Correct: B)
Explanation: The correct answer is B because increasing wing loading (typically via water ballast) shifts the speed polar to higher speeds while preserving approximately the same maximum glide ratio, allowing faster inter-thermal cruise and improving average cross-country speed according to MacCready theory. A is wrong because higher wing loading increases the stall speed rather than reducing it. C is wrong because higher wing loading forces the glider to fly faster, not slower. D is wrong because climb rate in thermals actually worsens with higher wing loading due to increased minimum sink speed.
Correct: D)
Explanation: The correct answer is D because the tail wheel is mounted at the extreme rear of the fuselage with a very large moment arm. Even though its mass is small, the large lever arm creates a significant aft moment that shifts the CG rearward, potentially beyond the aft limit, causing dangerous longitudinal instability. A is wrong because a tail wheel does not improve flight manoeuvrability. B is wrong because the tail wheel is behind the CG, shifting it aft, not forward. C ignores the amplifying effect of the large moment arm.
Correct: C)
Explanation: The correct answer is C because the maximum cockpit payload is a certification limit that must not be exceeded. The only correct action is to reduce the payload until it falls within limits. A and B are wrong because trim adjustments change elevator deflection and stick forces but do not alter the aircraft's actual mass or structural loading. D is wrong because reducing water ballast lowers total mass but does not address the cockpit load exceeding its own independent structural limit.
Correct: C)
Explanation: The correct answer is C because a motorless glider converts altitude into airspeed by descending along an inclined flight path. The component of the weight vector projected along the flight path direction balances drag and maintains airspeed. A is wrong because ascending currents slow the descent relative to the ground but do not propel the aircraft forward through the air. B is wrong because drag always opposes motion, never drives it. D is wrong because a tailwind increases ground speed but does not propel the aircraft through the airmass.
Correct: D)
Explanation: The correct answer is D: 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 CG = 47,300 / 600 = 78.833. Since the baggage was aft of the current CG (arm 150 > 80), removing it shifts the CG forward. A and C are too far forward. B implies a rearward shift, contradicting removal of aft-located mass.
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 |
*Water ballast in both wing tanks (kg). For empty mass 245 kg and ballast 184 kg: the maximum seat load is 90 kg.* - A) 100 kg - B) 110 kg - C) 90 kg - D) 80 kg
Correct: C)
Explanation: The correct answer is C because the loading table at 245 kg empty mass shows 184 kg ballast is achievable with seat loads of 70, 80, and 90 kg, but at 100 kg the maximum drops to 180 kg and at 110 kg to 170 kg. Since the full 184 kg of ballast is required, the highest permissible seat load is 90 kg. A and B would require reducing ballast below 184 kg. D is unnecessarily restrictive.
Correct: B)
Explanation: The correct answer is B because landing uphill provides strong deceleration from the slope gradient, dramatically shortening the ground roll, while a slightly higher approach speed compensates for potential wind shear and the altered visual perspective over unfamiliar terrain. A is wrong because full airbrakes depend on the situation. C is wrong because on significant slopes, the uphill direction takes priority over wind. D is wrong because flare height depends on the specific approach geometry.
Correct: D)
Explanation: The correct answer is D because heavy rain wets the wing surface, degrading the aerodynamic profile and potentially raising the effective stall speed. A higher approach speed provides the necessary safety margin. A is wrong because the critical concern is degraded lift, not a slower aircraft. B is wrong because rain requires specific adaptations. C is wrong because a shallower approach reduces obstacle clearance and extends exposure to poor visibility.
Correct: A)
Explanation: The correct answer is A because waterlogged grass increases rolling resistance from soft ground deformation and water drag on the wheel, slowing acceleration and lengthening the takeoff run. B is wrong because the runway condition, not the glider's wetness, is the primary concern. C is wrong because wet, soft grass increases resistance. D is wrong because aquaplaning occurs on hard, smooth surfaces with standing water, not soft grass.
ASK 21 Speed Polar:
] Two curves: G=470 kp and G=570 kp. - A) Regardless of the mass of the ASK21, the sink rate stays constant. - B) As the mass of the ASK21 rises, the sink rate increases. - C) As the mass of the ASK21 increases, the sink rate increases. - D) As the mass of the ASK21 decreases, the glide angle improves.
Correct: C)
Explanation: The correct answer is C because at 170 km/h, the heavier configuration (G=570 kp) shows a noticeably higher sink rate than the lighter one (G=470 kp) on the polar diagram. A heavier aircraft needs more lift at the same speed, producing greater induced drag and a higher rate of descent. A is wrong because the curves clearly show different sink rates. B states the same physical fact as C. D is wrong because at this high speed, the lighter aircraft does not necessarily have a significantly better ground glide angle.
