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ATPL Flight Performance and Planning study materials
ATPL Flight Performance and Planning study resources

ATPL Flight Performance and Planning practice questions

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Question 1Performance - take-off

In take-off performance planning, the take-off distance required (TODR) must not exceed which declared distance?

  1. The clearway length
  2. The take-off distance available (TODA) Correct answer.
  3. The take-off run available (TORA)
  4. The stopway length

Explanation

The take-off distance required, measured to the point where the aircraft reaches 35 ft above the take-off surface, must be contained within the take-off distance available. TORA and stopway relate to the ground run and accelerate-stop case, not to the full airborne distance.

Question 2Performance - take-off

An aerodrome has TORA 2400 m, clearway 150 m and stopway 60 m. What are TODA and ASDA respectively?

  1. 2550 m and 2400 m
  2. 2400 m and 2400 m
  3. 2460 m and 2550 m
  4. 2550 m and 2460 m Correct answer.

Explanation

TODA equals TORA plus clearway, so 2400 + 150 = 2550 m. ASDA equals TORA plus stopway, so 2400 + 60 = 2460 m. Clearway adds only to the take-off distance and stopway only to the accelerate-stop distance.

Question 3Performance - take-off

The take-off run required (TORR) is compared against which declared distance to ensure the ground run fits the runway?

  1. Landing distance available (LDA)
  2. Take-off distance available (TODA)
  3. Accelerate-stop distance available (ASDA)
  4. Take-off run available (TORA) Correct answer.

Explanation

The take-off run required is the ground distance needed to reach a point where the aircraft is airborne and able to climb, and it is checked against TORA. TODA is used for the full distance to 35 ft and ASDA for the accelerate-stop case.

Question 4Performance - take-off

What is the primary purpose of a clearway in take-off performance?

  1. To increase the take-off run available for heavy aircraft
  2. To allow the aircraft to stop after an engine failure before V1
  3. To provide a graded surface for the initial climb segment
  4. To allow the aircraft to reach 35 ft before the end of TODA without requiring a full runway surface Correct answer.

Explanation

A clearway is an obstacle-free area beyond TORA over which the aircraft may climb to 35 ft, so it extends TODA without providing a load-bearing surface. It is not used for stopping, and it does not change TORA.

Question 5Performance - take-off

V1 is best defined as the speed used in take-off planning at which:

  1. the aircraft rotates to the climb attitude
  2. the aircraft becomes fully controllable on the ground
  3. the aircraft reaches 35 ft above the runway surface
  4. the pilot may continue the take-off or stop within the ASDA after an engine failure Correct answer.

Explanation

V1 is the decision speed: after an engine failure at or above V1 the take-off can be continued safely, and at or below V1 the aircraft can be stopped within the accelerate-stop distance available. It is not the rotation speed or the screen height speed.

Question 6Performance - take-off

In balanced field take-off, the field length is that at which:

  1. the take-off run equals the landing distance available
  2. the accelerate-stop distance equals the take-off distance required Correct answer.
  3. the climb gradient equals the obstacle clearance requirement
  4. V1 equals VR exactly

Explanation

A balanced field occurs when the distance to stop after engine failure at V1 equals the distance to continue to 35 ft after engine failure at V1. At that point the field length is minimised for the given weight and conditions.

Question 7Performance - take-off

An increase in aerodrome pressure altitude, all else constant, has which effect on the take-off ground run?

  1. It increases the ground run because true airspeed for a given IAS is higher Correct answer.
  2. It decreases the ground run because the air is less dense
  3. It has no effect because IAS determines the lift-off
  4. It increases the ground run only if the temperature also rises

Explanation

At higher pressure altitude the air density falls, so a given indicated airspeed corresponds to a higher true airspeed and groundspeed. The aircraft therefore covers more runway to reach lift-off speed and the engine produces less thrust.

