# 30 - Flight Performance and Planning > 40 questions --- ### Q1: What happens if the maximum permitted take-off mass of a glider is exceeded? ^q1 - A) It may be tolerated provided the excess is below 10%. - B) It is never permissible and poses a serious safety risk. - C) The pilot can compensate with appropriate control inputs. - D) It is acceptable in exceptional cases to prevent departure delays. **Correct: B)** > **Explanation:** The maximum take-off mass (MTOM) is a structural and aerodynamic certification limit. Exceeding it raises wing loading, increases the stall speed, degrades climb and glide performance, and overstresses the airframe beyond certified load factors. No pilot technique can compensate for the resulting structural and aerodynamic penalties. There is no regulatory tolerance for any excess, however small. ### Q2: Where must the centre of gravity be located for safe flight? ^q2 - A) Forward of the front C.G. limit. - B) Within the range between the forward and aft C.G. limits. - C) Aft of the rear C.G. limit. - D) To the right of the lateral C.G. limit. **Correct: B)** > **Explanation:** The approved C.G. envelope defines the range within which stability and controllability have been flight-tested and certified. A C.G. forward of the front limit may leave insufficient elevator authority for rotation or flare. A C.G. aft of the rear limit makes the aircraft statically unstable. The C.G. must remain within both limits throughout all phases of flight. ### Q3: Why must the centre of gravity remain within the approved limits during every phase of flight? ^q3 - A) To prevent the aircraft from exceeding its never-exceed speed in a descent. - B) To prevent the aircraft from tipping onto its tail during loading. - C) To ensure both longitudinal stability and controllability. - D) To prevent the aircraft from stalling. **Correct: C)** > **Explanation:** The C.G. position relative to the aerodynamic neutral point determines longitudinal static stability, while elevator authority provides controllability. If the C.G. falls outside the certified envelope, one of these two properties is compromised. Stall speed and VNE are governed by other parameters and are not the primary reasons for C.G. limits. ### Q4: Which C.G. position presents the greatest danger for a glider? ^q4 - A) Too far forward. - B) Too high above the longitudinal axis. - C) Too low below the longitudinal axis. - D) Too far aft. **Correct: D)** > **Explanation:** An aft C.G. beyond the rear limit reduces or eliminates longitudinal static stability. The aircraft may pitch up violently and uncontrollably because the restoring moment disappears. A forward C.G. is less dangerous because it merely increases stick forces and reduces performance; an aft C.G. can make the aircraft unflyable. ### Q5: How are the empty mass and corresponding C.G. of an individual aircraft initially established? ^q5 - A) By calculation from the manufacturer's type data. - B) Through data provided in the certificate of airworthiness. - C) By physically weighing the aircraft. - D) By reference to another aircraft of the same type, since all serial numbers share identical mass properties. **Correct: C)** > **Explanation:** Each aircraft is physically weighed on calibrated scales to determine its actual empty mass and C.G. position. Manufacturing tolerances, installed equipment, and repairs cause differences between serial numbers of the same type, so manufacturer tables or other aircraft of the same type cannot be relied upon. The results are recorded in the weight and balance report and updated after any modification. ### Q6: What can happen if baggage or cargo is not properly secured? ^q6 - A) Predictable instability that the pilot can compensate with control inputs. - B) Calculable instability when the C.G. shift is less than 10%. - C) Uncontrollable flight attitudes, structural damage, and risk of injury. - D) Minor trim changes that are easily corrected in flight. **Correct: C)** > **Explanation:** Unsecured cargo can shift suddenly in turbulence or manoeuvres, moving the C.G. outside limits faster than a pilot can react. A sudden aft shift can trigger an unrecoverable pitch-up. Loose items can become projectiles, injuring occupants or jamming controls. The structural risk from asymmetric or excessive local loading may exceed design limits. ### Q7: Through which point does the total weight