Atmospheric Flight: Performance, Stability and Control
1. The standard atmosphere, altitudes and the four airspeeds
The International Standard Atmosphere (ISA) is a reference, not a forecast: a dry, still, perfect gas (R = 287 J/(kg·K), γ = 1.4) obeying the hydrostatic equation dp = −ρ g dh with a prescribed temperature profile. At sea level T0 = 288.15 K, p0 = 101 325 Pa and ρ0 = 1.225 kg/m³. In the troposphere temperature falls linearly at 6.5 K per km, so T = 288.15 − 0.0065 h, reaching 216.65 K at 11 km; from 11 km to 20 km the temperature is constant at 216.65 K. Combining the hydrostatic equation with p = ρRT gives the two working laws: in a gradient layer p/p1 = (T/T1)^(−g0/(aR)), with −g0/(aR) = 5.256 for a = −0.0065 K/m, and ρ/ρ1 = (T/T1)^4.256; in an isothermal layer p/p1 = ρ/ρ1 = exp[−g0(h − h1)/(RT)].
Altitude has several meanings. Geometric altitude is the tape-measure height; geopotential altitude is the height that would give the same potential energy with g held at g0, and it is the altitude the ISA tables are written in. Pressure altitude is the ISA altitude at which the standard pressure equals the pressure the aircraft is actually in, and density altitude is the same idea for density — a hot day raises density altitude and lengthens the take-off run. The density ratio σ = ρ/ρ0 links the airspeeds: indicated airspeed (IAS) is what the airspeed indicator shows; calibrated airspeed (CAS) is IAS corrected for instrument and position (installation) errors; equivalent airspeed (EAS) is CAS corrected for compressibility; and true airspeed (TAS) is the actual speed through the air, with EAS = TAS·√σ. EAS is the speed that gives the true dynamic pressure at sea-level density: q = ½ρV_TAS² = ½ρ0V_EAS².
| Airspeed | Obtained from | What it is good for |
|---|---|---|
| IAS | The reading of the airspeed indicator | What the pilot flies by; stall speed in IAS is almost independent of altitude |
| CAS | IAS corrected for instrument and position errors | Equals EAS at sea level in the standard atmosphere |
| EAS | CAS corrected for compressibility (EAS ≤ CAS at altitude) | Sets dynamic pressure, hence structural loads and the V-n diagram |
| TAS | EAS/√σ | Navigation, range, ground speed after wind correction |
2. The aeroplane, its instruments, forces and controls
Aircraft are classified first as lighter-than-air (balloons and airships, supported by buoyancy) or heavier-than-air (supported by aerodynamic force); the latter are fixed-wing (aeroplanes and gliders), rotary-wing (helicopters and autogyros) or flapping-wing. A fixed-wing aeroplane has a fuselage carrying payload and crew; a wing producing lift, fitted with ailerons, flaps and often slats and spoilers; an empennage — the horizontal stabiliser with its elevator and the vertical fin with its rudder; landing gear, tricycle or tail-wheel; and a powerplant. The resultant aerodynamic force is resolved into lift (normal to the free stream), drag (along it) and side force, and the resultant moment into rolling (about the longitudinal x-axis), pitching (about the lateral y-axis) and yawing (about the normal z-axis) moments. The angle of attack α is the angle between the relative wind and the body x-axis (or chord) in the plane of symmetry; the sideslip angle β is the angle between the relative wind and that plane, β = sin⁻¹(v/V).
- Altimeter — an aneroid capsule senses static pressure and the dial is graduated in ISA altitude; with the subscale set to 1013.25 hPa it reads pressure altitude.
- Airspeed indicator — measures pitot (total) minus static pressure and is calibrated with sea-level ISA density, so it reads IAS; only at sea level in the standard atmosphere does a perfect instrument read true airspeed.
- Vertical speed indicator — static pressure goes straight into a capsule and into the case through a calibrated leak; while the aircraft climbs or descends the two pressures differ in proportion to the rate of change of static pressure, i.e. the rate of climb. It lags a sudden change.
- Turn-and-bank (turn-and-slip) indicator — a rate gyroscope whose precession shows the rate of turn, plus a ball in a curved tube (an inclinometer) that shows whether the turn is balanced or the aircraft is slipping or skidding.
