Flight Performance and Planning

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Q61: Based on the polar below, which statement applies at a speed of 150 km/h? See attached sheet... ^t30q61

] - A) the sink rate of the ASK21 is independent of its mass - B) the ASK21 has a worse glide ratio at lower flying mass - C) the ASK21 has a higher sink rate at higher flying mass - D) the ASK21 has a better glide ratio at lower flying mass

Correct: A)

Explanation: The correct answer is A because at 150 km/h, the two polar curves for different masses intersect, meaning both configurations have the same sink rate at this specific speed. This intersection is a characteristic feature of speed polars. B is wrong because glide ratio is virtually identical for both mass configurations at 150 km/h. C is wrong at this particular speed because the curves cross here, making the sink rates equal. D is also wrong because neither mass has a better glide ratio at the intersection speed.

Q62: At Amlikon aerodrome, what is the maximum available landing distance heading East? ^t30q62

] - A) 700 ft. - B) 780m. - C) 780 ft - D) 700m.

Correct: B)

Explanation: The correct answer is B because the AIP Switzerland chart for Amlikon shows 780 m as the available landing distance for the eastward runway direction. A and C express values in feet, which are far too short (700 ft = 213 m, 780 ft = 238 m) and are the wrong unit for Swiss AIP runway distances. D (700 m) is close but does not match the published figure for the eastward heading. Always verify both the numerical value and the unit.

Q63: From what altitude must you request a transit clearance for the EMMEN TMA between Cham (approx. N47 deg 11' / E008 deg 28') and Hitzkirch (approx. N47 deg 14' / E008 deg 16')? ^t30q63

] - A) 2400 ft AMSL. - B) 3500 ft AMSL. - C) 2000ft GND. - D) 5000 ft AMSL.

Correct: B)

Explanation: The correct answer is B because the ICAO chart shows the EMMEN TMA floor at 3500 ft AMSL in the sector between Cham and Hitzkirch. Below this altitude, flight proceeds in uncontrolled airspace without clearance. A (2400 ft) and D (5000 ft) do not match any published boundary in this sector. C (2000 ft GND) uses an above-ground reference, which is not how Swiss TMA floors are defined -- they use AMSL altitudes.

Q64: The maximum permitted payload is exceeded. What action must be taken? ^t30q64

• A) Trim aft.
• B) Increase takeoff speed by 10%.
• C) Trim forward.
• D) Reduce the payload.

Correct: D)

Explanation: The correct answer is D because the maximum payload is a certification limit that cannot be circumvented by trim changes or speed adjustments. The only correct response is to reduce the payload until it falls within limits. A and C are wrong because trimming changes pitch forces but does not affect mass or structural loads. B is wrong because increasing takeoff speed would impose even greater aerodynamic loads on an already overloaded airframe.

Q65: Which is the effect of wind on the glide angle over the ground if the aircraft's true airspeed remains constant? ^t30q65

• A) With a tailwind, the glide angle increases.
• B) With a headwind, the glide angle decreases.
• C) Wind has no effect on the glide angle.
• D) With a headwind, the glide angle rises.

Correct: D)

Explanation: The correct answer is D because a headwind reduces groundspeed while the sink rate (determined by TAS through the airmass) remains unchanged, so the glider covers less horizontal distance per unit of height lost and the descent angle over the ground steepens. A is wrong because a tailwind flattens the glide angle rather than steepening it. B directly contradicts the physics. C is wrong because while wind does not change the aerodynamic polar, it clearly changes the ground-referenced trajectory.

Q66: How does indicated airspeed (IAS) compare to true airspeed (TAS) as altitude increases? ^t30q66

• A) It rises.
• B) It decreases.
• C) It cannot be measured.
• D) It stays identical.

Correct: B)

Explanation: The correct answer is B because IAS is proportional to the square root of dynamic pressure, which depends on air density. As altitude increases and density decreases, the same TAS produces less dynamic pressure, so IAS reads lower than TAS. The difference grows with altitude. A is wrong because IAS decreases relative to TAS. D is true only at sea level under standard conditions. C is wrong because IAS is always measurable via the pitot-static system.

