22-Mec-B7 Aero and Space Flight · December 2013

Question 5 of 7: Take-off distance to 15 m and landing distance from 15 m

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

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Mec-B7 Aero and Space Flight. Three hours, OPEN BOOK, any non-communicating calculator permitted. Seven questions are printed and all carry equal value; the paper states that any six constitute a complete examination and that only the first six appearing in the answer book will be marked, so each question is worth 20 of the 120 marks a candidate can attempt. The printed marking key splits every question into its sub-parts, and those weights are reproduced in the headings below. All seven questions are solved here, because this set is a study resource rather than a timed sitting. The paper also instructs the candidate to state any assumption made where a required quantity has been omitted; that instruction is used explicitly in Questions 4 and 5, where the thrust lapse with altitude and the ground-run averaging method are not given.

Reference texts. Solutions follow the conventions of the texts recommended for this examination code:

J. D. Anderson Jr., Introduction to Flight, 9th ed. — the standard atmosphere and the four altitudes (Ch. 3), incompressible aerodynamics and the Pitot-static tube (§3.4, §4.11), airplane performance (Ch. 6), atmospheric entry (§8.11). This is the primary reference throughout.
J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed. — finite-wing theory and induced drag (Ch. 5), transonic flow and drag divergence (Ch. 11).
B. N. Pamadi, Performance, Stability, Dynamics and Control of Airplanes, 3rd ed. — take-off and landing field lengths (Ch. 2), turning flight and the load factor (Ch. 2), static stability (Ch. 3).
W. F. Phillips, Mechanics of Flight, 2nd ed. — the parabolic drag polar, minimum-drag and maximum-climb speeds (Ch. 3).
P. H. Oosthuizen & W. E. Carscallen, Introduction to Compressible Fluid Flow, 2nd ed. — critical Mach number and compressibility effects (Ch. 1, Ch. 8).
H. J. Allen & 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) — the closed-form ballistic-entry solution used in Question 7.

SI units are used throughout, matching the paper. The International Standard Atmosphere constants are taken as $p_{0}=101.325\ \text{kPa}$, $T_{0}=288.15\ \text{K}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$, lapse rate $L=0.0065\ \text{K}\,\text{m}^{-1}$ through the troposphere, $R=287.05\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ and $g=9.80665\ \text{m}\,\text{s}^{-2}$ inside the atmosphere model. Aircraft weights use the rounded $g=9.81\ \text{m}\,\text{s}^{-2}$ that Canadian examination practice expects. All pressures are absolute.

Question 5: Take-off distance to 15 m and landing distance from 15 m (20 marks — take-off 10, landing 10)

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.

Quantity	Value
mass, hence weight	$m=10{,}000\ \text{kg}$, $W=98{,}100\ \text{N}$
wing area, sea-level thrust	$S=45\ \text{m}^{2}$, $T_{\max}=45\ \text{kN}$
in-flight polar	$C_{D}=0.025+0.035\,C_{L}^{2}$
$C_{L,\max}$: clean / take-off / landing	$1.3$ / $1.8$ / $2.2$
ground-run lift coefficients	take-off $0.25$; landing $-0.05$ (spoilers)
landing-run drag coefficient	$C_{D}=0.06$
thrust settings	approach $0.001T_{\max}$; landing run $-0.1T_{\max}$ (reversers)
rolling friction coefficients	take-off $\mu=0.02$; landing $\mu=0.08$ (braking)
reference speeds	$V_{TO}=1.25V_{\min,\text{TO}}$, $V_{LD}=1.2V_{\min,\text{LD}}$
screen height, sea-level density	$15\ \text{m}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$

Find. The total take-off distance from brake release to the point at which the aircraft has climbed to $15\ \text{m}$, and the total landing distance from the $15\ \text{m}$ screen to a full stop.

Check: assumed integration method. The paper does not prescribe how the ground run is to be integrated. The standard treatment is used here: the net accelerating force is evaluated once, at the speed for which $V^{2}$ equals the mean of its initial and final values (that is, at $V_{TO}/\sqrt{2}$), and the run is then obtained from constant-acceleration kinematics, $s=V^{2}/(2a)$. This is exact for the aerodynamic terms, which are linear in $V^{2}$, and introduces error only through the friction term's dependence on lift. The airborne segments are treated as steady, constant-speed energy exchanges — a climb at $V_{TO}$ for take-off and a steady glide at $V_{LD}$ for landing. No wind, no runway slope and no allowance for pilot reaction time or a certification safety factor are included; a certified field length would add both.

