89 questions
Correct: B)
Explanation: The maximum allowable mass (MTOM) is a structural and aerodynamic certification limit, not a guideline. Exceeding it increases wing loading, raises the stall speed, degrades climb performance, and overstresses the airframe — potentially beyond its certified load factors. No pilot input can compensate for a structurally compromised aircraft. There is no regulatory or safety margin that permits any excess, even temporarily.
Correct: B)
Explanation: The approved C.G. envelope defines the range within which the aircraft's stability and controllability have been certified. If the C.G. moves forward of the front limit, elevator authority may be insufficient to rotate at takeoff or flare on landing. If it moves aft of the rear limit, the aircraft becomes statically unstable and pitch oscillations can become uncontrollable. The C.G. must remain between both limits throughout the entire flight.
Correct: C)
Explanation: The C.G. position relative to the aerodynamic neutral point determines longitudinal static stability. A C.G. forward of the neutral point produces a restoring pitching moment (stability), while control authority provides maneuverability (controllability). If the C.G. is outside limits, one of these two properties is compromised — either the pilot cannot correct a pitch upset, or the aircraft does not naturally resist one. Stall speed and Vne are influenced by other parameters and are not the primary reasons for the C.G. requirement.
Correct: B)
Explanation: Each individual aircraft is physically weighed — typically on three-point scales — to determine its actual empty mass and C.G. position. Manufacturing tolerances, repairs, and installed equipment vary between serial numbers of the same type, so manufacturer tables alone are insufficient. The results are recorded in the aircraft's weight and balance report and must be updated after any modification that changes mass or mass distribution.
Correct: C)
Explanation: In turbulence or during aerobatics, unsecured cargo can shift suddenly and move the C.G. outside limits instantaneously — faster than a pilot can react. A sudden aft C.G. shift can cause an unrecoverable pitch-up; items becoming projectiles can injure occupants or jam controls. The structural risk arises from asymmetric loading exceeding design limits. No prior stability analysis can make unsecured cargo acceptable.
Correct: B)
Explanation: By definition, the center of gravity (C.G.) is the single point through which the resultant gravitational force (weight vector) acts on the entire aircraft. The center 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 is zero on the leading edge — none of these is where gravity acts.
Correct: D)
Explanation: The center of gravity is the point through which the total weight force of the aircraft acts. It is the mass-weighted average position of all individual mass elements of the aircraft. It is not the physically heaviest point, and it is distinct from the neutral point (an aerodynamic concept). All mass and balance calculations reference moments about the datum to locate this point.
Correct: C)
Explanation: The C.G. is the point through which gravity (weight) is considered to act on the entire aircraft as if all mass were concentrated there. This definition is fundamental to mass and balance calculations: moments of all individual masses are summed and divided by total mass to locate this point. The datum is a fixed reference point, not the C.G. itself, and moment is the product of mass times arm.
Correct: B)
Explanation: In mass and balance, moment = mass x balance arm (M = m x d), expressed in kg-m or lb-in. This follows the physical definition of a torque or moment of force. The total C.G. position is then found by: C.G. = (sum of all moments) / (total mass). Using a sum, difference, or quotient instead of a product would yield a dimensionally and physically incorrect result.
Correct: C)
Explanation: The balance arm (or moment arm) is the horizontal distance measured from the aircraft's datum line to the center of gravity of a particular mass item (e.g., pilot, ballast, equipment). It determines the leverage that mass exerts about the datum. Distances from the C.G. itself are not balance arms — the datum is always the reference point. The datum is defined in the aircraft's flight manual and is fixed for that aircraft type.
Correct: B)
Explanation: In mass and balance terminology, the balance arm (also called moment arm) is specifically the horizontal distance from the aircraft datum to any given point of interest — including the overall C.G. once calculated. Torque/moment is the product of mass and arm, not the distance itself. Span width is a geometric wing parameter unrelated to longitudinal mass and balance.
Correct: B)
Explanation: The datum is an arbitrary but fixed reference plane (often the firewall, wing leading edge, or nose) defined in the aircraft's flight manual. The balance arm of any mass is measured as the horizontal distance from this datum to the center of gravity of that specific mass. All moment calculations use this datum as the common reference, allowing moments to be summed algebraically to find the total C.G. position.
