Flight Performance and Planning

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Q1: Exceeding the maximum allowed aircraft mass is ^t30q1

DE · FR

• A) Not allowable and essentially dangerous
• B) Exceptionally allowable to avoid delays
• C) Compensated by the pilot's control inputs.
• D) Only relevant if the excess is more than 10 %.

Answer

A)

Explanation

The correct answer is A because the maximum takeoff mass (MTOM) is a hard certification limit set by the manufacturer based on structural strength, stall speed, and climb performance. Exceeding it increases wing loading, raises the stall speed, reduces climb performance, and may overstress the airframe beyond its certified load factors.

• B is wrong because no operational convenience justifies exceeding a safety limit.
• C is wrong because no pilot technique can compensate for structural overloading.
• D is wrong because there is no regulatory tolerance or percentage margin — any exceedance is prohibited.

Q2: The center of gravity has to be located ^t30q2

DE · FR

• A) Between the front and the rear C.G. limit.
• B) In front of the front C.G. limit.
• C) Right of the lateral C. G. limit.
• D) Behind the rear C.G. limit

Answer

A)

Explanation

The correct answer is A because the aircraft's stability and controllability are only certified within the approved C.G. envelope, which lies between the forward and aft C.G. limits.

• B is wrong because a C.G. ahead of the forward limit requires excessive elevator authority to flare or rotate, potentially making landing impossible.
• D is wrong because a C.G. behind the aft limit causes longitudinal instability and uncontrollable pitch-up.
• C is irrelevant — lateral C.G. limits are not the primary concern in standard mass-and-balance calculations for gliders.

Key Terms

• D — Drag D — Drag

Q3: An aircraft has to be loaded and operated in such a way that the center of gravity (CG) stays within the approved limits during all phases of flight. This is done to ensure ^t30q3

DE · FR

• A) That the aircraft does not stall.
• B) That the aircraft does not exceed the maximum allowable airspeed during a descent
• C) That the aircraft does not tip over on its tail while it is being loaded.
• D) Both stability and controllability of the aircraft.

Answer

D)

Explanation

The correct answer is D because the C.G. position relative to the neutral point determines longitudinal static stability (the tendency to return to equilibrium after a disturbance), while the elevator's ability to command pitch changes provides controllability. Both properties must be maintained throughout flight, and the C.G. envelope ensures this.

• A is wrong because stall speed depends primarily on wing loading and angle of attack, not C.G. position.
• B is wrong because Vne is an airframe limit unrelated to C.G.
• C describes a ground-handling issue, not an in-flight safety requirement.

Key Terms

CG = Centre of Gravity ### Q4: The empty weight and the corresponding center of gravity (CG) of an aircraft are initially determined ^t30q4

DE · FR

• A) For one aircraft of a type solely, since all aircraft of the same type have the same mass and CG position
• B) By calculation.
• C) By weighing.
• D) Through data provided by the aircraft manufacturer.

Answer

C)

Explanation

The correct answer is C because each individual airframe must be physically weighed — typically on calibrated scales at three support points — to determine its actual empty mass and C.G. position. Manufacturing tolerances, repairs, modifications, and installed equipment vary between serial numbers.

• A is wrong because no two aircraft of the same type are guaranteed to have identical mass and C.G.
• B is wrong because calculation alone cannot account for all variables.
• D is wrong because manufacturer data provides type-level reference values, not the specific values for each individual aircraft.

Key Terms

CG = Centre of Gravity ### Q5: Baggage and cargo has to be properly stowed and fastened, otherwise a shift of the cargo may cause ^t30q5

DE · FR

• A) Structural damage, angle of attack stability, velocity stability.
• B) Continuous attitudes which can be corrected by the pilot using the flight controls.
• C) Uncontrollable attitudes, structural damage, risk of injuries.
• D) Calculable instability if the C.G. is shifting by less than 10 %.

Answer

C)

Explanation

The correct answer is C because unsecured cargo can shift suddenly during turbulence or manoeuvres, moving the C.G. outside approved limits instantaneously — faster than a pilot can react. A sudden aft C.G. shift can cause an unrecoverable pitch-up, loose items can become projectiles injuring occupants or jamming controls, and asymmetric loading can overstress the structure.

• A is wrong because the terminology is inaccurate.

• B is wrong because a large sudden C.G. shift may be uncontrollable, not merely "continuous.

• D is wrong because no amount of prior analysis makes unsecured cargo acceptable.

Q6: The total weight of an aeroplane is acting vertically through the ^t30q6

DE · FR

• A) Center of gravity
• B) Stagnation point.
• C) Center of pressure.
• D) Neutral point.

Answer

A)

Explanation

The correct answer is A because the center of gravity is, by definition, the single point through which the resultant gravitational force (the weight vector) acts on the entire aircraft.

• B is wrong because the stagnation point is where airflow velocity reaches zero on the wing's leading edge — an aerodynamic concept unrelated to weight.
• C is wrong because the center of pressure is where the net aerodynamic force acts.
• D is wrong because the neutral point is the aerodynamic reference used for stability analysis.

Q7: The term "center of gravity" is described as ^t30q7

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• A) The heaviest point on an aeroplane.
• B) Half the distance between the neutral point and the datum line.
• C) Another designation for the neutral point.
• D) The distance between the leading edge and the trailing edge of the wing.

Answer

B)

Explanation

The correct answer is B. The center of gravity is the mass-weighted average position of all individual mass elements — the point where the total weight force is considered to act. It is found by summing all moments about the datum and dividing by total mass.

• A is wrong because the C.G. is not a "heaviest point" but a balance point.
• C is wrong because the neutral point is a separate aerodynamic concept relating to stability.
• D duplicates one of the other options and does not correctly define C.G. either.

Q8: The center of gravity (CG) defines ^t30q8

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• A) The point on the longitudinal axis or its extension from which the centers of gravity of all masses are referenced.
• B) The distance from the datum to the centre of gravity of an individual mass.
• C) The product of mass and balance arm
• D) The point through which the force of gravity is said to act on a mass.

Answer

D)

Explanation

The correct answer is D because the C.G. is the point through which the entire gravitational force (weight) acts as if all mass were concentrated there. This is the fundamental definition used in physics and aircraft mass-and-balance. A and B both describe the datum (reference point), not the C.G. itself. - C describes a moment (mass times arm), which is a calculation quantity, not the definition of the center of gravity.

Key Terms

CG = Centre of Gravity ### Q9: The term "moment" with regard to a mass and balance calculation is referred to as ^t30q9

DE · FR

• A) Sum of a mass and a balance arm.
• B) Difference of a mass and a balance arm.
• C) Product of a mass and a balance arm.
• D) Quotient of a mass and a balance arm.

Answer

C)

Explanation

The correct answer is C because in mass and balance, moment equals mass multiplied by balance arm (M = m x d), expressed in units such as kg-m or lb-in. The total C.G. position is then found by dividing the sum of all moments by the total mass.

• A is wrong because adding mass and arm has no physical meaning.
• B is wrong because subtracting them is equally meaningless.
• D is wrong because dividing mass by arm does not produce a moment — it would yield an incorrect dimension.

Q10: The term "balance arm" in the context of a mass and balance calculation defines the ^t30q10

DE · FR

• A) Point through which the force of gravity is said to act on a mass.
• B) Point on the longitudinal axis of an aeroplane or its extension from which the centers of gravity of all masses are referenced.
• C) Distance from the datum to the center of gravity of a mass.
• D) Distance of a mass from the center of gravity

Answer

C)

Explanation

The correct answer is C because the balance arm (or moment arm) is the horizontal distance measured from the aircraft's datum to the center of gravity of a specific mass item. This distance determines the leverage that mass exerts about the datum.

• A is wrong because that defines the center of gravity, not the arm.
• B is wrong because that defines the datum point itself.
• D is wrong because balance arms are measured from the datum, not from the aircraft's overall C.G.

Q11: The distance between the center of gravity and the datum is called ^t30q11

DE · FR

• A) Span width.
• B) Balance arm.
• C) Torque.
• D) Lever.

Answer

B)

Explanation

The correct answer is B because 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.

• A is wrong because span width is a wing geometric parameter.
• C is wrong because torque (or moment) is the product of force and distance, not the distance itself.
• D is wrong because "lever" is a general mechanical term, not the specific aviation mass-and-balance term used.

Key Terms

• D — Drag D — Drag

Q12: The balance arm is the horizontal distance between ^t30q12

DE · FR

• A) The C.G. of a mass and the rear C.G. limit.
• B) The front C.G. limit and the datum line
• C) The C.G. of a mass and the datum line.
• D) The front C.G. limit and the rear C.G. limit.

Answer

C)

Explanation

The correct answer is C because the balance arm of any mass item is measured as the horizontal distance from the aircraft's datum to that item's center of gravity. The datum is a fixed reference point defined in the flight manual.

• A is wrong because it references the rear C.G. limit, not the datum.
• B is wrong because it describes the distance between the forward C.G. limit and the datum.
• D describes the allowable C.G. range, not a balance arm.

Q13: The required data for a mass and balance calculation including masses and balance arms can be found in the ^t30q13

DE · FR

• A) Documentation of the annual inspection.
• B) Certificate of airworthiness
• C) Performance section of the pilot's operating handbook of this particular aircraft.
• D) Mass and balance section of the pilot's operating handbook of this particular aircraft.

Answer

D)

Explanation

The correct answer is D because 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 loading configurations.

• A is wrong because annual inspection documents record maintenance work, not loading data.
• B is wrong because the certificate of airworthiness merely certifies the aircraft type.
• C is wrong because the performance section covers speeds and climb rates, not mass-and-balance data.

