Hot, High and Uphill: A Take-off That Does Not Add Up
Cessna 172S at 1,150 kg departing Runway 07 from a 1,180 m dry-grass strip. Aerodrome elevation 1,200 ft, QNH 1005 hPa, OAT 30 °C. The runway is 1.5% upslope, with no clearway or stopway, so TORA = TODA = 1,180 m. Wind is 260°/12 kt, giving a 12 kt tailwind component and about 2 kt of crosswind. The AFM quotes a ground roll of 480 m at 1,200 ft pressure altitude, 30 °C, MTOW, level paved surface, still air. AFM corrections: add 15% for dry grass, add 10% for each 1% of upslope, add 10% for each 2 kt of tailwind component (no credit is given for headwind), and total distance to 50 ft is 1.65 × ground roll.
1. Compute the take-off distance required and decide whether the departure may be made.
Read the worked answer
Apply the corrections multiplicatively, to the ground roll, before the 1.65 factor.
| Step | Factor | Distance |
|---|---|---|
| AFM ground roll (1,200 ft PA, 30 °C, MTOW, paved, level) | - | 480 m |
| Dry grass | × 1.15 | 552 m |
| 1.5% upslope (+10% per 1%) | × 1.15 | 635 m |
| 12 kt tailwind (+10% per 2 kt = +60%) | × 1.60 | 1,016 m |
| Distance to 50 ft | × 1.65 | 1,676 m |
TODR ≈ 1,680 m against a TODA of 1,180 m - a shortfall of about 500 m (42%). The departure may not be made.
The trap is the ground roll: 1,016 m looks as though it fits inside 1,180 m. It does not, because the limiting figure is the distance to 50 ft, and the aeroplane must also be at 50 ft with a positive gradient at the end of the TODA. Even if the surface were paved and level, the tailwind correction alone would take the ground roll to 768 m and the TODR to 1,267 m, still outside the runway.
2. Is 1,200 ft the correct pressure altitude at which to enter the AFM data, and does it change the outcome?
Read the worked answer
No. Pressure altitude = elevation + 30 ft per hPa below 1013:
PA = 1,200 + (1013 − 1005) × 30 = 1,200 + 240 = 1,440 ft
QNH below 1013 means the pressure surface is low, so the aerodrome sits at a higher pressure altitude than its surveyed elevation. Entering the chart at 1,200 ft underestimates the distance. Using the rough rule of +10% ground roll per 1,000 ft of pressure altitude, add about 2.4%: 1,016 × 1.024 ≈ 1,040 m, so TODR ≈ 1,717 m.
The direction of the answer does not change - the take-off is still not possible - but the principle matters. In a case with 1,250 m available, entering at 1,200 ft instead of 1,440 ft would turn a legal take-off into an illegal one. Interpolating for both pressure altitude and temperature, rather than rounding towards the convenient line, is not optional.
3. Before start, the wind shifts to 070°/12 kt and the surface is confirmed dry. Recompute and state your decision with any conditions you would attach.
Read the worked answer
Runway 07 with wind 070°/12 kt is a pure headwind, and the AFM allows no credit for headwind, so the distance is computed as still air:
480 × 1.15 (grass) × 1.15 (slope) = 635 m ground roll 635 × 1.65 = 1,048 m to 50 ft
Applying the 2.4% pressure-altitude correction: 1,048 × 1.024 ≈ 1,073 m against 1,180 m - a margin of about 107 m (9%). The departure may be made.
Conditions I would attach:
- Confirm the wind is steady on 070°. If it veers towards the tail, the correction is not linear in the same way and the earlier numbers return very quickly - at 260°/12 kt the TODR was 1,680 m.
- Confirm the grass is dry. Wet grass typically adds a further 10%, giving 698 m ground roll and 1,152 m to 50 ft - inside 1,180 m, but by only 28 m, which is inside the noise of the chart.
- Confirm the aeroplane is actually at or below 1,150 kg. Every one of these figures is scaled to MTOW; 50 kg of extra baggage invalidates the chart entry, not just the margin.