Correct: C)
Explanation: The Pilot's Operating Handbook (POH) or Aircraft Flight Manual (AFM) contains a dedicated mass and balance section with the aircraft's empty mass, empty C.G. position, datum reference, C.G. limits, and approved loading configurations. The certificate of airworthiness merely certifies the aircraft type is approved; the annual inspection records maintenance history. Performance data (speeds, glide ratios) is in a different POH section.
Correct: D)
Explanation: The Weight and Balance section (Section 6 in EASA-standardized AFM/POH structure) contains the aircraft's basic empty mass, empty C.G. location, allowable C.G. range, and loading instructions. The Limitations section covers maximum speeds, load factors, and operating envelope. Normal Procedures covers checklists. Performance covers speeds, climb rates, and glide distances. Each section has a specific regulatory and operational purpose.
Correct: C)
Explanation: A headwind reduces groundspeed at touchdown for a given airspeed, so the aircraft arrives over the threshold with less kinetic energy to dissipate — shortening the ground roll. As a rule of thumb, a headwind component equal to 10% of approach speed reduces landing distance by approximately 19%. Conversely, high pressure altitude and high density altitude increase true airspeed at a given IAS, increasing groundspeed and lengthening landing distance. Heavy rain can reduce braking effectiveness, further increasing landing distance.
Correct: B)
Explanation: For aircraft not certified for flight into known icing (FIKI), operating in known or forecast icing conditions is a regulatory prohibition, not merely a performance consideration. Ice accretion on a glider's wings dramatically increases weight (shifting the C.G.), increases drag, reduces the maximum lift coefficient, and raises the stall speed — all simultaneously. If inadvertently encountered, the pilot must exit the icing environment immediately by changing altitude or heading, regardless of visual conditions.
Correct: D)
Explanation: The angle of descent (or glide angle) is geometrically defined as the angle between the horizontal and the actual flight path vector, measured in degrees. It is related to — but not the same as — the glide ratio: glide ratio = horizontal distance / height lost = 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.
Correct: B)
Explanation: Interception lines are prominent, linear geographic features — rivers, coastlines, railways, motorways — selected during pre-flight planning that run roughly perpendicular to the planned route. If a pilot becomes disoriented, flying toward the nearest interception line will produce an unmistakable landmark that allows position recovery. They do not extend permissions below VFR minima and are not range indicators; they are specifically a lost-procedure planning tool.
Note: This question originally references a chart excerpt (PFP-056) showing LO R 16 airspace boundaries.
Correct: D)
Explanation: Low-level restricted areas (LO R) published in national AIPs and on VFR charts typically express their vertical limits in feet MSL (above mean sea level) unless explicitly stated otherwise with GND/AGL. The notation "1 500 ft MSL" means the restriction applies from the surface (or a lower altitude boundary) up to 1,500 feet above mean sea level. Glider pilots must cross-check the AIP ENR section and current NOTAM for activation times and exact limits.
Note: This question originally references a chart excerpt (PFP-030) showing LO R 4 airspace boundaries.
Correct: B)
Explanation: As with Q19, restricted airspace limits are read directly from the relevant chart or NOTAM. The designation "4 500 ft MSL" indicates the upper vertical boundary is 4,500 feet above mean sea level — higher than a typical low-level restriction, reflecting terrain or operational considerations for that specific area. AGL (above ground level) would imply the limit varies with terrain; MSL is an absolute altitude referenced to a fixed datum.
Note: This question originally references a NOTAM excerpt (PFP-024).
Correct: C)
Explanation: NOTAM altitude references follow ICAO conventions: "Altitude" refers to height above MSL (mean sea level), "Height" refers to height above a local ground reference, and "Flight Level" is a pressure altitude reference (used above the transition altitude). The NOTAM in question prohibits overflight up to 9,500 ft MSL — a specific absolute altitude. 9,500 m MSL would be approximately 31,000 ft, clearly inconsistent with a typical VFR NOTAM restriction.
Correct: D)
Explanation: Under ICAO Annex 2 and national regulations, flight plans are mandatory for international flights crossing state borders, even for VFR glider flights. The flight plan is required for border control coordination, search and rescue alerting, and compliance with customs/immigration procedures. A filed and activated flight plan ensures that the relevant Air Traffic Services units and SAR services are aware of the flight. Hazard reports and location messages are separate AIREP/PIREP procedures.