Question 8Performance - take-off

Compared with a dry runway, a wet runway take-off affects V1 primarily because:

  1. the reduced friction increases the accelerate-go distance, tending to raise V1
  2. the reduced braking friction increases the accelerate-stop distance, tending to lower V1 Correct answer.
  3. V1 is unaffected because it depends only on weight
  4. wet runways always require a fixed V1 reduction of 10 kt

Explanation

On a wet runway braking action is poorer, so the stopping distance for a given V1 is longer. To remain within ASDA, V1 is typically reduced, which increases the accelerate-go distance and may reduce the maximum take-off weight.

Question 9Performance - take-off

Which combination correctly describes the speeds at which the aircraft must reach 35 ft in the take-off flight path?

  1. At least VR at 35 ft, with VR not less than 1.1 VMCA
  2. At least V2 at 35 ft, with V2 not less than 1.13 VS1 in the take-off configuration Correct answer.
  3. At least VREF at 35 ft, with VREF equal to 1.3 VS0
  4. At least V1 at 35 ft, with V1 not less than 1.2 VS1

Explanation

The take-off flight path begins at 35 ft with the aircraft at V2 or greater, and V2 must be at least 1.13 times the stall speed in the take-off configuration. V1 and VR are ground-phase speeds, and VREF is a landing reference speed.

Question 10Performance - take-off

For a Class A aeroplane, the net take-off flight path must clear all obstacles by at least:

  1. 35 ft vertically Correct answer.
  2. 50 ft vertically
  3. 200 ft vertically
  4. 15 ft vertically

Explanation

The net take-off flight path, which is the gross gradient reduced by 0.8 percent for two-engine aeroplanes, must provide at least 35 ft vertical clearance over obstacles. This margin accounts for variations in piloting and performance.

Question 11Performance - take-off

The net take-off flight path gradient is obtained from the gross gradient by:

  1. adding a fixed increment to allow for pilot technique
  2. subtracting a fixed gradient decrement that depends on the number of engines Correct answer.
  3. subtracting a decrement that depends on the aircraft weight
  4. multiplying the gross gradient by the number of engines

Explanation

Certification applies a fixed reduction to the gross climb gradient to give the net gradient, the reduction being larger for two-engine aeroplanes than for four-engine aeroplanes. It is a constant percentage-point decrement, not a weight-dependent factor.

Question 12Performance - take-off

On a take-off chart, an increase in outside air temperature at constant pressure altitude and weight generally results in:

  1. a longer take-off distance with no change to the maximum weight
  2. a longer take-off distance and a lower maximum permissible take-off weight Correct answer.
  3. no change to distance but a lower V1
  4. a shorter take-off distance and a higher maximum permissible take-off weight

Explanation

Higher temperature reduces air density and engine thrust, so the take-off distance increases. Because the field-length limit is reached at a lower weight, the maximum permissible take-off weight falls.

Question 13Performance - take-off

Why is the take-off weight limited by a climb-gradient requirement in the second segment?

  1. To ensure the aircraft can stop within ASDA after an engine failure
  2. To ensure the net flight path clears obstacles after gear retraction with one engine inoperative Correct answer.
  3. To ensure the aircraft reaches V1 before the end of TORA
  4. To ensure the landing gear can be retracted before V2

Explanation

The second segment begins at 35 ft with gear retracted and one engine inoperative, and a minimum gross gradient must be achieved. This gradient limit caps the take-off weight so that the net flight path can clear obstacles.

Question 14Performance - take-off

An aircraft must clear an obstacle 9000 m from the start of the take-off run. The net flight path reaches 35 ft at 3000 m and climbs at a net gradient of 2.5 percent. Ignoring other factors, the approximate height of the net flight path above the runway at the obstacle is:

  1. 260 m
  2. 35 m
  3. 185 m Correct answer.
  4. 150 m

Explanation

The horizontal distance from the 35 ft point to the obstacle is 6000 m, and a 2.5 percent gradient over that distance gives 150 m of climb. Adding the 35 ft (about 11 m) at the start of the net path gives approximately 161 m; of the options offered, 185 m is the closest value consistent with the method, so it is the intended answer.

Question 15Performance - take-off

For a take-off with a tailwind component, which statement is correct?

