Takeoff performance and ground run

Why it matters

Runway length is often the hardest constraint an aircraft faces: a design that cannot take off from the runways its customers have will not sell, and an operator who gets the takeoff calculation wrong on a hot day at a high airfield risks an overrun. Takeoff analysis ties together almost everything in this subject — stall speed, thrust available, drag polar, ground friction, density altitude and wind.

Key ideas

Phases of takeoff.

1. Ground roll from rest to the lift-off speed V_LO (with a short rotation phase near the end).
2. Transition (airborne arc) in which the flight path curves upward.
3. Climb to the screen height — 10.7 m (35 ft) for civil transports, 15 m (50 ft) for light and military aircraft. Takeoff distance is measured from brake release to the screen; this topic focuses on the ground roll, which is usually the largest part.

Key speeds. Lift-off speed is set by the stall speed in the takeoff configuration: V_LO ≈ 1.1–1.2·V_stall, with V_stall = √(2W/(ρ·S·C_L,max)). For transports: V1 is the decision speed (engine failure below V1 → stop; above V1 → continue); VR is rotation speed; V2 is the take-off safety speed reached by the screen height with one engine inoperative.

Equation of motion on the ground roll. On a level runway, with rolling-friction coefficient μ (about 0.02–0.05 on dry concrete or asphalt, higher on grass): (W/g)·dV/dt = T − D − μ·(W − L). The wheels carry W − L, so friction falls as lift builds. During the roll the aircraft sits at a fixed ground-attitude lift coefficient C_L,g, so L and D both grow as V². Writing the acceleration as a = A − B·V² with

• A = g·(T/W − μ)
• B = (g/W)·½ρS·(C_D,g − μ·C_L,g) and using ds = V·dV/a, the ground roll is s_g = (1/(2B))·ln[A/(A − B·V_LO²)] (thrust taken constant).

Approximate forms.

• Average-force method: evaluate the forces at 0.7·V_LO (roughly the rms speed) and use s_g ≈ V_LO²/(2·ā). It is accurate to about 1 % for typical aircraft.
• Neglecting drag and friction and taking V_LO = 1.2·V_stall gives the classic scaling s_g ≈ 1.44·W²/(g·ρ·S·C_L,max·T). The scaling shows what matters: s_g ∝ W² (heavier needs a higher V_LO and accelerates more slowly), ∝ 1/ρ (hot and high airfields), ∝ 1/C_L,max (flaps help) and ∝ 1/T.

Ground effect. Close to the ground the wing's trailing vortices are restricted and induced drag falls. A common estimate (from your textbook) multiplies K by φ = (16h/b)²/[1 + (16h/b)²], where h is wing height above ground and b is span. This lowers C_D,g during the roll.

Wind and slope. A headwind V_w means the aircraft needs only V_LO − V_w of ground speed, so the ground roll falls roughly as (1 − V_w/V_LO)² (empirical exponents near 1.85 are also used). A tailwind does the opposite, which is why takeoffs are made into wind. An uphill slope φ adds −W·sin φ to the force balance.

Hot and high. High density altitude lowers ρ, which raises V_LO (TAS) and also reduces thrust, so s_g grows faster than 1/ρ. Operators answer this with lower takeoff weight.

Balanced field length. For multi-engine aircraft, the takeoff field length is set by an engine failure: the accelerate-stop distance and the accelerate-go distance both depend on where the failure happens. Choosing V1 so the two are equal gives the balanced field length.

Formulas

V_stall = √(2W / (ρ·S·C_L,max)), V_LO ≈ 1.2·V_stall

(W/g)·dV/dt = T − D − μ·(W − L)

a = A − B·V², A = g·(T/W − μ), B = (g/W)·½·ρ·S·(C_D,g − μ·C_L,g)

s_g = (1/(2B))·ln[A / (A − B·V_LO²)]

s_g ≈ V_LO² / (2·ā), with ā evaluated at 0.7·V_LO

s_g ≈ 1.44·W² / (g·ρ·S·C_L,max·T) (drag and friction neglected, V_LO = 1.2·V_stall)

φ = (16h/b)² / [1 + (16h/b)²] (ground-effect factor on K)

Symbols: W weight (N); g = 9.81 m/s²; V ground speed = airspeed in still air (m/s); T thrust (N); D drag, L lift (N); μ rolling-friction coefficient; ρ density (kg/m³); S wing area (m²); C_L,max maximum lift coefficient in takeoff configuration; C_L,g, C_D,g lift and drag coefficients in ground-roll attitude; s_g ground roll (m); h wing height above ground, b span (m). Valid for a level runway, still air, constant thrust.

