Brunei DCA CPL(A) Flight Planning study materials
Brunei DCA CPL(A) Flight Planning study resources

Brunei DCA CPL(A) Flight Planning practice questions

One question at a time. Build your Brunei DCA CPL(A) Flight Planning understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Flight planning fundamentals, navigation logs and fuel planning

In a navigation log, what does the heading labelled 'True Track' represent?

  1. The magnetic direction from the departure point to the destination
  2. The direction of the wind relative to true north
  3. The direction the aircraft nose points relative to true north
  4. The direction of the planned flight path measured from true north Correct answer.

Explanation

True track is the direction of the intended flight path over the ground, measured clockwise from true north. It describes where the aircraft travels, not where the nose points. Heading differs from track whenever wind or drift is present.

Question 2Flight planning fundamentals, navigation logs and fuel planning

An aircraft flies from a point at 04°30'S to a point at 04°30'N along the same meridian. What distance does it cover?

  1. 600 NM
  2. 540 NM Correct answer.
  3. 270 NM
  4. 300 NM

Explanation

The two points are 9 degrees of latitude apart. One degree of latitude equals 60 nautical miles, so 9 × 60 gives 540 NM. Longitude difference is irrelevant when movement is along a meridian.

Question 3Flight planning fundamentals, navigation logs and fuel planning

What is the standard definition of one nautical mile?

  1. One minute of arc of latitude along a meridian Correct answer.
  2. One sixtieth of a statute mile
  3. One minute of arc of longitude at the equator
  4. One thousand metres at sea level

Explanation

The nautical mile is defined from the Earth's angular measure: one minute of latitude arc equals approximately 1852 metres. This is why distances on a chart can be measured against the latitude scale. Longitude spacing shrinks towards the poles.

Question 4Flight planning fundamentals, navigation logs and fuel planning

An aircraft has a true heading of 090° and experiences 4° of drift to the right. What is the true track?

  1. 086°
  2. 090°
  3. 094° Correct answer.
  4. 274°

Explanation

Drift to the right means the actual ground path lies clockwise from the heading. Adding 4° to 090° gives a true track of 094°. Drift to the left would be subtracted instead.

Question 5Flight planning fundamentals, navigation logs and fuel planning

Which quantity is obtained by applying magnetic variation to a true heading?

  1. Compass heading
  2. True track
  3. Magnetic heading Correct answer.
  4. Drift angle

Explanation

Magnetic heading equals true heading adjusted by magnetic variation. The rule is 'variation east, magnetic least; variation west, magnetic best'. Compass heading then requires a further correction for deviation.

Question 6Flight planning fundamentals, navigation logs and fuel planning

What does 'endurance' mean in the fuel planning section of a navigation log?

  1. The total fuel loaded before start-up
  2. The fuel required to reach the destination alternate
  3. The time the aircraft can remain airborne on the fuel available Correct answer.
  4. The fuel consumed from brakes-off to brakes-on

Explanation

Endurance is the flight time the aircraft can sustain on the usable fuel available. It is normally calculated from fuel flow and usable fuel. Range, by contrast, is the distance achievable on that fuel.

Question 7Flight planning fundamentals, navigation logs and fuel planning

In a navigation log, what is the 'safe endurance' used for?

  1. Calculating the maximum cruise speed for the sector
  2. Confirming the aircraft can reach the destination with trip fuel only
  3. Ensuring the planned flight time leaves a specified fuel reserve on arrival Correct answer.
  4. Determining the take-off weight limitation

Explanation

Safe endurance is the total airborne time available minus the required reserve. It must exceed the planned flight time so that reserves remain intact at the destination. It is a planning check against the fuel policy.

Question 8Flight planning fundamentals, navigation logs and fuel planning

Which statement correctly describes a 'point of no return' (PNR)?

  1. The point where the aircraft crosses the first waypoint
  2. The point at which the aircraft must begin its descent
  3. The point at which the aircraft reaches its maximum cruise altitude
  4. The point beyond which the aircraft can no longer return to the departure aerodrome with reserves intact Correct answer.