Q67: What has to be particularly observed when landing in heavy rain? ^t30q67

• A) Approach speed must be increased.
• B) Wing loading must be increased.
• C) The approach angle must be shallower than usual.
• D) Approach speed must be lower than usual.

Correct: A)

Explanation: The correct answer is A because rain on the wing surface increases roughness, disrupts the boundary layer, and can raise the effective stall speed, necessitating a higher approach speed as a safety margin. B is wrong because the pilot cannot change wing loading on approach. C is wrong because a shallower approach increases exposure to poor conditions. D is the opposite of what is needed and would bring the aircraft dangerously close to the elevated stall speed.

Q68: What must a glider pilot take into account at Bex aerodrome? ^t30q68

] - A) The traffic pattern for runway 33 is clockwise. - B) The traffic pattern for runway 15 is clockwise. - C) The traffic pattern for runway 33 is counter-clockwise. - D) Depending on wind, the traffic pattern for runway 33 may be either clockwise or counter-clockwise.

Correct: D)

Explanation: The correct answer is D because Bex aerodrome is in a narrow alpine valley with terrain constraints, so the traffic pattern for runway 33 cannot always follow a single fixed direction. The Visual Approach Chart specifies that circuit direction depends on the prevailing wind. A and C each assert a single fixed direction for runway 33, contradicting the published information. B refers to runway 15 rather than runway 33.

Q69: What is the maximum flying altitude above Biel Kappelen aerodrome (SE of Biel) if you wish to avoid requesting a transit clearance for TMA BERN 1? ^t30q69

] - A) 3500 ft AGL. - B) FL 100. - C) FL 35. - D) 3500 ft AMSL.

Correct: D)

Explanation: The correct answer is D because the ICAO chart shows TMA BERN 1 with a lower boundary of 3500 ft AMSL over Biel Kappelen. Staying below this keeps you in uncontrolled airspace. A (3500 ft AGL) uses an above-ground reference that could exceed the actual TMA floor over elevated terrain. B (FL 100) is the upper limit of lower Swiss airspace, not the TMA floor. C (FL 35) coincides with D only at standard pressure, but TMA floors are published in AMSL, not flight levels.

Q70: Which of these statements is correct? ^t30q70

• A) New C.G: 76.7, within approved limits.
• B) New C.G: 78.5, within approved limits.
• C) New C.G: 82.0, outside approved limits.
• D) New C.G: 75.5, outside approved limits.

Correct: A)

Explanation: The correct answer is A because performing the mass-and-balance calculation with the provided data yields a new CG of 76.7, which falls within the certified forward and aft CG limits. The procedure is: sum all individual moments (mass x arm), divide by total mass, then check against the published CG envelope. B (78.5), C (82.0), and D (75.5) do not result from the correct arithmetic using the given data.

Q71: What is the effect of a waterlogged grass runway on landing? ^t30q71

• A) Landing distance will be shorter.
• B) Landing distance will be longer.
• C) The glider risks running off the runway (groundloop).
• D) No effect.

Correct: A)

Explanation: The correct answer is A because waterlogged grass dramatically increases rolling resistance as the wheel sinks into the soft, saturated surface, decelerating the glider much more quickly than dry grass and shortening the ground roll. B describes what might happen on a flooded hard runway where aquaplaning reduces braking, not on soft grass. C is a concern on firm crosswind surfaces, not the primary effect here. D is clearly wrong since surface condition directly affects deceleration.

Q72: At Schanis aerodrome, what is the maximum available landing distance heading NNW? ^t30q72

] - A) 520 m. - B) 470m. - C) 520 ft. - D) 470 ft.

Correct: B)

Explanation: The correct answer is B because the AIP Switzerland chart for Schanis shows 470 m as the available landing distance for the NNW direction. A (520 m) corresponds to the opposite (SSE) direction. C and D are in feet (520 ft = 158 m, 470 ft = 143 m), far too short for any normal landing and clearly the wrong unit.