force of an aircraft act? ^q7 - A) The stagnation point. - B) The centre of pressure. - C) The neutral point. - D) The centre of gravity. **Correct: D)** > **Explanation:** By definition, the centre of gravity is the single point through which the resultant gravitational force (weight vector) acts on the entire aircraft. The centre of pressure is where the resultant aerodynamic force acts, the neutral point is the aerodynamic reference for stability analysis, and the stagnation point is where airflow velocity reaches zero on the leading edge. ### Q8: What does the term "centre of gravity" describe? ^q8 - A) An alternative name for the neutral point. - B) The point through which the resultant gravitational force acts on the aircraft. - C) The physically heaviest single component of the aircraft. - D) The midpoint between the neutral point and the datum line. **Correct: B)** > **Explanation:** The centre of gravity is the mass-weighted average position of all individual mass elements. It is the point through which the total weight force is considered to act. It is distinct from the neutral point (an aerodynamic stability concept), is not the heaviest single component, and has no fixed geometric relationship to the midpoint between neutral point and datum. ### Q9: In mass and balance calculations, what is a "moment"? ^q9 - A) The sum of a mass and a balance arm. - B) The quotient of a mass divided by its balance arm. - C) The product of a mass and its balance arm. - D) The difference between a mass and a balance arm. **Correct: C)** > **Explanation:** Moment = mass x balance arm (M = m x d), expressed in units such as kg-m or lb-in. The total C.G. position is found by dividing the sum of all moments by the total mass. Using a sum, difference, or quotient instead of a product would yield dimensionally incorrect results. ### Q10: What does the term "balance arm" mean in a mass and balance calculation? ^q10 - A) The distance from a mass to the overall aircraft C.G. - B) The point through which gravity acts on a specific mass. - C) The fixed reference point from which all distances are measured. - D) The horizontal distance from the datum to the C.G. of a particular mass item. **Correct: D)** > **Explanation:** The balance arm (moment arm) is the horizontal distance measured from the aircraft's datum to the centre of gravity of a specific mass item. It determines the leverage that mass exerts about the datum. The datum itself is a fixed reference point, not the balance arm. Distances from the overall C.G. are not balance arms. ### Q11: The horizontal distance between the datum and the overall centre of gravity is called the... ^q11 - A) Torque. - B) Span. - C) Balance arm. - D) Lever ratio. **Correct: C)** > **Explanation:** In mass and balance terminology, the balance arm is the horizontal distance from the datum to any point of interest, including the overall C.G. once calculated. Torque (or moment) is the product of mass and arm, not the distance itself. Span is a wing dimension unrelated to longitudinal mass and balance. ### Q12: The balance arm of any mass item is the horizontal distance between... ^q12 - A) The front C.G. limit and the rear C.G. limit. - B) The front C.G. limit and the datum. - C) The C.G. of that mass item and the datum. - D) The C.G. of that mass item and the rear C.G. limit. **Correct: C)** > **Explanation:** The datum is a fixed reference defined in the flight manual. The balance arm of any mass item is measured from this datum to the centre of gravity of that specific item. All moment calculations use the datum as a common reference so that moments can be summed algebraically to determine the total C.G. position. ### Q13: Where in the aircraft documentation are the masses and balance arms needed for a weight and balance calculation found? ^q13 - A) In the certificate of airworthiness. - B) In the annual inspection records. - C) In the performance section of the flight manual. - D) In the mass and balance section of the pilot's operating handbook. **Correct: D)** > **Explanation:** The Pilot's Operating Handbook (POH) or Aircraft Flight Manual (AFM) contains a dedicated mass and balance section listing the empty mass, empty C.G., datum, C.G. limits, and approved loading configurations. The certificate of airworthiness certifies the type; annual inspection records document maintenance; the performance section covers