High-lift devices raise C_L,max so that the aeroplane can fly slowly for take-off and landing. Trailing-edge flaps — plain, split, slotted and Fowler (which also extends the chord and so the area) — increase effective camber: the C_L-α curve shifts up and to the left, α_L0 becomes more negative and the stall comes at a slightly lower α. Leading-edge devices — slats, slots, Krueger flaps and droop — delay leading-edge separation, so the curve simply extends to a higher stalling angle. The primary controls are the elevator (pitch), the ailerons (roll) and the rudder (yaw). Aileron deflection produces adverse yaw: the wing with the down-going aileron makes more lift and therefore more induced drag, yawing the nose away from the intended turn — the reason for differential and Frise ailerons and for co-ordinated rudder.
3. The drag polar, level flight, climb, glide, ceilings, take-off and landing
Below the stall the lift curve is linear, C_L = a(α − α_L0). The drag polar writes total drag as parasite plus lift-dependent drag: C_D = C_D0 + K·C_L², with K = 1/(π e AR) where AR is the aspect ratio and e the Oswald efficiency. In steady level flight L = W and T = D, so the thrust required is T_R = W/(L/D) and is least where L/D is greatest. Differentiating C_L/C_D gives the three conditions that the whole of performance rests on.
| Maximise | C_L | Drag split | Used for |
|---|---|---|---|
| C_L/C_D | √(C_D0/K) | Induced = parasite; C_D = 2C_D0 | Minimum thrust (drag); min glide angle; prop range; jet endurance |
| C_L^(3/2)/C_D | √(3C_D0/K) | Induced = 3 × parasite | Minimum power; minimum sink rate; prop endurance |
| C_L^(1/2)/C_D | √(C_D0/(3K)) | Induced = parasite/3 | Jet range at constant altitude |
The maximum lift-to-drag ratio is (L/D)max = 1/(2√(K·C_D0)). In a steady climb at flight-path angle γ, T − D − W sin γ = 0, so the rate of climb is R/C = V sin γ = (TV − DV)/W — excess power per unit weight. In a power-off glide tan γ = D/L, so the flattest glide is at (L/D)max and the glide range from height h is h·(L/D); the smallest sink rate is at the minimum-power condition, which is a different, slower speed. As altitude rises the excess power shrinks: the absolute ceiling is where the maximum rate of climb falls to zero, the service ceiling where it falls to 100 ft/min (0.5 m/s). Stall speed is V_s = √(2W/(ρ S C_L,max)). For take-off, lift-off is taken at about 1.2 V_s and the ground roll is approximately s_LO ≈ 1.44 W²/(g ρ S C_L,max T) — it grows with the square of weight and falls with density, which is why a hot, high airfield is the hardest case. Landing approach is at about 1.3 V_s.
4. Range and endurance — the Breguet equations
Range and endurance come from integrating the fuel burn as weight falls from W0 (take-off) to W1 (empty tanks). For a jet the thrust-specific fuel consumption c_t is the fuel weight burned per unit thrust per unit time, so dW/dt = −c_t T = −c_t W/(L/D). Endurance is then E = (1/c_t)(L/D) ln(W0/W1), largest at (L/D)max. Range, flying at constant V and L/D (a cruise-climb), is R = (V/c_t)(L/D) ln(W0/W1); at constant altitude, where V must fall as W falls, the optimum is at maximum C_L^(1/2)/C_D. For a propeller aircraft the engine meters fuel per unit power, so with specific fuel consumption c (fuel weight per unit power per unit time) and propeller efficiency η_p, R = (η_p/c)(L/D) ln(W0/W1) — best at (L/D)max, independent of speed — while endurance is best at maximum C_L^(3/2)/C_D, i.e. minimum power.
5. Load factor, turning flight and the V-n diagram
The load factor is n = L/W. In a level, co-ordinated turn at bank angle φ the vertical component of lift supports the weight, L cos φ = W, so n = 1/cos φ — 2 at 60° of bank. The horizontal component supplies the centripetal force, giving the turn radius R = V²/(g√(n² − 1)) = V²/(g tan φ) and the turn rate ω = V/R = g√(n² − 1)/V. A tight, fast turn therefore needs a high load factor, and the stall speed in the turn rises to V_s√n. In a pull-up from level flight n = 1 + V²/(gR).