Correct: D)
Explanation: The Flight Information Service (FIS), reached on the published FIS frequency in each FIR, can accept an airborne flight plan (AFIL) during flight. This is the standard procedure when a flight plan was not filed before departure or when an extension is needed. SAR is a response service, not a flight planning authority. AIS distributes aeronautical information but does not accept real-time flight plans. Airport operators handle local arrivals and departures, not en-route plan filing.
Correct: C)
Explanation: Thermal convection depends on differential ground heating. Moist ground, water bodies, and marshes have high thermal inertia and specific heat capacity — they absorb solar radiation without heating up as quickly as dry land, suppressing thermal development above them. Flying over large water areas or wetlands thus means less lift and potentially a forced landing in unsuitable terrain. Conversely, dry fields, rocky areas, and built-up areas with dark surfaces (asphalt, concrete) generate strong thermals.
Correct: B)
Explanation: At a downwind turning point, the glider must turn and fly back into the wind (or at an angle into it), immediately losing tailwind assistance and gaining a headwind component. Arriving high provides the maximum altitude reserve for the subsequent upwind leg, where groundspeed is reduced and glide distance over ground is shortened. Arriving low with a turn ahead is tactically dangerous — any failure to find lift on the upwind leg leaves no margin for landing field selection.
Correct: B)
Explanation: When a glider turns through 90 or 180 degrees at a waypoint, the pilot's perspective of the sky changes dramatically — the sun appears to have "moved" relative to the aircraft heading, and cumulus clouds that were previously in the pilot's peripheral vision or behind may now appear in front, and vice versa. This perceptual shift can make the sky look completely different even if objectively unchanged. Pilots must re-orient their thermal assessment relative to the new heading rather than relying on their previous mental picture.
Correct: D)
Explanation: ICAO chart symbology for aeronautical charts (defined in ICAO Annex 4 and Document 8697) uses specific symbols to distinguish obstacle types: lit vs. unlit, single vs. group. A group of unlighted (unlit) obstacles is shown with a specific symbol (D in the referenced figure). Knowing these symbols is essential for cross-country flight planning to identify terrain and obstruction hazards that would not be illuminated at dusk or during poor visibility conditions.
Correct: C)
Explanation: ICAO aeronautical chart symbology differentiates airports by category: civil vs. military, international vs. domestic, and runway surface (paved vs. unpaved). A civil domestic airport with a paved runway is represented by a specific symbol (A in the referenced figure) — typically a circle with a line or specific fill pattern. Glider pilots use these symbols when planning outlanding fields or alternate airports, as paved runways are preferable to grass strips for emergency landings in many conditions.
Correct: D)
Explanation: ICAO chart symbols differentiate between spot elevations (general terrain high points), surveyed elevation points, and obstruction heights. A general spot elevation (symbol B in the referenced figure) marks a notable terrain elevation that may not be the highest peak but is charted for situational awareness. Cross-country glider pilots must be familiar with these symbols to identify terrain clearance requirements, especially when planning routes through valleys or near mountain ranges where minimum safe altitudes are critical.
Correct: D)
Explanation: Glide distance = glide ratio x height available. With a glide ratio of 1:30 (30 metres forward for every 1 metre of height lost) and 1,500 m of height: distance = 30 x 1,500 m = 45,000 m = 45 km. Note: 45 NM would be approximately 83 km, which would require a glide ratio of roughly 1:55 — far above this aircraft's performance. The calculation is straightforward in metric: ratio x altitude in metres gives distance in metres. Always verify units — mixing NM and metres is a common error.
Correct: D)
Explanation: In active thermal conditions with strong lift, the glider can fly faster between thermals to optimise the average cross-country speed (MacCready theory). A higher wing loading (achieved with water ballast) shifts the speed polar towards higher speeds, improving the glide ratio at high speed. The trade-off is a higher stall speed and a higher best-glide speed — acceptable when thermals are strong enough to compensate.
Correct: D)
Explanation: The tail wheel is mounted at the extreme rear of the fuselage, at a large distance aft of the nominal centre of gravity. Even though its mass is small in absolute terms, its large moment arm gives it a significant moment. Leaving the tail wheel installed during flight shifts the C.G. aftward — potentially beyond the aft C.G. limit — making the aircraft pitch-unstable and difficult to control.
Correct: C)
Explanation: The maximum pilot seat load is a certification limit that cannot be circumvented by any trim adjustment or ballast reduction. Exceeding the maximum payload may place the C.G. outside the forward limit and subjects the structure to uncertified loads. The only correct action is to reduce the payload (e.g. by removing ballast or equipment) until the limits are respected. Trimming does not alter mass and does not make the aircraft compliant with its limitations.
