Question 6 of 7: Range and Endurance Speeds, Landing Distance and Gust Load Factor

Question 6: Range and Endurance Speeds, Landing Distance and Gust Load Factor (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Speeds for maximum range and maximum endurance at 9000 m

Given. The Question 5 aircraft ($C_{D0} = 0.031$, $K = 0.037$, $W = 495\,405\ \text{N}$, $S = 120\ \text{m}^2$) at $h = 9000\ \text{m}$, where $T = 288.15 - 0.0065(9000) = 229.65\ \text{K}$ and $\rho = 0.4663\ \text{kg}/\text{m}^3$.

Find. The true airspeeds for maximum range and for maximum endurance.

Approach. For a jet, fuel flow is proportional to thrust, so endurance is maximised where thrust required (i.e. drag) is least, and range is maximised where $V/D$ is greatest — that is, where $C_L^{1/2}/C_D$ is a maximum. Each condition fixes a lift coefficient, and the lift equation then converts it to a speed.

1. Endurance condition. A jet burns fuel in proportion to thrust, and in level flight thrust equals drag, so maximum endurance is flown at minimum drag — the condition already established in Question 5(d), $C_L = \sqrt{C_{D0}/K} = 0.9153$. Hence $$V_{endurance} = \sqrt{\frac{2W}{\rho S\,C_{L,md}}} = \sqrt{\frac{2(495\,405)}{0.4663 \times 120 \times 0.9153}} = \boxed{139.1\ \text{m}/\text{s} = 501\ \text{km}/\text{h}}$$
2. Range condition. The distance flown per unit fuel is proportional to $V/D$, so range is maximised where $C_L^{1/2}/C_D$ is greatest. Setting $d\!\left(C_L^{1/2}/(C_{D0}+KC_L^2)\right)/dC_L = 0$ gives $C_{D0} = 3KC_L^2$, i.e. $$C_{L,range} = \sqrt{\frac{C_{D0}}{3K}} = \sqrt{\frac{0.031}{3(0.037)}} = 0.5285.$$ At this condition the induced drag is one third of the parasite drag, not equal to it.
3. Range speed. $$V_{range} = \sqrt{\frac{2W}{\rho S\,C_{L,range}}} = \sqrt{\frac{2(495\,405)}{0.4663 \times 120 \times 0.5285}} = \boxed{183.0\ \text{m}/\text{s} = 659\ \text{km}/\text{h}}$$

The two speeds are in the ratio $V_{range}/V_{endurance} = 3^{1/4} = 1.316$, which is a useful check: for a jet with a parabolic polar the best-range speed is always 31.6 % above the best-endurance speed, whatever the weight, the altitude or the polar constants. The physical reason is that the jet is paid in distance per unit fuel rather than time per unit fuel, so it is worth accepting some extra drag — here 38.7 kN against the minimum of 33.6 kN — in exchange for the higher speed. At 9000 m the range speed corresponds to $M = 0.603$, comfortably below the drag-divergence Mach number, so the incompressible analysis is defensible; a real transport cruises somewhat faster than the classical best-range speed for schedule reasons.

(b) Landing distance from 15 m

Given. The landing case has its own mass and configuration, so the data are restated in full.

Find. The total landing distance from the 15 m screen height to a full stop at sea level.

Approach. The landing distance splits into two segments handled separately. The airborne segment is a straight descent whose angle follows from the force balance $\sin\gamma = (D - T)/W$ at the approach speed; the ground run is a constant-deceleration problem in which the retarding force is evaluated once at the mean of $V^2$, i.e. at $V_L/\sqrt{2}$.