The V-n (manoeuvre) diagram is the flight envelope the structure is designed for, plotted against equivalent airspeed. On the left it is bounded by the stall curves n = ½ρ0V_E²S C_L,max/W, parabolas that reach the positive and negative limit load factors; the positive curve meets the limit load factor at the corner or manoeuvring speed V_A = V_s√n_max, the lowest speed at which the limit load can be pulled and the speed at which the smallest radius and highest rate of turn are obtained. The top and bottom are the limit load factors, the right edge is the design dive speed. Gust lines are superimposed on it: a sharp-edged vertical gust raises the angle of attack and so the load factor, the increment growing with speed and lift-curve slope and falling with wing loading W/S.
6. Longitudinal static stability — stick fixed and stick free
Static stability asks only which way the first moment acts after a disturbance. Longitudinally, the pitching-moment coefficient about the centre of gravity is C_m = C_m0 + C_mα·α. The aeroplane is statically stable if a nose-up disturbance produces a nose-down moment, C_mα < 0, and it can be trimmed (C_m = 0) at a positive angle of attack only if C_m0 > 0 as well. With the centre of gravity at h and the wing aerodynamic centre at h_ac (both as fractions of the mean chord, measured aft from the leading edge), the wing contributes C_L,w(h − h_ac) — destabilising when the c.g. is behind the a.c. — and the tail contributes −η V_H C_L,t, where V_H = l_t S_t/(S c̄) is the horizontal tail volume ratio. With tail lift slope a_t, wing-body slope a and downwash gradient dε/dα, C_mα = a(h − h_ac) − η V_H a_t (1 − dε/dα).
The neutral point h_n is the c.g. position that makes C_mα zero: h_n = h_ac + η V_H (a_t/a)(1 − dε/dα). The static margin is h_n − h, and C_mα = −a(h_n − h), so the c.g. must lie ahead of the neutral point. A bigger tail, a longer tail arm or a smaller downwash gradient moves h_n aft and widens the allowable c.g. range; this is the whole reasoning behind horizontal tail position and size. If the pilot lets go of the stick, a free elevator floats with the local flow and the tail loses part of its restoring power. With the hinge-moment coefficient C_h = b0 + b1 α_t + b2 δ_e (+ b3 δ_tab) set to zero, the floating angle is δ_float = −(b1/b2)α_t, and the tail lift slope is multiplied by (1 − τ b1/b2), where τ = dα_t/dδ_e is the elevator effectiveness. Since b1/b2 is normally positive, the stick-free neutral point lies ahead of the stick-fixed one and the stick-free static margin is the smaller.
The hinge moment H = C_h q S_e c_e is what the pilot must hold, and through the control gearing G the stick force is F = G·H. A trim tab (b3 term) is used to drive the hinge moment, and so the stick force, to zero at the chosen speed; away from trim the stick-force gradient dF/dV is proportional to the stick-free static margin, which is what gives the pilot a feel for speed. Aerodynamic balance — a set-back hinge, a horn balance or an internal seal — reduces b1 and b2 and so the forces, but over-balancing makes them reverse. The elevator’s control power is C_mδe = −η V_H a_t τ.
7. Directional and lateral static stability
Directional (weathercock) stability requires that a sideslip produce a yawing moment that turns the nose back into the relative wind: C_nβ > 0 in the usual sign convention. The fuselage ahead of the c.g. is destabilising; the vertical tail provides the restoring moment, its contribution C_nβ,v ≈ η_v V_v a_v (corrected for the sidewash the fuselage and wing induce) growing with the vertical tail volume ratio V_v = l_v S_v/(S b) — so the fin is sized by its area and its moment arm behind the c.g. The rudder’s control power C_nδr must be enough to hold the aircraft straight with an engine out and in a crosswind landing.