Q73: The current mass of an aircraft is 6400 lbs. Current CG: 80. CG limits: forward CG: 75.2, aft CG: 80.5. What mass can be moved from its current position to arm 150 without exceeding the aft CG limit? ^t30q73

• A) 27.82 lbs.
• B) 56.63 lbs.
• C) 39.45 lbs.
• D) 45.71 lbs.

Correct: D)

Explanation: The correct answer is D. The mass x currently sits at the aircraft CG (arm 80) and moves to arm 150. The equation is: (6400 x 80 + x x (150 - 80)) / 6400 = 80.5. Solving: 512,000 + 70x = 515,200, so x = 3200 / 70 = 45.71 lbs. A (27.82), B (56.63), and C (39.45) result from algebraic errors in setting up or solving the moment equation.

Q74: Correct loading of an aircraft depends on:... ^t30q74

• A) Only compliance with the maximum allowable mass.
• B) Only correct payload distribution.
• C) Correct payload distribution and compliance with the maximum allowable mass.
• D) The maximum allowable mass of baggage in the aft section of the aircraft.

Correct: C)

Explanation: The correct answer is C because safe loading requires simultaneously satisfying two conditions: total mass within the certified maximum, and payload distribution keeping the CG within forward and aft limits. A is wrong because correct mass alone says nothing about CG position. B is wrong because correct distribution is meaningless if total mass exceeds structural limits. D addresses only a subset of the problem.

Q75: What information can be read from this speed polar? (See attached sheet.)... ^t30q75

] - A) in the speed range up to 100 km/h, an increase in flying mass reduces the sink rate. - B) minimum speed is independent of flying mass. - C) both glide ratio and minimum speed are independent of flying mass. - D) only the maximum glide ratio is independent of flying mass, apart from a minor Reynolds number effect.

Correct: D)

Explanation: The correct answer is D because when mass increases, the polar shifts right and down, but the tangent from the origin touches the new curve at the same angle, preserving the maximum L/D ratio (with only negligible Reynolds number effects). A is wrong because increasing mass raises the sink rate at all speeds. B is wrong because minimum speed increases with mass (proportional to the square root of wing loading). C is half-correct on glide ratio but wrong on minimum speed.

Q76: At what indicated speed do you approach an aerodrome located at an altitude of 1800 m AMSL? ^t30q76

• A) At the same speed as at sea level.
• B) At a lower speed than at sea level.
• C) At the minimum sink rate speed.
• D) At a higher speed than at sea level.

Correct: A)

Explanation: The correct answer is A because IAS is a dynamic-pressure measurement that already accounts for density variations. The same IAS at 1800 m produces the same aerodynamic conditions as at sea level. B would reduce safety margins. D is unnecessary and would increase groundspeed. C is for thermal soaring, not circuit approach.

Q77: At what speed must you fly to achieve the best glide ratio for a flying mass of 450 kg? (See attached sheet.)... ^t30q77

] - A) 130km/h - B) 90km/h - C) 70km/h - D) 110km/h

Correct: B)

Explanation: The correct answer is B because the tangent from the origin to the 450 kg polar curve touches at approximately 90 km/h, which gives the maximum glide ratio. A (130 km/h) is in the high-drag region with poor glide ratio. C (70 km/h) is near the minimum sink speed, maximising endurance not distance. D (110 km/h) is between the two and does not correspond to the graphical tangent point.

Q78: The maximum aft CG limit is exceeded. What action must be taken? ^t30q78

• A) Trim aft.
• B) As long as the maximum takeoff mass is not exceeded, no particular action is required.
• C) Redistribute the useful load differently.
• D) Trim forward.

Correct: C)

Explanation: The correct answer is C because the only fix for an aft CG exceedance is to physically move mass forward by redistributing the payload. A (trim aft) would worsen the situation. D (trim forward) adjusts elevator deflection but does not move the actual CG. B is wrong because mass and CG limits are independent -- being within mass limits does not exempt the CG requirement.

Q79: Which factors increase the aerotow takeoff run distance? ^t30q79

• A) Low temperature, headwind.
• B) Grass runway, strong headwind.
• C) High atmospheric pressure.
• D) High temperature, tailwind.