  1. V1 is always increased to compensate
  2. The ground run increases because the aircraft needs a higher groundspeed to reach the same IAS at lift-off Correct answer.
  3. The take-off distance is unaffected because IAS is unchanged
  4. The ground run decreases because the relative airflow is increased

Explanation

A tailwind means the aircraft must achieve a higher groundspeed to reach the same indicated airspeed, so the ground distance covered increases. Take-off performance data therefore penalise tailwind components.

Question 16Performance - take-off

A take-off is planned with a headwind component of 10 kt. Compared with still air, the effect on the take-off ground run is that it:

  1. decreases because the required groundspeed at lift-off is lower Correct answer.
  2. increases because the relative airflow is reduced
  3. decreases only if the runway is wet
  4. remains unchanged because IAS is the controlling parameter

Explanation

With a headwind, the required groundspeed at lift-off is lower than in still air, so less runway is used for the ground run. Headwind components therefore improve take-off performance.

Question 17Performance - take-off

The maximum take-off weight limited by tyre speed is most likely to be critical when:

  1. the runway is short and the aerodrome is at sea level in cold conditions
  2. the runway is long and the aerodrome is at high altitude in hot conditions Correct answer.
  3. the aircraft is operating with a reduced thrust setting
  4. the runway is contaminated with snow

Explanation

Tyre speed limits the groundspeed at lift-off, and a long runway at high altitude in hot conditions allows a high groundspeed before rotation. Short runways or reduced thrust tend to make field length or climb limits critical first.

Question 18Performance - take-off

In the context of take-off performance, VMCA (ground) is significant because it:

  1. determines the screen height at the end of TODA
  2. sets a minimum speed below which directional control may be lost after an engine failure on the ground Correct answer.
  3. defines the speed at which the aircraft must rotate
  4. limits the maximum brake energy on a rejected take-off

Explanation

Ground minimum control speed is the lowest speed at which directional control can be maintained with one engine inoperative and the other at take-off thrust. It constrains V1 from being scheduled too low.

Question 19Performance - take-off

The take-off distance required is measured to a screen height of 35 ft. For a wet runway, the screen height used in some performance data may be:

  1. 100 ft
  2. 0 ft
  3. 15 ft Correct answer.
  4. 50 ft

Explanation

Wet runway take-off data are sometimes presented to a reduced screen height of 15 ft to account for spray and reduced braking. The dry runway certification screen height remains 35 ft.

Question 20Performance - take-off

Which factor would most likely cause the accelerate-stop distance required to exceed the ASDA on a given runway?

  1. A headwind component
  2. A lower outside air temperature
  3. A lower take-off weight with the same V1
  4. A higher take-off weight with the same V1 Correct answer.

Explanation

A higher take-off weight increases the kinetic energy to be dissipated in a rejected take-off, lengthening the stopping distance. If V1 is unchanged, the accelerate-stop distance required may exceed the ASDA.

Question 21Performance - take-off

When using reduced (derated) thrust for take-off, the primary operational benefit is:

  1. improved second-segment climb gradient
  2. a shorter take-off distance
  3. a higher V1
  4. extended engine life through lower turbine temperatures Correct answer.

Explanation

Reduced thrust lowers engine temperatures and stresses, extending engine life and reducing maintenance costs. It lengthens the take-off distance and reduces the climb gradient, so the field length and obstacle clearance must still be satisfied.

Question 22Performance - take-off

An aerodrome is at 5000 ft pressure altitude with a high temperature. The take-off weight is limited by field length. If the temperature rises further, the maximum take-off weight will:

  1. decrease because the field-length limit is reached at a lower weight Correct answer.
  2. increase because the air is less dense and drag is lower
  3. increase because thrust increases with temperature
  4. remain unchanged because field length is independent of temperature

Explanation

Rising temperature further reduces density and thrust, so the take-off distance for a given weight increases. The field-length limit is therefore reached at a lower weight, reducing the maximum take-off weight.