Lateral stability (the dihedral effect) requires that a sideslip produce a rolling moment that lifts the leading, into-wind wing away from the slip: C_lβ < 0. Geometric dihedral Γ raises the angle of attack of the leading wing and is stabilising; wing sweepback makes the leading wing more nearly perpendicular to the flow, and its effect grows with C_L; a high wing on a fuselage is stabilising and a low wing destabilising, from the cross-flow round the fuselage; and a vertical tail above the c.g. adds a stabilising rolling moment. Too much dihedral effect is as bad as too little, which is why many high-wing swept transports carry anhedral. The aileron’s control power is C_lδa.
| Stability | Derivative | Stable sign | Main contributor |
|---|---|---|---|
| Longitudinal | C_mα | Negative | Horizontal tail; c.g. ahead of neutral point |
| Directional | C_nβ | Positive | Vertical tail (fuselage opposes) |
| Lateral | C_lβ | Negative | Dihedral, sweepback, high wing, fin above c.g. |
8. Equations of motion, Euler angles and the dynamic modes
Dynamic stability asks what the motion does over time. The aeroplane is a rigid body with six degrees of freedom; in body axes the force equations are X − mg sin θ = m(u̇ + qw − rv) and its two companions, and the moment equations are Euler’s equations with the products of inertia that the plane of symmetry leaves (only I_xz). Attitude is described by Euler angles in the 3-2-1 sequence — yaw ψ, then pitch θ, then roll φ — whose rates are not the body rates: φ̇ = p + (q sin φ + r cos φ) tan θ, θ̇ = q cos φ − r sin φ, ψ̇ = (q sin φ + r cos φ) sec θ. The tan θ and sec θ terms blow up at θ = ±90°, the singularity (gimbal lock) of this description.
Linearising about steady, wings-level, symmetric flight splits the equations in two. Because the aeroplane is symmetric about its xz-plane, small symmetric perturbations (u, w or α, q, θ) produce no asymmetric forces to first order and vice versa, so the longitudinal set decouples from the lateral-directional set (v or β, p, r, φ, ψ). Each set has characteristic roots whose pattern is remarkably consistent across aircraft.
- Short period (longitudinal) — a fast, heavily damped oscillation in α and pitch rate at almost constant speed; period of a few seconds.
- Phugoid (longitudinal) — a slow, lightly damped exchange of kinetic and potential energy, speed and height oscillating at nearly constant α. Lanchester’s approximation gives ω_n ≈ √2·g/U0, a period T ≈ π√2·U0/g, and a damping ratio ζ ≈ 1/(√2·L/D).
- Roll subsidence (lateral) — a fast, non-oscillatory, heavily damped convergence of roll rate, set by the roll damping derivative.
- Spiral (lateral) — a very slow, non-oscillatory mode that may converge or diverge; strong directional stability with weak dihedral effect tends to make it divergent.
- Dutch roll (lateral-directional) — a lightly damped oscillation coupling yaw and roll; strong dihedral effect with weak directional stability worsens it, and a yaw damper is the usual cure.
Key takeaways
- ISA: 288.15 K, 101 325 Pa, 1.225 kg/m³ at sea level; −6.5 K/km to 11 km, then 216.65 K constant to 20 km. EAS = TAS·√σ, and the order of correction is IAS → CAS → EAS → TAS.
- With C_D = C_D0 + K C_L²: (L/D)max = 1/(2√(K C_D0)) at C_L = √(C_D0/K); minimum power at √(3C_D0/K); jet range at √(C_D0/(3K)).
- Breguet: jet range (V/c_t)(L/D) ln(W0/W1), endurance (1/c_t)(L/D) ln(W0/W1); propeller range (η_p/c)(L/D) ln(W0/W1). Convert TSFC to per second.
- Level turn: n = 1/cos φ, R = V²/(g√(n² − 1)); manoeuvring speed V_A = V_s√n_max; R/C = excess power/W; ceilings at R/C = 0 (absolute) and 0.5 m/s (service).
- Stable signs: C_mα < 0 (c.g. ahead of h_n = h_ac + ηV_H(a_t/a)(1 − dε/dα)), C_nβ > 0, C_lβ < 0; stick-free neutral point is ahead of stick-fixed.
