Landing Ground Roll Distance (Simplified) Formula Calculator

Q: What do the variables W, g, ρ, S, CLmax, and Davg represent?

W = aircraft weight (N)g = gravity (m/s²)ρ = air density (kg/m³)S = wing area (m²)CLmax = maximum lift coefficient (with flaps)Davg = average drag during landing (N) – includes aerodynamic and braking drag.

Q: Give a worked example.

W = 150,000 N, g = 9.81, ρ = 1.225, S = 50 m², CLmax = 2.0 (with flaps), Davg = 30,000 N. sL ≈ 1.69 × 150000² / (9.81 × 1.225 × 50 × 2.0 × 30000) = 1.69 × 2.25e10 / (9.81 × 1.225 × 3,000,000) = 3.8025e10 / (9.81 × 3.675e6) = 3.8025e10 / 36.05e6 ≈ 1055 m.

Q: What is the effect of spoilers on landing distance?

Spoilers increase drag (Davg) and reduce lift (effectively reducing CLmax for braking), shortening the distance.

Interpretation

Landing ground roll distance: s_L ≈ 1.69·W²/(g·ρ·S·C_Lmax·D_avg), where D_avg is average drag during braking. It estimates the distance to stop after touchdown. Example: W=20,000 N, ρ=1.225, S=20, C_Lmax=1.5, D_avg=2000 N → s_L ≈ 1.69×400e6/(9.81×1.225×20×1.5×2000) ≈ 676e6/(720,000) ≈ 939 m.

s_L ≈ 1.69 * W^2 / (g * ρ * S * C_Lmax * D_avg)

Landing Ground Roll Distance (Simplified)

Variables

Symbol	Quantity	Unit
s_L	Landing ground roll distance	m
W	Weight	N
g	Gravitational acceleration	m/s2
ρ	Air density	kg/m3
S	Wing area	m2
C_Lmax	Maximum lift coefficient
D_avg	Average deceleration force	N

What it means

The landing ground roll is the distance required to decelerate from touchdown speed to a stop. The formula assumes a constant deceleration force (braking + drag) and uses a factor of 1.69 (derived from energy considerations). It depends on landing weight, density, wing area, C_Lmax (which determines touchdown speed), and average drag (including braking). This estimate is used for field length certification and for design of braking systems. Understanding this relation is essential for ensuring safe landing performance.

Worked example

Landing Ground Roll – Two Examples

Real‑World

Scenario: W = 75,000 N, S = 20 m², C_Lmax = 2.0, D_avg = 30,000 N. Find landing roll.

Parameter	Value
W	75,000 N
S	20 m²
C_Lmax	2.0
D_avg	30,000 N

1s_L = 1.69 × W²/(g·ρ·S·C_Lmax·D_avg) = 1.69 × 5.625e9/(9.81×1.225×20×2×30000) = 9.506e9/(9.81×1.225×1,200,000) = 9.506e9/14,421,600 = 659 m

Result 659 m ✓ Typical

Scenario: W = 140,000, S = 30, C_Lmax = 2.2, D_avg = 50,000. Find s_L.

Parameter	Value
W	140,000
S	30
C_Lmax	2.2
D_avg	50,000

1s_L = 1.69 × 1.96e10/(9.81×1.225×30×2.2×50000) = 3.312e10/(9.81×1.225×3,300,000) = 3.312e10/39,659,000 = 835 m

Result 835 m ✓ Longer

Key insight: Landing roll depends on weight, wing loading, and average drag – brakes increase drag.

Common mistakes

Landing ground roll distance (simplified): s_L ≈ 1.69 · W² / (g·ρ·S·C_Lmax·D_avg).
D_avg: Average drag during landing (including braking).
1.69 factor accounts for deceleration.
Assumes no reverse thrust.
Braking effectiveness affects D_avg.

Applications

Landing ground roll distance, s_L ≈ 1.69·W²/(g·ρ·S·C_Lmax·D_avg), estimates the distance needed to stop after touchdown. This is essential for airfield design and for ensuring safe landing performance. The factor 1.69 accounts for the average deceleration during braking. Engineers use this to design braking systems, to evaluate landing gear performance, and to set approach speeds. By reducing landing distance through aerodynamic braking (spoilers) and efficient brakes, aerospace engineers can improve safety and enable operation on shorter runways.

Landing field length and certification analysis
Braking system design (wheel brakes, thrust reversers)
Landing gear and tire wear analysis
Spoiler and lift‑dump design for deceleration
Airfield compatibility and operational planning

Frequently Asked Questions

Q01What is the Landing Ground Roll Distance used for?

A01

It is a simplified estimate of the ground‑roll distance required to stop after touchdown, based on aerodynamic and braking drag. It is used for runway length design.

Q02What do the variables W, g, ρ, S, C_Lmax, and D_avg represent?

A02

W = aircraft weight (N)
g = gravity (m/s²)
ρ = air density (kg/m³)
S = wing area (m²)
C_Lmax = maximum lift coefficient (with flaps)
D_avg = average drag during landing (N) – includes aerodynamic and braking drag.

Q03Why is the landing ground roll important?

A03

It determines the required landing distance, which is often a limiting factor for airport operations.

Q04What are the assumptions of this formula?

A04

It assumes a constant deceleration, no thrust reversal, and that the landing speed is 1.3 times the stall speed. It also assumes a constant average drag.

Q05What are common mistakes when using this formula?

A05

Neglecting reverse thrust and spoiler effects, which substantially shorten real‑world landing distance versus wheel‑brake‑only estimates.
Using the wrong C_Lmax (e.g., without flaps).
Ignoring the effect of runway condition (wet, icy).

Q06Give a worked example.

A06

W = 150,000 N, g = 9.81, ρ = 1.225, S = 50 m², C_Lmax = 2.0 (with flaps), D_avg = 30,000 N. s_L ≈ 1.69 × 150000² / (9.81 × 1.225 × 50 × 2.0 × 30000) = 1.69 × 2.25e10 / (9.81 × 1.225 × 3,000,000) = 3.8025e10 / (9.81 × 3.675e6) = 3.8025e10 / 36.05e6 ≈ 1055 m.

Q07How does the landing distance vary with altitude?

A07

At higher altitude, ρ decreases, so the landing distance increases (since s_L ∝ 1/ρ).

Q08What is the effect of spoilers on landing distance?

A08

Spoilers increase drag (D_avg) and reduce lift (effectively reducing C_Lmax for braking), shortening the distance.

Q09How do you account for reverse thrust?

A09

Reverse thrust adds to the deceleration, effectively increasing D_avg. The formula would need to be modified to include it.

Q10What is the relationship between landing distance and landing speed?

A10

The distance is proportional to the square of the landing speed, so a lower landing speed reduces the distance significantly.