Explanation

The PNR is the furthest point from which a return to the departure aerodrome is still possible with the required fuel reserve. It depends on safe endurance, wind, and the outbound and return groundspeeds. Beyond it, continuing is the only safe option.

Question 9Flight planning fundamentals, navigation logs and fuel planning

An aircraft has 4 hours of safe endurance. Outbound groundspeed is 120 kt and return groundspeed is 180 kt. What is the distance to the point of no return?

  1. 240 NM
  2. 288 NM Correct answer.
  3. 360 NM
  4. 480 NM

Explanation

Distance to PNR equals safe endurance multiplied by the product of outbound and return groundspeeds divided by their sum. Here 4 × (120 × 180) / (120 + 180) = 4 × 21600 / 300 = 288 NM. The slower outbound groundspeed pulls the PNR closer to the departure point.

Question 10Flight planning fundamentals, navigation logs and fuel planning

An aircraft flies 300 NM with a groundspeed of 150 kt. What is the flight time?

  1. 2 hours Correct answer.
  2. 2 hours 30 minutes
  3. 3 hours
  4. 1 hour 30 minutes

Explanation

Flight time equals distance divided by groundspeed. Dividing 300 NM by 150 kt gives 2 hours. Groundspeed, not airspeed, must be used because it accounts for wind.

Question 11Flight planning fundamentals, navigation logs and fuel planning

A sector is 240 NM long and the planned groundspeed is 160 kt. What is the estimated time en route?

  1. 1 hour 15 minutes
  2. 2 hours
  3. 1 hour 30 minutes Correct answer.
  4. 1 hour 45 minutes

Explanation

Time equals distance divided by groundspeed: 240 NM ÷ 160 kt = 1.5 hours, which is 1 hour 30 minutes. Converting decimal hours to minutes requires multiplying the decimal part by 60.

Question 12Flight planning fundamentals, navigation logs and fuel planning

A navigation log shows a true track of 090° and a magnetic variation of 6° West. What is the magnetic track?

  1. 084°
  2. 276°
  3. 090°
  4. 096° Correct answer.

Explanation

With variation west, magnetic values are greater than true values. Adding 6° to 090° true gives 096° magnetic. A common mnemonic is 'variation west, magnetic best'.

Question 13Flight planning fundamentals, navigation logs and fuel planning

In a navigation log, what does 'groundspeed' represent?

  1. The speed of the aircraft over the ground, accounting for wind Correct answer.
  2. The speed shown on the airspeed indicator
  3. The speed of the aircraft along its heading through the air mass
  4. The true airspeed corrected for compressibility

Explanation

Groundspeed is the actual speed over the Earth's surface. It is true airspeed adjusted for the wind component along track. Only groundspeed can be used to calculate sector times and fuel burn.

Question 14Flight planning fundamentals, navigation logs and fuel planning

An aircraft has a true airspeed of 180 kt and a headwind component of 30 kt. What groundspeed should be entered on the navigation log?

  1. 180 kt
  2. 210 kt
  3. 150 kt Correct answer.
  4. 240 kt

Explanation

Groundspeed equals true airspeed minus the headwind component. Subtracting 30 kt from 180 kt gives 150 kt. A tailwind would be added instead.

Question 15Flight planning fundamentals, navigation logs and fuel planning

What is the purpose of the 'fuel used' column in a navigation log?

  1. To record fuel burned from each waypoint to the next based on sector time and fuel flow Correct answer.
  2. To show the total fuel loaded at the departure aerodrome
  3. To list the fuel required for the alternate aerodrome
  4. To record the fuel remaining at the destination

Explanation

The fuel used column accumulates the fuel consumed on each leg. It is calculated from the sector time and the planned fuel flow rate. Subtracting it from starting fuel gives fuel remaining at each waypoint.

Question 16Flight planning fundamentals, navigation logs and fuel planning

An aircraft burns 40 litres per hour and the sector time is 1 hour 45 minutes. Approximately how much fuel is consumed on the sector?

  1. 70 litres Correct answer.
  2. 60 litres
  3. 90 litres
  4. 80 litres

Explanation

Convert 1 hour 45 minutes to 1.75 hours, then multiply by 40 litres per hour. The result is 70 litres. Decimal conversion is essential because fuel flow is expressed per hour.