Q14: Which section of the flight manual describes the basic empty mass of an aircraft? ^t30q14

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• A) Normal procedures
• B) Performance
• C) Weight and balance
• D) Limitations

Answer

C)

Explanation

The correct answer is C because the Weight and Balance section of the flight manual contains the basic empty mass, empty C.G. location, allowable C.G. range, and loading instructions.

• A is wrong because Normal Procedures covers checklists and operational sequences.
• B is wrong because Performance covers speeds, climb rates, and glide distances.
• D is wrong because Limitations covers maximum speeds, load factors, and the operating envelope — not the basic empty mass data.

Q15: Which factor shortens landing distance? ^t30q15

DE · FR

• A) High pressure altitude
• B) Strong head wind
• C) Heavy rain
• D) High density altitude

Answer

B)

Explanation

The correct answer is B because a headwind reduces the groundspeed at touchdown for a given indicated airspeed, so the aircraft crosses the threshold with less kinetic energy relative to the ground, shortening the ground roll significantly.

• A is wrong because high pressure altitude means lower air density, higher true airspeed at the same IAS, and therefore longer landing distance.
• C is wrong because heavy rain can degrade braking effectiveness and contaminate the wing surface.
• D is wrong for the same reason as A — high density altitude increases groundspeed and lengthens the landing roll.

Key Terms

IAS = Indicated Airspeed ### Q16: Unless the aircraft is equipped and certified accordingly ^t30q16

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• A) Flight into forecast icing conditions is prohibited. Should the aircraft enter an area of icing conditions inadvertantly, the flight may be continued as long as visual meteorological conditions are maintained.
• B) Flight into known or forecast icing conditions is prohibited. Should the aircraft enter an area of icing conditions inadvertantly, it should be left without delay.
• C) Flight into known or forecast icing conditions is only allowed as long as it is ensured that the aircraft can still be operated without performance degradation.
• D) Flight into areas of precipitation is prohibited.

Answer

B)

Explanation

The correct answer is B because for non-FIKI certified aircraft, flying into known or forecast icing is a regulatory prohibition. If icing is inadvertently encountered, the pilot must exit immediately by changing altitude or heading.

• A is wrong because maintaining VMC does not make icing safe — ice accumulates regardless of visual conditions.
• C is wrong because it implies icing flight is permissible with performance monitoring, which is not the case.
• D is wrong because not all precipitation involves icing conditions.

Key Terms

VMC = Visual Meteorological Conditions ### Q17: The angle of descent is described as ^t30q17

DE · FR

• A) The ratio between the change in height and the horizontal distance distance travelled within the same time, expressed in degrees [°].
• B) The angle between a horizontal plane and the actual flight path, expressed in degrees [°].
• C) The ratio between the change in height and the horizontal distance travelled within the same time, expressed in percent [%].
• D) The angle between a horizontal plane and the actual flight path, expressed in percent [%].

Answer

B)

Explanation

The correct answer is B because the angle of descent (glide angle) is geometrically defined as the angle between the horizontal and the flight path vector, measured in degrees.

• A is wrong because a "ratio expressed in degrees" is contradictory — a ratio is dimensionless or expressed as a percentage, not in degrees.
• C describes a gradient (percentage), not an angle.
• D incorrectly expresses an angle in percent.
• For a glider with a 1:30 glide ratio, the glide angle is approximately 1.9 degrees.

Q18: Which is the purpose of "interception lines" in visual navigation? ^t30q18

DE · FR

• A) They help to continue the flight when flight visibility drops below VFR minima
• B) To visualize the range limitation from the departure aerodrome
• C) To mark the next available en-route airport during the flight
• D) They are used as easily recognizable guidance upon a possible loss of orientation

Answer

D)

Explanation

The correct answer is D because interception lines (also called catching lines) are prominent linear ground features — rivers, motorways, railways, coastlines — selected during pre-flight planning that the pilot can navigate toward if orientation is lost. Flying to the nearest interception line provides an unmistakable landmark for position recovery.

• A is wrong because nothing permits continuing flight below VFR minima.
• B is wrong because interception lines are not range indicators.
• C is wrong because they are geographic features, not airport markers.

Key Terms

VFR = Visual Flight Rules ### Q19: The upper limit of LO R 16 equals ^t30q19

DE · FR


• A) 1 500 m MSL.
• B) FL150.
• C) 1.500 ft GND.
• D) 1 500 ft MSL.

Answer

D)

Explanation

The correct answer is D because low-level restricted areas (LO R) on VFR charts typically express their vertical limits in feet MSL (above mean sea level). The value 1,500 ft MSL is a fixed, absolute altitude reference.

• A is wrong because 1,500 metres MSL would be approximately 4,900 ft — a different altitude entirely.
• B is wrong because FL150 (15,000 ft pressure altitude) is far too high for a typical low-level restriction.
• C is wrong because 1,500 ft GND (above ground level) would vary with terrain and is not the published limit.

Key Terms

• FL = Flight Level
• MSL = Mean Sea Level
• VFR = Visual Flight Rules ### Q20: The upper limit of LO R 4 equals ^t30q20

DE · FR


• A) 4.500 ft MSL
• B) 1.500 ft AGL
• C) 4.500 ft AGL.
• D) 1.500 ft MSL.

Answer

A)

Explanation

The correct answer is A because LO R 4 has its upper limit published at 4,500 ft MSL — a fixed altitude above mean sea level.

• B is wrong because 1,500 ft AGL references above ground level, which varies with terrain.
• C is wrong because 4,500 ft AGL would not be a fixed boundary.
• D is wrong because 1,500 ft MSL is too low and does not match the chart data for this particular restricted area.

Key Terms

• AGL = Above Ground Level
• MSL = Mean Sea Level ### Q21: Up to which altitude is an overflight prohibited according to the NOTAM? ^t30q21

DE · FR


• A) Flight Level 95
• B) Height 9500 ft
• C) Altitude 9500 ft MSL
• D) Altitude 9500 m MSL

Answer

C)

Explanation

The correct answer is C because NOTAM altitude references follow ICAO conventions where "altitude" refers to height above MSL. The NOTAM prohibits overflight up to 9,500 ft MSL.

• A is wrong because FL 95 is a pressure altitude reference (based on 1013.25 hPa), not the same as an MSL altitude.
• B is wrong because "height" implies above ground level (AGL).
• D is wrong because 9,500 m MSL would be approximately 31,000 ft — clearly inconsistent with a typical VFR restriction.

Key Terms

• AGL = Above Ground Level
• FL = Flight Level
• ICAO = International Civil Aviation Organization
• MSL = Mean Sea Level
• NOTAM = Notice to Air Missions
• VFR = Visual Flight Rules ### Q22: What must be considered for cross-border flights? ^t30q22

DE · FR

• A) Transmission of hazard reports
• B) Approved exceptions
• C) Requires flight plans
• D) Regular location messages

Answer

C)

Explanation

The correct answer is C because under ICAO Annex 2 and national regulations, a flight plan is mandatory for any international flight crossing state borders, even for VFR glider flights. This ensures coordination for border control, search and rescue alerting, and customs/immigration procedures.

• A is wrong because hazard reports (PIREPs) are a separate communication procedure.
• B is wrong because approved exceptions is too vague and not the primary requirement.
• D is wrong because regular position reports are separate from the flight plan requirement.

Key Terms

• ICAO = International Civil Aviation Organization
• VFR = Visual Flight Rules ### Q23: During a flight, a flight plan can be filed at the ^t30q23

DE · FR

• A) Next airport operator en-route.
• B) Flight Information Service (FIS).
• C) Aeronautical Information Service (AIS)
• D) Search and Rescue Service (SAR).

Answer

B)

Explanation

The correct answer is B because the Flight Information Service (FIS), reached on the published FIS frequency, can accept an airborne flight plan (AFIL) during flight. This is the standard procedure for filing when airborne.

• A is wrong because airport operators handle local ground operations, not en-route plan filing.
• C is wrong because AIS distributes aeronautical publications but does not accept real-time flight plans.
• D is wrong because SAR is a response service activated when an aircraft is overdue or in distress.

Q24: While planning a cross country gliding flight, what ground structure ought to be avoided enroute? ^t30q24

DE · FR

• A) Stone quarries and large sand areas
• B) Moist ground, water areas, marsh areas
• C) Highways, railroad tracks and channels.
• D) Areas with buildings, concrete and asphalt.

Answer

B)

Explanation

The correct answer is B because moist ground, water bodies, and marshes have high thermal inertia and specific heat capacity — they absorb solar radiation without heating quickly, suppressing thermal development above them. Flying over these areas means less lift and potentially a forced landing in unsuitable terrain.

• A is wrong because stone quarries and sandy areas heat up well and often produce good thermals.
• C is wrong because linear features like highways and railways are useful navigation aids.
• D is wrong because built-up areas with dark surfaces (asphalt, concrete) generate strong thermals.

Q25: During a cross-country flight, you approach a downwind turning point. The point ought to be taken ... (2,00 P.) ^t30q25

DE · FR

• A) As high as possible.
• B) With as less bank as possible
• C) As low as possible.
• D) As steep as possible.

Answer

A)

Explanation

The correct answer is A because at a downwind turning point, the glider must reverse direction and fly back into the wind. This immediately reduces groundspeed and shortens the achievable glide distance over the ground. Arriving high provides maximum altitude reserve for the subsequent upwind leg.

• B is wrong because bank angle is a secondary concern compared to altitude.
• C is wrong because arriving low with a turn ahead and headwind return is tactically dangerous.
• D is wrong because steep turns lose more altitude, compounding the problem.