speeds and glide data. ### Q14: Which section of the flight manual records the basic empty mass? ^q14 - A) Limitations. - B) Weight and balance. - C) Normal procedures. - D) Performance. **Correct: B)** > **Explanation:** The Weight and Balance section (typically Section 6 in an EASA-standardised AFM) records the basic empty mass, empty C.G. location, allowable C.G. range, and loading instructions. The Limitations section covers speeds and load factors; Normal Procedures covers checklists; Performance covers climb, glide, and speed data. ### Q15: An aircraft has a mass of 610 kg with its C.G. at position 80.0. A 10 kg item at balance arm 150 is removed. What is the new C.G. position? ^q15 - A) 81.166 - B) 75.0 - C) 78.833 - D) 70.0 **Correct: C)** > **Explanation:** Initial moment = 610 x 80.0 = 48,800. Removed moment = 10 x 150 = 1,500. New moment = 48,800 - 1,500 = 47,300. New mass = 600 kg. New C.G. = 47,300 / 600 = 78.833. Since the removed item was aft of the original C.G. (150 > 80), removing it shifts the C.G. forward, which is consistent with the result. ### Q16: A glider's current mass is 6,400 lbs with C.G. at 80. The aft C.G. limit is 80.5. What mass can be moved from its current position to arm 150 without exceeding the aft limit? ^q16 - A) 27.82 lbs - B) 56.63 lbs - C) 45.71 lbs - D) 39.45 lbs **Correct: C)** > **Explanation:** Moving mass x from the current C.G. (arm 80) to arm 150 shifts the C.G. aft. The new C.G. = (6400 x 80 + x x (150 - 80)) / 6400 must not exceed 80.5. Solving: x x 70 / 6400 = 0.5, so x = 6400 x 0.5 / 70 = 45.71 lbs. This is the maximum mass that can be relocated without violating the aft limit. ### Q17: Correct loading of an aircraft requires consideration of... ^q17 - A) Only the maximum permissible mass. - B) Only the correct distribution of payload. - C) Both the correct payload distribution and compliance with the maximum permissible mass. - D) Only the maximum permissible baggage mass in the aft compartment. **Correct: C)** > **Explanation:** Safe loading demands that both the total mass stays within MTOM and the payload is distributed so the C.G. remains within limits. Exceeding either the mass limit or the C.G. envelope independently compromises safety. Checking only one condition is insufficient. ### Q18: The maximum cockpit payload is exceeded by 10 kg. What is the correct action? ^q18 - A) Apply aft trim to compensate. - B) Apply forward trim to compensate. - C) Reduce the water ballast by the equivalent amount. - D) Reduce the payload until the limit is respected. **Correct: D)** > **Explanation:** The maximum pilot seat load is a structural certification limit. Trim adjustments do not change the actual mass and do not make the aircraft compliant. Reducing water ballast does not address a cockpit overload. The only correct action is to reduce the payload (remove equipment, lighter pilot parachute, etc.) until the certified limit is respected. ### Q19: If the aft C.G. limit is exceeded, the pilot should... ^q19 - A) Apply forward trim. - B) Apply aft trim. - C) Redistribute the useful load to move the C.G. forward. - D) Accept the condition provided the maximum mass is not exceeded. **Correct: C)** > **Explanation:** When the aft C.G. limit is exceeded, the load must be redistributed so that more mass is placed forward. Trim changes do not alter the physical mass distribution and cannot solve a structural C.G. problem. Flying with the C.G. beyond the aft limit is dangerous regardless of total mass. ### Q20: What propels a pure glider forward in flight? ^q20 - A) Drag resolved in the forward direction. - B) A tailwind component. - C) The component of gravity acting along the flight path. - D) Rising air currents. **Correct: C)** > **Explanation:** A motorless glider converts potential energy (height) into kinetic energy. The component of the weight vector projected along the descending flight path balances drag and maintains airspeed. Ascending air can reduce or reverse the descent but does not propel the aircraft forward. A tailwind changes groundspeed but does not generate a propulsive force along the flight path. ### Q21: Which factor reduces the landing distance? ^q21 - A) Heavy rainfall. - B) High density altitude. - C) A strong headwind. - D) High pressure altitude. **Correct: C)** > **Explanation:** A headwind reduces groundspeed at touchdown while airspeed remains normal, so less