Approach. Each of the two distances splits at the screen height into a ground segment, where Newton's second law along the runway is integrated, and an airborne segment, where the work done by the net streamwise force is equated to the change in potential energy.

Take-off — find the minimum and unstick speeds. The minimum speed uses the take-off configuration maximum lift coefficient, $1.8$: $$\begin{aligned} V_{\min,\text{TO}} &= \sqrt{\frac{2W}{\rho_{0}SC_{L,\max}}} \\ &= \sqrt{\frac{2\times98\,100}{1.225\times45\times1.8}} \\ &= 44.47\ \text{m}\,\text{s}^{-1} \end{aligned}$$ and the aircraft rotates and unsticks at $V_{TO}=1.25\times44.47=55.58\ \text{m}\,\text{s}^{-1}$.
Evaluate the ground-run forces at the mean of $V^{2}$. On the ground the wing is at $C_{L}=0.25$, so $C_{D}=0.025+0.035(0.25)^{2}=0.02719$. With $V^{2}=V_{TO}^{2}/2=1544.8\ \text{m}^{2}\,\text{s}^{-2}$ the dynamic pressure is $q=946.2\ \text{Pa}$, and $$\begin{aligned} D &= qSC_{D}=1158\ \text{N}, \\ L &= qSC_{L}=10\,645\ \text{N} \end{aligned}$$ The wheels therefore carry only $W-L=87.5\ \text{kN}$ of the weight, and the rolling resistance is $\mu(W-L)=0.02\times87\,455=1749\ \text{N}$.
Integrate the ground run. Newton's second law along the runway gives the acceleration $$\begin{aligned} a &= \frac{T-D-\mu(W-L)}{m} \\ &= \frac{45\,000-1158-1749}{10\,000} \\ &= 4.209\ \text{m}\,\text{s}^{-2} \end{aligned}$$ and with $s=V_{TO}^{2}/(2a)$, $$s_{\text{ground}}=\frac{(55.58)^{2}}{2\times4.209}=366.9\ \text{m}$$ Note how small the drag and friction terms are beside the thrust: the ground run of a jet is essentially a constant-acceleration problem.
Climb to the screen height by an energy balance. Once airborne the wing carries the full weight, so $C_{L}=W/(qS)=98\,100/(1892.4\times45)=1.152$ and $C_{D}=0.025+0.035(1.152)^{2}=0.0714$, giving $D=6084\ \text{N}$. Equating the work done by the excess thrust to the potential energy gained at constant speed, $$\begin{aligned} (T-D)\,s_{\text{air}} &= Wh \\ \Longrightarrow\quad s_{\text{air}} &= \frac{Wh}{T-D} \\ &= \frac{98\,100\times15}{45\,000-6084} \\ &= 37.8\ \text{m} \end{aligned}$$ so that $$\boxed{\;s_{\text{take-off}}=366.9+37.8=405\ \text{m}\ \text{to the }15\ \text{m} \ \text{screen}\;}$$ The airborne portion is only $9\%$ of the total, because the climb gradient at this thrust-to-weight ratio is nearly $0.40$.
Landing — find the minimum and threshold speeds. The landing configuration reaches $C_{L,\max}=2.2$, so $$\begin{aligned} V_{\min,\text{LD}} &= \sqrt{\frac{2\times98\,100}{1.225\times45\times2.2}} =40.22\ \text{m}\,\text{s}^{-1}, \\ V_{LD} &= 1.2\times40.22=48.27\ \text{m}\,\text{s}^{-1} \end{aligned}$$
Descend from the screen as a steady glide. At $V_{LD}$ the wing carries the weight, so $q=1426.9\ \text{Pa}$, $C_{L}=1.528$, $C_{D}=0.025+0.035(1.528)^{2}=0.1067$ and $D=6851\ \text{N}$. The approach thrust is only $0.001T_{\max}=45\ \text{N}$, so the flight-path angle follows from the streamwise equilibrium $W\sin\gamma=D-T$: $$\begin{aligned} \sin\gamma &= \frac{6851-45}{98\,100}=0.06938 \\ \Longrightarrow\quad \gamma &= 3.98^{\circ}, \\ s_{\text{air}} &= \frac{h}{\tan\gamma} \\ &= \frac{15}{0.06955} \\ &= 215.7\ \text{m} \end{aligned}$$ A four-degree approach is a little steeper than the three degrees of a normal ILS glideslope, which is consistent with the engines being all but closed.
Integrate the landing ground run. After touchdown the spoilers give $C_{L}=-0.05$, so the wing presses the aircraft onto its wheels and the normal load exceeds the weight. Evaluating again at $V^{2}=V_{LD}^{2}/2$, that is $q=713.5\ \text{Pa}$: $$\begin{aligned} D &= qS(0.06)=1926\ \text{N}, \\ L &= qS(-0.05)=-1605\ \text{N}, \\ W-L &= 99\,705\ \text{N} \end{aligned}$$ The three retarding contributions are the aerodynamic drag, the braking friction $0.08\times99\,705=7976\ \text{N}$, and the reverse thrust $4500\ \text{N}$, totalling $14\,403\ \text{N}$, so $a=1.440\ \text{m}\,\text{s}^{-2}$ and $$s_{\text{ground}}=\frac{(48.27)^{2}}{2\times1.440}=808.8\ \text{m}$$
Add the two landing segments. $$\boxed{\;s_{\text{landing}}=215.7+808.8=1024\ \text{m}\ \text{from the }15\ \text{m} \ \text{screen}\;}$$ The landing distance is two and a half times the take-off distance, which is the usual result: an engine can put out four times as much accelerating force as brakes and reversers can put out retarding force at these speeds, and the landing case starts with a $215\ \text{m}$ air distance that the take-off case does not pay.