Correct: D)
Explanation: A motorless glider is propelled exclusively by the component of the weight vector (gravity) projected in the direction of the flight path. In steady gliding flight, the aircraft is in equilibrium between lift (perpendicular to the flight path), drag (opposing motion) and weight. The component of weight along the flight path axis balances drag and maintains airspeed. Ascending air currents can slow or cancel the descent but do not propel the aircraft forward.
Correct: C)
Explanation: Calculation of the new C.G.: Initial moment = 610 x 80.0 = 48,800. Removed moment = 10 x 150 = 1,500. New total moment = 48,800 - 1,500 = 47,300. New mass = 610 - 10 = 600 kg. New C.G. = 47,300 / 600 = 78.833. Since the baggage was located aft of the current C.G. (150 > 80), its removal shifts the C.G. forward, which is consistent with the result obtained (78.833 < 80.0).
Extract from the Discus B Flight Manual -- Loading table with water ballast
] Max. permitted all-up weight including water ballast: 525 kg Lever arm of water ballast: 203 mm aft of datum (BE)
Table of water ballast loads at various empty weights and seat loads:
| Empty mass (kg) | Seat load 70 kg | 80 kg | 90 kg | 100 kg | 110 kg | |---|---|---|---|---|---| | 220 | 184 | 184 | 184 | 184 | 184 | | 225 | 184 | 184 | 184 | 184 | 184 | | 230 | 184 | 184 | 184 | 184 | 184 | | 235 | 184 | 184 | 184 | 184 | 180 | | 240 | 184 | 184 | 184 | 184 | 175 | | 245 | 184 | 184 | 184 | 180 | 170 | | 250 | 184 | 184 | 184 | 175 | 165 |
Correct: D)
Explanation: According to the Discus B loading table (extract from the flight manual): with an empty mass of 245 kg and 184 kg of water ballast in both wing tanks, the maximum seat load is 90 kg. The maximum permitted all-up weight with ballast is 525 kg; according to the row of the table corresponding to 245 kg / 184 kg, the seat load is limited to 90 kg in order to remain within the approved C.G. envelope.
Correct: D)
Explanation: On sloping terrain, the fundamental rule is to land uphill, which considerably shortens the landing roll — deceleration is assisted by gravity. An approach speed slightly above normal is recommended to maintain manoeuvrability and safety in the face of possible wind shear or turbulence on low final over unknown terrain. Landing downhill would be extremely dangerous as deceleration would be insufficient.
Correct: B)
Explanation: In heavy rain, the wing surface is wet, which can degrade aerodynamic characteristics (surface roughness, modification of the effective aerofoil profile). The stall speed may be slightly higher and the airbrakes less effective due to water on the surface. A higher approach speed therefore provides an appropriate safety margin. A shallower approach angle would be dangerous as it reduces obstacle clearance margins and extends the final approach.
Correct: D)
Explanation: A waterlogged grass runway offers greater rolling resistance (friction and soft ground deformation), which increases ground drag during acceleration. In addition, long or rain-flattened grass can create extra resistance. The takeoff distance is therefore longer compared to a dry grass runway. Aquaplaning is possible on hard runways with standing water but does not apply directly to wet grass — and wet grass offers more resistance, not less.
ASK 21 Speed Polar:
] Two curves: G=470 kp (light mass) and G=570 kp (heavy mass).
Correct: C)
Explanation: The ASK21 speed polar is shown for two masses: G=470 kp and G=570 kp. At 170 km/h, reading both curves, the sink rate is higher for the greater mass (570 kp). This is physically logical: a higher mass requires more lift to fly, which results in a higher angle of attack (at the same speed), greater induced drag and therefore a higher sink rate at that speed. The best L/D ratio remains approximately the same as both polars are nearly geometrically similar, but the absolute sink rate increases with mass.
Speed Polar (AIRSPEED):
] Two curves: 450 kg and 580 kg.
Correct: C)
Explanation: The speed at minimum sink rate (V min sink) corresponds to the top of the speed polar curve — the point where the curve is highest (lowest sink rate). Reading the polar for a mass of 450 kg, this point is at approximately 75 km/h. This is the optimum speed for maximising endurance in still air and for centring thermals. It differs from the best glide speed (which corresponds to the tangent from the origin to the polar).