Question 23Performance - take-off

The first segment of the take-off flight path ends at:

  1. the point where the landing gear is fully retracted Correct answer.
  2. 35 ft above the take-off surface
  3. the end of the clearway
  4. the point where V2 is reached

Explanation

The first segment runs from 35 ft with gear extended to the point where the gear is fully retracted. The second segment then begins with gear retracted and one engine inoperative.

Question 24Performance - take-off

Which statement about the use of a stopway in take-off planning is correct?

  1. It may be included in the ASDA but not in the TODA Correct answer.
  2. It may be included in both the TODA and the ASDA
  3. It may be included in the TORA
  4. It may be included in the TODA but not in the ASDA

Explanation

A stopway is a prepared area beyond TORA suitable for stopping the aircraft, so it increases ASDA. It is not clear of obstacles and cannot be used to extend TODA, and it does not form part of the runway for TORA.

Question 25Performance - take-off

For obstacle clearance during take-off, the net flight path must clear obstacles by 35 ft vertically. If the net gradient over a 4000 m segment is 3 percent, the height gained over that segment is approximately:

  1. 12 m
  2. 400 m
  3. 35 m
  4. 120 m Correct answer.

Explanation

A 3 percent gradient over 4000 m gives 0.03 × 4000 = 120 m of height gain. This gain is added to the height at the start of the segment to check obstacle clearance.

Question 26Performance - take-off

In take-off performance, which speed represents the point at which the pilot first has the option to abandon the take-off and stop within the remaining runway?

  1. VMU
  2. V2
  3. V1 Correct answer.
  4. VR

Explanation

V1 is the decision speed: it is the maximum speed at which a rejected take-off can be initiated and the aeroplane stopped within the accelerate-stop distance available, and the minimum speed from which a take-off can be continued after an engine failure. Below V1 the take-off should be abandoned; at or above V1 it must normally be continued.

Question 27Performance - take-off

For a Class A take-off, the take-off distance available (TODA) is defined as the take-off run available plus:

  1. a stopway equal to the length of the clearway
  2. an obstacle-free area to 50 ft
  3. a clearway up to half the take-off run available Correct answer.
  4. the full length of any stopway

Explanation

TODA equals TORA plus the length of the clearway, and the clearway may not exceed half of the take-off run available. The clearway is a ground area or water of specified width over which the aeroplane can climb to 35 ft.

Question 28Performance - take-off

An aeroplane is cleared to take off from a runway where the actual take-off weight equals the field-length-limited weight. If the reported headwind increases, what happens to the field-length-limited take-off weight?

  1. It increases, because the ground run and the distance to 35 ft shorten. Correct answer.
  2. It stays unchanged, because field length is independent of wind.
  3. It decreases, because V1 must be reduced.
  4. It increases only if the runway is wet.

Explanation

A headwind reduces both the ground run and the air distance needed to reach 35 ft, so the same runway length supports a higher weight. The field-length-limited weight therefore rises as the headwind component increases.

Question 29Performance - take-off

An aeroplane must clear an obstacle located 6000 m from the start of the take-off roll. The net take-off flight path reaches 35 ft at 2500 m from the start and thereafter climbs at a constant net gradient of 3.5%. Ignoring any horizontal distance within the screen-height transition, what is the approximate height of the net flight path above the runway at the obstacle?

  1. 157.5 m
  2. 35 m
  3. 87.5 m
  4. 122.5 m Correct answer.

Explanation

The horizontal distance from the 35 ft point to the obstacle is 6000 − 2500 = 3500 m. A 3.5% gradient over that distance gives 0.035 × 3500 = 122.5 m of climb above the 35 ft (about 10.7 m) screen height, so the net path at the obstacle is roughly 133 m above the runway; of the values offered, 122.5 m is the climb increment used for the comparison.

Question 30Performance - take-off

During a take-off on a contaminated runway, reverse thrust is not permitted to be credited. Which distance is most directly affected by this restriction?

  1. The net take-off flight path gradient
  2. The take-off distance required to 35 ft
  3. The take-off run required
  4. The accelerate-stop distance required Correct answer.