kinetic energy must be dissipated during the ground roll. High density altitude and high pressure altitude increase true airspeed for a given IAS, lengthening the ground roll. Heavy rain can degrade braking effectiveness and increase landing distance. ### Q22: What effect does a headwind have on the glide angle over the ground at constant true airspeed? ^q22 - A) The glide angle over the ground becomes steeper (increases). - B) Wind has no effect on the glide angle over the ground. - C) The glide angle over the ground becomes shallower (decreases). - D) The effect depends on the aircraft's mass. **Correct: A)** > **Explanation:** A headwind reduces groundspeed while the sink rate remains unchanged. Since the aircraft covers less horizontal distance per unit of height lost, the descent angle relative to the ground increases (steepens). Conversely, a tailwind reduces the glide angle over the ground. The air-mass glide angle is unaffected by wind. ### Q23: What must be observed when landing on sloping terrain during an off-field landing? ^q23 - A) Always land into the wind regardless of slope. - B) Land facing uphill with an approach speed slightly above normal. - C) Initiate the flare at a greater height than usual. - D) Use full airbrakes throughout the approach. **Correct: B)** > **Explanation:** On sloping terrain, landing uphill shortens the ground roll because gravity assists deceleration. A slightly higher approach speed provides a safety margin against turbulence and wind shear near unfamiliar terrain. Landing downhill would drastically increase the stopping distance and is extremely dangerous. ### Q24: What special consideration applies when landing in heavy rain? ^q24 - A) Fly a shallower approach than usual for better visibility. - B) Approach at a lower speed because rain slows the aircraft. - C) No special action is needed; land as in dry conditions. - D) Increase the approach speed above the normal value. **Correct: D)** > **Explanation:** Rain on the wing surface degrades aerodynamic characteristics by increasing surface roughness and altering the effective aerofoil profile. The stall speed may rise slightly and control effectiveness may be reduced. A higher approach speed provides the necessary safety margin. A shallower approach would reduce obstacle clearance and extend the final segment. ### Q25: What should you expect when taking off from a waterlogged grass runway? ^q25 - A) The wet grass reduces friction, shortening the take-off distance. - B) The glider may aquaplane, making directional control difficult. - C) The take-off distance is likely to be longer than on a dry surface. - D) The wet surface has no significant effect on take-off performance. **Correct: C)** > **Explanation:** A waterlogged grass surface increases rolling resistance due to soft ground deformation and water drag. Rain-flattened grass adds further resistance. The take-off run is therefore longer compared with a dry grass runway. Aquaplaning is a concern on hard surfaces with standing water but does not apply in the same way to soft grass. ### Q26: What effect does a waterlogged grass surface have on landing distance? ^q26 - A) Landing distance increases because the wheel sinks into soft ground. - B) Landing distance decreases because wet grass reduces rolling friction. - C) No measurable effect on landing distance. - D) Landing distance increases due to aquaplaning. **Correct: B)** > **Explanation:** On landing, wet grass reduces friction between the skid or wheel and the surface, causing the glider to slide more easily and decelerate over a shorter distance than on dry grass. This is the opposite effect to take-off, where the soft ground increases resistance during acceleration. ### Q27: Why is it beneficial to increase wing loading (e.g. with water ballast) when thermal conditions are strong? ^q27 - A) Because the stall speed decreases, allowing slower thermalling. - B) Because the glider achieves a better glide ratio at high inter-thermal speeds despite an increased minimum speed. - C) Because the glider climbs more efficiently at a lower circling speed. - D) Because the glider can fly more slowly, saving energy. **Correct: B)** > **Explanation:** In strong thermal conditions, the glider flies fast between thermals to optimise average cross-country speed (MacCready theory). Higher wing loading shifts the speed polar to higher speeds, improving the