Q26: After getting around a turning point, what should a glider pilot be prepared for? (2,00 P.) ^t30q26

DE · FR

• A) For weakening thermals due to the progressing time
• B) For a changed horizontal picture due to lower cloud bases
• C) For increased cloud dissipation due to the progressing time
• D) For a changed cloud picture due to the apparently changed position of the sun

Answer

D)

Explanation

The correct answer is D because when a glider turns 90 or 180 degrees at a waypoint, the pilot's entire visual perspective of the sky shifts dramatically. The sun appears to have moved relative to the heading, and cumulus clouds that were behind or beside the aircraft now appear in different positions. This perceptual shift can make the sky look completely different.

• A is wrong because thermal weakening is a time-of-day issue, not a turning-point issue.
• B is wrong because cloud bases do not change from turning.
• C is wrong because cloud dissipation is unrelated to heading changes.

Q27: According ICAO, what symbol indicates a group of unlighted obstacles? ^t30q27

DE · FR

• A) Single lighted obstacle
• B) Single unlighted obstacle
• C) Group of lighted obstacles
• D) Group of unlighted obstacles

Answer

D)

Explanation

The correct answer is D because ICAO Annex 4 chart symbology uses distinct symbols to differentiate between single obstacles versus groups, and lighted versus unlighted. The symbol for a group of unlighted obstacles is shown as D in the figure — two filled circles side by side with no light rays. Knowing these symbols is critical for cross-country planning and obstacle avoidance.

• Option A — represents a single lighted obstacle (filled circle with light rays).
• Option B — represents a single unlighted obstacle (filled circle only).
• Option C — represents a group of lighted obstacles (two filled circles with light rays).

Key Terms

ICAO = International Civil Aviation Organization ### Q28: According ICAO, what symbol indicates a civil airport (not international airport) with paved runway? ^t30q28

DE · FR

• A) Civil airport, paved runway
• B) Military airport, paved runway
• C) Civil airport, unpaved runway
• D) Heliport

Answer

A)

Explanation

The correct answer is A because ICAO aeronautical chart symbology differentiates airports by civil versus military status and runway surface type. A civil airport with a paved runway is shown as symbol A in the figure — a circle with a solid filled runway bar through the centre. Glider pilots use these symbols when planning outlanding fields or alternate airports.

• Option B — represents a military airport with paved runway (circle with runway bar and crossbar).
• Option C — represents a civil airport with unpaved runway (circle with open/outlined runway bar).
• Option D — represents a heliport (square with H).

Key Terms

ICAO = International Civil Aviation Organization ### Q29: According ICAO, what symbol indicates a general spot elevation? ^t30q29

DE · FR

• A) General spot elevation
• B) Highest spot elevation on chart
• C) Mountain peak / summit
• D) Trigonometric point

Answer

A)

Explanation

The correct answer is A because ICAO charts use specific symbols to differentiate between general spot elevations, the highest elevation on a chart, mountain peaks, and trigonometric points. A general spot elevation is shown as symbol A — a small dot with a plain elevation number beside it. Familiarity with these symbols is essential for terrain clearance planning.

• Option B — represents the highest spot elevation on the chart (larger bold dot with bold underlined number).
• Option C — represents a mountain peak or summit (filled triangle with elevation number).
• Option D — represents a trigonometric point (open triangle with centre dot and elevation number).

Key Terms

ICAO = International Civil Aviation Organization ### Q30: What distance can be covered during a glide in a glider plane with glide ratio 1/30 from a height of 1500 m? (Neglect wind and thermal effects) ^t30q30

DE · FR

• A) 45 NM
• B) 30 km
• C) 45 km
• D) 81 NM

Answer

C)

Explanation

The correct answer is C because glide distance equals glide ratio multiplied by height: 30 x 1,500 m = 45,000 m = 45 km. The glide ratio of 1:30 means the glider covers 30 metres horizontally for every 1 metre of height lost.

• A is wrong because 45 NM equals approximately 83 km, which would require a glide ratio of about 1:55.
• B is wrong because 30 km would correspond to a glide ratio of only 1:20.
• D is wrong because 81 NM (150 km) would require a glide ratio of 1:100.
• Always verify that units are consistent — mixing nautical miles and metres is a common exam trap.

Key Terms

NM = Nautical Mile(s) ### Q31: Why can wing loading be increased when soaring conditions are good? ^t30q31

DE · FR

• A) Because the stall speed diminishes.
• B) Because the glider achieves a better glide ratio at high speed even though the minimum speed rises.
• C) Because the glider can fly more slowly and achieves a better glide ratio.
• D) Because the glider has a better climb rate even though it must fly more slowly.

Answer

B)

Explanation

The correct answer is B because in strong thermal conditions, the glider benefits from flying faster between thermals (MacCready theory). Adding water ballast increases wing loading, which shifts the speed polar to the right — improving the glide ratio at high cruising speeds while accepting a higher stall and minimum sink speed.

• A is wrong because increasing wing loading raises the stall speed.
• C is wrong because higher wing loading means the glider must fly faster, not slower.
• D is wrong because a heavier glider has a worse climb rate in thermals due to its higher minimum sink speed.

Q32: The tail wheel of a glider was not removed before departure. What will be the consequence? ^t30q32

DE · FR

• A) Better manoeuvrability at departure.
• B) The centre of gravity shifts forward.
• C) No consequence. The wheel represents only a tiny fraction of the total weight of the glider and has no effect on the centre of gravity.
• D) The centre of gravity will be further aft and possibly too far aft, which is dangerous.

Answer

D)

Explanation

The correct answer is D because the tail wheel is mounted at the extreme rear of the fuselage, far aft of the nominal C.G. Even though its absolute mass is small, its large moment arm produces a significant moment that shifts the C.G. aftward — potentially beyond the aft limit, making the aircraft pitch-unstable and difficult to control.

• A is wrong because the tail wheel does not improve manoeuvrability.
• B is wrong because the tail wheel is aft of the C.G., so its presence shifts the C.G. backward, not forward.
• C is wrong because the long arm amplifies the effect of even a small mass.

Q33: The pilot exceeds the maximum cockpit payload by 10 kg. What has to be done? ^t30q33

DE · FR

• A) Trim aft.
• B) Trim forward.
• C) Reduce the payload.
• D) Compensate by reducing the water ballast slightly.

Answer

C)

Explanation

The correct answer is C because the maximum seat load is a certification limit that cannot be circumvented. Exceeding it may place the C.G. outside the forward limit and subjects the structure to loads beyond what was tested. The only remedy is to reduce the payload until the limits are respected. A and B are wrong because trimming changes the aerodynamic forces on the elevator but does not alter the aircraft's mass or C.G. position.

• D is wrong because reducing water ballast changes total mass but does not address the specific seat load limitation.

Q34: What propels a pure glider forward? ^t30q34

DE · FR

• A) Ascending air currents.
• B) Drag directed forward.
• C) The component of gravity acting in the direction of the flight path.
• D) A tailwind.

Answer

C)

Explanation

The correct answer is C because in steady gliding flight, the weight vector can be resolved into two components: one perpendicular to the flight path (balanced by lift) and one along the flight path. This along-path component of gravity provides the forward-driving force that balances drag and maintains airspeed.

• A is wrong because ascending air can reduce the descent rate but does not propel the glider forward through the air.
• B is wrong because drag always opposes the direction of motion.
• D is wrong because a tailwind affects groundspeed but does not propel the aircraft through the airmass.

Q35: The current mass of an aircraft is 610 kg and the centre of gravity (C.G.) position is at 80.0. You remove a 10 kg item of baggage located at a moment arm of 150. Which is the new centre of gravity? ^t30q35

DE · FR

• A) 75.0
• B) 81.166
• C) 70.0
• D) 78.833

Answer

D)

Explanation

The correct answer is D. The calculation proceeds as follows: 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. (arm 150 > 80), removing it shifts the C.G. forward — consistent with the result (78.833 < 80.0).

• Option A (75.0) and C (70.0) are too far forward.
• Option B (81.166) incorrectly shows a rearward shift.

Q36: The empty mass of the Discus B is 245 kg. You are planning to carry 184 kg of water ballast. What is the maximum load at the pilot's seat? ^t30q36

DE · FR

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 (column 90 kg → value 184, but column 100 kg → 180 and column 110 kg → 170; with ballast=184 required, read the 245 kg row and find the seat load corresponding to ballast=184, i.e. max 90 kg permitted according to the table).*

• A) 100 kg
• B) 110 kg
• C) 90 kg
• D) 80 kg

Answer

C)

Explanation

The correct answer is C (90 kg). Reading the Discus B loading table at the row for empty mass 245 kg: with a seat load of 90 kg the permitted water ballast is 184 kg (matching our requirement), but at 100 kg seat load only 180 kg of ballast is permitted, and at 110 kg only 170 kg. Since we need the full 184 kg of ballast, the maximum seat load that still allows this is 90 kg.

• Option A (100 kg) and B (110 kg) would require reducing the water ballast below 184 kg.
• Option D (80 kg) is unnecessarily restrictive — the table shows 184 kg is still permitted at 90 kg.

Q37: What important principle must be observed when making an off-field landing on sloping terrain? ^t30q37

DE · FR

• A) Only land with airbrakes fully extended.
• B) Land facing uphill with an approach speed slightly above normal.
• C) Always land into wind regardless of the slope.
• D) The landing flare must be initiated at a greater height than usual.

Answer

B)

Explanation

The correct answer is B because landing uphill uses the slope to decelerate the glider — gravity assists braking, dramatically shortening the ground roll. A slightly higher approach speed provides a safety margin against wind shear and turbulence near unfamiliar terrain.

