Navigation
Q1: Through which points does the Earth's rotational axis pass? ^t60q1
DE · FR
- A) The geographic North Pole and the magnetic south pole.
- B) The magnetic north pole and the geographic South Pole.
- C) The geographic North Pole and the geographic South Pole.
- D) The magnetic north pole and the magnetic south pole.
Answer
C)
Explanation
The Earth's rotational axis is the physical axis around which the planet spins, and it passes through the geographic (true) poles — not the magnetic poles. The geographic poles are fixed points defined by the rotational axis, while the magnetic poles are offset from them and drift over time due to changes in the Earth's molten core.
Source
Q2: Which statement correctly describes the polar axis of the Earth? ^t60q2
DE · FR
- A) It passes through the geographic South Pole and the geographic North Pole and is tilted 23.5° relative to the equatorial plane.
- B) It passes through the magnetic south pole and the magnetic north pole and is tilted 66.5° relative to the equatorial plane.
- C) It passes through the magnetic south pole and the magnetic north pole and is perpendicular to the equatorial plane.
- D) It passes through the geographic South Pole and the geographic North Pole and is perpendicular to the equatorial plane.
Answer
D)
Explanation
The polar axis passes through the geographic poles and is perpendicular (90°) to the plane of the equator by definition. The Earth's axis is indeed tilted 23.5° relative to the plane of its orbit around the sun (the ecliptic), but it is perpendicular to the equatorial plane — those two facts are consistent and not contradictory.
- Option A confuses the tilt to the ecliptic with the relationship to the equator.
Source
Q3: For navigation systems, which approximate geometrical shape best represents the Earth? ^t60q3
DE · FR
- A) A flat plate.
- B) An ellipsoid.
- C) A sphere of ecliptical shape.
- D) A perfect sphere.
Answer
B)
Explanation
The Earth is not a perfect sphere — it is slightly flattened at the poles and bulges at the equator due to its rotation. This shape is called an oblate spheroid or ellipsoid. Modern navigation systems (including GPS) use the WGS-84 ellipsoid as the reference model, which accurately accounts for this flattening in coordinate calculations.
Source
- [?] Source PDF non identifiée (original: B)
### Q4: Which of the following statements about a rhumb line is correct? ^t60q4
DE · FR
- A) The shortest path between two points on the Earth follows a rhumb line.
- B) A rhumb line crosses each meridian at an identical angle.
- C) The centre of a complete rhumb line circuit is always the centre of the Earth.
- D) A rhumb line is a great circle that meets the equator at 45°.
Answer
B)
Explanation
A rhumb line (also called a loxodrome) is defined as a line that crosses every meridian of longitude at the same constant angle.
Why is this useful? A pilot can fly a rhumb line simply by maintaining a fixed compass heading — no course corrections needed. On a Mercator chart, a rhumb line appears as a straight line, making it easy to plot and follow. This is why Mercator projection is the standard for aviation charts.
Rhumb line (loxodrome) vs great circle (orthodrome):
| | Loxodrome (rhumb line) | Orthodrome (great circle) |
|---|---|---|
| Definition | Crosses all meridians at the same angle | Arc of the largest circle on the sphere |
| On Mercator chart | Straight line | Curved line (bows toward the poles) |
| Heading | Constant — no corrections needed | Changes continuously |
| Distance | Longer than great circle | Shortest possible path |
| Use case | Easy to fly with compass; ideal for short/medium distances | Long-distance routes (airlines); plotted, then flown as rhumb-line segments |
For glider cross-country flights, the difference is negligible. For long-distance flights, pilots break the orthodrome into short rhumb-line segments with periodic heading updates.
- A is wrong: the shortest path is an orthodrome (great circle), not a loxodrome.
- C is wrong: a loxodrome spirals toward the poles — its centre is not the Earth's centre (that defines a great circle).
- D is wrong: a loxodrome is not a great circle.
Source
- [?] Source PDF non identifiée (original: B)
### Q5: The shortest route between two points on the Earth's surface follows a segment of ^t60q5
DE · FR
- A) A small circle
- B) A great circle.
- C) A rhumb line.
- D) A parallel of latitude.
Answer
B)
Explanation
A great circle (orthodrome) is any circle whose plane passes through the centre of the Earth, and the arc of a great circle between two points is the shortest possible path along the Earth's surface (the geodesic).
The figure shows the difference: a rhumb line (loxodrome) crosses all meridians at a constant angle but is longer; a great circle (orthodrome) is the shortest path but requires constantly changing heading. Long-haul aircraft routes follow great circle tracks to minimize fuel and time, broken into rhumb-line segments for practical navigation.
- A (small circle) — a small circle does not pass through the Earth's centre and is never the shortest path.
- C (rhumb line) — a rhumb line is longer than the great circle (except along meridians or the equator where they coincide).
- D (parallel of latitude) — parallels (except the equator) are small circles, not great circles.
Source
- [?] Source PDF non identifiée (original: B)
### Q6: What is the approximate circumference of the Earth measured along the equator? ^t60q6
DE · FR
- A) 40000 NM.
- B) 21600 NM.
- C) 10800 km.
- D) 12800 km.
Answer
B)
Explanation
The equator spans 360 degrees of longitude, and each degree of longitude on the equator equals 60 NM (since 1 NM = 1 arcminute on a great circle). Therefore: 360° x 60 NM = 21,600 NM. In kilometers, the Earth's equatorial circumference is approximately 40,075 km — so option A has the right number but wrong unit. Knowing this relationship (1° = 60 NM on the equator) is fundamental to navigation calculations.
Key Terms
NM = Nautical Mile(s)
Source
Q7: What is the latitude difference between point A (12°53'30''N) and point B (07°34'30''S)? ^t60q7
DE · FR
- A) .20°28'00''
- B) .05°19'00''
- C) .20,28°
- D) .05,19°
Answer
A)
Explanation
When two points are on opposite sides of the equator, the difference in latitude is the sum of their respective latitudes. Here: 12°53'30''N + 07°34'30''S = 20°28'00''. Converting minutes: 53'30'' + 34'30'' = 88'00'' = 1°28'00'', so 12° + 7° + 1°28' = 20°28'00''. Always add latitudes when they are in opposite hemispheres (N and S).
Key Terms
S — Wing Area — total planform area of the wings
Source
Q8: At what positions are the two polar circles located? ^t60q8
DE · FR
- A) 23.5° north and south of the equator
- B) At a latitude of 20.5°S and 20.5°N
- C) 20.5° south of the poles
- D) 23.5° north and south of the poles
Answer
D)
Explanation
The Arctic Circle lies at approximately 66.5°N and the Antarctic Circle at 66.5°S — which is 90° - 23.5° = 66.5°, placing them 23.5° away from their respective geographic poles. This 23.5° offset directly corresponds to the axial tilt of the Earth. The Tropics of Cancer and Capricorn (option A) are the ones located 23.5° from the equator.
Source
Q9: Along a meridian, what is the distance between the 48°N and 49°N parallels of latitude? ^t60q9
DE · FR
- A) 111 NM
- B) 10 NM
- C) 60 NM
- D) 1 NM
Answer
C)
Explanation
Along any meridian (line of longitude), 1 degree of latitude always equals 60 nautical miles. This is because meridians are great circles and 1 NM is defined as 1 arcminute of arc along a great circle. The 111 km figure (option A) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations.