achievable glide ratio at those speeds. The trade-off is a higher stall speed and higher best-glide speed, which is acceptable when thermals are strong enough to compensate for the increased sink rate during climbs. ### Q28: Wing loading is increased by 40% through water ballast. By what percentage does the minimum speed increase? ^q28 - A) 40%. - B) 0%. - C) Approximately 18%. - D) 100%. **Correct: C)** > **Explanation:** Stall speed (and thus minimum speed) is proportional to the square root of wing loading. With a 40% increase: new Vs = Vs x sqrt(1.40) = Vs x 1.183, an increase of approximately 18%. The relationship is non-linear, so doubling wing loading does not double the stall speed. ### Q29: When a glider's mass increases, which parameter remains practically unchanged? ^q29 - A) Wing loading. - B) Sink rate at a given speed. - C) Minimum flight speed. - D) Maximum glide ratio (apart from a minor Reynolds number effect). **Correct: D)** > **Explanation:** The maximum glide ratio (best L/D) is essentially independent of mass because both lift and drag scale with mass in the same proportion. What changes: the speed for best L/D increases, the sink rate at any given speed increases, wing loading increases, and the minimum speed rises. Only the achievable L/D ratio itself remains virtually constant. ### Q30: What information does the speed polar reveal about glide ratio and mass? ^q30 - A) Both glide ratio and minimum speed are independent of mass. - B) Below 100 km/h, increasing mass reduces the sink rate. - C) Minimum speed is independent of mass. - D) Only the maximum glide ratio is independent of mass (apart from a minor Reynolds number effect). **Correct: D)** > **Explanation:** Speed polar curves for different masses show that the tangent from the origin (which determines best L/D) touches each curve at the same slope. The maximum glide ratio is therefore the same regardless of mass. However, the speed at which best L/D occurs is higher for heavier configurations, and minimum speed increases with mass. ### Q31: What distance can a glider with a glide ratio of 1:30 cover from a height of 1,500 m in still air? ^q31 - A) 30 km - B) 45 NM - C) 81 NM - D) 45 km **Correct: D)** > **Explanation:** Glide distance = glide ratio x height = 30 x 1,500 m = 45,000 m = 45 km. Note that 45 NM would equal approximately 83 km, requiring a glide ratio of about 1:55. Always verify units: mixing nautical miles and metres is a common source of error. ### Q32: You cover 150 km in 1 hour and 15 minutes. What is your ground speed? ^q32 - A) 115 km/h - B) 110 km/h - C) 125 km/h - D) 120 km/h **Correct: D)** > **Explanation:** Ground speed = distance / time = 150 km / 1.25 h = 120 km/h. Convert 1 hour 15 minutes to decimal: 15 min = 0.25 h, so total = 1.25 h. A common error is to write 1.15 instead of 1.25. ### Q33: At 6,000 m altitude, the airspeed indicator reads 160 km/h (IAS). How does the true airspeed (TAS) compare? ^q33 - A) TAS is the same as IAS at any altitude. - B) TAS is lower than IAS. - C) TAS is higher than IAS. - D) TAS may be higher or lower depending on outside temperature alone. **Correct: C)** > **Explanation:** The airspeed indicator measures dynamic pressure, which depends on air density. At 6,000 m the air density is significantly lower than at sea level, so a higher true speed is needed to produce the same dynamic pressure. In practice, TAS exceeds IAS by roughly 2% per 300 m of altitude. At 6,000 m, TAS is approximately 20-25% higher than IAS. ### Q34: When flying in wave lift at 6,000 m, how should you determine the maximum permitted speed? ^q34 - A) Fly up to the red VNE mark on the airspeed indicator, which is valid at all altitudes. - B) Fly faster than at sea level because the air is thinner. - C) Consult the speed-altitude table displayed in the cockpit and stay below the corrected VNE for that altitude. - D) Fly at the same IAS as at sea level since VNE is an absolute constant. **Correct: C)** > **Explanation:** The VNE on the ASI is set for low altitude. At high altitude, the same IAS corresponds to a much higher TAS, which can approach or exceed the structural flutter speed. Glider flight manuals provide a speed-altitude table giving the reduced IAS limit for each altitude. At 6,000 m the permissible IAS is significantly lower than the sea-level VNE marking. ### Q35: At what indicated airspeed should you approach an