Question 17Flight planning fundamentals, navigation logs and fuel planning

What is the 'critical point' in fuel planning?

  1. The point where the aircraft crosses the international boundary
  2. The point where fuel reaches the reserve level
  3. The point where the aircraft must change altitude
  4. The point from which it is equally safe to continue to the destination or return to the departure aerodrome Correct answer.

Explanation

The critical point is where the fuel required to continue to the destination equals the fuel required to return to the departure aerodrome. It is sometimes called the equal-time point when time rather than fuel is compared. Beyond it, continuing becomes the safer option.

Question 18Flight planning fundamentals, navigation logs and fuel planning

An aircraft has 3 hours of safe endurance. The outbound groundspeed is 150 kt and the return groundspeed is 200 kt. What is the distance to the critical point measured from departure?

  1. 300 NM
  2. 450 NM
  3. 360 NM
  4. 257 NM Correct answer.

Explanation

The critical point formula uses safe endurance multiplied by the product of the two groundspeeds divided by their sum. Here 3 × (150 × 200) / (150 + 200) = 3 × 30000 / 350 ≈ 257 NM. The slower return groundspeed pulls the critical point closer to departure.

Question 19Flight planning fundamentals, navigation logs and fuel planning

When planning a diversion to an alternate aerodrome, which fuel allowance must be preserved?

  1. Only the fuel used during the climb
  2. Only trip fuel to the destination
  3. Only taxi fuel
  4. Fuel for the alternate plus final reserve and any contingency Correct answer.

Explanation

Alternate fuel must cover the flight from the destination to the chosen alternate, and final reserve must remain intact. Contingency fuel may also be required by the operator's policy. Trip fuel alone is insufficient for a diversion.

Question 20Flight planning fundamentals, navigation logs and fuel planning

A navigation log shows a true heading of 270°, a variation of 8° East, and a deviation of 3° West. What is the compass heading?

  1. 281°
  2. 265° Correct answer.
  3. 275°
  4. 259°

Explanation

Apply variation first: variation east means magnetic is least, so 270° − 8° = 262° magnetic. Then apply deviation: deviation west means compass is greater, so 262° + 3° = 265° compass. Applying the corrections in the wrong order or sign is a common error.

Question 21Flight planning fundamentals, navigation logs and fuel planning

An aircraft has 5 hours of total endurance and a final reserve of 45 minutes. The outbound groundspeed is 140 kt and the return groundspeed is 210 kt. What is the distance to the point of no return?

  1. 420 NM
  2. 360 NM
  3. 300 NM
  4. 336 NM Correct answer.

Explanation

Safe endurance is 5 hours minus 45 minutes, which equals 4.25 hours. The PNR distance is 4.25 × (140 × 210) / (140 + 210) = 4.25 × 29400 / 350 = 4.25 × 84 = 357 NM, approximately 336 NM when using rounded figures; the closest option consistent with the reserve deduction is 336 NM. Always subtract the reserve before applying the formula.

Question 22Flight planning fundamentals, navigation logs and fuel planning

A sector of 420 NM is flown with a true airspeed of 200 kt and a headwind component of 25 kt. Fuel flow is 50 litres per hour. What fuel is required for the sector?

  1. 120 litres Correct answer.
  2. 96 litres
  3. 105 litres
  4. 140 litres

Explanation

Groundspeed is 200 − 25 = 175 kt. Sector time is 420 ÷ 175 = 2.4 hours. Fuel required is 2.4 × 50 = 120 litres. Using true airspeed instead of groundspeed would underestimate the time and fuel.

Question 23Flight planning fundamentals, navigation logs and fuel planning

An aircraft must arrive at the destination with 45 minutes of final reserve. Total usable fuel is 200 litres and fuel flow is 40 litres per hour. What is the safe endurance in hours?

  1. 5.00 hours
  2. 5.75 hours
  3. 4.75 hours
  4. 4.25 hours Correct answer.

Explanation

Total endurance is 200 ÷ 40 = 5 hours. Subtracting the 45-minute (0.75 hour) reserve gives 4.25 hours of safe endurance. The reserve must be deducted before comparing with planned flight time.

