22-Mec-B7 Aero and Space Flight · December 2017

Question 5 of 7: Landing Distance and Gust Load Factor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2017 — 16-Mec-B7 Aero and Space Flight. Three hours, OPEN BOOK, any non-communicating calculator permitted. Seven questions, all of equal value; any SIX constitute a complete paper, so full marks are 120 and the percentage grade is $[(\text{mark obtained}/120)\times 100]$. Some questions require an essay answer, and clarity and organisation of the answer are explicitly marked. All seven questions are solved below, and every sub-part is addressed.

Reference texts. J. D. Anderson, Introduction to Flight, 9th ed. (standard atmosphere, altitude definitions, airplane performance, take-off and landing, atmospheric entry, rocket staging, elliptical orbits); J. D. Anderson, Fundamentals of Aerodynamics, 6th ed. (airfoil stall, high-lift devices, critical Mach number and drag divergence, wave drag and area ruling); W. F. Phillips, Mechanics of Flight, 2nd ed. (parabolic drag polar, minimum-drag speed, climb and glide angles, jet range and endurance, static and dynamic stability); H. J. Allen and A. J. Eggers, A Study of the Motion and Aerodynamic Heating of Ballistic Missiles Entering the Earth’s Atmosphere at High Supersonic Speeds, NACA Report 1381 (1958) (ballistic entry, maximum deceleration); G. P. Sutton and O. Biblarz, Rocket Propulsion Elements, 9th ed. (liquid-propellant engine architecture, solid-propellant grain design).

Check: standing assumptions. The paper’s page-1 note invites the candidate to “submit with their answer paper a clear statement of any assumptions made.” Four assumptions are used throughout and are stated once here. (i) The International Standard Atmosphere with $T_0 = 288.15\ \text{K}$, $p_0 = 101.325\ \text{kPa}$, $\rho_0 = 1.225\ \text{kg}/\text{m}^3$, tropospheric lapse rate $L = 0.0065\ \text{K}/\text{m}$ to $11\ \text{km}$, $R = 287.05\ \text{J}/(\text{kg}\cdot\text{K})$ and $\gamma = 1.4$, giving the exponents $g/(LR) = 5.2559$ for pressure and $4.2559$ for density. (ii) Where a question needs the variation of thrust with altitude but does not state it, the fixed-geometry jet assumption $T = T_{SL}\,(\rho/\rho_0)$ is used. (iii) Ground-run accelerations are evaluated once at $V/\sqrt{2}$, the speed at which $V^2$ takes its mean value, so that $s = V^2/(2a)$. (iv) Aircraft weights use $g = 9.81\ \text{m}/\text{s}^2$; the atmosphere model itself uses the defining value $9.80665\ \text{m}/\text{s}^2$. Earth data for Question 7 use $\mu = GM = 3.986\times 10^{14}\ \text{m}^3/\text{s}^2$.

Question 5: 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.

Given. One aircraft, configured for landing at sea level in the Standard Atmosphere, with the full data set reproduced below.

Given data
Quantity	Value
Mass $m$; weight $W$	$13\,500\ \text{kg}$; $132\,435\ \text{N}$
Wing area $S$; mean chord $c$	$72\ \text{m}^2$; $3.8\ \text{m}$ (span $18.95\ \text{m}$, $AR = 4.99$)
Maximum sea-level thrust $T_{max}$	$59\ \text{kN}$
In-flight drag polar	$C_D = 0.031 + 0.039\,C_L^2$
Maximum $C_L$, clean / landing configuration	$1.3$ / $2.3$
$C_L$ during the landing run (spoilers deployed)	$-0.07$
Thrust on approach / during the landing run	$0.0015\,T_{max} = 88.5\ \text{N}$ / $-0.15\,T_{max} = -8850\ \text{N}$
$C_D$ during the landing run	$0.15$
Wheel–runway friction coefficient $\mu$	$0.085$
Landing speed	$1.15 \times$ the minimum speed in landing configuration
Lift-curve slope $a = dC_L/d\alpha$	$6$ per radian
Screen height; gust case	$15\ \text{m}$; $400\ \text{km}/\text{hr}$ with a $50\ \text{km}/\text{hr}$ upward gust

Find. (a) the total landing distance measured from the point at which the aircraft is $15\ \text{m}$ above the runway until it comes to rest; (b) the load factor produced by a sharp-edged vertical gust of $50\ \text{km}/\text{hr}$ at $400\ \text{km}/\text{hr}$.

Approach. The landing is treated in two segments. In the airborne segment the aircraft flies a straight, steady approach path whose angle follows from the equilibrium of drag, the small residual thrust and the component of weight along the path, and the horizontal distance is the screen height divided by the tangent of that angle. In the ground run the decelerating force — aerodynamic drag, wheel friction on a normal load increased by the spoilers, and reverse thrust — is evaluated once at $V_L/\sqrt{2}$, the speed at which $V^2$ takes its mean value, so that $s = V_L^2/(2a)$. Part (b) uses the sharp-edged gust model, in which the gust changes the angle of attack instantaneously by $U/V$.

Part (a) — landing distance from the 15 m screen height.

