NCAA CPL Principles of Flight study materials
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Engine failure after takeoff with an obstacle and a marginal single-engine gradient

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Engine failure after takeoff with an obstacle and a marginal single-engine gradient

A light twin (MTOW 2,100 kg, Vmc 65 KCAS, Vr 80 KCAS, Vxse 82 KCAS, Vyse 88 KCAS) departs a 1,400 m runway at elevation 3,800 ft with OAT 28°C, giving a density altitude of about 6,200 ft. The aircraft is 60 kg below MTOW. At 150 ft AGL, 78 KCAS, gear down and 10° flap, the right engine fails. The pilot applies left rudder and 5° bank into the failed engine. The published single-engine climb gradient is 2.0% at 82 KCAS and 1.5% at 88 KCAS at sea level, ISA. A 15 m obstacle lies 300 m beyond the runway end. The pilot must decide within seconds whether to continue the takeoff or reject. The first officer calls 'engine failure, right engine' and reaches for the gear handle. The captain says 'continue, we'll make the obstacle'.

1. Assess the controllability margin at 78 KCAS with gear down, 10° flap and 5° bank into the failed engine. Is Vmc a fixed speed? List the factors in this scenario that change it and the direction of change.

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At 78 KCAS the aircraft is 13 kt above the published Vmc of 65 KCAS, but that does not give a comfortable margin because Vmc is not a fixed number. It is a certification minimum measured under defined conditions: maximum takeoff power, gear up, takeoff flap, and 5° bank toward the operating engine.

In this scenario, gear down increases Vmc, and 10° flap also increases Vmc. Banking into the failed engine is the wrong direction: it increases Vmc because the horizontal component of lift adds to the yaw from the live engine rather than opposing it. Banking into the live engine reduces Vmc. The high density altitude reduces engine power, which tends to reduce Vmc in IAS terms, but it also severely reduces the excess thrust available for climb.

The pilot is also below Vxse and only 13 kt above Vmc, so there is little margin for yaw or turbulence. The first action should have been to bank into the live engine, not the failed engine, and to lower the nose toward Vxse.

2. Using the obstacle and gradient data, evaluate the captain's decision to continue. Show the required gradient and the likely effect of density altitude on the available gradient. What would you expect to happen?

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The required climb gradient to clear the obstacle is 15 m / 300 m = 5.0%. The published sea-level ISA gradients are only 2.0% at 82 KCAS and 1.5% at 88 KCAS. At a density altitude of about 6,200 ft, a normally aspirated engine loses roughly 3% of power per 1,000 ft, so it produces only about 81% of sea-level power. The single-engine gradient will therefore be well below the published figures, perhaps around 1.2% to 1.6% at best, and worse with gear down. It cannot clear a 5% obstacle gradient.

The captain's decision to continue is therefore unsound. At 150 ft AGL the practical options are to reject and land straight ahead if sufficient runway remains, or to accept a forced landing with minimum descent rate. Turning back is not appropriate at that height. The takeoff should not have been commenced without performance data showing the obstacle could be cleared on one engine.

3. If the decision is to reject or to continue, state the immediate actions and the target speed/configuration. Explain the aerodynamic reason for banking into the live engine.

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If continuing, the immediate actions are: maintain directional control with rudder, bank 5° into the live engine, pitch to Vxse (82 KCAS) for obstacle clearance, retract the gear, identify and verify the failed engine, feather it, apply maximum continuous power, and check climb performance. If the aircraft is not climbing, do not attempt to stretch the glide; prepare to land ahead.

If rejecting, close both throttles, keep the wings level, land straight ahead or on the remaining runway, and do not attempt a turn back. Lower the gear only when landing is assured.

Banking into the live engine produces a horizontal component of lift toward the live engine. That component opposes the yaw produced by asymmetric thrust, reducing the rudder force required and lowering Vmc. Banking into the failed engine has the opposite effect and increases the control problem.