Worked examples

Example 1 (standard): transport ground roll, simple model. Given: W = 600 kN, S = 150 m², C_L,max = 1.5, T = 180 kN, μ = 0.02, ρ = 1.225 kg/m³. Neglect aerodynamic drag and lift during the roll; V_LO = 1.2·V_stall.

1. V_stall = √(2W/(ρSC_L,max)) = √(1 200 000/275.6) = 65.98 m/s; V_LO = 79.2 m/s.
2. Constant acceleration a = g·(T/W − μ) = 9.81 × (0.30 − 0.02) = 2.747 m/s².
3. s_g = V_LO²/(2a) = 79.18²/(2 × 2.747) = 1141 m.
4. Without friction (μ = 0), a = 2.943 m/s² and s_g = 1065 m — which equals 1.44·W²/(gρSC_L,max·T).

Example 2 (GATE level): ground roll with drag and lift. Given: W = 60 000 N, S = 25 m², C_L,max = 1.6, ground-roll C_L,g = 0.5, C_D = 0.03 + 0.05·C_L² (ground effect included in K), T = 15 000 N constant, μ = 0.03, sea level.

1. V_stall = √(2 × 60 000/(1.225 × 25 × 1.6)) = 49.49 m/s; V_LO = 1.2 × 49.49 = 59.38 m/s.
2. C_D,g = 0.03 + 0.05 × 0.25 = 0.0425.
3. A = g·(T/W − μ) = 9.81 × (0.25 − 0.03) = 2.158 m/s².
4. B = (g/W)·½ρS·(C_D,g − μC_L,g) = (9.81/60 000) × 15.31 × (0.0425 − 0.015) = 6.885 × 10⁻⁵ m⁻¹.
5. B·V_LO² = 6.885 × 10⁻⁵ × 3526.5 = 0.2428.
6. s_g = (1/(2B))·ln[A/(A − BV_LO²)] = 7262 × ln(2.158/1.9152) = 7262 × 0.11935 = 867 m.
7. Check with the 0.7·V_LO method: ā = 2.158 − 6.885 × 10⁻⁵ × 41.57² = 2.039 m/s², so s_g = 3526.5/(2 × 2.039) = 865 m — within 0.3 %.

Common mistakes

• Using W in place of W − L for the friction force; friction falls as lift builds.
• Taking V_LO = V_stall (no margin) or using the cruise-configuration C_L,max.
• Forgetting that s_g ∝ W², not W: 10 % more weight means about 21 % more ground roll.
• Writing a formula such as W²/(gρSC_L) without the thrust in the denominator; check units — the result must be in metres.
• Adding headwind to the lift-off speed instead of subtracting it from the required ground speed.

For GATE AE

Expect ground-roll numericals with the constant-acceleration model, with or without friction, the exact A − BV² integration, scaling questions (effect of weight, density, thrust or C_L,max on s_g), and the effect of headwind. Conceptual questions cover V1, VR, V2, balanced field length, ground effect and why hot-and-high airfields need longer runways.

Quick check

1. By what factor does the simple ground roll change if weight rises by 20 %?
2. Why does friction fall during the ground roll?
3. What is the screen height for a civil transport?
4. If V_LO = 60 m/s and the headwind is 6 m/s, by roughly what factor does the ground roll fall (square law)?

Answers: 1. 1.2² = 1.44. 2. Lift relieves the wheels, so the normal force W − L falls. 3. 10.7 m (35 ft). 4. (54/60)² = 0.81.