Key Terms
NM = Nautical Mile(s)
Source
Q10: Along any line of longitude, what distance corresponds to one degree of latitude? ^t60q10
DE · FR
- A) 30 NM
- B) 1 NM
- C) 60 km
- D) 60 NM
Answer
D)
Explanation
One degree of latitude = 60 arcminutes, and since 1 NM equals exactly 1 arcminute of latitude along a meridian, 1° of latitude = 60 NM. This relationship holds along any meridian because all meridians are great circles. In SI units, 1° of latitude ≈ 111 km, not 60 km as stated in option C.
Key Terms
- e — Oswald Efficiency Factor — wing efficiency factor (1.0 for ideal elliptical lift distribution)
- NM = Nautical Mile(s)
Source
Q11: Point A lies at exactly 47°50'27''N latitude. Which point is precisely 240 NM north of A? ^t60q11
DE · FR
- A) 49°50'27''N
- B) 43°50'27''N
- C) 53°50'27''N
- D) 51°50'27'N'
Answer
D)
Explanation
Converting 240 NM to degrees of latitude: 240 NM / 60 NM per degree = 4°. Adding 4° to 47°50'27''N gives 51°50'27''N. Moving north increases the latitude value.
- Option C would require 6° (360 NM), and option A would require only 2° (120 NM).
Key Terms
NM = Nautical Mile(s)
Source
- [ ] ~ VV Q7 p.147 (clé: B, original: B)
### Q12: Along the equator, what is the distance between the 150°E and 151°E meridians? ^t60q12
DE · FR
- A) 1 NM
- B) 60 NM
- C) 60 km
- D) 111 NM
Answer
B)
Explanation
On the equator, meridians of longitude are separated by great circle arcs, and 1° of longitude along the equator equals 60 NM — the same as 1° of latitude along any meridian, because the equator is also a great circle. At higher latitudes, the distance between meridians decreases (multiplied by cos(latitude)), but at the equator it is exactly 60 NM per degree.
Key Terms
NM = Nautical Mile(s)
Source
Q13: When two points A and B on the equator are separated by exactly one degree of longitude, what is the great circle distance between them? ^t60q13
DE · FR
- A) 216 NM
- B) 120 NM
- C) 60 NM
- D) 400 NM
Answer
C)
Explanation
The equator itself is a great circle, so the great circle distance between two points on the equator separated by 1° of longitude is simply 60 NM (1° x 60 NM/degree). This is the same principle as measuring along a meridian. Any confusion arises if one tries to calculate using km instead — 1° ≈ 111 km on the equator, but the question asks for NM.
Key Terms
NM = Nautical Mile(s)
Source
Q14: Consider two points A and B on the same parallel of latitude (not the equator). A is at 010°E and B at 020°E. The rhumb line distance between them is always ^t60q14
DE · FR
- A) More than 600 NM.
- B) More than 300 NM.
- C) Less than 300 NM.
- D) Less than 600 NM.
Answer
D)
Explanation
The rhumb line distance between points on the same parallel of latitude is: 10° x 60 NM x cos(latitude). Since cos(latitude) is always less than 1 for any latitude other than the equator (where it equals exactly 60 NM x 10 = 600 NM), the rhumb line distance is always strictly less than 600 NM. At the equator it would equal 600 NM, but since they are specifically "not on the equator," the distance is always less than 600 NM.
Key Terms
- e — Oswald Efficiency Factor — wing efficiency factor (1.0 for ideal elliptical lift distribution)
- NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: B)
### Q15: How much time elapses as the sun traverses 20° of longitude? ^t60q15
DE · FR
- A) 0:20 h
- B) 1:20 h
- C) 0:40 h
- D) 1:00 h
Answer
B)
Explanation
The Earth rotates 360° in 24 hours, so it rotates 15° per hour, or 1° every 4 minutes. For 20° of longitude: 20 x 4 minutes = 80 minutes = 1 hour 20 minutes. Alternatively: 20° / 15°/h = 1.333 h = 1:20 h. This relationship (15°/hour or 4 min/degree) is essential for time zone calculations and solar noon determination.
Source
- [?] Source PDF non identifiée (original: B)
### Q16: How much time passes as the sun crosses 10° of longitude? ^t60q16
DE · FR
- A) 0:30 h
- B) 0:40 h
- C) 1:00 h
- D) 0:04 h
Answer
B)
Explanation
Using the same principle as Q15: the Earth rotates 15° per hour, so 10° corresponds to 10/15 hours = 2/3 hour = 40 minutes = 0:40 h.
- Option D (4 minutes) would be the time for only 1° of longitude.
- Option A (30 minutes) would correspond to 7.5° of longitude.
Key Terms
D — Drag
Source
- [?] Source PDF non identifiée (original: D)
### Q17: The sun traverses 10° of longitude. What is the corresponding time difference? ^t60q17
DE · FR
- A) 0.33 h
- B) 1 h
- C) 0.4 h
- D) 0.66 h
Answer
D)
Explanation
This is the same calculation as Q16 but expressed as a decimal fraction of an hour: 10° / 15°/h = 0.6667 h ≈ 0.66 h (40 minutes in decimal hours). Note that Q16 and Q17 appear to ask the same question but expect different answer formats — Q16 expects 0:40 h (40 minutes) while Q17 expects 0.66 h (the decimal equivalent). Both represent the same 40-minute time difference.
Source
- [?] Source PDF non identifiée (original: A)
### Q18: If Central European Summer Time (CEST) is UTC+2, what is the UTC equivalent of 1600 CEST? ^t60q18
DE · FR
- A) 1400 UTC.
- B) 1600 UTC.
- C) 1500 UTC.
- D) 1700 UTC.
Answer
A)
Explanation
UTC+2 means CEST is 2 hours ahead of UTC. To convert from local time to UTC, subtract the offset: 1600 CEST - 2 hours = 1400 UTC. A simple mnemonic: "to get UTC, subtract the positive offset." This is critical in aviation as all flight plans, ATC communications, and NOTAMs use UTC regardless of local time zone.
Key Terms
ATC = Air Traffic Control
Source
- [?] Source PDF non identifiée (original: D)
### Q19: What is UTC? ^t60q19
DE · FR
- A) A local time in Central Europe.
- B) Local mean time at a specific point on Earth.
- C) A zonal time
- D) The mandatory time reference used in aviation.
Answer
D)
Explanation
Coordinated Universal Time (UTC) is the mandatory time reference for all international aviation operations — flight plans, ATC communications, weather reports (METARs/TAFs), and NOTAMs all use UTC to eliminate confusion from time zone differences. It is not a zonal or local time, and it is not referenced to any geographic location (though it closely tracks Greenwich Mean Time).
Key Terms
ATC = Air Traffic Control
Source
- [?] Source PDF non identifiée (original: C)
### Q20: If Central European Time (CET) is UTC+1, what is the UTC equivalent of 1700 CET? ^t60q20
DE · FR
- A) 1800 UTC.
- B) 1500 UTC.
- C) 1600 UTC.
- D) 1700 UTC.
Answer
C)
Explanation
CET is UTC+1, meaning it is 1 hour ahead of UTC. To convert to UTC, subtract the offset: 1700 CET - 1 hour = 1600 UTC. Switzerland uses CET (UTC+1) in winter and CEST (UTC+2) in summer — knowing the current offset is essential when filing flight plans or reading NOTAMs.