Correct: C)
Explanation: The PAYERNE TMA has a lower limit that varies by sector. On the route between Murten and Neuchatel, the lower limit of the relevant TMA is at 700 m AMSL (2300 ft). Below this altitude, flight may be conducted without authorisation in the lower airspace (Class E or G depending on the area). Above 700 m AMSL, authorisation from the responsible ATC unit is required to cross the Class D TMA. This information is found on the Swiss ICAO aeronautical chart 1:500,000 or the gliding chart 1:300,000.
Correct: D)
Explanation: Birrfeld aerodrome lies within Class E airspace above the local CTR/ATZ. At 1400 m AMSL in this sector, you are in Class E. VFR minima in Class E are: horizontal visibility 5 km, cloud clearance 1500 m horizontally and 300 m vertically. Class E provides an air traffic service for IFR; VFR flights are permitted without a clearance but must comply with these meteorological minima.
DABS -- Daily Airspace Bulletin Switzerland (extract)
]
| Firing-Nr D-/R-Area NOTAM-Nr | Validity UTC | Lower Limit AMSL or FL | Upper Limit AMSL or FL | Location | Center Point | Covering Radius | Activity / Remarks | |---|---|---|---|---|---|---|---| | B0685/14 | 0000-2359 | 900m / 3000ft | FL 130 | SION TMA SECT 1 | 461610N 0072940E | 4.7 KM / 2.5 NM | TMA SECT 1 ACT HX ONLY | | W0912/15 | 1145-1300 | GND | FL 120 | MORGARTEN | 470507N 0083758E | 10.0 KM / 5.4 NM | R-AREA ACT. ENTRY PROHIBITED. FOR INFO CTC ZURICH INFO 124.7 | | W0957/15 | 1400-1700 | 2150m / 7000ft | FL 120 | HINWIL | 471721N 0084859E | 7.0 KM / 3.8 NM | TEMPO R-AREA ACTIVE. ENTRY PROHIBITED. CTC 118.975 | | W0960/15 | 0800-1700 | GND | 1200m / 4050ft | 1.7 KM SE CERNIER | 470352N 0065442E | 1.5 KM / 0.8 NM | D-AREA ACT |
Correct: B)
Explanation: Consulting the DABS extract provided: zone W0957/15 is active from 1400 to 1700 UTC. On 20 June 2015 (summer time CEST = UTC+2), 1515-1545 LT corresponds to 1315-1345 UTC. Zone W0957/15 is therefore not yet active at this time (it starts at 1400 UTC). Zone W0912/15 is active from 1145 to 1300 UTC — already expired. The route can therefore be flown without coordination between 1500 and 1600 LT (i.e. 1300-1400 UTC), just before W0957/15 becomes active. The DABS applies to all airspace users, including gliders.
Correct: C)
Explanation: Over Schwyz, the Swiss ICAO aeronautical chart 1:500,000 shows the lower limit of Class C airspace at FL 130. Below this, the airspace is Class E (or D depending on the area). Entering Class C requires an ATC clearance regardless of flight rules. Glider pilots flying wave or cross-country flights at high altitude over the Swiss central Alps must therefore contact the competent ATC unit (Zurich Information or Zurich ACC) before reaching FL 130.
AD INFO 1 -- LA COTE / LSGP
]
| Data | Value | |--------|--------| | Hours MON-FRI | 0700-1200 LT / 1400-ECT -30 min | | Hours SAT/SUN | 0800-1200 LT / 1400-ECT -30 min |
ECT = End of Civil Twilight.
Correct: D)
Explanation: According to the AD INFO 1 sheet for LSGP La Cote, the afternoon opening hours are shown as "1400-HRH -30 min" where HRH denotes "end of civil twilight" (Swiss notation). The aerodrome therefore closes 30 minutes before the end of civil twilight (not before sunset, which is an earlier moment). This applies on weekdays (MON-FRI) and at weekends (SAT-SUN). PPR (Prior Permission Required) also applies.
Visual Approach Chart -- GRUYERES / LSGT
]
| Data | Value | |--------|--------| | AD Frequency | 124.675 MHz | | Elevation | 2,257 ft (688 m) | | Winch launches | Intensive SAT/SUN |
Correct: D)
Explanation: According to the Visual Approach Chart for LSGT Gruyeres, the aerodrome frequency is shown in the top right: AD 124.675. This is the frequency on which local traffic information is broadcast, including information on intensive winch launches at weekends ("Intense winch launching SAT/SUN"). Frequencies 110.85 and 113.9 correspond to the VOR/DME SPR (Saanen/Pringy) shown on the chart, and 119.175 is the GENEVA DELTA frequency.