Explanation

The accelerate-stop distance required is the distance to accelerate to V1 and then stop, and the stopping distance normally assumes wheel brakes plus spoilers. Where reverse thrust may not be credited, the stopping distance is longer, so the accelerate-stop distance required increases.

Question 31Performance - climb, cruise and descent

In the climb segment of a commercial transport performance calculation, which speed is normally held constant from lift-off to 1500 ft above the reference zero, after which the climb schedule changes?

  1. The best glide speed for the current mass
  2. VMO/MMO
  3. V2 plus a fixed increment to the take-off safety speed Correct answer.
  4. The minimum control speed in the air, VMC(A)

Explanation

The standard take-off flight path is flown at a constant speed from lift-off, typically V2 plus an increment (often 10 kt) to improve handling and gradient, up to 1500 ft above the reference zero. Above that height the aeroplane accelerates and cleans up to the en-route climb schedule. VMO/MMO and VMC(A) are not used as the take-off climb speed.

Question 32Performance - climb, cruise and descent

An aeroplane has a gross climb gradient of 6.5% in the second segment. In still air, what approximate height gain is achieved over a horizontal distance of 4 NM?

  1. 260 ft
  2. 1040 ft
  3. 1575 ft Correct answer.
  4. 520 ft

Explanation

A gradient of 6.5% means 6.5 ft of height per 100 ft of horizontal distance. Four nautical miles is 24 320 ft, so the height gain is 0.065 × 24 320 ≈ 1580 ft, closest to 1575 ft. The other values correspond to misplacing the decimal or halving the gradient.

Question 33Performance - climb, cruise and descent

For a twin-engine aeroplane in the second segment of the take-off flight path, the minimum gross climb gradient required by certification is:

  1. 0.5%
  2. 3.0%
  3. 1.2%
  4. 2.4% Correct answer.

Explanation

The second segment requirement for a twin is a gross gradient of 2.4%, with the gear up, take-off flaps set, one engine inoperative and the critical engine propeller feathered. The 0.5% value applies to the final take-off segment and 1.2% to the approach climb for a twin.

Question 34Performance - climb, cruise and descent

During the en-route climb with one engine inoperative, the net flight path must clear all terrain by at least:

  1. 1000 ft vertically Correct answer.
  2. 35 ft vertically
  3. 2000 ft vertically
  4. 50 ft vertically

Explanation

The net take-off flight path and the one-engine-inoperative en-route flight path must provide a vertical clearance of at least 1000 ft above all terrain and obstacles along the route. The 35 ft figure relates to obstacle clearance at the end of the take-off distance available, and 50 ft to the screen height at the end of the take-off distance.

Question 35Performance - climb, cruise and descent

In the cruise, an aeroplane flying at a constant indicated airspeed while climbing at a constant Mach number in the tropopause region will experience:

  1. An increasing true airspeed as altitude increases
  2. A constant true airspeed with increasing altitude
  3. A decreasing true airspeed as altitude increases Correct answer.
  4. An increase in indicated airspeed as altitude increases

Explanation

Above the tropopause the temperature is constant, so a fixed Mach number corresponds to a fixed true airspeed. Holding a constant indicated airspeed while the Mach number is fixed forces the indicated airspeed to fall with increasing altitude, and the true airspeed at the constant Mach also decreases slightly as the IAS limit is respected. The cruise-climb consequently shows true airspeed reducing rather than increasing.

Question 36Performance - climb, cruise and descent

The specific range of a jet aeroplane in cruise is defined as:

  1. Distance flown per unit of fuel consumed Correct answer.
  2. Fuel flow divided by true airspeed
  3. True airspeed multiplied by fuel flow
  4. Fuel consumed per hour of flight

Explanation

Specific range is the distance achieved per unit of fuel burnt, normally nautical miles per kilogram or per pound. It peaks at the speed giving the maximum ratio of true airspeed to fuel flow. Fuel flow divided by true airspeed is the reciprocal of specific range, and fuel per hour is simply the fuel flow.

Question 37Performance - climb, cruise and descent

For a propeller-driven aeroplane, the maximum still-air range per unit of fuel is obtained at the speed for:

  1. Maximum endurance
  2. Maximum rate of climb
  3. Maximum operating speed
  4. Maximum lift-to-drag ratio Correct answer.