Speed Polar (AIRSPEED):
] - A) 75 km/h - B) 95 km/h - C) 50 km/h - D) 140 km/h
Correct: A)
Explanation: The correct answer is A because the minimum sink rate corresponds to the apex (highest point) of the speed polar curve. For the 450 kg configuration, this point is at approximately 75 km/h. This is the optimal speed for maximum endurance and thermal centring. B is closer to the min-sink speed for the heavier 580 kg configuration. C is below the stall speed. D is deep in the high-drag region.
Correct: C)
Explanation: The correct answer is C because the lower limit of the relevant Payerne TMA sector on this route is 700 m AMSL (2300 ft) per the Swiss ICAO chart. Below this altitude, flight proceeds without clearance in Class E or G airspace. A applies to a different TMA sector. B (FL 100) is the upper boundary, not the entry point. D is wrong because a TMA does not extend to the ground -- that would be a CTR.
Correct: A)
Explanation: The correct answer is A because at 1400 m AMSL over Birrfeld, the Swiss chart shows Class E airspace with VFR minima of 5 km visibility, 1500 m horizontal and 300 m vertical cloud clearance. B is wrong because Birrfeld is not within a Class D CTR at this altitude. C is wrong because Class G applies only in the lowest airspace layer. D is wrong because Class C starts at FL 130.
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. | | W0957/15 | 1400-1700 | 2150m / 7000ft | FL 120 | HINWIL | 471721N 0084859E | 7.0 KM / 3.8 NM | TEMPO R-AREA ACTIVE. ENTRY PROHIBITED. | | W0960/15 | 0800-1700 | GND | 1200m / 4050ft | 1.7 KM SE CERNIER | 470352N 0065442E | 1.5 KM / 0.8 NM | D-AREA ACT | - A) It is not possible to fly the planned route that day. - B) You can ignore the DABS as it only applies to commercial aviation. - C) You can pass through all relevant danger and restricted areas below 1000 ft AGL or above 12,000 ft AMSL. - D) The route can be flown without coordination between 1500 and 1600 LT.
Correct: D)
Explanation: The correct answer is D because 1515-1545 LT CEST (UTC+2) converts to 1315-1345 UTC. Zone W0912/15 expired at 1300 UTC. Zone W0957/15 becomes active at 1400 UTC but its lower limit is 7000 ft, above the planned 6500 ft. The route is clear. A is wrong because the route is flyable. B is wrong because the DABS applies to all airspace users. C is wrong because the cited altitudes do not match the published restrictions.
Correct: C)
Explanation: The correct answer is C because the Swiss ICAO chart shows Class C airspace beginning at FL 130 over Schwyz. Below FL 130, the airspace is Class E. A and B are within Class E. D marks the upper boundary of Swiss controlled airspace, not the Class C entry point.
AD INFO 1 -- LA COTE / LSGP
] - A) Until half an hour before the start of civil twilight. - B) Until half an hour before sunset. - C) Until half an hour before the end of civil twilight. - D) Until the end of civil twilight.
Correct: C)
Explanation: The correct answer is C because the AD INFO shows afternoon hours ending at "ECT -30 min," meaning the aerodrome closes 30 minutes before the end of civil twilight. A references the start of civil twilight, which is incorrect. B references sunset, which occurs earlier than the end of civil twilight. D omits the 30-minute advance closure.
Visual Approach Chart -- GRUYERES / LSGT
] - A) 113.9 - B) 124.675 - C) 119.175 - D) 110.85
Correct: B)
Explanation: The correct answer is B because 124.675 MHz is the aerodrome frequency for LSGT, shown on the Visual Approach Chart. All local traffic information including intensive winch launches at weekends is broadcast on this frequency. A (113.9) is the VOR/DME SPR navigation frequency. C (119.175) is the Geneva Delta sector frequency. D (110.85) does not appear on the chart.
Correct: B)
Explanation: The correct answer is B because distance = speed x time = 90 km/h x 1.5 h (90 min / 60) = 135 km. A results from treating 90 minutes as 1 hour. C and D do not match any correct calculation. Always convert minutes to decimal hours before multiplying.
Correct: D)
Explanation: The correct answer is D because at 6000 m, air density is significantly reduced. The ASI measures dynamic pressure; to produce the same reading (160 km/h IAS) in thinner air, the aircraft must fly at a much higher true airspeed -- approximately 20-25% higher. A is wrong because TAS is always greater than IAS above sea level. B is wrong because equality holds only at sea level under standard conditions. C is wrong because at altitude, TAS is always higher, not variable in direction.
Correct: B)
Explanation: The correct answer is B because VNE is fundamentally a TAS limit. At high altitude, the same IAS produces a much higher TAS, so the permissible IAS must be reduced according to the flight manual's speed-altitude table placarded in the cockpit. A is wrong because you must fly at a lower IAS. C is wrong because VNE IAS must be reduced at altitude. D is wrong because the green arc boundary may exceed the altitude-corrected VNE.