- Longitudinal modes: short period (fast, damped) and phugoid (slow, speed–height exchange, T ≈ π√2·U0/g); lateral: roll subsidence, spiral, Dutch roll.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In the International Standard Atmosphere, what is the temperature at a geopotential altitude of 5 km, in K (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 255.65
The troposphere lapse rate is 6.5 K/km from 288.15 K, so T = 288.15 − 6.5 × 5 = 288.15 − 32.5 = 255.65 K. Using 288 K for sea level, or the isothermal value 216.65 K that only applies above 11 km, is the usual slip.An aircraft flies at a true airspeed of 200 m/s at an altitude where the density ratio σ = ρ/ρ0 = 0.25. What is its equivalent airspeed, in m/s?
Numerical answer — type the value.
Show answer
Answer: 100
Equal dynamic pressure: ½ρV_TAS² = ½ρ0V_EAS², so V_EAS = V_TAS·√σ = 200 × 0.5 = 100 m/s. Dividing by √σ instead gives 400 m/s, which would make the equivalent airspeed larger than the true airspeed at altitude — impossible.With its subscale set to the standard sea-level pressure of 1013.25 hPa, an altimeter reads:
Show answer
Answer: B — Pressure altitude
An altimeter is an aneroid barometer graduated in ISA altitude, so on the standard setting it reports the ISA altitude having the measured static pressure — the pressure altitude. It reads geometric altitude only when the real atmosphere happens to match ISA, and density altitude needs the temperature as well.Which of the following statements about high-lift devices are correct?
Show answer
Answer: A — A trailing-edge flap increases effective camber and makes the zero-lift angle more negative; B — A leading-edge slat mainly extends the lift curve to a higher stalling angle; C — A Fowler flap increases both camber and wing area
Trailing-edge flaps add camber (the C_L-α line shifts up and left, α_L0 more negative), Fowler flaps also slide aft to add area, and slats delay leading-edge separation so the curve continues to a higher α. The fourth is false: with trailing-edge flaps the stall usually comes at a slightly LOWER angle of attack, even though C_L,max is higher.An aircraft has the drag polar C_D = 0.02 + 0.045 C_L². What is its maximum lift-to-drag ratio (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 16.67
(L/D)max = 1/(2√(K C_D0)) = 1/(2√(0.045 × 0.02)) = 1/(2 × 0.03) = 16.67. Check: at C_L* = √(0.02/0.045) = 0.667, C_D = 0.02 + 0.045 × 0.444 = 0.04 = 2C_D0, and 0.667/0.04 = 16.67. Forgetting the factor 2 gives 33.3.An aircraft of weight 50 kN flies at 60 m/s with thrust 10 kN and drag 5 kN. What is its steady rate of climb, in m/s?
Numerical answer — type the value.
Show answer
Answer: 6
R/C = (T − D)V/W = (10 − 5) × 60/50 = 300/50 = 6 m/s — the excess power 300 kW divided by the weight 50 kN. Using thrust alone (10 × 60/50 = 12 m/s) ignores the power already spent against drag.A glider with a maximum lift-to-drag ratio of 20 starts gliding from a height of 2 km in still air. What is the greatest horizontal distance it can cover, in km?
Numerical answer — type the value.
Show answer
Answer: 40
In a steady glide tan γ = D/L, so the flattest glide is at (L/D)max and the range is h × (L/D)max = 2 × 20 = 40 km. The minimum-sink speed keeps the glider up longest but covers less distance.A jet aircraft cruise-climbs at a constant 250 m/s and constant L/D = 16. Its TSFC is 0.6 per hour and the ratio of initial to final weight is 1.25. Using the Breguet equation, what is its range in km (nearest integer)?
Numerical answer — type the value.
Show answer
Answer: 5355
c_t = 0.6/3600 = 1.667 × 10⁻⁴ s⁻¹. R = (V/c_t)(L/D) ln(W0/W1) = (250/1.667 × 10⁻⁴) × 16 × ln 1.25 = 1.5 × 10⁶ × 16 × 0.22314 = 5.355 × 10⁶ m ≈ 5355 km. Leaving c_t per hour gives about 1.5 km, and using log10 instead of ln gives about 2326 km.For a propeller-driven aircraft with a parabolic drag polar, maximum range and maximum endurance are obtained respectively at the maximum of:
Show answer
Answer: A — C_L/C_D and C_L^(3/2)/C_D
A piston-propeller engine burns fuel per unit power, so range R = (η_p/c)(L/D) ln(W0/W1) is best at (L/D)max, and endurance, which needs minimum power, is best at maximum C_L^(3/2)/C_D. The second option is the jet pair (range at C_L^(1/2)/C_D, endurance at C_L/C_D) — the classic swap.An aircraft makes a level, co-ordinated turn at 100 m/s with a bank angle of 60°. Taking g = 9.81 m/s², what is the turn radius, in m (nearest integer)?