• A is wrong because full airbrakes may not always be appropriate on short or steep fields.
• C is wrong because on significant slopes, landing uphill takes priority over landing into wind.
• D is wrong because the flare height should be adapted to the terrain, but this is not the primary principle.

Q38: You must land in heavy rain. What must you pay particular attention to? ^t30q38

DE · FR

• A) The approach speed is lower than usual because rain slows the aircraft.
• B) The landing is performed as in dry conditions.
• C) Due to poor visibility, the approach angle must be shallower than usual.
• D) A higher approach speed must be used.

Answer

D)

Explanation

The correct answer is D because heavy rain on the wing surface degrades the aerodynamic profile through increased roughness, potentially raising the stall speed. A higher approach speed provides an adequate safety margin.

• A is wrong because rain does not lower the safe approach speed — if anything, the stall speed increases.
• B is wrong because rain significantly changes conditions (reduced visibility, wet surfaces, degraded aerodynamics).
• C is wrong because a shallower approach reduces obstacle clearance margins and extends the final approach in poor visibility.

Q39: You are taking off from a grass runway that has become waterlogged after several days of rain. What should you expect? ^t30q39

DE · FR

• A) The takeoff distance is likely to be longer.
• B) The glider is wet and has reduced performance.
• C) The wet grass offers less resistance, which is why the takeoff distance will be shorter.
• D) The glider may skid sideways (aquaplaning).

Answer

A)

Explanation

The correct answer is A because a waterlogged grass runway creates greater rolling resistance due to soft ground deformation and water drag on the wheels, slowing acceleration and increasing the takeoff distance.

• B is wrong because while a wet glider has slightly degraded performance, the primary issue is the runway condition.
• C is wrong because wet, soft grass increases resistance rather than reducing it.
• D is wrong because aquaplaning occurs on hard surfaces with standing water, not on soft grass — and the question asks about takeoff distance, not directional control.

Q40: Which of these statements is correct at a speed of 170 km/h, taking into account the following speed polar? ^t30q40

DE · FR

ASK 21 Speed Polar:

]

Two curves: G=470 kp (light mass, min sink rate ~0.657 m/s at ~75 km/h) and G=570 kp (heavy mass, min sink rate ~0.724 m/s). The best glide ratio is read from the tangent from the origin. At 170 km/h, the sink rate is higher for G=570 kp than for G=470 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.

Answer

C)

Explanation

The correct answer is C because at 170 km/h, reading both polar curves, the heavier configuration (570 kp) shows a higher sink rate than the lighter one (470 kp). A heavier glider requires more lift to maintain flight, producing greater induced drag and therefore a higher sink rate at any given speed.

• A is wrong because the two curves clearly show different sink rates at 170 km/h.
• B and C state the same thing — sink rate increases with mass — which is correct.
• D is wrong because at high speeds the glide angle is not necessarily better at lower mass.

Q41: Which is the speed at the minimum sink rate in still air for a mass of 450 kg? ^t30q41

DE · FR

Speed Polar (AIRSPEED):

]

Two curves: 450 kg and 580 kg. The minimum sink rate (top of the curve) for 450 kg is at approximately 75 km/h. The 580 kg curve is shifted to the right (higher speeds) and downward (greater sink rate).

• A) 75 km/h
• B) 95 km/h
• C) 50 km/h
• D) 140 km/h

Answer

A)

Explanation

The correct answer is A because the minimum sink rate speed corresponds to the highest point on the speed polar curve — where the sink rate is smallest. For 450 kg, this peak occurs at approximately 75 km/h. This speed maximises flight endurance in still air and is optimal for centring thermals.

• Option B (95 km/h) would be closer to the best-glide speed or the minimum-sink speed at higher mass.
• Option C (50 km/h) is below the stall speed.
• Option D (140 km/h) is far into the high-speed range where sink rate is much greater.

Q42: From what altitude on the route between Murten (approx. N46°56'/E007°07') and Neuchâtel aerodrome (approx. N46°57'/E006°52') are you required to request permission to cross the PAYERNE TMA? ^t30q42

DE · FR

• A) 950 m AMSL (3100 ft).
• B) 3050 m AMSL (FL 100).
• C) 700 m AMSL (2300 ft).
• D) At any altitude since the lower limit of the TMA is represented by the ground surface (GND).

Answer

C)

Explanation

The correct answer is C because on the route between Murten and Neuchatel, the relevant sector of the PAYERNE TMA has a lower limit at 700 m AMSL (2300 ft). Below this altitude, flight can proceed in uncontrolled airspace without clearance. Above 700 m AMSL, ATC authorisation is required. A (950 m) does not match the published boundary. B (FL 100) is far too high — that is the upper limit of some TMAs, not the lower limit here.

• D is wrong because the TMA does not extend to the ground in this sector.

Key Terms

• AMSL = Above Mean Sea Level
• ATC = Air Traffic Control
• FL = Flight Level
• TMA = Terminal Manoeuvring Area ### Q43: In which airspace class are you flying at 1400 m AMSL (QNH 1013 hPa) over Birrfeld aerodrome (47°25'36"N/007°14'02"E), and what are the visibility and cloud distance minima in that airspace? ^t30q43

DE · FR

• A) Airspace class E, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.
• B) Airspace class D, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.
• C) Airspace class G, horizontal visibility 1.5 km, clear of cloud with permanent ground contact.
• D) Airspace class C, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.

Answer

A)

Explanation

The correct answer is A because at 1400 m AMSL over Birrfeld, you are in Class E airspace. VFR minima in Class E require 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance.

• B is wrong because Class D applies within specific CTRs or TMAs, not over Birrfeld at this altitude.
• C is wrong because Class G applies below a certain altitude and has reduced minima.
• D is wrong because Class C begins at a higher altitude in this area (typically FL 130 in Switzerland).

Key Terms

• AMSL = Above Mean Sea Level
• FL = Flight Level
• QNH = Pressure adjusted to mean sea level
• VFR = Visual Flight Rules ### Q44: The route shown below towards SCHWYZ (dotted line) is planned for 20 June 2015 (summer time) between 1515–1545 LT at 6500 ft AMSL. Which of the following statements is correct? ^t30q44

DE · FR

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 |

• 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.

Answer

D)

Explanation

The correct answer is D. On 20 June 2015 (CEST = UTC+2), the planned time of 1515-1545 LT corresponds to 1315-1345 UTC. Zone W0912/15 (MORGARTEN) was active 1145-1300 UTC and has already expired. Zone W0957/15 (HINWIL) activates at 1400 UTC (1600 LT) — it is not yet active. The route can therefore be flown without coordination between 1500 and 1600 LT.

• A is wrong because the route is flyable during the given time window.
• B is wrong because the DABS applies to all airspace users including gliders.
• C is wrong because it incorrectly suggests blanket altitude-based exemptions.

Key Terms

• AGL = Above Ground Level
• AMSL = Above Mean Sea Level
• FL = Flight Level
• NM = Nautical Mile(s)
• NOTAM = Notice to Air Missions
• TMA = Terminal Manoeuvring Area ### Q45: According to the ICAO aeronautical chart at 1:500,000, at what altitude over Schwyz (approx. 47°01' N, 8°39' E) must you request permission to enter Class C airspace? ^t30q45

DE · FR

• A) FL 90
• B) 4500 ft
• C) FL 130
• D) FL 195

Answer

C)

Explanation

The correct answer is C because over Schwyz, the Swiss ICAO 1:500,000 chart shows Class C airspace beginning at FL 130. Below FL 130, the airspace is Class E. Entering Class C requires ATC clearance regardless of flight rules.

• Option A (FL 90) is below the actual boundary.
• Option B (4500 ft) is far too low and in uncontrolled airspace.
• Option D (FL 195) is the upper limit of Swiss controlled airspace, not the lower limit of Class C over Schwyz.

Key Terms

• ATC = Air Traffic Control
• FL = Flight Level
• ICAO = International Civil Aviation Organization ### Q46: Until what time is La Côte aerodrome (LSGP) open in the evening? ^t30q46

DE · FR

AD INFO 1 — LA CÔTE / LSGP

]

| Data | Value | |--------|--------| | ICAO | LSGP | | Elevation | 1352 ft (412 m) | | ARP | 46°24'23"N / 006°15'28"E | | Runway | 04 / 22 — true/mag: 041°/040° and 221°/220° | | Dimensions | 560 x 30 m — GRASS | | LDG distance available | 490 m | | TKOF distance available | 490 m | | SFC strength | 0.25 MPa | | Status | Private — Airfield, PPR | | Location | 25 km NE Geneva | | Hours MON–FRI | 0700–1200 LT / 1400–ECT –30 min | | Hours SAT/SUN | 0800–1200 LT / 1400–ECT –30 min | | ECT reference | → VFG RAC 1-1 |

ECT = End of Civil Twilight. The aerodrome closes 30 minutes before end of civil twilight.

• 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.

Answer

C)

Explanation

The correct answer is C because the AD INFO sheet for LSGP shows afternoon hours as "1400-ECT -30 min," meaning the aerodrome closes 30 minutes before the end of civil twilight.

• A is wrong because it references the start of civil twilight, not the end.
• B is wrong because sunset occurs earlier than the end of civil twilight.
• D is wrong because the aerodrome closes 30 minutes before ECT, not at ECT itself.