Source
- [?] Source PDF non identifiée (original: D)
### Q21: Vienna (LOWW) is at 016°34'E and Salzburg (LOWS) at 013°00'E, both at approximately the same latitude. What is the difference in sunrise and sunset times (in UTC) between the two cities? ^t60q21
DE · FR
- A) In Vienna sunrise is 14 minutes earlier and sunset is 14 minutes later than in Salzburg
- B) In Vienna sunrise and sunset are about 14 minutes earlier than in Salzburg
- C) In Vienna sunrise is 4 minutes later and sunset is 4 minutes earlier than in Salzburg
- D) In Vienna sunrise and sunset are about 4 minutes later than in Salzburg
Answer
B)
Explanation
The difference in longitude is 016°34' - 013°00' = 3°34' ≈ 3.57°. At 4 minutes per degree, this gives approximately 14.3 minutes ≈ 14 minutes. Vienna is east of Salzburg, so the sun reaches Vienna earlier — both sunrise and sunset occur about 14 minutes earlier in Vienna (as seen in UTC). Local time zones disguise this difference, but in UTC the eastern location always sees solar events first.
Source
- [?] Source PDF non identifiée (original: A)
### Q22: How is "civil twilight" defined? ^t60q22
DE · FR
- A) The interval before sunrise or after sunset when the sun's centre is no more than 6° below the true horizon.
- B) The interval before sunrise or after sunset when the sun's centre is no more than 12° below the apparent horizon.
- C) The interval before sunrise or after sunset when the sun's centre is no more than 6° below the apparent horizon.
- D) The interval before sunrise or after sunset when the sun's centre is no more than 12° below the true horizon.
Answer
A)
Explanation
Civil twilight is the period when the sun's center is between 0° and 6° below the true (geometric) horizon — there is still sufficient natural light for most outdoor activities without artificial lighting. The true horizon (geometric) is used in the formal definition, not the apparent horizon (which is affected by refraction). Nautical twilight uses 12°, and astronomical twilight uses 18° below the true horizon. In aviation regulations, civil twilight often defines the boundary for day/night VFR operations.
Key Terms
VFR = Visual Flight Rules
Source
Q23: Given: WCA: -012°; TH: 125°; MC: 139°; DEV: 002°E. Determine TC, MH, and CH. ^t60q23
DE · FR
- A) TC: 113°. MH: 139°. CH: 125°.
- B) TC: 137°. MH: 127°. CH: 125°.
- C) TC: 137°. MH: 139°. CH: 125°.
- D) TC: 113°. MH: 127°. CH: 129°.
Answer
B)
Explanation
The heading chain works as follows: TC → (apply WCA) → TH → (apply VAR) → MH → (apply DEV) → CH. Given TH = 125° and WCA = -12°, then TC = TH - WCA = 125° - (-12°) = 137°. For MH: MC = MH + WCA, so MH = MC - WCA = 139° - 12° = 127°. For CH: DEV = 002°E means compass reads 2° high, so CH = MH - DEV = 127° - 2° = 125°. Negative WCA means wind from the right, requiring a left correction in heading.
Key Terms
- CH = Compass Heading
- DEV = Compass Deviation
- MC = Magnetic Course
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- VAR = Magnetic Variation
- WCA = Wind Correction Angle
Source
- [?] Source PDF non identifiée (original: A)
### Q24: Given: TC: 179°; WCA: -12°; VAR: 004° E; DEV: +002°. What are MH and MC? ^t60q24
DE · FR
- A) MH: 163°. MC: 175°.
- B) MH: 167°. MC: 175°.
- C) MH: 167°. MC: 161°
- D) MH: 163°. MC: 161°.
Answer
A)
Explanation
TH = TC + WCA = 179° + (-12°) = 167°. Then MH = TH - VAR (E is subtracted): MH = 167° - 4° = 163°. For MC: MC = TC - VAR = 179° - 4° = 175°. Alternatively: MC = MH + WCA = 163° + (-12°) = 151° — wait, that doesn't match; MC is measured from magnetic north to the course line, so MC = TC - VAR = 179° - 4° = 175°. East variation is subtracted when converting from True to Magnetic ("East is least").
Key Terms
- DEV = Compass Deviation
- MC = Magnetic Course
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- VAR = Magnetic Variation
- WCA = Wind Correction Angle
Source
- [ ] ~ VV Q104 p.168 (clé: B, original: B)
### Q25: The angular difference between the true course and the true heading is known as the ^t60q25
DE · FR
- A) Variation.
- B) WCA.
- C) Deviation.
- D) Inclination.
Answer
B)
Explanation
The Wind Correction Angle (WCA) is the angular difference between the true course (the direction of intended track over the ground) and the true heading (the direction the aircraft's nose points). A crosswind requires the pilot to angle the nose into the wind, creating a difference between heading and track — this offset angle is the WCA. It is neither variation (true-to-magnetic difference) nor deviation (magnetic-to-compass difference).
Key Terms
WCA = Wind Correction Angle
Source
Q26: The angular difference between the magnetic course and the true course is called ^t60q26
DE · FR
- A) Deviation.
- B) WCA.
- C) Variation
- D) Inclination.
Answer
C)
Explanation
Magnetic variation (also called declination) is the angle between true north (geographic) and magnetic north at any given location, which creates a difference between the true course and the magnetic course. Variation changes with location and over time as the magnetic poles shift. Deviation is the error introduced by the aircraft's own magnetic field on the compass, affecting the difference between magnetic north and compass north.
Key Terms
WCA = Wind Correction Angle
Source
Q27: How is "magnetic course" (MC) defined? ^t60q27
DE · FR
- A) The angle between true north and the course line.
- B) The direction from any point on Earth toward the geographic North Pole.
- C) The direction from any point on Earth toward the magnetic north pole.
- D) The angle between magnetic north and the course line.
Answer
D)
Explanation
The magnetic course is the direction of the intended flight path (course line) measured clockwise from magnetic north. It differs from the true course by the local magnetic variation. Pilots use magnetic course because aircraft compasses point to magnetic north, making magnetic references more directly usable for navigation without additional corrections.
Key Terms
MC = Magnetic Course
Source
Q28: How is "True Course" (TC) defined? ^t60q28
DE · FR
- A) The angle between true north and the course line.
- B) The direction from any point on Earth toward the magnetic north pole.
- C) The angle between magnetic north and the course line.
- D) The direction from any point on Earth toward the geographic North Pole.
Answer
A)
Explanation
The True Course is the angle measured clockwise from true (geographic) north to the intended flight path (course line). It is determined from aeronautical charts, which are oriented to true north. To fly a true course, pilots must apply magnetic variation to get the magnetic course, then apply wind correction angle to get the true heading they must fly.
Key Terms
TC = True Course
Source
Q29: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and VAR? ^t60q29
DE · FR
- A) TH: 172°. VAR: 004° W
- B) TH: 194°. VAR: 004° W
- C) TH: 194°. VAR: 004° E
- D) TH: 172°. VAR: 004° E
Answer
B)
Explanation
TH = TC + WCA = 183° + 11° = 194°. For variation: VAR is the difference between TC and MC, or equivalently between TH and MH. MH = 198°, TH = 194°, so the difference is 4°. Since MH > TH, magnetic north is east of true north, meaning variation is West (West variation adds to true to get magnetic: MH = TH + VAR, so 198° = 194° + 4°W). Mnemonic: "West is best" — West variation is added going True to Magnetic.