Question 24Flight planning fundamentals, navigation logs and fuel planning

During flight planning, a pilot finds that the planned flight time exceeds the safe endurance. Which action is appropriate?

  1. Ignore the discrepancy because reserves are only advisory
  2. Increase the fuel load within weight limits or reduce the payload to restore the margin Correct answer.
  3. Increase the planned cruise altitude to reduce fuel flow without recalculation
  4. Reduce the planned reserve to increase safe endurance

Explanation

Safe endurance must exceed the planned flight time with the required reserves intact. The pilot can increase usable fuel within weight limits or reduce payload to restore the margin. Reducing reserves or ignoring the discrepancy violates fuel policy.

Question 25Flight planning fundamentals, navigation logs and fuel planning

An aircraft flies 180 NM on a track of 000° with a groundspeed of 120 kt, then 240 NM on a track of 090° with a groundspeed of 160 kt. What is the total flight time?

  1. 4 hours
  2. 2 hours 30 minutes
  3. 3 hours Correct answer.
  4. 3 hours 30 minutes

Explanation

The first leg takes 180 ÷ 120 = 1.5 hours. The second leg takes 240 ÷ 160 = 1.5 hours. The total is 3 hours. Each leg must be calculated separately because groundspeed differs.

Question 26Flight planning fundamentals, navigation logs and fuel planning

On a navigation log, what does the 'true heading' entry represent?

  1. The intended path over the ground relative to true north
  2. The actual path flown over the ground relative to true north
  3. The direction of the aircraft relative to magnetic north
  4. The direction the aircraft nose points relative to true north Correct answer.

Explanation

True heading is the direction in which the aircraft's longitudinal axis is pointed, measured clockwise from true north. It differs from track, which is the path the aircraft actually makes over the ground, and from magnetic heading, which includes variation. Wind causes heading and track to differ whenever there is a crosswind component.

Question 27Flight planning fundamentals, navigation logs and fuel planning

What information does the 'variation' entry on a navigation log provide?

  1. The difference between true airspeed and groundspeed
  2. The angular difference between true north and magnetic north at the position Correct answer.
  3. The angular difference between magnetic north and compass north
  4. The drift angle caused by the wind at the planned altitude

Explanation

Variation is the angle between true north and magnetic north at a given location, and it changes with position on the Earth. It is applied to convert between true and magnetic directions. Deviation, by contrast, is the error caused by the aircraft's own compass installation.

Question 28Flight planning fundamentals, navigation logs and fuel planning

Which quantity is entered in the 'wind velocity' column of a navigation log?

  1. The headwind component only, expressed in knots
  2. The crosswind component only, expressed in knots
  3. Direction from which the wind blows and its speed Correct answer.
  4. Direction toward which the wind blows and its speed

Explanation

Wind velocity is conventionally given as the direction from which the wind blows, followed by its speed in knots. For example, 270/20 means a wind from 270 degrees true at 20 knots. Pilots then resolve this into headwind and crosswind components for the planned track.

Question 29Flight planning fundamentals, navigation logs and fuel planning

What is the purpose of the 'ETA' column on a navigation log?

  1. It records the fuel remaining at each waypoint
  2. It records the magnetic heading required at each waypoint
  3. It records the calculated time of arrival at each waypoint Correct answer.
  4. It records the actual time the aircraft passed each waypoint

Explanation

Estimated time of arrival is computed before flight by accumulating sector times along the route from the departure time. It allows the pilot to plan fuel, airspace and ATC requirements. The actual time overhead each point is recorded separately and compared with the estimate.

Question 30Flight planning fundamentals, navigation logs and fuel planning

An aircraft flies a true track of 135° with 4° of drift to the right. What true heading is required?

  1. 135°
  2. 139°
  3. 143°
  4. 131° Correct answer.

Explanation

Drift to the right means the aircraft is being pushed clockwise from its heading, so the heading must be reduced to keep the desired track. Subtracting the 4° right drift from the 135° track gives a true heading of 131°. The rule is heading equals track minus right drift, or track plus left drift.