Numerical answer — type the value.
Show answer
Answer: 589
n = 1/cos 60° = 2, so R = V²/(g√(n² − 1)) = 10 000/(9.81 × √3) = 10 000/16.99 = 588.5 m. The same comes from R = V²/(g tan φ) with tan 60° = 1.732. Using n in place of √(n² − 1) gives 510 m.An aircraft with a 1-g stall speed V_s is designed for a positive limit load factor n_max. Its manoeuvring (corner) speed is:
Show answer
Answer: B — V_s·√n_max
On the stall boundary n = (V/V_s)² because lift at C_L,max grows with V². Setting n = n_max gives V_A = V_s√n_max. Below V_A the wing stalls before the limit load is reached; above it the pilot can overstress the structure.For an aircraft, the wing aerodynamic centre is at 0.25c̄, the horizontal tail volume ratio is 0.6, tail efficiency is 1, the wing-body and tail lift slopes are 5 and 4 per rad, and dε/dα = 0.4. With the c.g. at 0.30c̄, what is the stick-fixed static margin, as a fraction of c̄ (to three decimal places)?
Numerical answer — type the value.
Show answer
Answer: 0.238
h_n = h_ac + ηV_H(a_t/a)(1 − dε/dα) = 0.25 + 0.6 × (4/5) × 0.6 = 0.25 + 0.288 = 0.538. Static margin = h_n − h = 0.538 − 0.30 = 0.238. Leaving out the (1 − dε/dα) factor puts h_n at 0.73 and overstates the margin.Compared with the stick-fixed neutral point, the stick-free neutral point of a conventional aircraft (with b1/b2 > 0) is:
Show answer
Answer: B — Forward, so the stick-free static margin is smaller
A free elevator floats at δ = −(b1/b2)α_t, in the direction that unloads the tail, so the tail lift slope is reduced by the factor (1 − τb1/b2). A less effective tail puts the neutral point further forward, and the stick-free margin is the smaller one — which is why stick-free stability is the more demanding requirement.Which of the following make the rolling-moment derivative C_lβ more negative, that is, increase lateral static stability?
Show answer
Answer: A — Positive geometric dihedral; B — Wing sweepback; C — A high-wing position on the fuselage
Dihedral raises the leading wing’s angle of attack in a slip, sweepback makes the leading wing more normal to the flow, and a high wing gains from the cross-flow round the fuselage — all three roll the aircraft away from the slip, so C_lβ becomes more negative. A low wing does the opposite and is destabilising, which is why low-wing aircraft carry more geometric dihedral.Which of the following statements about the longitudinal dynamic modes of a conventional aircraft are correct?
Show answer
Answer: A — The short-period mode is a fast, heavily damped oscillation at nearly constant speed; B — The phugoid is a slow exchange of kinetic and potential energy at nearly constant angle of attack; C — By Lanchester’s approximation, the phugoid period is proportional to the flight speed
Short period: α and q oscillate in seconds with strong damping while speed barely changes. Phugoid: speed and height trade at nearly constant α, with T ≈ π√2·U0/g — directly proportional to U0 (about 45 s at 100 m/s). The Dutch roll couples yaw and roll and belongs to the lateral-directional set, which decouples from the longitudinal one.In the 3-2-1 Euler angle description of aircraft attitude, the kinematic equations relating Euler-angle rates to body rates become singular when:
Show answer
Answer: B — The pitch angle θ is ±90°
φ̇ and ψ̇ contain tan θ and sec θ, which are infinite at θ = ±90°: the yaw and roll axes line up and the description loses a degree of freedom (gimbal lock). A 90° bank is an ordinary attitude in this sequence, since φ enters only through sin φ and cos φ.