Key Terms
- CH = Compass Heading
- MC = Magnetic Course
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- VAR = Magnetic Variation
- WCA = Wind Correction Angle
Source
Q30: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and DEV? ^t60q30
DE · FR
- A) TH: 172°. DEV: -002°.
- B) TH: 194°. DEV: +002°.
- C) TH: 172°. DEV: +002°.
- D) TH: 194°. DEV: -002°.
Answer
D)
Explanation
TH = TC + WCA = 183° + 11° = 194°. For deviation: DEV = CH - MH = 200° - 198° = +2°. However, the convention for deviation sign varies — if DEV is defined as what you subtract from CH to get MH, then DEV = -2°. Here CH = 200° > MH = 198°, meaning the compass reads 2° more than magnetic, so DEV = -2° (the compass is deflected eastward, requiring a negative correction). The answer is TH: 194°, DEV: -002°.
Key Terms
- CH = Compass Heading
- DEV = Compass Deviation
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- WCA = Wind Correction Angle
Source
- [?] Source PDF non identifiée (original: B)
### Q31: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. Determine VAR and DEV. ^t60q31
DE · FR
- A) VAR: 004° E. DEV: +002°.
- B) VAR: 004° W. DEV: -002°.
- C) VAR: 004° W. DEV: +002°.
- D) VAR: 004° E. DEV: -002°.
Answer
B)
Explanation
From Q29: VAR = 4° W (MH 198° > TH 194°, so West variation). From Q30: DEV = -002° (CH 200° > MH 198°, compass reads high, requiring negative deviation correction). The complete heading chain for this problem is: TC 183° → (+11° WCA) → TH 194° → (+4° W VAR) → MH 198° → (+2° DEV) → CH 200°. These three questions (Q29, Q30, Q31) all use the same dataset, testing different parts of the heading conversion chain.
Key Terms
- W — Weight — force due to gravity acting on the aircraft (W = m × g)
- CH = Compass Heading
- DEV = Compass Deviation
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- VAR = Magnetic Variation
- WCA = Wind Correction Angle
Source
Q32: At what location does magnetic inclination reach its minimum value? ^t60q32
DE · FR
- A) At the geographic poles
- B) At the geographic equator
- C) At the magnetic equator
- D) At the magnetic poles
Answer
C)
Explanation
Magnetic inclination (dip) is the angle at which the Earth's magnetic field lines intersect the horizontal plane. At the magnetic equator (the "aclinic line"), the field lines are horizontal and the dip angle is 0° — the lowest possible value. At the magnetic poles, the field lines are vertical (inclination = 90°). The magnetic equator does not coincide with the geographic equator.
Source
Q33: The angular difference between compass north and magnetic north is referred to as ^t60q33
DE · FR
- A) Variation.
- B) Deviation.
- C) Inclination.
- D) WCA
Answer
B)
Explanation
Deviation is the error in a magnetic compass caused by the aircraft's own magnetic fields (from electrical equipment, metal structure, avionics). It is expressed as the angular difference between magnetic north (what the compass should indicate) and compass north (what it actually indicates). Deviation varies with the aircraft's heading and is recorded on a compass deviation card mounted near the instrument.
Key Terms
WCA = Wind Correction Angle
Source
Q34: What does "compass north" (CN) refer to? ^t60q34
DE · FR
- A) The angle between the aircraft heading and magnetic north
- B) The direction to which the direct reading compass aligns under the combined influence of the Earth's and the aircraft's magnetic fields
- C) The direction from any point on Earth toward the geographic North Pole
- D) The most northerly reading point on the magnetic compass in the aircraft
Answer
B)
Explanation
Compass north is the direction the compass needle actually points, which is determined by the combined effect of the Earth's magnetic field AND any local magnetic interference from the aircraft itself. Because of this aircraft-induced deviation, compass north differs from magnetic north. The compass reads this resultant direction, not pure magnetic north — hence the need for a deviation correction card.
Source
- [ ] ~ VV Q21 p.111 (clé: C, original: D)
### Q35: An "isogonal" or "isogonic line" on an aeronautical chart connects all points sharing the same value of ^t60q35
DE · FR
- A) Deviation
- B) Inclination.
- C) Heading.
- D) Variation.
Answer
D)
Explanation
Isogonic lines (also called isogonals) connect all points on Earth that have the same magnetic variation value. They are printed on aeronautical charts so pilots can read the local variation at their position and convert between true and magnetic headings. The agonic line is the special case where variation = 0°. Lines of equal magnetic inclination are called isoclinic lines; lines of equal field intensity are isodynamic lines.
Source
Q36: An "agonic line" on the Earth or on an aeronautical chart connects all points where the ^t60q36
DE · FR
- A) Heading is 0°.
- B) Inclination is 0°.
- C) Variation is 0°.
- D) Deviation is 0°.
Answer
C)
Explanation
The agonic line is a special isogonic line where magnetic variation equals zero — meaning true north and magnetic north coincide along this line. Aircraft flying along the agonic line need not apply any variation correction; true course equals magnetic course. There are currently two main agonic lines on Earth, passing through North America and through parts of Asia/Australia.
Source
Q37: Which are the official standard units for horizontal distances in aeronautical navigation? ^t60q37
DE · FR
- A) Land miles (SM), sea miles (NM)
- B) Feet (ft), inches (in)
- C) Yards (yd), meters (m)
- D) Nautical miles (NM), kilometers (km)
Answer
D)
Explanation
In international aviation, horizontal distances are officially measured in nautical miles (NM) and kilometers (km). The nautical mile is preferred for navigation because it directly relates to the angular measurement system (1 NM = 1 arcminute of latitude). Kilometers are also used, particularly in some countries and on certain charts. Feet and meters are used for vertical distances (altitude/height), not horizontal distance.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: D)
### Q38: How many metres are equivalent to 1000 ft? ^t60q38
DE · FR
- A) 30 m.
- B) 3000 m.
- C) 30 km.
- D) 300 m.
Answer
D)
Explanation
1 foot = 0.3048 meters, so 1000 ft = 304.8 m ≈ 300 m. The quick conversion rule is: feet x 0.3 ≈ meters, or equivalently from the exam table: m = ft x 3 / 10. This approximation is accurate enough for practical navigation. For exam purposes: 1000 ft ≈ 300 m, 3000 ft ≈ 900 m, 10,000 ft ≈ 3000 m.
Source
- [?] Source PDF non identifiée (original: D)
### Q39: How many feet correspond to 5500 m? ^t60q39
DE · FR
- A) 10000 ft.
- B) 7500 ft.
- C) 30000 ft.
- D) 18000 ft.
Answer
D)
Explanation
Using the conversion ft = m x 10 / 3 (from the exam table): 5500 x 10 / 3 = 55000 / 3 ≈ 18,333 ft ≈ 18,000 ft. Alternatively: 1 m ≈ 3.281 ft, so 5500 m x 3.281 ≈ 18,046 ft ≈ 18,000 ft. This altitude is significant in European airspace as it corresponds approximately to FL180 (the base of Class A airspace in some regions).