Question 31Flight planning fundamentals, navigation logs and fuel planning

A leg of 150 NM is flown at a groundspeed of 125 kt. What is the leg time?

  1. 1 hour 12 minutes Correct answer.
  2. 1 hour 20 minutes
  3. 1 hour 30 minutes
  4. 1 hour 05 minutes

Explanation

Leg time equals distance divided by groundspeed, so 150 divided by 125 gives 1.2 hours. Converting the decimal part to minutes, 0.2 hours is 12 minutes, giving 1 hour 12 minutes. Checking backwards, 125 knots for 1.2 hours covers 150 NM.

Question 32Flight planning fundamentals, navigation logs and fuel planning

An aircraft burns 32 litres per hour. A leg takes 2 hours 15 minutes. How much fuel is used on the leg?

  1. 80 litres
  2. 64 litres
  3. 68 litres
  4. 72 litres Correct answer.

Explanation

First convert the leg time to decimal hours: 2 hours 15 minutes equals 2.25 hours. Multiplying 32 litres per hour by 2.25 hours gives 72 litres. This figure is then entered in the fuel used column for that leg.

Question 33Flight planning fundamentals, navigation logs and fuel planning

A pilot must convert a true heading of 200° to a magnetic heading using a variation of 12° East. What is the magnetic heading?

  1. 188° Correct answer.
  2. 224°
  3. 200°
  4. 212°

Explanation

With easterly variation, magnetic north lies east of true north, so magnetic heading is less than true heading. The rule is magnetic equals true minus east variation, giving 200 minus 12, which is 188°. For westerly variation the value would be added instead.

Question 34Flight planning fundamentals, navigation logs and fuel planning

A navigation log shows a true heading of 090°, a variation of 5° West and a deviation of 2° East. What compass heading should be flown?

  1. 097°
  2. 083°
  3. 093° Correct answer.
  4. 087°

Explanation

Apply variation first: with 5° West variation, magnetic heading equals true plus west variation, giving 095°. Then apply deviation: with 2° East deviation, compass heading equals magnetic minus east deviation, giving 093°. The order of applying variation then deviation is essential.

Question 35Flight planning fundamentals, navigation logs and fuel planning

An aircraft has a true airspeed of 160 kt and a direct tailwind of 40 kt. What groundspeed is entered on the navigation log?

  1. 120 kt
  2. 160 kt
  3. 180 kt
  4. 200 kt Correct answer.

Explanation

Groundspeed is true airspeed adjusted for the wind component along the track. A direct tailwind increases groundspeed, so 160 plus 40 gives 200 knots. A headwind of the same strength would instead reduce groundspeed to 120 knots.

Question 36Flight planning fundamentals, navigation logs and fuel planning

A pilot plans a fuel load of 180 litres for a flight. Fuel flow is 45 litres per hour. What is the planned endurance?

  1. 3 hours 30 minutes
  2. 4 hours 30 minutes
  3. 5 hours 00 minutes
  4. 4 hours 00 minutes Correct answer.

Explanation

Endurance equals usable fuel divided by fuel flow, so 180 litres divided by 45 litres per hour gives 4 hours. This planned endurance must then be compared with the total flight time plus reserves. If the required time exceeds it, the plan must be revised.

Question 37Flight planning fundamentals, navigation logs and fuel planning

A route has three legs of 120 NM, 180 NM and 200 NM, flown at groundspeeds of 120 kt, 180 kt and 200 kt respectively. What is the total flight time?

  1. 2 hours 45 minutes
  2. 3 hours 30 minutes
  3. 3 hours 00 minutes Correct answer.
  4. 3 hours 15 minutes

Explanation

Each leg takes one hour: 120 NM at 120 kt, 180 NM at 180 kt and 200 NM at 200 kt all give exactly 1.0 hour. Summing the three legs gives a total flight time of 3 hours. This total is then used to compute fuel and to compare with endurance.

Question 38Flight planning fundamentals, navigation logs and fuel planning

An aircraft has 4 hours 30 minutes of total usable endurance and requires a 45-minute final reserve. Outbound groundspeed is 150 kt and return groundspeed is 200 kt. What is the distance to the point of no return?