Key Terms

• ICAO = International Civil Aviation Organization
• ECT = End of Civil Twilight
• HRH = Heure de Référence Horaire (Swiss AD INFO term for ECT, see VFG RAC 1-1)
• PPR = Prior Permission Required
• ARP = Aerodrome Reference Point
• LDG = Landing
• TKOF = Takeoff
• SFC = Surface
• LT = Local Time ### Q47: On which frequency do you receive information about winch launches at Gruyères aerodrome (LSGT) at weekends? ^t30q47

DE · FR

Visual Approach Chart — GRUYÈRES / LSGT

]

AD 124.675 — PPR — ELEV 2257 ft (688 m)

Key chart data (altitudes in ft, magnetic headings):

| Data | Value | |--------|--------| | ICAO | LSGT | | AD Frequency | 124.675 MHz | | Elevation | 2257 ft (688 m) | | Status | PPR | | Minimum AD overfly altitude (MNM ALT) | 4000 ft | | Glider ARR/DEP sector W (GLD ARR/DEP W) | MAX 3100 ft | | Glider ARR/DEP sector E (GLD ARR/DEP E) | MAX 3600 ft | | HEL ARR/DEP | 3000 ft | | Preferred ARR sectors | WEST and EAST | | CTN (cross-country traffic) | 3000 ft | | MNM AD overfly | 4000 ft | | Class C airspace above | FL 100 / 119.175 GENEVA DELTA | | Winch launches | Intensive SAT/SUN (CTN: Intense winch launching SAT/SUN) | | Nearby VOR/DME | SPR R076, 113.9 MHz |

Noise-sensitive areas (yellow) around Bulle/Broc. Avoid overflying the field during PJE (parachute dropping). Contact RTF 5 min before ETA.

• A) 113.9
• B) 124.675
• C) 119.175
• D) 110.85

Answer

B)

Explanation

The correct answer is B (124.675 MHz) because this is the aerodrome frequency shown on the Visual Approach Chart for LSGT Gruyeres. Local traffic information, including intensive winch launching activity on weekends, is broadcast on this frequency.

• Option A (113.9) is the VOR/DME SPR navigation frequency.
• Option C (119.175) is the Geneva Delta sector frequency for Class C airspace above.
• Option D (110.85) is not shown on this chart and does not relate to LSGT operations.

Key Terms

• ETA = Estimated Time of Arrival
• FL = Flight Level
• ICAO = International Civil Aviation Organization ### Q48: What distance do you cover in 90 minutes at a ground speed of 90 km/h? ^t30q48

DE · FR

• A) 90 km
• B) 135 km
• C) 100 km
• D) 120 km

Answer

B)

Explanation

The correct answer is B because distance = speed x time. Ground speed = 90 km/h, time = 90 minutes = 1.5 hours. Distance = 90 x 1.5 = 135 km. Remember to convert minutes to hours before multiplying: 90 minutes = 1.5 hours, not 0.9 hours.

• Option A (90 km) results from incorrectly using 1 hour instead of 1.5 hours.
• Option C (100 km) and D (120 km) do not correspond to any correct calculation.

Q49: At an altitude of 6000 m, the airspeed indicator shows 160 km/h (IAS). The true airspeed (TAS) ^t30q49

DE · FR

• A) is lower than the IAS.
• B) is also 160 km/h.
• C) can be higher or lower than the IAS depending on atmospheric pressure and temperature.
• D) is higher than the IAS.

Answer

D)

Explanation

The correct answer is D because the airspeed indicator measures dynamic pressure, which depends on air density. At 6000 m, air density is significantly lower than at sea level. For the pitot tube to register the same dynamic pressure (same IAS), the aircraft must be moving faster through the thinner air. TAS increases by approximately 2% per 300 m of altitude gain, so at 6000 m, TAS is roughly 40% higher than IAS.

• A is wrong because TAS is always higher than IAS at altitude.
• B is wrong because they only equal each other at sea level in ISA conditions.
• C is wrong because at any altitude above sea level, TAS is always higher than IAS.

Key Terms

• IAS = Indicated Airspeed
• ISA = International Standard Atmosphere
• TAS = True Airspeed ### Q50: You are flying in wave lift at 6000 m altitude. Which is the maximum speed you may fly? ^t30q50

DE · FR

• A) In the low-density air, at a higher speed than usual.
• B) Below the red V_NE mark on the airspeed indicator, according to the speed-altitude table displayed on the instrument panel.
• C) At the same speed as at sea level since V_NE is an absolute value.
• D) Maximum within the green arc.

Answer

B)

Explanation

The correct answer is B because at high altitude the true airspeed corresponding to a given IAS is much higher, and it is the TAS that determines aerodynamic loads on the structure. Glider flight manuals provide a speed-altitude table (or VNE reduction curve) displayed in the cockpit, giving the corrected maximum IAS at each altitude. At 6000 m, the allowable IAS is lower than the sea-level VNE mark.

• A is wrong because you must fly slower (lower IAS), not faster.
• C is wrong because V_NE as indicated must be reduced with altitude.
• D is wrong because the green arc alone does not account for altitude corrections.

Key Terms

• IAS = Indicated Airspeed
• TAS = True Airspeed ### Q51: 1235 lbs (rounded) correspond to (1 kg = approx. 2.2 lbs): ^t30q51

DE · FR

• A) approx. 620 kg.
• B) approx. 2720 kg.
• C) approx. 560 kg.
• D) approx. 2470 kg.

Answer

C)

Explanation

The correct answer is C because to convert pounds to kilograms, divide by 2.2: 1235 / 2.2 = 561.4 kg, which rounds to approximately 560 kg. The key formula is: mass in kg = weight in lbs / 2.2.

• Option A (620 kg) would correspond to about 1364 lbs.
• Option B (2720 kg) results from multiplying instead of dividing.
• Option D (2470 kg) is also the result of a multiplication error.

Q52: What has to be particularly observed when landing on an upsloping field with a tailwind? ^t30q52

DE · FR

• A) Fly final a little faster than usual.
• B) Flare higher than usual.
• C) Fly at the normal approach speed (yellow triangle).
• D) You must land with all airbrakes fully extended.

Answer

C)

Explanation

The correct answer is C because on an upsloping field with a tailwind, the competing effects partially cancel each other: the upslope shortens the ground roll while the tailwind lengthens it. The normal approach speed (yellow triangle on the ASI) provides the correct balance of energy management.

• A is wrong because a faster approach would result in excessive float on the upslope.
• B is wrong because flaring higher risks ballooning on the slope.
• D is wrong because full airbrakes may cause an excessively steep descent on short final.

Q53: In which airspace class are you above Langenthal aerodrome (47 deg 10'58''N / 007 deg 44'29''E) at an altitude of 2000 m AMSL (QNH 1013 hPa), and what are the minimum visibility and cloud distance requirements? ^t30q53

DE · FR

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

Answer

A)

Explanation

The correct answer is A because at 2000 m AMSL above Langenthal, you are in Class E airspace. VFR flight in Class E requires 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance.

• B is wrong because Class G with its reduced minima applies only at very low altitudes.
• C is wrong because there is no Class D TMA at this location and altitude.
• D is wrong because Class C begins at FL 130 in this region, far above 2000 m AMSL.

Key Terms

• AMSL = Above Mean Sea Level
• FL = Flight Level
• QNH = Pressure adjusted to mean sea level
• TMA = Terminal Manoeuvring Area
• VFR = Visual Flight Rules ### Q54: Which center of gravity position is the most dangerous for a glider? ^t30q54

DE · FR

• A) Too far forward.
• B) Too low.
• C) Too far aft.
• D) Too high.

Answer

C)

Explanation

The correct answer is C because when the C.G. is too far aft, the glider loses longitudinal static stability — the nose tends to pitch up without returning to equilibrium, potentially leading to uncontrollable divergent oscillations or a stall/spin.

• Option A (too far forward) is less dangerous because the aircraft remains stable, though elevator authority may be insufficient for landing.
• Option B and D are wrong because vertical C.G. displacement is not the primary concern in standard glider mass-and-balance analysis.

Q55: How does the indicated VNE (never-exceed speed) change as altitude increases? ^t30q55

DE · FR

• A) It rises.
• B) It decreases.
• C) It stays the same; the airspeed indicator accounts for this automatically.
• D) It diminishes.

Answer

C)

Explanation

The correct answer is C because the airspeed indicator measures dynamic pressure, which inherently accounts for air density. The V_NE marking on the ASI (red line) represents a fixed IAS value that corresponds to the structural limit. However, note that the allowable maximum IAS must actually be reduced at high altitude per the flight manual's speed-altitude table — the ASI marking itself does not change, but the pilot must observe a lower limit. The subtlety is that while the ASI reading mechanism inherently accounts for density, glider pilots must consult the altitude-correction table for the actual limit at high altitude.

• Option A and B/D are wrong because the physical mark on the instrument does not move.

Key Terms

• IAS = Indicated Airspeed
• VNE = Never Exceed Speed ### Q56: You have covered a distance of 150 km in 1 hour and 15 minutes. Your calculated ground speed is: ^t30q56

DE · FR

• A) 125 km/h.
• B) 115 km/h.
• C) 120 km/h.
• D) 110 km/h.

Answer

C)

Explanation

The correct answer is C because ground speed = distance / time = 150 km / 1.25 hours = 120 km/h. The key step is converting 1 hour 15 minutes to decimal hours: 15 minutes = 0.25 hours, so total time = 1.25 hours.

• Option A (125 km/h) results from dividing by 1.2 hours.
• Option B (115 km/h) and D (110 km/h) do not correspond to any correct calculation with these inputs.

Q57: The following NOTAM was published on 18 August (summer time). Which of the following statements is correct? ^t30q57

DE · FR

]

• 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.