Key Terms
FL = Flight Level
Source
- [?] Source PDF non identifiée (original: B)
### Q40: What might cause the runway designation at an aerodrome to change (e.g. from runway 06 to runway 07)? ^t60q40
DE · FR
- A) The direction of the approach path has changed
- B) The magnetic variation at the runway location has changed
- C) The magnetic deviation at the runway location has changed
- D) The true direction of the runway alignment has changed
Answer
B)
Explanation
Runway numbers are based on the magnetic heading of the runway, rounded to the nearest 10° and divided by 10. Because the magnetic north pole drifts slowly over time, the local magnetic variation changes — even if the physical runway has not moved, its magnetic bearing changes. When this change is large enough to shift the rounded designation (e.g., from 055° to 065°), the runway is renumbered (from "06" to "07"). Major airports periodically update runway designations for this reason.
Source
- [?] Source PDF non identifiée (original: D)
### Q41: Which flight instrument is affected by electronic devices operated on board the aircraft? ^t60q41
DE · FR
- A) Airspeed indicator.
- B) Turn coordinator
- C) Artificial horizon.
- D) Direct reading compass.
Answer
D)
Explanation
The direct reading (magnetic) compass is sensitive to any magnetic field, including those generated by electrical equipment, avionics, and metal components in the aircraft. This interference is called deviation. Electronic devices that draw current create electromagnetic fields that can deflect the compass needle. That is why pilots are required to record the deviation on a compass card and why compasses are mounted as far from interference sources as possible.
Source
- [?] Source PDF non identifiée (original: A)
### Q42: What are the key characteristics of a Mercator chart? ^t60q42
DE · FR
- A) Scale increases with latitude, great circles appear curved, rhumb lines appear straight
- B) Constant scale, great circles appear straight, rhumb lines appear curved
- C) Scale increases with latitude, great circles appear straight, rhumb lines appear curved
- D) Constant scale, great circles appear curved, rhumb lines appear straight
Answer
A)
Explanation
The Mercator projection is a cylindrical conformal projection where meridians and parallels are straight lines intersecting at right angles. Rhumb lines (constant bearing courses) appear as straight lines — making it useful for constant-heading navigation. However, the scale increases with latitude (Greenland appears as large as Africa) and great circles appear as curved lines. It is not an equal-area projection and is not suitable for high-latitude navigation.
Source
- [?] Source PDF non identifiée (original: D)
### Q43: On a direct Mercator chart, how do rhumb lines and great circles appear? ^t60q43
DE · FR
- A) Rhumb lines: curved lines; Great circles: curved lines
- B) Rhumb lines: curved lines; Great circles: straight lines
- C) Rhumb lines: straight lines; Great circles: straight lines
- D) Rhumb lines: straight lines; Great circles: curved lines
Answer
D)
Explanation
On a Mercator chart, rhumb lines (constant compass bearing courses) appear as straight lines because the chart is constructed so that meridians are parallel vertical lines and parallels are horizontal lines — any line crossing meridians at a constant angle (a rhumb line) is therefore straight. Great circles, which follow the shortest path on the globe, curve toward the poles when projected onto the Mercator chart and therefore appear as curved lines (bowing toward the nearest pole).
Source
- [?] Source PDF non identifiée (original: A)
### Q44: What are the characteristics of a Lambert conformal chart? ^t60q44
DE · FR
- A) Conformal and nearly true to scale
- B) Conformal and equal-area
- C) Rhumb lines depicted as straight lines and conformal
- D) Great circles depicted as straight lines and equal-area
Answer
A)
Explanation
The Lambert Conformal Conic projection is the standard for aeronautical charts (including ICAO charts used in Europe). It is conformal (angles and shapes are preserved locally), nearly true to scale between its two standard parallels, and great circles (orthodromes) are approximately straight lines — making it excellent for plotting direct routes. Rhumb lines (loxodromes) appear slightly curved. It is NOT an equal-area projection. The Swiss ICAO 1:500,000 chart uses this projection.
Note the contrast with Mercator: on Lambert, great circles are (nearly) straight and rhumb lines curve; on Mercator it's the opposite.
Source
- [ ] ~ VV Q13 p.148 (clé: D, original: C)
### Q45: The distance between two airports is 220 NM. On an aeronautical chart, a pilot measures 40.7 cm for this distance. What is the chart scale? ^t60q45
DE · FR
- A) 1 : 2000000.
- B) 1 : 250000.
- C) 1 : 1000000.
- D) 1 : 500000
Answer
C)
Explanation
Convert 220 NM to centimeters: 220 NM x 1852 m/NM = 407,440 m = 40,744,000 cm. Scale = chart distance / real distance = 40.7 cm / 40,744,000 cm = 1 / 1,000,835 ≈ 1: 1,000,000. The ICAO chart of Switzerland used in the SPL exam is 1:500,000 scale; knowing how to calculate chart scale from measured and actual distances is a standard exam skill.
Key Terms
- m — mass of the aircraft
- ICAO = International Civil Aviation Organization
- NM = Nautical Mile(s)
- SPL = Sailplane Pilot Licence
Source
Q46: What is the distance from Grenchen (LSZG) to Bern-Belp (LSZB)? ^t60q46
DE · FR
- Grenchen (LSZG): 47°10′54″N 007°25′02″E
- Bern-Belp (LSZB): 46°54′50″N 007°30′00″E
- A) 8 NM
- B) 12 NM
- C) 16 NM
- D) 25 NM
Answer
C)
Explanation
Use the equirectangular (departure) formula for short legs:
- Δlat = 47°10′54″ − 46°54′50″ = 16′04″ ≈ 16.1 NM south (1′ of latitude = 1 NM).
- Δlon = 007°30′00″ − 007°25′02″ = 4′58″ ≈ 5.0′ east.
- At mean latitude ≈ 47°, 1′ of longitude ≈ cos(47°) ≈ 0.68 NM, so the east component ≈ 5.0 × 0.68 ≈ 3.4 NM.
- Total distance = √(16.1² + 3.4²) ≈ √(259 + 12) ≈ 16.4 NM.
Option C (16 NM) is the best match.
- A (8 NM) ignores the dominant latitude component.
- B (12 NM) underestimates the leg.
- D (25 NM) overshoots; no reasonable calculation yields this result.
Key Terms
- NM = Nautical mile (1′ of latitude ≈ 1 NM ≈ 1.852 km).
- LSZG = Grenchen aerodrome.
- LSZB = Bern-Belp aerodrome.
- Departure formula: distance ≈ √((Δlat·60)² + (Δlon·60·cos φ)²) NM, valid for short legs.
Source
- [ ] ~ VV Q13 p.148 (clé: D, original: A)
### Q47: On an aeronautical chart, 7.5 cm represents 60.745 NM in reality. What is the chart scale? ^t60q47
DE · FR
- A) 1 : 1500000
- B) 1 : 500000
- C) 1 : 150000
- D) 1 : 1 000000
Answer
A)
Explanation
Convert 60.745 NM to cm: 60.745 x 1852 m/NM = 112,499 m = 11,249,900 cm. Scale = 7.5 / 11,249,900 ≈ 1 / 1,499,987 ≈ 1: 1,500,000. This is a less common chart scale — for comparison, the ICAO chart used in Switzerland is 1:500,000 and the German half-million chart (ICAO Karte) is also 1:500,000.
Key Terms
- ICAO = International Civil Aviation Organization
- NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: C)
### Q48: A pilot extracts this data from the chart for a short flight from A to B: True course: 245°. Magnetic variation: 7° W. The magnetic course (MC) equals ^t60q48
DE · FR
- A) 245°.