Correct: D)

Explanation: The correct answer is D because high temperature reduces air density (decreasing engine power and lift), while a tailwind requires higher groundspeed to reach lift-off TAS, both lengthening the takeoff roll. A is wrong because low temperature increases density and headwind shortens the run. B is wrong because strong headwind dominates, shortening takeoff despite grass surface. C is wrong because high pressure increases density, improving performance.

Q80: The following NOTAM was published for 18 November. Which of these statements is correct? ^t30q80

] - A) On 18 November, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: Class E airspace, upper limit: max. FL150. - B) On 18 November from 1800 LT to 2100 LT, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. - C) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise with helicopters will take place. - D) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: GND, upper limit: max. 15,000 ft AMSL.

Correct: D)

Explanation: The correct answer is D because the NOTAM specifies times in UTC, identifies the areas as ZUGERSEE, SUSTEN and TICINO, with vertical limits from GND to 15,000 ft AMSL. A incorrectly states FL150 and "Class E" as the lower limit. B uses local time instead of UTC. C introduces "helicopters" as the exclusive aircraft type, which is not stated in the NOTAM.

Q81: What is the maximum permitted flying altitude within the CTR of Bern-Belp airport? ^t30q81

] - A) 5500 ft GND. - B) 4500 ft AMSL. - C) 5000 ft AMSL - D) 3000 ft AMSL.

Correct: D)

Explanation: The correct answer is D because the Bern-Belp CTR has a published upper ceiling of 3000 ft AMSL as shown on the ICAO chart. A (5500 ft GND), B (4500 ft AMSL), and C (5000 ft AMSL) are all above the actual CTR ceiling and would place you in the overlying TMA.

Q82: In which airspace class are you above BEX aerodrome at an altitude of 1700 m AMSL, and what are the minimum visibility and cloud distance requirements? ^t30q82

] - A) Class G airspace, horizontal visibility 1.5 km, clear of cloud with continuous sight of the ground. - B) Class C airspace, horizontal visibility 8 km, cloud clearance 1.5 km horizontally, 300 m vertically. - C) Class C airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically. - D) Class E airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically.

Correct: D)

Explanation: The correct answer is D because above Bex at 1700 m AMSL, the airspace is Class E with standard VFR minima: 5 km visibility, 1500 m horizontal and 300 m vertical cloud clearance. A is wrong because Class G applies only below the Class E floor. B and C are wrong because Class C begins at FL 130 in this area, well above 1700 m AMSL.

Q83: Which is the sink rate at 160 km/h for this glider at a flying mass of 580 kg? (See attached sheet.) ^t30q83

] - A) 1,6m/s - B) 0,8m/s - C) 2,0m/s - D) 1,2m/s

Correct: C)

Explanation: The correct answer is C because reading the 580 kg curve at 160 km/h on the speed polar, the sink rate is approximately 2.0 m/s. At 160 km/h the glider is well beyond best-glide speed, in the region of rapidly increasing drag. B (0.8 m/s) corresponds to the minimum sink rate at much lower speed. A (1.6 m/s) and D (1.2 m/s) are intermediate values that do not match the polar reading.

Q84: 550 kg (rounded) correspond to (1 kg = approx. 2.2 lbs):... ^t30q84

• A) approx. 12,100 lbs.
• B) approx. 1210 lbs.
• C) approx. 2500 lbs.
• D) approx. 250 lbs.

Correct: B)

Explanation: The correct answer is B because 550 kg x 2.2 = 1210 lbs. A (12,100 lbs) results from a decimal error (multiplying by 22 instead of 2.2). C (2500 lbs) implies a conversion factor of about 4.5. D (250 lbs) implies dividing instead of multiplying. The factor 2.2 lbs per kg is the standard aviation approximation.

Q85: At what speed must a glider fly in calm air to cover the maximum possible distance? ^t30q85

• A) At the minimum sink rate speed.
• B) At the maximum allowed speed.
• C) At minimum flying speed.
• D) At the best glide ratio speed.