1. Minimum and landing speeds. In the landing configuration at sea level, $$V_{\min} = \sqrt{\frac{2W_L}{\rho_0 S\,C_{L\max}}} = \sqrt{\frac{2(392\,400)}{1.225(120)(2.4)}} = 47.16\ \text{m}/\text{s},$$ $$V_L = 1.15\,V_{\min} = \boxed{54.24\ \text{m}/\text{s} = 195\ \text{km}/\text{h}}$$
2. Approach angle. At $V_L$ the dynamic pressure is $q = \tfrac{1}{2}(1.225)(54.24)^2 = 1802\ \text{Pa}$, so the aircraft must fly at $$C_L = \frac{W_L}{qS} = \frac{392\,400}{1802(120)} = 1.815, \qquad C_D = 0.031 + 0.037(1.815)^2 = 0.1529,$$ giving $D = qSC_D = 1802(120)(0.1529) = 33.05\ \text{kN}$. With the approach thrust $T = 0.002(160\,000) = 320\ \text{N}$, the descent angle follows from the along-path balance $W\sin\gamma = D - T$: $$\sin\gamma = \frac{33\,051 - 320}{392\,400} = 0.08341, \qquad \gamma = 4.79^\circ.$$
3. Airborne distance from the screen height. Treating the approach as a straight line from 15 m to the runway, $$s_{air} = \frac{h}{\sin\gamma} = \frac{15}{0.08341} = \boxed{180\ \text{m}}$$
4. Retarding forces during the ground run. Evaluating once at $V = V_L/\sqrt{2}$, i.e. at $q = \tfrac{1}{2}\rho_0 V_L^2/2 = 901.0\ \text{Pa}$. With $C_L = -0.05$ the wing pushes down: $$L = qSC_L = 901.0(120)(-0.05) = -5.41\ \text{kN},$$ $$C_D = 0.031 + 0.037(-0.05)^2 = 0.03109, \qquad D = 901.0(120)(0.03109) = 3.36\ \text{kN}.$$ The normal reaction is therefore $W_L - L = 392.4 + 5.41 = 397.8\ \text{kN}$ and the friction force is $\mu(W_L - L) = 0.08(397\,806) = 31.82\ \text{kN}$. Adding the reversed thrust of 24 kN, $$F_{decel} = 24\,000 + 3362 + 31\,824 = 59\,186\ \text{N}.$$
5. Ground run. The deceleration is $a = F_{decel}/m = 59\,186/40\,000 = 1.480\ \text{m}/\text{s}^2$, and with constant deceleration from $V_L$ to rest, $$s_{ground} = \frac{V_L^2}{2a} = \frac{(54.24)^2}{2(1.4797)} = \boxed{994\ \text{m}}$$
6. Total landing distance. $$s_{landing} = s_{air} + s_{ground} = 180 + 994 = \boxed{1174\ \text{m}}$$

The breakdown is instructive. Reverse thrust supplies 41 % of the retarding force, wheel friction 54 % and aerodynamic drag only 6 %; and the negative ground-run lift coefficient, small as it looks, adds 5.4 kN to the normal reaction and hence 0.43 kN of extra braking force. That is the quantitative justification for the ground spoilers described in Question 3(d): dumping the lift at touchdown is what lets the brakes work at their full authority immediately. Note also that the ground run is 85 % of the total — the airborne segment is short precisely because the approach is comparatively steep, the engines being at flight idle.

(c) Load factor in a horizontal frontal gust

Given. Level flight at 500 m, $V = 350\ \text{km/h} = 97.22\ \text{m}/\text{s}$, suddenly encountering a horizontal head-on gust of $U = 60\ \text{km/h} = 16.67\ \text{m}/\text{s}$.

Find. The load factor $n = L/W$ immediately after the encounter.

Approach. A horizontal gust does not change the angle of attack — it changes the airspeed. The lift coefficient is therefore unchanged at its trimmed value, and the load factor is simply the ratio of the new dynamic pressure to the old.

1. Trimmed condition before the gust. In level flight $W = \tfrac{1}{2}\rho V^2 S C_L$, so $C_L$ takes whatever value that equation requires. At 500 m ($\rho = 1.1673\ \text{kg}/\text{m}^3$, $q = 5517\ \text{Pa}$) this is $C_L = 495\,405/(5517 \times 120) = 0.748$.
2. Immediately after the gust. A head-on gust adds directly to the airspeed, $V' = V + U$, while the aircraft's attitude — and hence its angle of attack and $C_L$ — has not yet had time to change. The new lift is $$L' = \tfrac{1}{2}\rho (V+U)^2 S\,C_L.$$
3. Load factor. Dividing by $W = \tfrac{1}{2}\rho V^2 S C_L$, everything except the speeds cancels — the density, the wing area and the lift coefficient all drop out: $$n = \frac{L'}{W} = \left(\frac{V + U}{V}\right)^2 = \left(1 + \frac{U}{V}\right)^2 = \left(1 + \frac{16.67}{97.22}\right)^2 = (1.1714)^2 = \boxed{1.372}$$

Because the density cancels, the stated altitude of 500 m does not enter the answer at all; it is needed only if one wants the trimmed lift coefficient quoted in step 1. That is a deliberate feature of the question, not an omission, and it distinguishes the horizontal gust case from the more familiar sharp-edged vertical gust, for which $n = 1 + \rho V a U /(2W/S)$ and both the density and the wing loading matter a great deal. Physically, a horizontal gust of 17 % of the flight speed produces a 37 % overload because lift goes as the square of speed; the aircraft will then pitch and decelerate back towards trim, so 1.372 is the instantaneous peak rather than a sustained value. It is comfortably inside the 2.5 g manoeuvre envelope required of a transport category aeroplane, but a gust of this relative size at a lower flight speed — on approach, say — would be a much larger fraction of the airspeed and correspondingly more severe.