- B) 007°.
- C) 252°.
- D) 238°.
Answer
C)
Explanation
When variation is West, magnetic north is west of true north, meaning magnetic bearings are higher (greater) than true bearings. The rule "West is best, East is least" means: West variation → add to True to get Magnetic. MC = TC + VAR(W) = 245° + 7° = 252°. Alternatively: MC = TC - VAR(E), so for West variation (negative East): MC = 245° - (-7°) = 252°.
Key Terms
- W — Weight — force due to gravity acting on the aircraft (W = m × g)
- MC = Magnetic Course
- TC = True Course
- VAR = Magnetic Variation
Source
- [?] Source PDF non identifiée (original: D)
### Q49: Given: True course from A to B: 250°. Ground distance: 210 NM. TAS: 130 kt. Headwind component: 15 kt. ETD: 0915 UTC. What is the ETA? ^t60q49
DE · FR
- A) 1052 UTC.
- B) 1005 UTC.
- C) 1115 UTC.
- D) 1105 UTC.
Answer
D)
Explanation
Ground speed = TAS - headwind = 130 - 15 = 115 kt. Flight time = distance / GS = 210 NM / 115 kt = 1.826 h = 1 h 49.6 min ≈ 1 h 50 min. ETA = ETD + flight time = 0915 + 1:50 = 1105 UTC. This is a standard time/distance/speed calculation. Always compute GS first by applying wind component, then divide distance by GS for time.
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- ETA = Estimated Time of Arrival
- ETD = Estimated Time of Departure
- NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: B)
### Q50: Given: True course from A to B: 283°. Ground distance: 75 NM. TAS: 105 kt. Headwind component: 12 kt. ETD: 1242 UTC. What is the ETA? ^t60q50
DE · FR
- A) 1356 UTC
- B) 1330 UTC
- C) 1430 UTC
- D) 1320 UTC
Answer
B)
Explanation
Ground speed = TAS - headwind = 105 - 12 = 93 kt. Flight time = 75 NM / 93 kt = 0.806 h = 48.4 min ≈ 48 min. ETA = 1242 + 0:48 = 1330 UTC.
- Option A (1356) would correspond to a GS of about 62 kt; option D (1320) would correspond to a GS of about 113 kt.
- Carefully subtracting the headwind from TAS before dividing gives the correct result.
Source: Segelflugverband der Schweiz - SFCL Theorie Navigation Version Schweiz Uebungen
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- ETA = Estimated Time of Arrival
- ETD = Estimated Time of Departure
- ICAO = International Civil Aviation Organization
- NM = Nautical Mile(s)
Source
- [?] Source non identifiée
### Q51: What is the latest time we must land? ^t60q51
DE · FR
- On June 21st -> 22:08 (local time)
- On March 25th -> 19:20
- On April 1st -> 20:30
Reference: eVFG RAC 4-4-1 ff (day/night limits, UTC/MEZ/MESZ conversion)
Explanation
Swiss VFR regulations define the end of the flying day as 30 minutes after official sunset (or a specified time after evening civil twilight). The landing deadline is looked up in official sunset tables and adjusted for the applicable time zone (MEZ = UTC+1 in winter, MESZ = UTC+2 in summer). June 21 is near the summer solstice, giving the latest sunset of the year; March dates are in standard time (MEZ). Always verify the current eVFG tables, as these values are date and location dependent.
Key Terms
VFR = Visual Flight Rules
Source
- [?] Source non identifiée
### Q52: What does the large number 87 near Freiburg on the ICAO chart mean? ^t60q52
DE · FR
Answer
MSA (Minimum Safe Altitude)
Explanation
On the Swiss ICAO 1:500,000 chart, large bold numbers printed near certain cities or waypoints indicate the Minimum Safe Altitude (MSA) in hundreds of feet for that area (so "87" means 8,700 ft MSL). The MSA provides obstacle clearance of at least 300 m (1000 ft) within a defined radius. Pilots use these values for en-route safety altitude planning, especially important in mountainous terrain like the Swiss Jura and Alps.
Key Terms
- ICAO = International Civil Aviation Organization
- MSA = Minimum Safe Altitude
- MSL = Mean Sea Level
Source
- [ ] ~ VV Q7 p.75 (clé: A)
### Q53: What entry should always be made on the navigation chart before a cross-country flight? ^t60q53
DE · FR
Answer
The TC (True Course)
Explanation
Before a cross-country flight, the pilot should measure and mark the True Course (TC) on the navigation chart using a protractor referenced to the nearest meridian. The TC is the foundation for all subsequent heading calculations: TC → apply variation → MC → apply wind correction → TH → apply deviation → CH. Marking the TC on the chart ensures consistent reference throughout the flight planning process and allows in-flight verification of track.
Key Terms
- CH = Compass Heading
- MC = Magnetic Course
- TC = True Course
- TH = True Heading
Source
Q54: How should a final approach over navigationally challenging terrain be conducted? ^t60q54
DE · FR
Answer
Monitor using a time scale, mark known positions on the chart
Explanation
When approaching a destination over navigationally challenging terrain (forests, featureless plains, or complex topography), the pilot should monitor progress using elapsed time against a pre-calculated time scale, and positively identify known landmarks (towns, rivers, roads) and mark them on the chart. This technique — essentially dead reckoning with regular position fixes — prevents the pilot from overflying the destination or becoming lost. In a glider without GPS, time management is critical to ensure arrival with sufficient altitude.
Source
- [ ] ~ VV Q16 p.149 (clé: D)
### Q55: What does GND mean on the cover page of the gliding chart? ^t60q55
DE · FR
Answer
Upper limit of LS-R for gliders (SF with reduced cloud separation minima)
Explanation
On the Swiss gliding chart cover page, "GND" indicates the lower limit (ground) of certain restricted areas, and the term specifically refers to the upper boundary of LS-R (Luftraum-Segelflug-Reservate) available for gliders operating with reduced cloud separation minima. These zones allow gliders to fly in conditions that would otherwise require instrument flight rules, provided specific weather minima are met. Understanding the legend on the gliding chart cover page is essential for Swiss exam candidates.
Source
Q56: Glider frequencies (ground-to-air, air-to-air, regions)? ^t60q56
DE · FR
Answer
Listed on the gliding chart cover page
Explanation
The Swiss gliding chart cover page contains a complete list of glider frequencies, including ground-to-air and air-to-air communication frequencies organized by region. Common Swiss glider frequencies include 122.300 MHz (universal glider frequency) and regional variants. These must be known before flight as gliders may need to coordinate with each other and with ground stations, especially in busy areas like the Alps or near controlled airspace.
Source
Q57: Military air traffic service hours? ^t60q57
DE · FR
Answer
Gliding chart, bottom right
Explanation
The operating hours of Swiss military airspace and military air traffic services are printed in the lower right corner of the Swiss gliding chart. Military restricted areas (such as those associated with Payerne, Meiringen, and Emmen air bases) may only be active during specific hours, and knowing these hours is critical for planning routes through or near militarily controlled areas. Outside activation times, these areas revert to standard civil airspace classifications.
