Take-off Run: Two Small Changes That Cancel the Margin
A trainer aircraft of mass 1,100 kg begins its take-off run from rest with an average engine thrust of 3,200 N. Average aerodynamic drag during the ground roll is 900 N and the coefficient of rolling friction is 0.04. Use g = 9.81 m/s2 and treat thrust, drag and friction as constant. Rotation speed is 28 m/s. The available runway is 400 m long, and squadron procedure requires the take-off distance to be no more than 400/1.15 m, so that a 15 per cent margin always remains. The crew is considering two independent departures: one with two extra occupants totalling 240 kg, and one from a runway with a uniform 2 per cent upslope.
1. Calculate the take-off distance required on a level runway at 1,100 kg, and state whether it satisfies the procedure.
Read the worked answer
Step 1 - forces.
- Rolling friction = mu x m x g = 0.04 x 1,100 x 9.81 = 431.6 N
- Net accelerating force = 3,200 - 900 - 431.6 = 1,868.4 N
Step 2 - acceleration.
- a = 1,868.4 / 1,100 = 1.699 m/s2
Step 3 - distance from rest to 28 m/s.
- s = v^2 / (2a) = 784 / (2 x 1.699) = 784 / 3.397 = 230.8 m
Step 4 - procedure check.
- Permitted maximum = 400 / 1.15 = 347.8 m
- 230.8 m is well inside 347.8 m, so the departure is legal with about 117 m of the permitted distance unused.
The key move is to subtract friction and drag before dividing by mass: a candidate who divides 3,200 N by 1,100 kg gets 2.91 m/s2 and a distance of 135 m, which understates the run by nearly 100 m.
2. Recompute the distance for the departure carrying 240 kg extra and judge whether the flight still meets the 1/1.15 rule.
Read the worked answer
Step 1 - new mass. 1,100 + 240 = 1,340 kg
Step 2 - forces.
- Rolling friction = 0.04 x 1,340 x 9.81 = 525.8 N
- Net force = 3,200 - 900 - 525.8 = 1,774.2 N
Step 3 - acceleration. a = 1,774.2 / 1,340 = 1.324 m/s2
Step 4 - distance. s = 784 / (2 x 1.324) = 784 / 2.648 = 296.1 m
Judgement: 296.1 m against a permitted 347.8 m. The departure is still legal, but the margin has fallen from 117 m to about 52 m. The extra 240 kg cost 65 m of runway - roughly a quarter of the original run - for only a 22 per cent increase in mass, because friction grows with weight while thrust does not.
3. Now consider the 2 per cent upslope on its own, then both effects together. Decide whether the flight should depart, and say which factor the crew tends to underestimate.
Read the worked answer
Upslope alone (1,100 kg).
- Gravity component along the slope = m x g x sin(theta) = 1,100 x 9.81 x 0.02 = 215.8 N
- Net force = 1,868.4 - 215.8 = 1,652.6 N
- a = 1.502 m/s2
- s = 784 / 3.005 = 260.9 m - legal, costing about 30 m
Both together (1,340 kg, 2 per cent slope).
- Friction = 525.8 N; slope component = 1,340 x 9.81 x 0.02 = 262.9 N
- Net force = 3,200 - 900 - 525.8 - 262.9 = 1,511.3 N
- a = 1,511.3 / 1,340 = 1.128 m/s2
- s = 784 / 2.256 = 347.6 m against a permitted 347.8 m
Decision: do not depart. A margin of 0.2 m is not a margin; it is the arithmetic cancelling to zero. Either offload the 240 kg or use a level runway, then re-check.
What the crew underestimates: weight. Each factor alone is legal, and it is tempting to treat 'legal + legal' as safe. But friction and slope both scale with mass on the same run, so they compound rather than add independently - the mass penalty is about twice the slope penalty, and together they consume the entire 15 per cent margin.