Asymmetric climb gradient after engine failure
A light twin (MTOM 2,000 kg) departs a 1,400 m asphalt runway, dry, at sea level, OAT +30 °C, QNH 1013. Obstacle departure procedure requires a net climb gradient of 4.8% to 1,200 ft. The AFM gives, for the actual mass of 1,950 kg: single-engine rate of climb 210 ft/min at Vyse, single-engine climb gradient 6.1% in ISA, and a 1% gradient reduction per 10 °C above ISA. The pilot plans a flap-up, Vyse climb.
1. Determine the achievable single-engine climb gradient and whether the obstacle procedure can be flown.
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ISA at sea level is +15 °C; actual is +30 °C, so +15 °C above ISA. Gradient reduction = 1.5 × 1% = 1.5%. Achievable gradient = 6.1% − 1.5% = 4.6%. Required net gradient = 4.8%. 4.6% < 4.8%, so the aircraft cannot meet the published net gradient at this temperature and mass. Options: reduce mass (offload), wait for cooler temperature, or use a different departure procedure/runway. Note the AFM figure is gross; the net gradient is gross minus 0.9% for twins, so if the 6.1% is gross, net = 6.1 − 1.5 − 0.9 = 3.7%, even further below 4.8%.
2. The crew elects to reduce mass. About how much must be offloaded to restore the required gradient, given the AFM shows roughly 0.9% gradient gain per 100 kg reduction?
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Deficit (using net figures) = 4.8% − 3.7% = 1.1%. At 0.9% per 100 kg: 1.1 / 0.9 × 100 ≈ 122 kg. Offload approximately 125 kg (to about 1,825 kg) and re-check the AFM, since the gradient/mass relationship is not perfectly linear. Also re-check the accelerate-stop and take-off distances at the new mass on the dry runway - both improve with lower mass.