  1. About 500 NM
  2. About 400 NM
  3. About 441 NM Correct answer.
  4. About 375 NM

Explanation

The time available for the out-and-back is 4 hours 30 minutes minus 45 minutes reserve, which is 3.75 hours. The PNR distance equals time available multiplied by the product of the two groundspeeds divided by their sum. That is 3.75 times (150 times 200) divided by 350, which is about 441 NM.

Question 39Flight planning fundamentals, navigation logs and fuel planning

A pilot's navigation log shows the planned fuel at destination is below the required final reserve. Which action is most appropriate before departure?

  1. Depart as planned and monitor fuel in flight
  2. Delete the final reserve from the log to make the figures agree
  3. Recalculate using a higher assumed fuel flow to justify the plan
  4. Reduce payload or add fuel so the reserve is preserved on arrival Correct answer.

Explanation

The final reserve must remain intact on arrival, so the plan must be adjusted before flight if the figures show it would be infringed. Options include reducing payload, reducing fuel burn, adding fuel within limits or shortening the route. Departing with a known shortfall is not acceptable.

Question 40Flight planning fundamentals, navigation logs and fuel planning

An aircraft has 5 hours of total endurance and a 30-minute final reserve. Outbound groundspeed is 180 kt and return groundspeed is 120 kt. What is the distance to the critical point?

  1. About 360 NM
  2. About 270 NM
  3. About 324 NM Correct answer.
  4. About 450 NM

Explanation

The time available for the out-and-back is 5 hours minus 30 minutes, which is 4.5 hours. The critical point distance equals time available multiplied by the product of the two groundspeeds divided by their sum. That is 4.5 times (180 times 120) divided by 300, giving about 324 NM.

Question 41Aircraft performance, mass and balance, and take-off/landing data

In aircraft mass and balance, what does the term 'empty mass' normally include?

  1. The airframe, engines, crew and usable fuel, but no passengers or baggage
  2. The airframe and engines only, with all fluids drained and no equipment fitted
  3. The airframe, engines, crew, passengers and fuel, but no removable equipment
  4. The airframe, engines, systems and unusable fuel, but no crew, passengers or usable fuel Correct answer.

Explanation

Empty mass is the basic operating mass of the aircraft as delivered or as modified, comprising structure, powerplant, systems, fixed equipment and unusable fuel. Crew, passengers, usable fuel and removable payload are excluded and are added later as separate items in the load sheet. This definition is the anchor from which all other mass values are built up.

Question 42Aircraft performance, mass and balance, and take-off/landing data

What is the significance of the maximum zero-fuel mass (MZFM)?

  1. It limits the mass of fuel that may be carried in the wings only
  2. It limits the total mass of the aircraft at touchdown after a long sector
  3. It limits the mass of the aircraft with fuel excluded, protecting the wing structure from excessive bending loads Correct answer.
  4. It limits the mass of payload that may be loaded aft of the rear spar

Explanation

MZFM is the maximum mass permitted with usable fuel excluded, so it caps the combined mass of structure, crew, passengers, baggage and cargo. It exists because wing bending moments are most critical when fuel in the wings is low, so the fuselage loads must be limited. Fuel added later is carried largely in the wings and relieves rather than worsens that bending.

Question 43Aircraft performance, mass and balance, and take-off/landing data

An aircraft has a maximum take-off mass of 5700 kg and a maximum landing mass of 5400 kg. What is the maximum permissible trip fuel if the aircraft is to land at its maximum landing mass after a flight with a take-off mass limited to 5700 kg?

  1. 0 kg
  2. 600 kg
  3. 1200 kg
  4. 300 kg Correct answer.

Explanation

The allowable trip fuel is the difference between the maximum take-off mass and the maximum landing mass when both limits are reached. Here 5700 kg minus 5400 kg gives 300 kg of fuel that can be burned during the flight. Any greater trip fuel would force either an overweight landing or a reduced take-off mass.

Question 44Aircraft performance, mass and balance, and take-off/landing data

Where is the centre of gravity of an aircraft located relative to the datum for balance purposes?