Correct: C)
Explanation: Distance = speed x time. Ground speed = 90 km/h, duration = 90 minutes = 1.5 hours. Distance = 90 km/h x 1.5 h = 135 km. This is a basic navigation calculation: remember to convert minutes to a fraction of an hour before multiplying. 90 minutes represents one and a half hours, i.e. 1.5 h — not 0.9 h (a common error when confusing minutes with decimal hours).
Correct: D)
Explanation: The airspeed indicator measures dynamic pressure, which depends on air density. At 6000 m altitude, the air density is significantly lower than at sea level (standard ISA atmosphere). For the same dynamic pressure (same IAS), the TAS must be higher because less dense air requires a greater true speed to produce the same indicated pressure. In practice, TAS increases by approximately 2% per 300 m of altitude gain. At 6000 m, TAS is approximately 20-25% higher than IAS.
Correct: D)
Explanation: The VNE (never-exceed speed) displayed on the airspeed indicator is an IAS reference value at sea level (or low altitude). At high altitude, the TAS corresponding to the same IAS is higher, but it is the true airspeed (TAS) that determines structural aerodynamic loads. For gliders, the flight manual provides a **speed-altitude table** (or curve) giving the corrected VNE IAS as a function of altitude. At 6000 m, the V_NE IAS to be observed is lower than that shown at ground level — hence the reference to the table displayed in the cockpit.
Correct: C)
Explanation: 1235 lbs / 2.2 = 561.4 kg, approximately 560 kg. Formula: mass (kg) = weight (lbs) / 2.2.
Correct: C)
Explanation: On an upsloping field with tailwind, fly the normal approach speed (yellow triangle). The upslope shortens the flare distance and tailwind reduces effective landing distance. Normal speed is critical to avoid stall.
Correct: C)
Explanation: Langenthal at 2000 m AMSL is in Class E airspace (between 1500 ft AMSL and the TMA/CTA floor). In Class E, VMC requires: visibility 5 km, cloud clearance 1500 m horizontally and 300 m vertically.
Correct: B)
Explanation: A center of gravity too far aft is the most dangerous position as it makes the glider longitudinally unstable. Longitudinal stability disappears and the glider may pitch violently without possible correction.
Correct: B)
Explanation: VNE remains the same on the airspeed indicator (IAS) because IAS is already corrected for density by design. True airspeed (TAS) increases with altitude, but IAS remains constant.
Correct: C)
Explanation: GS = distance / time = 150 km / (1h15 min) = 150 / 1.25 = 120 km/h.
]
Correct: B)
Explanation: The NOTAM describes activation of extended CTR/TMA Payerne and zone LS-R4 from 2 to 6 September between 0600-1500 UTC as holding and demonstration areas. The region must be strictly avoided during these periods.
]
Correct: C)
Explanation: For a flying mass of 450 kg, best glide speed is read from the polar (attached sheet) where the tangent from the origin touches the curve. For 450 kg, this speed is approximately 75 km/h.
]
Correct: D)
Explanation: According to the DABS for 19 March 2013 (winter time) between 1205-1255 LT, the route can be flown without coordination between 1200-1300 LT as the zones are not active during this specific period.
Correct: C)
Explanation: With a 40% wing loading increase, minimum speed increases by the square root of 1.4 = 1.183, approximately 18%. Stall speed is proportional to the square root of wing loading.
]
Correct: C)
Explanation: At 150 km/h, the ASK21's sink rate is independent of its mass because the two polar curves (different masses) intersect at this speed. This is an aerodynamic property of the polar curve.
]
Correct: D)
Explanation: At Amlikon, the maximum available landing distance heading East is 780 m according to the AIP Switzerland chart.
]
Correct: C)
Explanation: Between Cham and Hitzkirch, the EMMEN TMA begins at 3500 ft AMSL. Below this you are in uncontrolled airspace. Above this you enter the TMA and must obtain clearance.
Correct: B)
Explanation: If the maximum allowed payload is exceeded, the only correct action is to reduce the payload. Trimming or increasing takeoff speed does not solve an excessive mass problem.
Correct: C)
Explanation: With a headwind, the angle of descent relative to the ground increases (the aircraft descends more steeply over the ground). With a tailwind, the angle decreases. Wind does not change the sink rate in m/s, but it changes the ground descent angle.