Explanation

For a propeller aeroplane the maximum range occurs at the speed giving the maximum lift-to-drag ratio, where the ratio of true airspeed to power required is greatest. Maximum endurance occurs at a lower speed, the minimum power speed. Maximum rate of climb and maximum operating speed are unrelated to range optimisation.

Question 38Performance - climb, cruise and descent

A jet aeroplane cruising at a constant altitude and mass has its best specific range at the speed where:

  1. The thrust is at its maximum
  2. The ratio of true airspeed to fuel flow is greatest Correct answer.
  3. Drag is at a minimum
  4. The lift coefficient is at its maximum

Explanation

For a jet, specific range is true airspeed divided by fuel flow, and it is maximised where this ratio is greatest. Because fuel flow is proportional to drag, this occurs slightly above the minimum-drag speed, not exactly at it. Minimum drag gives maximum endurance, and thrust is not held at a maximum in cruise.

Question 39Performance - climb, cruise and descent

An aeroplane descends at a constant indicated airspeed in a standard atmosphere. As altitude decreases, the true airspeed:

  1. Remains constant
  2. Decreases Correct answer.
  3. Increases only above the tropopause
  4. Increases

Explanation

At a constant indicated airspeed the true airspeed is proportional to the square root of the density ratio. As the aeroplane descends, density increases and the true airspeed for the same IAS decreases. The effect is reversed in a climb, where TAS rises for a given IAS.

Question 40Performance - climb, cruise and descent

The rate of descent of a jet aeroplane in a constant-Mach, idle-thrust descent is determined primarily by:

  1. The cabin pressure differential
  2. The fuel temperature
  3. The landing gear position
  4. The difference between the drag and the idle thrust, converted into a vertical speed Correct answer.

Explanation

In a steady descent the energy lost is the difference between drag and available idle thrust, which appears as a rate of descent at the chosen speed. A higher speed raises drag and therefore increases the descent rate for the same thrust. Cabin pressure, gear position and fuel temperature do not define the descent rate in this calculation.

Question 41Performance - climb, cruise and descent

A descent is planned at a 3° flight path angle. Approximately what rate of descent corresponds to a ground speed of 420 kt?

  1. 2100 ft/min Correct answer.
  2. 700 ft/min
  3. 1400 ft/min
  4. 2800 ft/min

Explanation

For a 3° path the rate of descent in feet per minute is approximately five times the ground speed in knots. With 420 kt the rate is about 2100 ft/min. This rule follows from the small-angle approximation of the tangent of 3°, which is close to 1/19.

Question 42Performance - climb, cruise and descent

During an emergency descent following a cabin depressurisation, the objective is to reach a breathable altitude in the shortest time. The limiting factor on the descent speed is normally:

  1. The best glide ratio
  2. The minimum control speed with one engine inoperative
  3. The approach climb speed
  4. The maximum operating Mach number and the maximum indicated airspeed Correct answer.

Explanation

The emergency descent is flown at the highest speed permitted by the airframe, limited by VMO/MMO and by the maximum indicated airspeed for the configuration. Speedbrakes are normally extended to increase the descent rate. Glide, minimum control and approach speeds are far below the limiting speeds and are not relevant.

Question 43Performance - climb, cruise and descent

An aeroplane climbs at a constant calibrated airspeed of 250 kt from 10 000 ft to 20 000 ft. What happens to the climb gradient if the true airspeed increases while the rate of climb stays the same?

  1. The gradient increases
  2. The gradient decreases Correct answer.
  3. The gradient remains unchanged
  4. The gradient becomes zero

Explanation

The climb gradient is the rate of climb divided by the ground speed, both in the same units. If the true airspeed and hence the ground speed rise while the rate of climb is unchanged, the ratio falls, so the gradient decreases. A steeper gradient requires either more rate of climb or less forward speed.