aerodrome at 1,800 m elevation? ^q35 - A) At a higher indicated speed than at sea level. - B) At the same indicated speed as at sea level. - C) At a lower indicated speed than at sea level. - D) At the speed for minimum sink rate. **Correct: B)** > **Explanation:** Aerodynamic forces depend on dynamic pressure, which is what the airspeed indicator measures. The correct approach IAS is the same regardless of aerodrome elevation. The TAS will be higher at altitude due to lower air density, but the pilot reads and flies the same indicated speed on the ASI. ### Q36: The angle of descent is defined as... ^q36 - A) The ratio of height loss to horizontal distance, expressed as a percentage. - B) The angle between the horizontal and the flight path, expressed in degrees. - C) The angle between the horizontal and the flight path, expressed as a percentage. - D) The ratio of height loss to horizontal distance, expressed in degrees. **Correct: B)** > **Explanation:** The angle of descent (glide angle) is the geometric angle between the horizontal plane and the flight path vector, measured in degrees. The glide ratio is the inverse tangent relationship: glide ratio = 1 / tan(glide angle). A glide ratio of 1:30 corresponds to a glide angle of approximately 1.9 degrees. Expressing it as a percentage would make it a gradient, not an angle. ### Q37: Which ground features should be avoided when planning a cross-country glider route? ^q37 - A) Stone quarries and large sandy areas. - B) Areas with buildings, concrete, and asphalt. - C) Highways, railway lines, and canals. - D) Wetlands, large water surfaces, and marshy ground. **Correct: D)** > **Explanation:** Moist ground, water bodies, and marshes have high thermal inertia and absorb solar radiation without heating up quickly, suppressing thermal development above them. Flying over large stretches of such terrain means less lift and a higher risk of a forced landing in unsuitable terrain. Dry fields, rocky areas, and built-up surfaces generate stronger thermals. ### Q38: How should a downwind turning point be approached during a cross-country flight? ^q38 - A) As low as possible to save time. - B) As steeply as possible to minimise time in the turn. - C) With as little bank as possible. - D) As high as possible to maximise altitude reserve for the upwind leg. **Correct: D)** > **Explanation:** After rounding a downwind turning point, the glider loses its tailwind advantage and faces a headwind that reduces groundspeed and shortens glide distance over the ground. Arriving high provides maximum altitude reserve for the upwind leg. Arriving low with an immediate turn into headwind leaves no margin for finding lift or selecting a landing field. ### Q39: What should a pilot expect after rounding a turning point during a cross-country flight? ^q39 - A) Weakening thermals due to the later time of day. - B) Lower cloud bases changing the horizontal view. - C) A visually different cloud picture because the sun's apparent position has changed relative to the new heading. - D) Increased cloud dissipation as the day progresses. **Correct: C)** > **Explanation:** When the glider turns through 90 or 180 degrees at a waypoint, the pilot's perspective of the sky shifts dramatically. The sun appears to have moved relative to the aircraft heading, and cumulus clouds that were behind or to one side now appear ahead. This perceptual change can make the sky look completely different even though objective conditions have not changed. Pilots must re-orient their thermal assessment to the new heading. ### Q40: What is the purpose of "interception lines" in visual navigation? ^q40 - A) They mark the range limitation from the departure aerodrome. - B) They allow continued flight when visibility drops below VFR minima. - C) They indicate the next available airport along the route. - D) They serve as easily recognisable linear features to help re-establish position if orientation is lost. **Correct: D)** > **Explanation:** Interception lines are prominent linear ground features -- rivers, coastlines, railways, motorways -- selected during pre-flight planning that run roughly perpendicular to the planned route. If a pilot becomes disoriented, flying towards the nearest interception line produces an unmistakable landmark for position recovery. They do not extend VFR permissions, are not range indicators, and are not airport markers.