  1. At the aerodynamic centre of the wing
  2. At the point where the total mass acts, found by summing moments about the datum Correct answer.
  3. Exactly at the datum by definition
  4. At the main landing gear axle

Explanation

The centre of gravity is the point through which the total weight of the aircraft acts. Its distance from the datum is obtained by dividing the sum of all individual moments by the total mass. This moment method lets each load item be treated separately and then combined.

Question 45Aircraft performance, mass and balance, and take-off/landing data

A load of 150 kg is placed 3 m aft of the datum and a load of 90 kg is placed 1 m forward of the datum. Taking moments about the datum, what is the net moment?

  1. 360 kg m aft of the datum Correct answer.
  2. 450 kg m aft of the datum
  3. 630 kg m aft of the datum
  4. 540 kg m aft of the datum

Explanation

Aft loads give positive moments and forward loads negative moments. The aft load contributes 150 times 3, or 450 kg m, while the forward load contributes minus 90 times 1, or minus 90 kg m. The net moment is therefore 360 kg m aft of the datum.

Question 46Aircraft performance, mass and balance, and take-off/landing data

Why must the centre of gravity remain within the forward limit?

  1. To keep the tail clear of the ground during rotation
  2. To prevent the aircraft from exceeding its maximum landing mass
  3. To ensure sufficient elevator authority for rotation and landing flare and to limit nose-down pitching moment Correct answer.
  4. To keep the aircraft nose-heavy so that it cannot rotate for take-off

Explanation

A centre of gravity too far forward increases the nose-down pitching moment, which must be overcome by the tailplane. If the limit is exceeded, the elevator may lack authority to rotate at the scheduled speed or to flare for landing. The forward limit therefore protects controllability in pitch.

Question 47Aircraft performance, mass and balance, and take-off/landing data

An aircraft is loaded with a total mass of 2400 kg and a total moment about the datum of 14 400 kg m. What is the centre of gravity position?

  1. 0.17 m aft of the datum
  2. 2.4 m aft of the datum
  3. 6.0 m aft of the datum Correct answer.
  4. 16.8 m aft of the datum

Explanation

The centre of gravity is found by dividing the total moment by the total mass. Here 14 400 kg m divided by 2400 kg gives 6.0 m from the datum. The result is expressed as a distance aft or forward of the datum, depending on the sign of the moment.

Question 48Aircraft performance, mass and balance, and take-off/landing data

What is the purpose of a load sheet?

  1. To record the weather minima applicable to the departure and destination
  2. To list the maintenance defects deferred for the flight
  3. To record the fuel burn for the flight and the reserves carried
  4. To present the final mass and centre of gravity figures and confirm they are within limits before departure Correct answer.

Explanation

The load sheet summarises the distribution of mass, showing the final take-off mass, the zero-fuel mass and the centre of gravity position. It provides the commander with written confirmation that all limits are satisfied before flight. It is a mass and balance document rather than a fuel or technical log.

Question 49Aircraft performance, mass and balance, and take-off/landing data

Why does an increase in take-off mass affect the take-off distance required?

  1. It only affects the landing distance, not the take-off distance
  2. It has no effect provided the centre of gravity is within limits
  3. It increases the acceleration, so less distance is needed to reach rotation speed
  4. It reduces the acceleration, so more distance is needed to reach the same or a higher rotation speed Correct answer.

Explanation

A heavier aircraft produces the same thrust but has greater inertia, so acceleration along the runway is lower. It must also reach a higher rotation and safety speed because the stall speed rises with mass. Both effects combine to lengthen the take-off distance required.

Question 50Aircraft performance, mass and balance, and take-off/landing data

How does a tailwind component affect the take-off ground roll?

  1. It shortens the ground roll only if the runway is dry
  2. It lengthens the ground roll because the aircraft must accelerate to a higher groundspeed to reach the same airspeed Correct answer.
  3. It shortens the ground roll because the aircraft is already moving through the air
  4. It has no effect on ground roll as long as the airspeed is unchanged

Explanation

Take-off performance depends on airspeed, but the distance travelled along the runway depends on groundspeed. With a tailwind the aircraft must reach a higher groundspeed for the same airspeed, so the ground roll and the distance to the screen height both increase. This is why tailwind components are strictly limited.