Minimum speed in the landing configuration. With flaps and slats fully deployed the wing reaches $C_{L,max} = 2.3$, so $V_{stall} = \sqrt{\dfrac{2W}{\rho_0 S\,C_{L,max}}} = \sqrt{\dfrac{2(132\,435)}{1.225 \times 72 \times 2.3}} = 36.13\ \text{m}/\text{s}$.
Landing (approach and touchdown) speed. The paper specifies $V_L = 1.15\,V_{stall} = 1.15 \times 36.13$, i.e. $V_L = 41.55\ \text{m}/\text{s} = 149.6\ \text{km}/\text{hr}$.
Lift and drag coefficients on the approach. Flying at $1.15$ times the stalling speed means the dynamic pressure is $1.15^2$ times the stalling value, so $C_{L,app} = C_{L,max}/1.15^2 = 2.3/1.3225 = 1.739$ and $C_{D,app} = 0.031 + 0.039(1.739)^2 = 0.1490$.
Drag on the approach. Because $\tfrac{1}{2}\rho V^2 S = W/C_L$ on a shallow path, the drag can be written without recomputing $q$: $D = W\,C_{D,app}/C_{L,app} = 132\,435 \times 0.1490/1.739 = 11\,343\ \text{N}$.
Approach path angle. Resolving along the descent path, $W\sin\gamma + T_{app} = D$, so $\sin\gamma = (D - T_{app})/W = (11\,343 - 88.5)/132\,435 = 0.08498$ and $\gamma = 4.88^\circ$. The residual thrust is only $88.5\ \text{N}$, so this is very nearly a glide.
Air distance from the screen height. Descending $15\ \text{m}$ on that straight path covers $$s_{air} = \frac{15}{\tan 4.88^\circ} = \boxed{\ 175.9\ \text{m}\ }$$
Set up the ground run. After touchdown at $V_L$ the decelerating forces are aerodynamic drag $D$, rolling friction $\mu N$ on the normal load $N = W - L$, and reverse thrust. Following the standing assumption, all speed-dependent terms are evaluated once at $\bar V = V_L/\sqrt{2} = 29.38\ \text{m}/\text{s}$, where $\bar q = \tfrac{1}{2}\rho_0 \bar V^2 = 528.8\ \text{Pa}$.
Aerodynamic terms with spoilers deployed. The spoilers make the lift coefficient negative, which is their purpose: $L = \bar q S C_{L,run} = 528.8 \times 72 \times (-0.07) = -2665\ \text{N}$, so the wheels carry $N = W - L = 132\,435 + 2665 = 135\,100\ \text{N}$, more than the weight. The drag is $D = \bar q S C_{D,run} = 528.8 \times 72 \times 0.15 = 5711\ \text{N}$.
Total decelerating force and the deceleration. $F = D + \mu N + |T_{rev}| = 5711 + 0.085(135\,100) + 8850 = 5711 + 11\,484 + 8850 = 26\,045\ \text{N}$, hence $a = F/m = 26\,045/13\,500 = 1.929\ \text{m}/\text{s}^2$.
Ground run and total landing distance. With constant deceleration from $V_L$ to rest, $s_{ground} = V_L^2/(2a) = 41.55^2/(2 \times 1.929) = 447.5\ \text{m}$, so $$\boxed{\ s_{landing} = 175.9 + 447.5 = 623\ \text{m}\ }$$

The airborne segment is a straight, essentially unpowered descent at 4.88°; the ground run is computed from a single mean deceleration of 1.93 m/s². Spoilers make the wing lift negative, so the wheels carry 135.1 kN — more than the aircraft's own weight — and the friction term is the largest of the three retarding forces.

Check: modelling assumptions in the landing calculation. No flare is modelled: the aircraft is taken to fly the straight approach path all the way to touchdown and to touch down at the full landing speed. A real flare lengthens the air distance and reduces the touchdown speed slightly, so the figure above is the standard textbook estimate rather than a certification landing distance, which would in addition include a delay before braking and a large regulatory factor. Full reverse thrust and spoilers are also assumed to be effective from the instant of touchdown, and brakes are represented solely by the given friction coefficient of 0.085.

The breakdown is instructive. Of the $26.0\ \text{kN}$ of retarding force, wheel friction contributes 44 per cent, reverse thrust 34 per cent and aerodynamic drag 22 per cent. Because the spoilers dump the lift and press the aircraft onto its wheels, the friction term is 2 per cent larger than it would be on a weightless-wing basis, and far larger than it would be if the wing were still lifting; deploying spoilers on touchdown is worth more than the drag they themselves generate.

Part (b) — load factor from a sharp-edged vertical gust.