Correct: C)
Explanation: The correct answer is C because 1235 / 2.2 = approximately 561 kg, rounding to 560 kg. A (620 kg) is too high. B (2720 kg) and D (2470 kg) result from multiplying instead of dividing. To convert pounds to kilograms, divide by 2.2.
Correct: C)
Explanation: The correct answer is C because the upslope deceleration compensates for the tailwind's tendency to extend the ground roll, making the normal approach speed (yellow triangle) appropriate. A is wrong because flying faster increases float distance. B is wrong because flare height is not universally higher. D is wrong because full airbrakes are not always required.
Correct: A)
Explanation: The correct answer is A because at 2000 m AMSL above Langenthal, the Swiss chart shows Class E airspace with standard VFR minima: 5 km visibility, 1500 m horizontal and 300 m vertical cloud clearance. B applies only in the lowest airspace layer. C is wrong because there is no Class D CTR here. D is wrong because Class C starts at FL 130.
Correct: C)
Explanation: The correct answer is C because an aft CG reduces longitudinal stability. Beyond the aft limit, the glider becomes unstable in pitch and can enter an uncontrollable divergent oscillation or pitch-up. A (too far forward) increases stability excessively but is generally survivable. B and D are not standard CG concerns for gliders.
Correct: C)
Explanation: The correct answer is C because the ASI measures dynamic pressure, which inherently accounts for density changes. The VNE IAS reading remains the same because at altitude, lower density requires higher TAS to produce the same dynamic pressure and aerodynamic loads. A is wrong because VNE IAS does not rise. B is less precise than C. D is wrong for the general case, though some gliders do require VNE IAS reduction at very high altitudes for flutter reasons.
Correct: C)
Explanation: The correct answer is C because ground speed = 150 km / 1.25 hours = 120 km/h. Converting 1 hour 15 minutes: 15 min = 0.25 h, total = 1.25 h. A (125), B (115), and D (110) do not match this calculation. A common mistake is treating 1h15m as 1.15 hours instead of 1.25.
]
- A) The extended CTR/TMA Payerne and restricted zone LS-R4 must be strictly avoided every day from 02 to 06 September 2013, between sunrise and sunset.
- B) An airshow is taking place in the Payerne area from 02 to 06 September 2013. The TMA Payerne and restricted zone LS-R4 are active each day during this period between 0600 UTC and 1500 UTC as holding areas and airshow demonstration sectors.
- C) Due to an airshow from 02 to 06 September 2013, the extended CTR/TMA Payerne is active each day between 0600 UTC and 1500 UTC. The TMA is used as a holding area, the restricted zone LS-R4 as a demonstration and holding area. The area must be strictly avoided.
- D) Due to an airshow, a transit clearance for the extended CTR/TMA Payerne and restricted zone LS-R4 must be requested on frequency 135.475 (Payerne TWR) from 02 to 06 September 2013.
Correct: C)
Explanation: The correct answer is C because the NOTAM describes activation of the extended CTR/TMA Payerne and LS-R4 from 02 to 06 September, 0600-1500 UTC daily, designating them as holding and demonstration areas with mandatory avoidance. A is wrong because the times are 0600-1500 UTC, not sunrise to sunset. B omits the mandatory avoidance requirement. D is wrong because no transit clearance is offered -- complete avoidance is required.
]
- A) 95km/h
- B) 75km/h
- C) 55km/h
- D) 135km/h
Correct: B)
Explanation: The correct answer is B because the best glide speed is found by drawing a tangent from the origin to the 450 kg polar curve. The tangent point falls at approximately 75 km/h. A (95 km/h) is too fast. C (55 km/h) is near or below stall speed. D (135 km/h) produces a very poor glide ratio.
]
- A) The DABS can be ignored as it solely applies to military aircraft.
- B) You may pass through all relevant danger and restricted zones below 1000 ft AGL or above 10,000 ft AMSL.
- C) The route can be flown without coordination between 1200 and 1300 LT.
- D) It is not possible to fly the planned route that day.
Correct: C)
Explanation: The correct answer is C because for 19 March 2013 (CET = UTC+1), 1205-1255 LT converts to 1105-1155 UTC. During this window, the relevant zones are not active, so the flight can proceed without coordination. A is wrong because the DABS applies to all airspace users. B cites incorrect altitude exemptions. D is wrong because the route is flyable during the planned window.
Correct: A)
Explanation: The correct answer is A because stall speed scales with the square root of wing loading. With 40% increase: sqrt(1.40) = 1.183, meaning an 18.3% speed increase. B (40%) incorrectly assumes a linear relationship. C (100%) implies a speed doubling. D (0%) wrongly suggests mass has no effect. The square-root relationship is fundamental to the lift equation.