Correct: C)
Explanation: Indicated airspeed (IAS) decreases relative to TAS as altitude increases, because air density decreases. At high altitude, IAS is less than TAS. At low altitude, they are close.
]
Correct: C)
Explanation: At Bex, the traffic pattern for runway 33 can be either direction depending on the wind, due to terrain constraints. The correct answer is that direction depends on wind conditions.
]
Correct: D)
Explanation: Above Biel Kappelen, the BERN 1 TMA begins at 3500 ft AMSL. By staying below 3500 ft AMSL, you do not need a transit clearance.
Correct: C)
Explanation: CG calculation question: with the attached sheet data, the new CG is calculated at 76.7, within approved limits.
Correct: D)
Explanation: A wet grass runway reduces rolling friction and shortens landing distance. Wet grass decreases braking, so the glider stops faster (sliding effect).
]
Correct: D)
Explanation: At Schanis, the maximum available landing distance heading NNW is 470 m according to AIP Switzerland.
Correct: C)
Explanation: CG calculation: current mass 6400 lbs, current CG 80, aft limit 80.5. Moving mass x from current position to arm 150 without exceeding 80.5: (6400x80 + x*(150-80))/(6400+x) = 80.5. Solution: x is approximately 45.71 lbs.
Correct: B)
Explanation: Correct loading depends on both respecting the maximum allowable mass AND correct payload distribution (to keep CG within limits). Both conditions are necessary.
]
Correct: C)
Explanation: On the speed polar, maximum glide ratio is independent of flying mass (apart from minor Reynolds number effects). Polar curves for different masses have the same maximum L/D but at different speeds.
Correct: D)
Explanation: At 1800 m AMSL, air is less dense. To maintain the same aerodynamic lift, TAS is higher, but IAS (what is read on the ASI) remains the same as at sea level. Therefore approach at the same indicated speed.
]
Correct: D)
Explanation: For 450 kg, best glide speed is read from the polar (attached sheet) at the tangent from the origin. For this glider type at 450 kg, it is approximately 90 km/h.
Correct: B)
Explanation: If the aft CG limit is exceeded, redistribute the useful load forward. Trimming is not a structural solution to the CG problem.
Correct: D)
Explanation: High temperature and tailwind lengthen the aerotow takeoff roll. High temperature reduces air density (less lift), tailwind increases the takeoff distance.
]
Correct: C)
Explanation: The NOTAM for 18 November shows a military night flying exercise from 1800 to 2100 UTC in the ZUGERSEE, SUSTEN and TICINO regions, between GND and 15,000 ft AMSL.
]
Correct: D)
Explanation: The CTR of Bern-Belp airport has an upper limit of 3000 ft AMSL.
]
Correct: C)
Explanation: Above Bex aerodrome at 1700 m AMSL: we are in Class E airspace (between 1500 ft AMSL and the TMA). VMC in Class E: visibility 5 km, cloud clearance 1500 m / 300 m.
]
Correct: C)
Explanation: At 160 km/h for 580 kg, sink rate is read from the polar (attached sheet), approximately 2.0 m/s.
Correct: D)
Explanation: 550 kg x 2.2 = 1210 lbs. Formula: lbs = kg x 2.2.
Correct: C)
Explanation: In calm air, to cover the maximum distance, fly at the best glide speed (best L/D). This is the optimal speed for maximising range in still air.
Correct: D)
Explanation: When glider mass increases, maximum glide ratio remains practically unchanged (mass-independent, apart from Reynolds effects). What changes: minimum speed increases, wing loading increases, sink rate increases.
Correct: C)
Explanation: Time = distance / speed = 150 km / 100 km/h = 1.5 h = 1 hour 30 minutes.
]
Correct: C)
Explanation: According to the DABS, when zones LS-R8 and LS-R8A are active, this alpine route cannot be flown as these restricted zones cover the itinerary.
Correct: B)
Explanation: For Zurich TMA transit: first radio contact on 124.7 MHz, at least 10 minutes before entering the TMA.
Correct: B)
Explanation: Stall speed in turn with load factor n=2.0: Vsturn = Vsnormal x sqrt(n) = 60 kts x sqrt(2) = 60 x 1.414, approximately 85 kts. Increase = (85-60)/60 x 100%, approximately 41%, rounded to 40%.