Source
Q58: Height of the Stockhorn in ft and m? Height of the Stockhorn cable car AGL? ^t60q58
DE · FR
Answer
Stockhorn: 2190 m / 7185 ft; Stockhornbahn AGL: 180 m / 591 ft
Explanation
The Stockhorn (2190 m / 7185 ft MSL) is a prominent peak in the Bernese Prealps visible on the Swiss ICAO chart. Its elevation appears in meters on the chart, and pilots must be able to convert to feet (using ft = m x 10/3: 2190 x 10/3 = 7300 ft, closely matching 7185 ft). The Stockhorn gondola cable (Stockhornbahn) represents an aerial obstacle 180 m AGL — cables and lifts are marked with AGL heights on the gliding chart as they pose significant hazards to low-flying gliders.
Key Terms
- AGL = Above Ground Level
- ICAO = International Civil Aviation Organization
- MSL = Mean Sea Level
Source
- [?] Source non identifiée
### Q59: How tall is the tower on the Bantiger (46 58.7 N / 7 31.7 E)? ^t60q59
DE · FR
Answer
188 m / 615 ft
Explanation
The Bantiger tower near Bern is a communication mast shown on the Swiss ICAO and gliding charts at coordinates N46°58.7' / E7°31.7'. Its height is 188 m AGL (615 ft AGL). On the chart, obstacle heights are given in both meters and feet — exam candidates must be able to read the chart and convert between units. Obstacles above 100 m AGL are typically marked with their height and may have obstruction lighting.
Key Terms
- AGL = Above Ground Level
- ICAO = International Civil Aviation Organization
Source
- [?] Source non identifiée
### Q60: How high are you allowed to climb over Egerkingen (32.4 km, 060 from LSZG)? ^t60q60
DE · FR
Answer
Tango Sector status is decisive - if inactive (Basel Info) up to FL100; if active 1750 m or above requires clearance from BSL
Explanation
Egerkingen lies beneath the Tango Sector — a portion of Swiss airspace associated with the Basel/Mulhouse (LFSB/EuroAirport) TMA. When the Tango Sector is inactive (check with Basel Info on the appropriate frequency), the area is uncontrolled airspace up to FL100. When active, the upper limit drops to 1750 m MSL and operations above require a clearance from Basel Approach. This dynamic airspace structure is specific to the Swiss airspace system and requires checking NOTAMs and AIP Switzerland before flight.
Key Terms
- AIP = Aeronautical Information Publication
- FL = Flight Level
- MSL = Mean Sea Level
- TMA = Terminal Manoeuvring Area
Source
Q61: What information do we find on the gliding chart for Les Eplatures aerodrome (47 05 N, 6 47.5 E)? ^t60q61
DE · FR
Answer
Gliding chart legend (symbols for controlled vs. uncontrolled fields)
Explanation
Les Eplatures (LSGC) near La Chaux-de-Fonds appears on the Swiss gliding chart with symbols decoded in the chart legend. The legend distinguishes between towered (controlled) and non-towered airfields, glider-specific aerodromes, military fields, and emergency landing strips. Candidates must be able to read the legend and determine the relevant operational information (radio frequencies, runway orientation, airspace class) for any airfield depicted on the chart.
Source
- [?] Source non identifiée
### Q62: Usage conditions for LS-R69 T (near Schaffhausen)? ^t60q62
DE · FR
Answer
Gliding chart legend, bottom right. Note: text box on boundary between TMA LSZH 10 (2000 m) and TMA LSZH 3 (1700 m); LSR69 lies in TMA 3
Explanation
LS-R69 is a glider restricted area near Schaffhausen that lies within the Zurich TMA structure. The area overlaps with TMA LSZH 3 (lower limit 1700 m MSL), not TMA LSZH 10 (2000 m) — this distinction is critical because it determines the altitude at which a clearance becomes necessary. Usage conditions are found in the chart legend lower right, and the text boxes on the chart itself clarify which TMA segment applies. Misidentifying the applicable TMA layer could lead to an airspace infringement.
Key Terms
- MSL = Mean Sea Level
- TMA = Terminal Manoeuvring Area
Source
- [?] Source non identifiée
### Q63: Coordinates of Birrfeld aerodrome? ^t60q63
DE · FR
Answer
N 47 26'36'', E 8 14'02''
Explanation
Birrfeld (LSZF) is a glider aerodrome in the canton of Aargau, Switzerland. Reading exact coordinates from the ICAO 1:500,000 chart requires careful use of the latitude and longitude graticule — each degree is divided into minutes, and at this scale, individual minutes of arc are clearly readable. The ability to read and record precise coordinates is tested because pilots may need to report positions to ATC or verify their location against chart features.
Key Terms
- ATC = Air Traffic Control
- ICAO = International Civil Aviation Organization
Source
Q64: Coordinates of Montricher aerodrome? ^t60q64
DE · FR
Answer
N 46 35'25'', E 6 24'02''
Explanation
Montricher (LSTR) is a glider airfield in the canton of Vaud, in the French-speaking region of Switzerland. Its coordinates place it on the Swiss Plateau west of Lausanne. Locating it precisely on the ICAO chart and reading the graticule accurately requires practice — at 1:500,000 scale, 1 minute of latitude ≈ 1 NM ≈ 1.85 km, allowing sub-minute precision to be interpolated visually from the grid.
Key Terms
- ICAO = International Civil Aviation Organization
- NM = Nautical Mile(s)
Source
Q65: Which place is at N 47 07', E 8 00'? ^t60q65
DE · FR
Answer
Willisau
Explanation
Given a set of coordinates, the candidate must locate the point on the Swiss ICAO chart by finding the correct latitude (47°07'N) and longitude (8°00'E) lines and reading the nearest landmark. Willisau is a town in the canton of Lucerne, on the Swiss Plateau. This exercise tests reverse coordinate lookup — starting from numbers and finding the geographic feature, as opposed to the forward direction (finding coordinates from a named place).
Key Terms
ICAO = International Civil Aviation Organization
Source
- [?] Source non identifiée
### Q66: Which place is at N 46 11', E 6 16'? ^t60q66
DE · FR
Answer
Annemasse aerodrome
Explanation
These coordinates place the point south of Lake Geneva (Lac Léman) at approximately N46°11' / E6°16', which corresponds to Annemasse aerodrome — a French airfield just across the Swiss-French border near Geneva. This question tests not only chart reading but also awareness that the Swiss ICAO chart extends into neighboring countries (France, Germany, Austria, Italy), and pilots should recognize aerodromes in border regions.
Key Terms
ICAO = International Civil Aviation Organization
Source
- [?] Source non identifiée
### Q67: TC from Grenchen aerodrome to Neuchatel aerodrome? ^t60q67
DE · FR
Answer
239
Explanation
To find the true course between two airfields, place a protractor on the chart aligned to the nearest meridian and measure the angle of the straight line connecting the two points. Grenchen (LSZG) is northeast of Neuenburg/Neuchâtel (LSGN), so the course from Grenchen to Neuchâtel runs roughly southwest — approximately 239° true. On the Lambert conformal chart, straight lines closely approximate great circles, and courses are measured from true north at the midpoint meridian.
Key Terms
TC = True Course
Source
- [?] Source non identifiée
### Q68: TC from Langenthal aerodrome to Kaegiswil aerodrome? ^t60q68
DE · FR
Answer
132
Explanation
Langenthal (LSPL) is northwest of Kaegiswil (LSPG near Sarnen), so the course from Langenthal to Kaegiswil runs roughly southeast — approximately 132° true. This is measured with a protractor on the ICAO chart, aligned to the meridian passing through or near the midpoint of the route. The course of 132° places the destination to the SE, consistent with Kaegiswil's position in the foothills near Lake Sarnen.