Convert the speeds. $V = 400/3.6 = 111.11\ \text{m}/\text{s}$ and the gust velocity $U = 50/3.6 = 13.889\ \text{m}/\text{s}$.
Model the gust. In the sharp-edged gust model the aircraft enters the gust instantaneously and its flight path has no time to respond, so the whole of the upward gust velocity appears as an increment in angle of attack: $\Delta\alpha = \tan^{-1}(U/V) \approx U/V = 13.889/111.11 = 0.125$ radian ($7.13^\circ$).
Convert the angle change into a lift increment. With the lift-curve slope $a = dC_L/d\alpha = 6$ per radian, $\Delta L = \tfrac{1}{2}\rho_0 V^2 S\,a\,\Delta\alpha = \tfrac{1}{2}\rho_0 V^2 S\,a\,(U/V) = \tfrac{1}{2}\rho_0 V S\,a\,U$.
Form the load factor. The load factor is total lift divided by weight, and before the gust $L = W$, so $$n = 1 + \frac{\Delta L}{W} = 1 + \frac{\rho_0 V a U}{2(W/S)}.$$
Substitute. The wing loading is $W/S = 132\,435/72 = 1839.4\ \text{N}/\text{m}^2$, so $n = 1 + \dfrac{1.225 \times 111.11 \times 6 \times 13.889}{2 \times 1839.4} = 1 + 3.083$ and $$\boxed{\ n = 4.08\ }$$

A cross-check confirms the result: the trimmed lift coefficient at $111.11\ \text{m}/\text{s}$ is $C_L = 2W/(\rho_0 V^2 S) = 0.2432$, the gust adds $a\,\Delta\alpha = 6(0.125) = 0.750$, and the ratio of the new lift coefficient to the old is $0.9932/0.2432 = 4.08$, identical to the boxed value because the dynamic pressure is unchanged.

Three features of the formula deserve comment. Wing loading appears in the denominator, which is why a lightly loaded aircraft such as this one — only $1839\ \text{N}/\text{m}^2$, roughly $187\ \text{kg}/\text{m}^2$ — is thrown about by turbulence that a heavily loaded airliner would barely register. Airspeed appears in the numerator, which is the reason for the turbulence-penetration speed published for every transport aircraft: slowing down is the pilot’s direct lever on gust loads. Finally, a load factor of 4.08 is well beyond the usual $+2.5$ limit manoeuvre load factor of a transport aircraft, which is exactly why certification does not use the sharp-edged model unmodified: the standards apply a gust-alleviation factor of roughly 0.7–0.8 to allow for the finite time the aircraft takes to penetrate the gust and for the response of the flight path, which would bring the figure here down to about 3.2–3.5. Even so, the message stands that a strong gust at high speed is a structurally significant event.

Final results
Quantity	Value
Minimum speed in landing configuration	$36.13\ \text{m}/\text{s}$
Landing speed $V_L$	$41.55\ \text{m}/\text{s}\ (149.6\ \text{km}/\text{hr})$
Approach $C_L$ / $C_D$	$1.739$ / $0.1490$
Drag on the approach	$11\,343\ \text{N}$
Approach path angle	$4.88^\circ$
Air distance from 15 m	$175.9\ \text{m}$
Wheel normal load during the run	$135\,100\ \text{N}$
Retarding force (drag + friction + reverse thrust)	$5711 + 11\,484 + 8850 = 26\,045\ \text{N}$
Mean deceleration	$1.929\ \text{m}/\text{s}^2$
Ground run	$447.5\ \text{m}$
Total landing distance from 15 m	$623\ \text{m}$
Gust-induced change in angle of attack	$0.125\ \text{rad}\ (7.13^\circ)$
Load factor in the gust	$n = 4.08$

Topic. Landing performance from a screen height, and the sharp-edged gust load factor.

Key relations. $V_{stall} = \sqrt{2W/(\rho S C_{L,max})}$; $V_L = 1.15\,V_{stall}$; $C_{L,app} = C_{L,max}/1.15^2$; $\sin\gamma = (D-T)/W$; $s_{air} = h/\tan\gamma$; ground run $F = D + \mu(W-L) + |T_{rev}|$ evaluated at $V_L/\sqrt{2}$, $s = V_L^2/(2a)$; sharp-edged gust $n = 1 + \rho V a U /(2W/S)$.

Why this works. The ground run is the interesting piece of modelling. Drag and lift both vary as $V^2$, so the deceleration is not constant and the exact answer needs an integral. Evaluating the net force once at $V_L/\sqrt{2}$ works because $V^2$ takes its mean value there, so the mean of the $V^2$-dependent forces is captured exactly and only the constant terms are treated approximately — the result is within a few per cent of the integrated answer for a fraction of the effort. The gust formula is equally economical: since the aircraft has no time to change its flight path, the gust is purely an angle-of-attack step of $U/V$, and because the dynamic pressure is unchanged the load factor is simply the ratio of the new lift coefficient to the trimmed one.

Common pitfall. Forgetting that spoilers make $C_L$ negative and therefore increase the wheel normal load above the weight — subtracting a positive lift here would reduce the friction term by 4 per cent and lengthen the ground run. In part (b), using the trimmed $C_L$ in place of the lift-curve slope, or leaving the gust and flight speeds in km/hr, which changes the answer even though the ratio $U/V$ does not, because $\rho V$ appears alone in the numerator.

Source. Anderson, Introduction to Flight, 9th ed., Ch. 6.15 (landing performance) and Ch. 6.16–6.17 (V–n diagram and gust loads); Phillips, Mechanics of Flight, 2nd ed., Ch. 3.11 (take-off and landing).

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