Answer

C)

Explanation

The correct answer is C because the NOTAM establishes that from 2 to 6 September 2013, between 0600 and 1500 UTC, the extended CTR/TMA Payerne is activated as a holding area, while LS-R4 serves as both a demonstration and holding area for an airshow. These areas must be strictly avoided during the active period.

• A is wrong because the times are 0600-1500 UTC, not sunrise to sunset.
• B incorrectly states both areas serve as holding and demonstration areas.
• D is wrong because transit is not permitted — the area must be avoided entirely, not transited with clearance.

Key Terms

• CTR = Control Zone
• NOTAM = Notice to Air Missions
• TMA = Terminal Manoeuvring Area ### Q58: Which is the best glide speed in calm air for a flying mass of 450 kg? See attached sheet. ^t30q58

DE · FR

]

• A) 95km/h
• B) 75km/h
• C) 55km/h
• D) 135km/h

Answer

B)

Explanation

The correct answer is B (75 km/h) because the best glide speed is found by drawing a tangent from the origin to the speed polar curve for 450 kg. The point where this tangent touches the curve gives the speed for maximum lift-to-drag ratio (best glide).

• Option A (95 km/h) is too fast and would correspond to a heavier mass or a different polar.
• Option C (55 km/h) is near the stall speed.
• Option D (135 km/h) is deep in the high-speed range where the glide ratio is significantly reduced.

Key Terms

• D — Drag D — Drag

Q59: A VFR flight will follow the route shown on the map below (dotted line) from APPENZELL towards MUOTATHAL. The route is planned for 19 March 2013 (winter time) between 1205 and 1255 LT. Answer using the DABS below. Which of these answers is correct? ^t30q59

DE · FR

]

• 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.

Answer

C)

Explanation

The correct answer is C because checking the DABS for 19 March 2013 (winter time, CET = UTC+1), the planned time of 1205-1255 LT converts to 1105-1155 UTC. During this period, the relevant danger and restricted zones along the route are not active, allowing the route to be flown without coordination.

• A is wrong because the DABS applies to all airspace users, including gliders.
• B is wrong because altitude-based exemptions do not automatically apply to all restricted areas.
• D is wrong because the route is flyable during the specified time window.

Key Terms

• AGL = Above Ground Level
• AMSL = Above Mean Sea Level
• VFR = Visual Flight Rules ### Q60: Wing loading is increased by 40% by water ballast. By what percentage does the glider's minimum speed increase? ^t30q60

DE · FR

• A) 18%.
• B) 40%.
• C) 100%.
• D) 0%.

Answer

A)

Explanation

The correct answer is A because stall speed (and therefore minimum speed) is proportional to the square root of wing loading. If wing loading increases by 40% (factor 1.4), the new minimum speed is the original multiplied by the square root of 1.4, which equals approximately 1.183 — an increase of about 18.3%.

• B is wrong because the speed does not increase linearly with wing loading.
• C is wrong because a 100% increase would mean doubling the speed.
• D is wrong because any mass increase raises the minimum speed.

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

DE · FR

]

• 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

Answer

A)

Explanation

The correct answer is A because at 150 km/h, the two polar curves for different masses of the ASK21 intersect, meaning both configurations have the same sink rate at this particular speed. This is an aerodynamic property of the polar: the curves cross at one speed where mass has no effect on sink rate.

• B is wrong because at 150 km/h the glide ratio is equal for both masses.
• C is wrong because the sink rates are identical at this intersection point.
• D is also wrong because neither mass has a better glide ratio at this specific speed.

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

DE · FR

]

• A) 700 ft.
• B) 780m.
• C) 780 ft
• D) 700m.

Answer

B)

Explanation

The correct answer is B (780 m) because the AIP chart for Amlikon aerodrome shows a maximum landing distance available of 780 metres in the eastward direction. Always verify the unit and the specific runway direction when reading aerodrome charts.

• Option A and C are wrong because landing distances in Switzerland are given in metres, not feet.
• Option D (700 m) does not match the published data for the eastward heading.

Key Terms

AIP = Aeronautical Information Publication ### 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

DE · FR

]

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

Answer

B)

Explanation

The correct answer is B because the EMMEN TMA lower boundary between Cham and Hitzkirch is at 3500 ft AMSL. Below this altitude, you remain in uncontrolled airspace and no clearance is needed. Above 3500 ft AMSL, you enter the TMA and must obtain an ATC clearance.

• Option A (2400 ft) is too low and does not correspond to the published limit.
• Option C (2000 ft GND) references above ground level, which is not how this TMA boundary is expressed.
• Option D (5000 ft) is too high.

Key Terms

• AMSL = Above Mean Sea Level
• ATC = Air Traffic Control
• TMA = Terminal Manoeuvring Area ### Q64: The maximum permitted payload is exceeded. What action must be taken? ^t30q64

DE · FR

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

Answer

D)

Explanation

The correct answer is D because when the maximum permitted payload is exceeded, the only correct action is to reduce the payload until it complies with the limit. The maximum payload is a certification limit based on structural strength and C.G. envelope. A and C are wrong because trimming adjusts aerodynamic forces on the tail but does not change the aircraft's mass or C.G. — it cannot make an overloaded aircraft safe.

• B is wrong because increasing takeoff speed does not solve an overweight condition and may actually overstress the structure further.

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

DE · FR

• 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.

Answer

D)

Explanation

The correct answer is D because a headwind reduces groundspeed while the sink rate in the airmass remains unchanged. Since the glider covers less horizontal ground distance per unit of altitude lost, the descent angle relative to the ground steepens (increases).

• A is wrong because a tailwind decreases (flattens) the glide angle over the ground by increasing groundspeed.
• B is wrong because a headwind increases, not decreases, the ground glide angle.
• C is wrong because wind significantly affects the ground track glide angle, even though it does not affect the airmass glide angle.

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

DE · FR

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

Answer

B)

Explanation

The correct answer is B because as altitude increases, air density decreases. For the same true airspeed, the pitot tube measures less dynamic pressure, so the IAS reading is lower than TAS. Conversely, to maintain the same IAS at altitude, the aircraft must fly at a higher TAS. The relationship is approximately TAS = IAS x square root of (sea-level density / actual density).

• A is wrong because IAS does not rise relative to TAS with altitude.
• C is wrong because IAS can always be measured.
• D is wrong because IAS and TAS diverge increasingly with altitude.

Key Terms

• IAS = Indicated Airspeed
• TAS = True Airspeed ### Q67: What has to be particularly observed when landing in heavy rain? ^t30q67

DE · FR

• 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.

Answer

A)

Explanation

The correct answer is A because heavy rain on the wing surface increases roughness and can degrade the boundary layer, potentially raising the stall speed and reducing maximum lift coefficient. A higher approach speed provides a safety margin against these effects.

• B is wrong because deliberately increasing wing loading in rain would require adding ballast, which is impractical and counterproductive.
• C is wrong because a shallower approach reduces obstacle clearance in poor visibility.
• D is wrong because a lower approach speed reduces the safety margin when aerodynamic degradation is already a risk.

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

DE · FR

]

• 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.

Answer

D)

Explanation

The correct answer is D because at Bex aerodrome, terrain constraints (the Rhone valley and surrounding mountains) mean the traffic pattern direction for runway 33 depends on the prevailing wind conditions. The chart shows that either a left or right circuit may be used.

• A is wrong because it limits the pattern to clockwise only.
• B relates to runway 15, not 33.
• C is wrong because it limits the pattern to counter-clockwise only.
• Pilots must check the local procedures and wind conditions before joining the circuit.

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

DE · FR

]

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

Answer

D)

Explanation

The correct answer is D because the lower limit of TMA BERN 1 over Biel Kappelen is 3500 ft AMSL. By staying below this altitude, you remain in uncontrolled airspace and do not need a transit clearance.

• Option A (3500 ft AGL) is wrong because TMA boundaries are referenced to MSL, not AGL.
• Option B (FL 100) is far above the relevant boundary.
• Option C (FL 35) converts to approximately 3500 ft in standard atmosphere, but flight levels use the standard pressure setting (1013.25 hPa), not QNH, so this is not the correct way to express the limit.

Key Terms

• AGL = Above Ground Level
• AMSL = Above Mean Sea Level
• FL = Flight Level
• MSL = Mean Sea Level
• QNH = Pressure adjusted to mean sea level
• TMA = Terminal Manoeuvring Area ### Q70: Which of these statements is correct? ^t30q70

DE · FR

• 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.

Answer

A)

Explanation

The correct answer is A because applying the mass-and-balance calculation with the data provided (from the attached sheet), the new C.G. position computes to 76.7, which falls within the approved forward and aft C.G. limits. Always verify your calculation by checking whether the result is between the published forward and aft limits.

• Option B (78.5) is an incorrect calculation result.
• Option C (82.0) is too far aft and would be outside limits.
• Option D (75.5) is incorrectly calculated and would also fall outside the forward limit.

Key Terms

• D — Drag D — Drag

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

DE · FR

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

Answer

A)

Explanation

The correct answer is A because a waterlogged grass surface creates greater friction and drag on the landing gear during the ground roll, causing the glider to decelerate faster and stop in a shorter distance. The water acts as a braking medium.

• B is wrong because wet grass increases, not decreases, rolling resistance for a glider.
• C is wrong because while directional control may be slightly affected, the primary effect is shortened stopping distance.
• D is wrong because surface conditions always affect landing distance.

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

DE · FR

]

• A) 520 m.
• B) 470m.
• C) 520 ft.
• D) 470 ft.

Answer

B)

Explanation

The correct answer is B (470 m) because the AIP chart for Schanis aerodrome shows a maximum landing distance available of 470 metres in the NNW direction. Always read the correct runway direction and corresponding distance from the aerodrome chart.

