How to Calculate Aircraft Take-Off Ground Run: A Flight Mechanics Tutorial
An aircraft does not become airborne simply because its engines are producing thrust. During the take-off ground run, thrust must overcome aerodynamic drag and rolling resistance. On a sloping runway, it must also overcome a component of the aircraft’s weight.
This tutorial derives a useful ground-run equation from Newton’s second law. It is an engineering model for understanding aircraft performance, not a substitute for approved Aircraft Flight Manual or Pilot’s Operating Handbook data.
What is the take-off ground run?
The ground run is the distance travelled from brake release to lift-off. It is only one part of total take-off distance; the airborne distance needed to clear a specified obstacle is calculated separately.
As the aircraft accelerates, its wings progressively generate more lift. This reduces the load carried by the wheels and therefore reduces rolling resistance. At lift-off, the available lift is sufficient for the aircraft to leave the runway.
For an overview of take-off speeds and phases, including rotation and lift-off speed, see Breaking Down Take-Off: Critical Flight Parameters.
Step 1: Draw the free-body diagram
Start with an aircraft on a level runway and take the forward direction as positive. The important forces are:
- T — thrust, acting forward;
- D — aerodynamic drag, acting opposite to motion;
- R — rolling resistance at the tyres;
- W — weight, equal to mg, acting vertically downward;
- L — aerodynamic lift, acting vertically upward; and
- N — normal reaction from the runway, acting vertically upward.
Step 2: Write the vertical force balance
During the ground run, the aircraft remains in contact with the runway. There is no significant vertical acceleration, so the vertical forces balance:
Therefore:
As lift increases, the normal reaction N falls. The wheels carry less of the aircraft’s weight.
Step 3: Express rolling resistance
Rolling resistance can be modelled as the coefficient of rolling resistance multiplied by the normal reaction:
Substituting N = W − L gives:
Where μ is the coefficient of rolling resistance. This shows why rolling resistance is not perfectly constant during a take-off run. At low speed, lift is small and the wheels carry almost all the aircraft weight. Near lift-off, lift is much greater and rolling resistance is lower.
Step 4: Apply Newton’s second law along the runway
Newton’s second law states that the resultant force equals mass multiplied by acceleration:
For the level-runway ground run:
Substituting the rolling-resistance expression gives the fundamental take-off ground-run equation:
This states that acceleration depends on the force remaining after thrust has overcome drag and rolling resistance.
Step 5: Allow for changing lift and drag
Lift and drag both vary approximately with the square of airspeed:
Where:
- ρ is air density;
- V is airspeed;
- S is wing reference area;
- CL is lift coefficient; and
- CD is drag coefficient.
Substituting these terms into the force equation gives:
This is more realistic than assuming fixed acceleration because aerodynamic lift and drag rise as the aircraft accelerates.
Step 6: Convert acceleration into distance
Acceleration is the rate of change of velocity with time:
However, the quantity required is runway distance, s. Using:
The ground-run equation becomes:
Rearranging:
The ground-run distance from rest to lift-off speed VLO is therefore:
This integral recognises that drag, lift, rolling resistance and potentially thrust change during acceleration.
A simpler constant-acceleration approximation
For an introductory estimate, assume the net accelerating force is approximately constant:
Using the standard kinematic relationship:
If the aircraft starts from rest, U = 0, so:
This approximation is useful for showing trends, but it is not a full aircraft-performance calculation. In reality, drag increases with speed, lift reduces rolling resistance, and available thrust may change as airspeed increases.
Worked example: estimating a ground run
Consider a simplified aircraft model operating from a level, dry runway in still air:
- Mass, m = 6,000 kg
- Weight, W = 6,000 × 9.81 = 58,860 N
- Wing area, S = 30 m2
- Air density, ρ = 1.225 kg/m3
- Take-off thrust, T = 26,000 N
- Rolling-resistance coefficient, μ = 0.02
- Ground-run lift coefficient, CL = 0.70
- Drag coefficient, CD = 0.040
- Lift-off speed, VLO = 60 m/s
First, combine the speed-dependent drag and rolling-resistance terms:
Substituting the values:
The ground-run integral becomes:
Solving gives:
In this deliberately simplified example, the aircraft requires approximately 451 m of ground roll to reach its assumed lift-off speed.
The result does not include rotation, obstacle clearance, wind, runway contamination, engine-transient behaviour, certification margins or the detailed aerodynamic data used in approved performance manuals.
What makes the ground run longer?
Higher aircraft mass
A heavier aircraft increases both W and m. More force is needed to accelerate it, and a higher lift-off speed may also be required.
Higher temperature or altitude
Lower air density reduces lift at a given true airspeed. The aircraft must accelerate to a higher true airspeed to generate the required lift, while engine or propeller performance may also be reduced.
Higher drag
High-lift devices can reduce lift-off speed, but they also alter drag. Engineers must balance the lift benefit against the drag penalty.
A poor runway surface
A higher value of μ increases rolling resistance. Soft, rough or contaminated surfaces can therefore have a substantial effect on acceleration.
Extending the model for runway slope
For an uphill runway with slope angle θ, part of the aircraft’s weight acts down the slope:
The longitudinal equation becomes:
An uphill slope therefore reduces acceleration and increases ground-run distance. On a downhill runway, the weight component acts in the direction of travel and assists acceleration.
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