Correct: D)

Explanation: The correct answer is D because maximum distance in calm air is achieved at the speed giving the best lift-to-drag ratio (best glide ratio), found graphically where a tangent from the origin touches the speed polar. A (minimum sink) maximises endurance, not distance. B (maximum speed) produces very poor glide ratio in the high-drag region. C (minimum speed, near stall) has high induced drag and poor glide ratio.

Q86: The mass of a glider is increased. Which parameter will NOT be affected by this increase? ^t30q86

• A) Maximum glide ratio (apart from a minor Reynolds number effect).
• B) Wing loading.
• C) Sink rate.
• D) Indicated airspeed (IAS).

Correct: A)

Explanation: The correct answer is A because maximum glide ratio is an aerodynamic property of the airfoil and planform that is preserved when mass changes -- the polar shifts right and down but the tangent angle from the origin remains the same. B is wrong because wing loading increases directly with mass. C is wrong because sink rate increases at any given speed. D is wrong because stall speed and all characteristic IAS values increase with the square root of wing loading.

Q87: How long does it take to cover a distance of 150 km at an average ground speed of 100 km/h? ^t30q87

• A) 1 hour 50 minutes.
• B) 1 hour 40 minutes.
• C) 2 hours.
• D) 1 hour 30 minutes.

Correct: D)

Explanation: The correct answer is D because time = 150 km / 100 km/h = 1.5 hours = 1 hour 30 minutes. A (1h50m) would mean 183 km at 100 km/h. B (1h40m) gives 167 km. C (2h) implies only 75 km/h average. A common error is treating 1.5 hours as 1 hour 50 minutes instead of 1 hour 30 minutes.

Q88: When preparing an alpine VFR flight along the route shown on the map below (dotted line) between MUNSTER and AMSTEG, you consult the DABS. You intend to fly this route on a summer weekday between 1445-1515 LT. According to the DABS, zones R-8 and R-8A are active during this period. Answer using the DABS map below and the ICAO aeronautical chart 1:500,000 Switzerland. Which of these answers is correct? ^t30q88

] - A) The route can be flown without restriction after contacting 128.375 MHz. - B) Restricted zones LS-R8 and LS-R8A may be transited below 28,000 ft AMSL. - C) It is not possible to fly this route while the restricted zones are active. - D) Restricted zones LS-R8 and LS-R8A may be overflown at 9200 ft AMSL or above.

Correct: C)

Explanation: The correct answer is C because when LS-R8 and LS-R8A are active, they prohibit all entry without exception, and the zones cover the planned route at relevant altitudes. A is wrong because radio contact does not grant automatic clearance into active restricted zones. B is wrong because the zones extend from low altitude, covering typical VFR flight levels. D is wrong because no published overfly exemption exists at 9200 ft.

Q89: You wish to obtain clearance to transit the ZURICH TMA. What must you do? ^t30q89

• A) First radio contact on frequency 124.7, at least 10 minutes before entering the TMA.
• B) First radio contact on frequency 124.7, at least 5 minutes before entering the TMA.
• C) First radio contact on frequency 118.975, at least 10 minutes before entering the TMA.
• D) First radio contact on frequency 118.1, at least 5 minutes before entering the TMA.

Correct: A)

Explanation: The correct answer is A because Swiss VFR procedures require contacting Zurich Information on 124.7 MHz at least 10 minutes before the TMA boundary. B uses the correct frequency but only 5 minutes lead time, which is insufficient. C uses the correct lead time but the wrong frequency (118.975 is Zurich Approach). D combines an incorrect frequency with insufficient lead time.

Q90: The minimum speed of your glider is 60 kts in straight flight. By what percentage would it increase in a steep turn with a bank angle of 60 deg (load factor n = 2.0)? ^t30q90

• A) approx. 40%.
• B) 0%.
• C) approx. 5%.
• D) approx. 20%.

Correct: A)

Explanation: The correct answer is A because stall speed in a turn increases by the square root of the load factor: Vs_turn = 60 x sqrt(2.0) = 60 x 1.414 = 84.9 kts. The increase is (84.9 - 60) / 60 x 100% = 41%, approximately 40%. B (0%) wrongly implies no effect from load factor. C (5%) and D (20%) underestimate the sqrt(2) factor. Only very shallow banks produce such small increases.