• Option A (520 m) does not match the published data for this heading.
• Option C and D are wrong because Swiss aerodrome distances are given in metres, not feet.

Key Terms

AIP = Aeronautical Information Publication ### 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

DE · FR

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

Answer

D)

Explanation

The correct answer is D (45.71 lbs). The calculation uses the shift formula: when mass x is moved from the current C.G. position (80) to arm 150, the C.G. shifts aft. The new C.G. must not exceed 80.5. Using the formula: delta CG = (x × delta arm) / total mass, we get: 0.5 = (x × 70) / 6400, therefore x = (0.5 × 6400) / 70 = 45.71 lbs.

• Option A (27.82), B (56.63), and C (39.45) result from incorrect calculations using wrong distances or mass values.

Key Terms

CG = Centre of Gravity ### Q74: Correct loading of an aircraft depends on: ^t30q74

DE · FR

• 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.

Answer

C)

Explanation

The correct answer is C because correct loading requires satisfying two independent conditions simultaneously: the total mass must not exceed the maximum allowable mass (MTOM), and the payload must be distributed so that the C.G. remains within the approved envelope.

• A is wrong because respecting the mass limit alone does not guarantee the C.G. is within limits.
• B is wrong because correct distribution alone does not ensure the total mass is within limits.
• D is wrong because it addresses only one specific baggage compartment rather than the complete loading requirements.

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

DE · FR

]

• 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.

Answer

D)

Explanation

The correct answer is D because when comparing polar curves for different masses, the tangent from the origin touches each curve at the same angle, meaning the maximum lift-to-drag ratio (best glide ratio) is essentially unchanged by mass, apart from minor Reynolds number effects. However, the speed at which this best glide ratio occurs increases with mass.

• A is wrong because increasing mass always increases the sink rate at any given speed.
• B is wrong because minimum speed increases with mass (proportional to the square root of mass ratio).
• C is wrong because while glide ratio is mass-independent, minimum speed is not.

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

DE · FR

• 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.

Answer

A)

Explanation

The correct answer is A because the airspeed indicator measures dynamic pressure, which directly relates to aerodynamic forces regardless of altitude. At 1800 m AMSL, air density is lower, so the TAS will be higher for the same IAS — but the aerodynamic forces (lift, stall characteristics) depend on IAS, not TAS. Therefore, the same indicated approach speed provides the same safety margins as at sea level.

• B is wrong because flying at a lower IAS would reduce the stall margin.
• D is wrong because a higher IAS is unnecessary and would result in excessive float.
• C is wrong because the minimum sink speed is not the correct approach speed.

Key Terms

• AMSL = Above Mean Sea Level
• IAS = Indicated Airspeed
• TAS = True Airspeed ### Q77: At what speed must you fly to achieve the best glide ratio for a flying mass of 450 kg? (See attached sheet.) ^t30q77

DE · FR

]

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

Answer

B)

Explanation

The correct answer is B (90 km/h) because the best glide ratio speed is found where the tangent from the origin touches the speed polar curve for 450 kg. For this glider type at 450 kg, this occurs at approximately 90 km/h.

• Option A (130 km/h) is too fast — at this speed the glide ratio is significantly reduced.
• Option C (70 km/h) is closer to the minimum sink speed, which maximises endurance but not distance.
• Option D (110 km/h) would give a reduced glide ratio compared to the optimum.

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

DE · FR

• 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.

Answer

C)

Explanation

The correct answer is C because when the aft C.G. limit is exceeded, the useful load must be redistributed to move mass forward — for example, adding nose ballast, repositioning equipment, or adjusting the pilot's seating position. This physically moves the C.G. within approved limits.

• A is wrong because trimming aft would worsen the situation aerodynamically.
• B is wrong because being within mass limits does not compensate for a C.G. out of limits — both must be satisfied independently.
• D is wrong because trim adjusts aerodynamic forces but does not change the actual C.G. position.

Key Terms

CG = Centre of Gravity ### Q79: Which factors increase the aerotow takeoff run distance? ^t30q79

DE · FR

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

Answer

D)

Explanation

The correct answer is D because high temperature reduces air density, decreasing the lift generated at any given groundspeed, requiring a longer acceleration to reach flying speed. A tailwind reduces the headwind component, meaning the aircraft needs a higher groundspeed to achieve the same airspeed, further lengthening the takeoff run.

• A is wrong because low temperature increases air density (more lift) and headwind shortens the run.
• B is wrong because a strong headwind shortens the takeoff distance.
• C is wrong because high atmospheric pressure increases density, which helps rather than hinders takeoff performance.

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

DE · FR

]

• 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.

Answer

D)

Explanation

The correct answer is D because the NOTAM specifies a military night flying exercise on 18 November from 1800 to 2100 UTC in the ZUGERSEE, SUSTEN, and TICINO areas, with vertical limits from GND to 15,000 ft AMSL.

• A is wrong because the lower limit is GND, not Class E airspace, and the upper limit is 15,000 ft AMSL, not FL150.
• B is wrong because the times are in UTC, not local time.
• C is wrong because it incorrectly specifies helicopter-only operations and omits the geographic areas.

Key Terms

• AMSL = Above Mean Sea Level
• FL = Flight Level
• NOTAM = Notice to Air Missions ### Q81: What is the maximum permitted flying altitude within the CTR of Bern-Belp airport? ^t30q81

DE · FR

]

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

Answer

D)

Explanation

The correct answer is D because the CTR (Control Zone) of Bern-Belp airport has an upper limit of 3000 ft AMSL. Above this altitude, you exit the CTR and enter different airspace. VFR flight within the CTR requires a clearance from Bern Tower and must remain below the published upper limit.

• Option A (5500 ft GND) does not match the published limit.
• Option B (4500 ft AMSL) is too high.
• Option C (5000 ft AMSL) is also too high.

Key Terms

• AMSL = Above Mean Sea Level
• CTR = Control Zone
• VFR = Visual Flight Rules ### 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

DE · FR

]

• 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.

Answer

D)

Explanation

The correct answer is D because at 1700 m AMSL above Bex aerodrome, you are in Class E airspace. VFR minima in Class E require 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance.

• A is wrong because Class G applies at lower altitudes with reduced requirements.
• B is wrong because Class C has the right visibility minimum (5 km in Switzerland, not 8 km) but starts at a much higher altitude.
• C is wrong for the same airspace classification reason — Class C begins at FL 130, well above 1700 m.

Key Terms

• AMSL = Above Mean Sea Level
• FL = Flight Level
• VFR = Visual Flight Rules ### Q83: Which is the sink rate at 160 km/h for this glider at a flying mass of 580 kg? (See attached sheet.) ^t30q83

DE · FR

]

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

Answer

C)

Explanation

The correct answer is C (2.0 m/s) because reading the speed polar curve for a flying mass of 580 kg at 160 km/h, the sink rate is approximately 2.0 m/s. When reading a speed polar, always identify the correct curve for the given mass before reading the value at the specified speed.

• Option A (1.6 m/s) would correspond to a lighter mass or lower speed.
• Option B (0.8 m/s) is near the minimum sink rate at much lower speed.
• Option D (1.2 m/s) is also too low for this speed and mass combination.

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

DE · FR

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

Answer

B)

Explanation

The correct answer is B because to convert kilograms to pounds, multiply by 2.2: 550 x 2.2 = 1,210 lbs. The key formula is: weight in lbs = mass in kg x 2.2.

• Option A (12,100 lbs) results from multiplying by 22 instead of 2.2. C (2,500 lbs) does not correspond to any correct calculation.
• Option D (250 lbs) results from dividing instead of multiplying.

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

DE · FR

• 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.

Answer

D)

Explanation

The correct answer is D because the best glide ratio speed (also called best L/D speed) maximises the horizontal distance covered per unit of altitude lost in still air. This speed is found on the polar curve where the tangent from the origin touches the curve.

• A is wrong because minimum sink speed maximises endurance (time aloft), not distance.
• B is wrong because maximum speed produces the worst glide ratio due to high parasite drag.
• C is wrong because minimum flying speed is near the stall and gives a poor glide ratio due to high induced drag.

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

DE · FR

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

Answer

A)

Explanation

The correct answer is A because the maximum glide ratio (best L/D) is essentially independent of mass — both the lift coefficient and drag coefficient at the optimal angle of attack remain the same, so their ratio is unchanged. Only a minor Reynolds number effect exists.

• B is wrong because wing loading = mass / wing area, which directly increases with mass.
• C is wrong because sink rate increases with mass at any given speed.
• D is wrong because the speeds corresponding to best glide and minimum sink both increase with mass.

Key Terms

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

DE · FR

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

Answer

D)

Explanation

The correct answer is D because time = distance / speed = 150 km / 100 km/h = 1.5 hours = 1 hour 30 minutes. The calculation is straightforward: 150 / 100 = 1.5 hours. Convert the decimal 0.5 hours to 30 minutes.

• Option A (1 hour 50 minutes) would correspond to a distance of about 183 km.
• Option B (1 hour 40 minutes = 1.667 hours) would correspond to about 167 km.
• Option C (2 hours) would correspond to 200 km.

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

DE · FR

]

• 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.

Answer

C)

Explanation

The correct answer is C because when restricted zones LS-R8 and LS-R8A are active, they cover the planned alpine route between Munster and Amsteg, making it impossible to fly through them. Restricted zones with "entry prohibited" status cannot be transited, regardless of altitude or radio contact.

