032 Performance topic guide
Take-Off Climb Segments and Gradients
After an engine fails at or after V1 on a multi-engine take-off, the climb to a safe height is flown in four defined segments, and each one changes the aeroplane's configuration, thrust setting, and target speed at a fixed boundary. Knowing which segment you are in tells you the gear position, the flap setting, the thrust rating, and the target speed without having to reason it out from scratch under exam time pressure.
Each segment also carries its own minimum climb gradient, because the whole point of the segmented climb is to guarantee obstacle clearance even with one engine inoperative. The gradients tested most often are the second segment's 2.4 per cent for a twin and the final segment's 1.2 per cent, and the exam expects both the number and the segment it belongs to.
The four segments in order
Each segment is defined by what the aeroplane is doing, not by a fixed altitude band, which is why the boundaries matter more than any single number.
- First segment: from lift-off to gear fully retracted, one engine inoperative, take-off thrust, speed building towards V2. A positive gradient is required, but no specific minimum percentage applies to a twin at this stage.
- Second segment: gear up, take-off thrust, climbing at V2 from 35 ft, the certification screen height, to the acceleration altitude. This is where the numerical minimum gross gradient is set, 2.4 per cent for a twin.
- Third segment: level acceleration at the acceleration altitude, flaps retracting from the take-off setting to clean, speed increasing from V2 towards the final segment speed. No height is gained here; it is traded for configuration and speed.
- Final segment: clean configuration, maximum continuous thrust, climbing from the acceleration altitude to 1500 ft at the final take-off speed. Its minimum gross gradient for a twin is 1.2 per cent.
Gross versus net gradient
Every gradient quoted above is a gross figure, the gradient the aeroplane actually demonstrates. Obstacle clearance calculations do not use the gross figure directly; they use a net gradient, the gross figure reduced by a defined margin, commonly 0.8 per cent for a twin, to build in a safety buffer.
An aeroplane that meets the gross certification minimum exactly still clears obstacles on the reduced net path, not on the gross one. A question about the certification minimum wants the gross gradient, and a question about obstacle clearance wants the net gradient, and the two are not interchangeable.
Converting a gradient into a rate of climb
Charts and stems sometimes state a requirement as a gradient and sometimes as a rate of climb, so converting between the two is a recurring skill. The link runs through groundspeed: a 1 per cent gradient at a groundspeed of 100 kt gives a rate of climb of approximately 100 ft per minute, so in general, gradient in per cent multiplied by groundspeed in knots gives an approximate rate of climb in feet per minute. Three per cent at 200 kt, for instance, is approximately 600 ft per minute.
The word approximately matters. The relationship uses groundspeed, so any wind component has to be applied to TAS first, and the true conversion factor is closer to 101 than 100, which is why this is always treated as a close estimate rather than an exact value.
Worked example
Worked example: gradient to rate of climb in the second segment
During the second segment, an aeroplane's true airspeed is 160 kt with a 10 kt headwind component, and it is achieving exactly the certification minimum gross gradient for a twin. Using the standard approximation, what rate of climb does this represent?
- A384 ft per minute
- B360 ft per minute
- C240 ft per minute
- D180 ft per minute
Show the answer and walkthrough
Correct answer: B
- A. This uses the 160 kt true airspeed directly. The gradient is a ratio against distance over the ground, so the headwind must be subtracted to get groundspeed first.
- B. Correct. Groundspeed is 160 minus 10, or 150 kt, and 2.4 per cent, the second segment minimum, multiplied by 150 gives 360.
- C. This uses 1.6 per cent, the net gradient (the 2.4 per cent gross figure reduced by the 0.8 per cent twin margin), instead of the gross figure the certification minimum refers to.
- D. This uses 1.2 per cent, the final segment's minimum gradient, instead of the second segment's 2.4 per cent.
Step by step
- Find groundspeed first: 160 kt TAS minus the 10 kt headwind gives 150 kt.
- Identify the correct gradient: the question specifies the second segment gross minimum for a twin, which is 2.4 per cent.
- Apply the approximation: rate of climb in feet per minute is approximately gradient in per cent multiplied by groundspeed in knots, so 2.4 multiplied by 150 gives 360.
- Sanity check: at a reference groundspeed of 100 kt, 2.4 per cent would be about 240 ft per minute; scaling that up for the faster 150 kt groundspeed pushes the figure higher still, and 360 ft per minute is a sensible increase over that reference.
Common mistakes
Feeding TAS into the gradient conversion instead of groundspeed
The approximation is built on distance covered over the ground per minute, so any wind component changes the answer, and this is exactly the kind of TAS-only distractor the exam plants.
Mixing up which segment a gradient belongs to
2.4 per cent and 1.2 per cent are both twin-engine minimums, one for the second segment and one for the final segment. Quoting the right number for the wrong segment still loses the mark.
Using net gradient where gross was asked, or the reverse
Certification minimums are gross figures, obstacle clearance margins are net figures, and a stem that names one while expecting the other is testing exactly this confusion.
Related topic guides
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Last reviewed July 2026