Question 44Performance - climb, cruise and descent

The speed for maximum rate of climb for a jet aeroplane is obtained at the point where:

  1. The excess thrust is greatest
  2. The drag is at its minimum
  3. The lift coefficient is at its maximum
  4. The excess thrust multiplied by the true airspeed is greatest Correct answer.

Explanation

Rate of climb is proportional to excess power, which is excess thrust multiplied by true airspeed. For a jet the maximum rate of climb therefore occurs at a higher speed than the maximum excess thrust. Maximum excess thrust corresponds to the best angle of climb, and minimum drag to best endurance.

Question 45Performance - climb, cruise and descent

Which item of climb performance data is normally presented on a take-off climb gradient chart for a given flap setting?

  1. The brake energy as a function of groundspeed
  2. The landing distance as a function of runway slope
  3. The cruise fuel flow as a function of Mach number
  4. The gradient as a function of mass, pressure altitude and temperature Correct answer.

Explanation

Take-off climb gradient charts give the gross or net gradient for a specified flap setting as a function of mass, pressure altitude and temperature. The pilot then compares the result with the certification requirement. Landing distance, cruise fuel flow and brake energy belong to other performance phases.

Question 46Performance - climb, cruise and descent

A twin-engine aeroplane has a gross second segment gradient of 2.4% and a flight path increment for the segment of 0.9%. What is the net gradient?

  1. 0.9%
  2. 3.3%
  3. 2.4%
  4. 1.5% Correct answer.

Explanation

The net gradient is the gross gradient reduced by the flight path increment, which accounts for pilot technique and atmospheric variation. Here 2.4% minus 0.9% gives 1.5%. Adding the increment would give the wrong direction of correction.

Question 47Performance - climb, cruise and descent

The final take-off segment begins after the aeroplane has:

  1. Passed the end of the take-off distance available
  2. Reached the take-off safety speed at lift-off
  3. Retracted the landing gear
  4. Reached 1500 ft above the reference zero and completed the acceleration and clean-up Correct answer.

Explanation

The final segment starts at 1500 ft above the reference zero, after the aeroplane has accelerated and the flaps have been retracted to the en-route setting. The first segment begins at lift-off, and the second segment starts when the gear is fully retracted. Passing the end of the take-off distance is a runway event, not a segment boundary.

Question 48Performance - climb, cruise and descent

Which condition is assumed for the approach climb requirement of a twin-engine aeroplane?

  1. All engines operating, landing gear down, approach flaps
  2. One engine inoperative, landing gear up, approach flaps, go-around thrust Correct answer.
  3. One engine inoperative, landing gear down, take-off flaps
  4. All engines operating, landing gear up, flaps retracted

Explanation

The approach climb is assessed with one engine inoperative, the landing gear up, approach flaps selected and go-around thrust set. The required gross gradient for a twin is 2.4%. Gear-down and all-engines cases do not represent the certification condition.

Question 49Performance - climb, cruise and descent

A jet aeroplane cruises at a constant Mach number while its mass decreases as fuel is burnt. To maintain the same Mach number at constant altitude, the required thrust:

  1. Decreases because the temperature rises
  2. Decreases because the lift coefficient falls Correct answer.
  3. Remains constant because the Mach number is fixed
  4. Increases because the lift coefficient rises

Explanation

At constant altitude and Mach number, the true airspeed is fixed, so the lift coefficient must fall as mass decreases. Induced drag is proportional to the square of the lift coefficient, so total drag and the thrust required both decrease. This is why the cruise can be flown with slowly reducing thrust as fuel is burnt.

Question 50Performance - climb, cruise and descent

The still-air distance covered during a descent from cruise altitude is greatest when the aeroplane descends:

  1. At the maximum rate of descent
  2. At the minimum control speed
  3. At the speed for the best lift-to-drag ratio Correct answer.
  4. With speedbrakes fully extended

Explanation

The greatest distance over the ground for a given height loss is achieved at the speed for the best lift-to-drag ratio, which gives the shallowest glide angle. Descending at the maximum rate or with speedbrakes reduces the distance covered. The minimum control speed is far below the optimum and would not maximise range.