 1 + 0.02715 × 310 × 0.5144 = 5.26 (positive value)

n = 1 – 0.02715V = 1 – 0.02715 × 310 × 0.5144 = -3.26 (negative value)

When the aircraft is flying with dive speed (V_D), the gust speed should be ±25 ft/s (i.e., ±7.5 m/s). Hence, the load factor is

n=1+\frac{0.733\times (\pm 7.5)\times V\times 6.3\times 0.9\times 19.33}{2\times 2300\times 9.81}\Rightarrow n=1\pm 0.01315V \quad \quad \quad \quad (9.206)

The dive speed (V_D) is given as 480.5 KEAS; therefore,

n = 1 + 0.01315V = 1 + 0.01315 × 480.5 × 0.5144 = 1 + 3.25 = 4.25 (positive value)

n = 1 – 0.01315V = 1 – 0.01315 × 480.5 × 0.5144 = 1 – 3.25 = -2.25 (negative value)

By comparison between the results of sections (a) and (b), we see that the maximum load factor at sea level is higher than the load factor at 10,000 ft (as we expected). Recall that equivalent airspeeds are independent of altitude.
Therefore, we conclude

+n_{\max} = 6.48
–n_{\max}= -4.48

Thus, the coordinates of points D and I are, respectively, (6.48, 310) and (−4.48, 310). The gust V-n diagram is sketched in Figure 9.31.

3. Combined V-n diagram

Now, we have sufficient data to plot the combined V-n diagram. We first place both diagrams (maneuver and gust) in one plot. Then, we identify and mark the intersection points or corner points for each individual diagram (points A, B, D, F, H, I, K in Figure 9.32). Now, we observe that the line BF and the gust line (n = 1 + 0.03436V) have an intersection. This is marked as point C. In addition, the line BF has an intersection with the line between two positive gust lines. This is marked as point E. The last step is to connect all outer intersection points with straight lines; the combined V-n diagram is created. As a check for sanity, we make sure that no intersection point or corner point is outside the envelope. Also, no point is in the stall regions.
Figure 9.32 demonstrates the final V-n diagram that includes the gust effect. As a conclusion, we observe that the FAR requires that the maximum allowable load factor should be 6. However, due to the gust, this parameter is increased to 6.48.

(3.9):        AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}

(2.49):     V_s=\sqrt{\frac{2W}{\rho SC_{L_{\max}}}}

(9.200):   V_D\geq 1.55V_C   (acrobatic aircraft)

(9.195):    n=\frac{V^2\rho SC_{L_{\max}}}{2mg}

(9.206):   n=1+\frac{k_gV_{gE}V_Ea\rho S}{2W}   where
W denotes the aircraft weight (in N)
ρ is the air density (in kg/m³)
S is the wing area (in m²)
V_E is the aircraft equivalent airspeed (in m/s)
V_{gE}is the gust equivalent speed (in m/s)
a is wing lift curve slope during the gust encounter (in 1/rad)

(9.208):    \mu_g=\frac{2m}{\rho Cas}     where
C is the wing mean geometric chord (in m)
m denotes the aircraft mass (in kg)

Table 9.9 Gus speed for constructing gust-induced load