Hot Day, Six Knots Tailwind, Trees Ahead
You are departing a private strip near Makurdi: elevation 1,250 ft, QNH 1,013, OAT 32°C, runway 09, 1,200 m of dry asphalt with a 1% upslope in the take-off direction. Wind is 270°/06 kt. The aircraft's POH chart, for the actual take-off mass, gives 480 m ground roll and 860 m to 50 ft at these conditions, and there is no clearway or stopway. Your operator's SOP corrections are +10% for each 2 kt of tailwind and +5% for each 1% of upslope; the chart assumes a smooth, dry, hard surface, a serviceable engine and average pilot technique. A colleague suggests using runway 09 anyway because the aircraft is about 100 kg below MTOW and 'the book figures are always conservative'. There are trees on the extended centreline and the strip has no fire cover. You are the pilot-in-command and must decide.
1. Calculate the take-off distance required to 50 ft on runway 09 and the margin over the available distance.
Read the worked answer
QNH equals 1,013 hPa, so pressure altitude = aerodrome elevation = 1,250 ft.
Wind 270° on runway 09 gives a full tailwind component of 6 kt.
Corrections, applied multiplicatively:
- Tailwind: 6 kt ÷ 2 × 10% = +30%
- Upslope: 1% × 5% = +5%
Distance to 50 ft = 860 × 1.30 = 1,118 m; × 1.05 = 1,174 m. Ground roll = 480 × 1.30 × 1.05 = 655 m.
Available TODA = 1,200 m, so the screen height is reached 26 m before the end of the runway - about 2% margin. That is not a margin; it is the difference between the chart and reality. Any degradation (soft asphalt edge, late rotation, a 1 kt stronger wind) puts the aircraft into the trees. The take-off is not acceptable on runway 09.
2. State your decision and justify it against the three 'fixes' likely to be proposed: use runway 27, wait, or offload fuel.
Read the worked answer
Decision: reject runway 09. Use runway 27, or wait for the wind to align.
Runway 27 gives a 6 kt headwind. Applying no headwind credit at all, the distance becomes 860 × 1.05 = 903 m, a margin of about 297 m - the tailwind penalty alone was 30%. The upslope also becomes downhill, an unclaimed bonus.
'Offload fuel' is only defensible if you have a documented mass-correction chart. Without one, you are guessing at a few percent, and the chart numbers you have are already at a lighter mass than MTOW; inventing a further credit is exactly the error the colleague is making. It also leaves less fuel for the diversion you may need on a hot day.
'Wait' is legitimate if the wind is thermally driven and forecast to drop, but you must set a go/no-go time and revert to runway 27 if it does not. Departing on 09 with 26 m of margin is a gamble on perfect technique and a perfect engine.
3. Quantify the density altitude and explain how it degrades the numbers in the chart.
Read the worked answer
ISA temperature at 1,250 ft = 15 − (2 × 1.25) = 12.5°C. OAT 32°C is therefore ISA +19.5°C. Using the standard approximation of 120 ft per °C of ISA deviation: DA ≈ 1,250 + (120 × 19.5) ≈ 3,590 ft.
At that density altitude the engine produces less power and the propeller is less efficient, so acceleration is slower and the ground roll grows. Because true airspeed for the same indicated airspeed is higher, the aircraft covers more runway before reaching flying speed, and the climb gradient after lift-off is reduced - which is why the 50 ft point moves down the runway and why obstacle clearance suffers most.
The chart was already computed at 32°C, so the density altitude is embedded in the 860 m figure; the danger is adding further penalties the chart does not contain - wet or soft grass, under-inflated tyres, a tailwind gusting above the mean, or a contaminant on the surface. On a day like this, treat every correction as a minimum and keep a hard abort point at 70% of the calculated ground roll.