• A is wrong because radio contact does not grant transit rights through active restricted zones.
• B is wrong because a 28,000 ft ceiling does not help a glider.
• D is wrong because overflying at 9,200 ft may still be within the zone's vertical limits.

Key Terms

• AMSL = Above Mean Sea Level
• ICAO = International Civil Aviation Organization
• VFR = Visual Flight Rules ### Q89: You wish to obtain clearance to transit the ZURICH TMA. What must you do? ^t30q89

DE · FR

• 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.

Answer

A)

Explanation

The correct answer is A because to transit the Zurich TMA, the pilot must make first radio contact on frequency 124.7 MHz (Zurich Information) at least 10 minutes before entering the controlled airspace. This provides ATC sufficient time to assess traffic, issue a clearance or alternative instructions, and ensure separation.

• B is wrong because 5 minutes is insufficient lead time.
• C is wrong because 118.975 is not the correct frequency for Zurich TMA transit requests.
• D is wrong on both the frequency and the lead time.

Key Terms

• ATC = Air Traffic Control
• TMA = Terminal Manoeuvring Area ### 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

DE · FR

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

Answer

A)

Explanation

The correct answer is A because in a turn, the stall speed increases by the square root of the load factor: Vsturn = Vsstraight x sqrt(n). With n = 2.0: Vs_turn = 60 x sqrt(2) = 60 x 1.414 = 84.85 kts. The increase is (84.85 - 60) / 60 x 100 = 41.4%, which rounds to approximately 40%.

• B is wrong because the stall speed always increases in a turn.
• C (5%) and D (20%) significantly underestimate the effect.
• This relationship between bank angle, load factor, and stall speed is fundamental to safe manoeuvring flight.

Key Terms

• n — Load Factor (ratio of lift to weight: n = L/W) n — Load Factor (ratio of lift to weight: n = L/W)

Q91: The upper limit of LO R 16 equals ^t30q91

DE · FR


• A) 1 500 m MSL.
• B) FL150.
• C) 1 500 ft MSL.
• D) 1.500 ft GND.

Answer

C)

Explanation

The correct answer is C because restricted airspace areas (LO R) on aeronautical charts express their limits using standard altitude references. LO R 16 has an upper limit of 1,500 ft MSL (mean sea level), which is a fixed, absolute altitude.

• A is wrong because 1,500 m MSL would be approximately 4,900 ft — a completely different altitude that confuses feet with metres.
• B is wrong because FL150 (15,000 ft pressure altitude) is far too high for a typical low-level restriction.
• D is wrong because 1,500 ft GND (above ground level) would vary with terrain elevation and is not the published reference.

Key Terms

FL = Flight Level; MSL = Mean Sea Level

Q92: The upper limit of LO R 4 equals ^t30q92

DE · FR


• A) 4.500 ft AGL.
• B) 4.500 ft MSL
• C) 1.500 ft AGL
• D) 1.500 ft MSL.

Answer

B)

Explanation

The correct answer is B because LO R 4 has its upper limit at 4,500 ft MSL, a fixed altitude above mean sea level.

• A is wrong because 4,500 ft AGL (above ground level) would vary with terrain, which is inappropriate for a fixed regulatory boundary.
• C is wrong because 1,500 ft AGL is both the wrong altitude value and the wrong reference.
• D is wrong because 1,500 ft MSL is too low and corresponds to a different restricted area (LO R 16).

Key Terms

AGL = Above Ground Level; MSL = Mean Sea Level

Q93: Up to which altitude is an overflight prohibited according to the NOTAM? ^t30q93

DE · FR


• A) Height 9500 ft
• B) Altitude 9500 ft MSL
• C) Flight Level 95
• D) Altitude 9500 m MSL

Answer

B)

Explanation

The correct answer is B because the NOTAM prohibits overflight up to an altitude of 9,500 ft MSL, following ICAO convention where "altitude" refers to height above mean sea level.

• A is wrong because "height" in aviation terminology means above a local ground reference (AGL), which is not what the NOTAM specifies.
• C is wrong because FL 95 is a pressure altitude reference based on 1013.25 hPa, which differs from an MSL altitude depending on actual atmospheric conditions.
• D is wrong because 9,500 m MSL would be approximately 31,000 ft — clearly inconsistent with a typical VFR NOTAM.

Key Terms

AGL = Above Ground Level; FL = Flight Level; ICAO = International Civil Aviation Organization; MSL = Mean Sea Level; NOTAM = Notice to Air Missions; VFR = Visual Flight Rules

Q94: According ICAO, what symbol indicates a group of unlighted obstacles? (2,00 P.) ^t30q94

DE · FR


• A) D
• B) C
• C) B
• D) A

Answer

B)

Explanation

The correct answer is B (symbol C in the annex) because ICAO aeronautical chart symbology (defined in ICAO Annex 4) uses specific symbols to distinguish between single and grouped obstacles, and between lighted and unlighted ones. Symbol C represents a group of unlighted obstacles. Correct identification of these symbols is essential for cross-country flight planning and obstacle avoidance.

• Option A (symbol D), C (symbol B), and D (symbol A) represent other obstacle categories such as single obstacles, lighted groups, or lighted single obstacles.

Key Terms

ICAO = International Civil Aviation Organization ### Q95: According ICAO, what symbol indicates a civil airport (not international airport) with paved runway? (2,00 P.) ^t30q95

DE · FR


• A) D
• B) A
• C) C
• D) B

Answer

B)

Explanation

The correct answer is B (symbol A in the annex) because ICAO chart symbology uses distinct depictions for different aerodrome types — civil versus military, international versus domestic, and paved versus unpaved. Symbol A represents a civil (non-international) airport with a paved runway. Glider pilots must recognise these symbols when identifying potential emergency landing options.

• Option A (symbol D), C (symbol C), and D (symbol B) represent other aerodrome categories such as international airports, military aerodromes, or grass-strip airfields.

Key Terms

ICAO = International Civil Aviation Organization ### Q96: According ICAO, what symbol indicates a general spot elevation? (2,00 P.) ^t30q96

DE · FR


• A) A
• B) B
• C) D
• D) C

Answer

D)

Explanation

The correct answer is D (symbol C in the figure) because on ICAO aeronautical charts, a general spot elevation is indicated by a specific symbol showing a terrain point of known height, used for situational awareness and terrain clearance planning.

• Option A (symbol A), B (symbol B), and C (symbol D) represent other elevation-related markings such as maximum elevation figures, surveyed points, or obstruction elevations defined in ICAO Annex 4.

Key Terms

ICAO = International Civil Aviation Organization ### Q97: The term center of gravity is defined as ^t30q97

DE · FR

• A) Half the distance between the neutral point and the datum line.
• B) Another designation for the neutral point.
• C) The distance between the leading edge and the trailing edge of the wing.
• D) The heaviest point on an aeroplane.

Answer

A)

Explanation

The correct answer is A. The center of gravity is the single point through which the resultant of all gravitational forces acts on the aircraft — it is the mass-weighted average position of all components.

• B is wrong because the neutral point is a distinct aerodynamic concept used for stability analysis, not another name for C.G.
• C duplicates the same incorrect description as A's wording, but the C.G. is defined by mass distribution, not as a geometric midpoint.
• D is wrong because the C.G. is not the heaviest point — it is where the total weight effectively acts.

Q98: The term moment with regard to a mass and balance calculation is referred to as ^t30q98

DE · FR

• A) Sum of a mass and a balance arm.
• B) Product of a mass and a balance arm.
• C) Quotient of a mass and a balance arm.
• D) Difference of a mass and a balance arm.

Answer

B)

Explanation

The correct answer is B because in mass-and-balance calculations, moment is defined as the product of mass and balance arm: Moment = Mass x Arm (e.g., in kg-m or lb-in). This follows the physical definition of a torque. The total C.G. is found by summing all moments and dividing by total mass.

• A is wrong because adding mass and arm is dimensionally meaningless.
• C is wrong because dividing mass by arm does not produce a moment.
• D is wrong because subtracting them is equally incorrect.

Key Terms

• D — Drag D — Drag

Q99: The term balance arm in the context of a mass and balance calculation defines the ^t30q99

DE · FR

• A) Point on the longitudinal axis of an aeroplane or its extension from which the centers of gravity of all masses are referenced.
• B) Distance of a mass from the center of gravity
• C) Distance from the datum to the center of gravity of a mass.
• D) Point through which the force of gravity is said to act on a mass.

Answer

C)

Explanation

The correct answer is C because the balance arm (moment arm) is the horizontal distance measured from the aircraft's datum reference point to the center of gravity of a specific mass item.

• A is wrong because that describes the datum itself, not the balance arm.
• B is wrong because balance arms are measured from the datum, not from the overall aircraft C.G.
• D is wrong because that is the definition of the center of gravity of a mass item, not the balance arm.

Q100: Which is the purpose of interception lines in visual navigation? ^t30q100

DE · FR

• A) To mark the next available en-route airport during the flight
• B) To visualize the range limitation from the departure aerodrome
• C) They help to continue the flight when flight visibility drops below VFR minima
• D) They are used as easily recognizable guidance upon a possible loss of orientation

Answer

D)

Explanation

The correct answer is D because interception lines (also called catching lines or line features) are prominent linear ground features — motorways, rivers, coastlines, railways — that a pilot selects during pre-flight planning to navigate toward if orientation is lost. By flying toward a known interception line, the pilot can re-establish position and resume navigation.

• A is wrong because interception lines are geographic features, not airport markers.
• B is wrong because they are not range indicators.
• C is wrong because nothing authorises continuing flight below VFR minima — interception lines are a lost-procedure tool, not a visibility workaround.

Key Terms

VFR = Visual Flight Rules