Each certified distance splits into an airborne segment governed by energy and a ground segment governed by the net accelerating or retarding force.

Segment	Distance
take-off ground run	$366.9\ \text{m}$
take-off airborne to $15\ \text{m}$	$37.8\ \text{m}$
total take-off distance to $15\ \text{m}$	$\mathbf{405\ m}$
landing air distance from $15\ \text{m}$	$215.7\ \text{m}$ (approach angle $3.98^{\circ}$)
landing ground run	$808.8\ \text{m}$
total landing distance from $15\ \text{m}$	$\mathbf{1024\ m}$

Topic. Take-off and landing field lengths from first principles.

Key relations. $V_{\min}=\sqrt{2W/(\rho SC_{L,\max})}$; ground run $s=V^{2}/(2a)$ with $ma=T-D-\mu(W-L)$ evaluated at $V/\sqrt{2}$; airborne climb $s=Wh/(T-D)$; steady glide $\sin\gamma=(D-T)/W$ and $s=h/\tan\gamma$.

Why this works. Both ground runs are one-dimensional accelerations under a force that varies only through $V^{2}$, and the aerodynamic part of that force is exactly linear in $V^{2}$; evaluating it once at the mean of $V^{2}$ therefore integrates the aerodynamic terms exactly and only approximates the friction term, whose variation is itself small. The airborne segments are handled by energy rather than by geometry because the flight path in a transition is not known without a further assumption about load factor, whereas the work-energy statement needs only the net streamwise force. The sign conventions carry all the physics of the landing case: negative lift from the spoilers increases the normal load and hence the braking force, and negative thrust from the reversers adds directly to the retarding force, which is why a modern jet can stop in roughly twice the distance it took to become airborne rather than five times.

Common pitfall. Using $W$ rather than $W-L$ in the friction term, which overstates take-off friction by about $12\%$ here and, more seriously, understates landing friction because the spoiler lift is negative. Another is using the clean $C_{L,\max}=1.3$ for the take-off or landing minimum speeds; the paper supplies three different maxima and each belongs to one configuration. A third is forgetting that the approach thrust, though tiny, is not zero, and that reverse thrust enters the landing balance with a positive retarding sign.

Source. Anderson, Introduction to Flight, 9th ed., §6.14–6.15 (take-off and landing performance); Pamadi, Performance, Stability, Dynamics and Control of Airplanes, 3rd ed., §2.11 for the ground-run averaging method used here.

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