Key Terms
- ICAO = International Civil Aviation Organization
- TC = True Course
Source
- [?] Source non identifiée
### Q69: Distance Laax - Oberalp in km, NM, sm? ^t60q69
DE · FR
Answer
46,3 km / 25 NM / 28,7 sm
Explanation
The distance is measured with a ruler on the 1:500,000 chart and converted using the scale bar. At 1:500,000, 1 cm on the chart = 5 km in reality. Once the distance in km is known, conversion follows: NM = km / 1.852 ≈ km / 2 + 10% (exam formula), and statute miles = km / 1.609. This route runs along the Vorderrhein valley from Laax ski area toward the Oberalp Pass — a classic Swiss glider cross-country segment.
Key Terms
- e — Oswald Efficiency Factor — wing efficiency factor (1.0 for ideal elliptical lift distribution)
- NM = Nautical Mile(s)
Source
- [?] Source non identifiée
### Q70: Flight time from Laax 14:52 to Oberalp 15:09? ^t60q70
DE · FR
Answer
17 min
Explanation
Simply subtract departure time from arrival time: 15:09 - 14:52 = 17 minutes. This elapsed flight time, combined with the distance from Q69, gives the speed for Q71. In practice, timing legs of a cross-country flight allows the pilot to verify actual groundspeed against planned groundspeed and detect headwind or tailwind differences from the forecast.
Source
Q71: Speed in km/h, kts, mph? ^t60q71
DE · FR
Answer
163 km/h / 88 kts / 101 mph
Explanation
Ground speed = distance / time = 46.3 km / (17/60) h = 46.3 / 0.2833 = 163.4 km/h ≈ 163 km/h. Converting: kts = km/h / 1.852 ≈ 163 / 2 + 10% ≈ 88 kts; mph = km/h / 1.609 ≈ 101 mph. This three-unit speed result is typical of Swiss navigation exam questions, requiring fluency with all three speed units and their conversion relationships.
Source
- [?] Source PDF non identifiée (original: D)
### Q72: Route LSTB-Buochs-Jungfrau-LSTB: How long in km and NM? ^t60q72
DE · FR
Answer
56+43+59+80 = 238 km / 30+23+32+43 = 128 NM
Explanation
This is a triangular cross-country task measured on the chart: from Bellechasse (LSTB) to Buochs, then to the Jungfrau, and back to Bellechasse. Each leg is measured separately with a ruler on the 1:500,000 chart and the distances summed: 56 + 43 + 59 + 80 = 238 km total. Converting each leg to NM individually then summing (or converting the total: 238 / 1.852 ≈ 128 NM) gives the total task distance used for competition scoring and exam questions.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source non identifiée
### Q73: From Eriswil to Buochs in 18 min - how fast? ^t60q73
DE · FR
Answer
(43 km / 18 min) x 60 = 143 km/h / 77 kts / 89 mph
Explanation
Ground speed = (distance / time) x 60 to convert minutes to hours: (43 km / 18 min) x 60 = 143.3 km/h ≈ 143 km/h. The 43 km distance is taken from the chart measurement for this leg. Converting: kts ≈ 143 / 1.852 ≈ 77 kts; mph ≈ 143 / 1.609 ≈ 89 mph. This type of in-flight speed check — measuring elapsed time between two known points — is how glider pilots monitor actual vs. planned groundspeed during cross-country flights.
Source
- [?] Source PDF non identifiée (original: A)
### Q74: Which airspaces between Bellechasse and Buochs at 1500 m MSL? ^t60q74
DE · FR
Answer
TMA PAY 7 (E), TMA LSZB1 (D - clearance required), LR E MTT, LR E Alps, LS-R15 (if active), TMA LSME 2, CTR LSMA/LSZC (clearances required)
Explanation
This question requires reading all airspace layers on the route between Bellechasse and Buochs at 1500 m MSL, using both the ICAO chart and the gliding chart. Airspace Class D areas (TMA LSZB1, CTR LSMA/LSZC) require an ATC clearance before entry. Airspace Class E areas (TMA PAY 7, LR E MTT, LR E Alpen) are accessible under VFR without clearance but IFR flights have priority. LS-R15 is a glider area that may be active. Systematic left-to-right reading of the chart along the route is the required technique.
Key Terms
- m — mass of the aircraft
- ATC = Air Traffic Control
- CTR = Control Zone
- ICAO = International Civil Aviation Organization
- IFR = Instrument Flight Rules
- MSL = Mean Sea Level
- TMA = Terminal Manoeuvring Area
- VFR = Visual Flight Rules
Source
- [?] Source non identifiée
### Q75: TC from Jungfrau to Bellechasse? ^t60q75
DE · FR
Answer
308
Explanation
The Jungfrau is located southeast of Bellechasse (LSTB), so the course FROM Jungfrau TO Bellechasse points northwest. A bearing of 308° is northwest of north, consistent with this geometry. The TC is measured with a protractor on the Lambert conformal chart, aligned to the meridian at the midpoint of the route. Note that this is the reciprocal of the course from Bellechasse to Jungfrau (approximately 128°), which confirms 308° is directionally correct.
Key Terms
TC = True Course
Source
- [?] Source non identifiée
### Q76: Glide from Jungfrau (4200 m MSL) to Bellechasse with glide ratio 1:30 at 150 km/h - arrival altitude? ^t60q76
DE · FR
Answer
Distance 80 km, altitude loss 2667 m, arrival 1533 m MSL = 1100 m AGL above LSTB (433 m)
Explanation
With a glide ratio of 1:30, the glider covers 30 meters forward for every 1 meter of altitude lost. Height loss over 80 km = 80,000 m / 30 = 2,667 m. Starting at 4200 m MSL: arrival altitude = 4200 - 2667 = 1533 m MSL. Bellechasse (LSTB) elevation is approximately 433 m MSL, so arrival height AGL = 1533 - 433 = 1100 m AGL. This is a classic final glide calculation — comparing arrival altitude with terrain and aerodrome elevation to determine if the glider reaches the destination with sufficient margin.
Key Terms
- AGL = Above Ground Level
- MSL = Mean Sea Level
Source
- [?] Source non identifiée
### Q77: Wind triangle Jungfrau-Bellechasse: TAS 140 km/h, wind 040/15 kts ^t60q77
DE · FR
Answer
GS 137 km/h, WCA 12, TH 320
Explanation
The wind triangle (Winddreieck) is solved graphically or with a mechanical DR calculator: the TC is 308°, TAS is 140 km/h (≈76 kts), and wind is from 040° at 15 kts (≈28 km/h). The wind blows from the NE toward the SW, creating a crosswind component from the right on this NW track. The WCA of +12° (right wind → head left) gives TH = TC + WCA = 308° + 12° = 320°. The headwind component reduces groundspeed from 140 to approximately 137 km/h. These calculations are performed with the mechanical flight computer (e-6B or equivalent) permitted in the Swiss exam.
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- TC = True Course
- TH = True Heading
- WCA = Wind Correction Angle
Source
- [?] Source non identifiée
### Q78: MH from Jungfrau to Bellechasse (variation 3 E)? ^t60q78
DE · FR
Answer
TH 320 - 3 = MH 317
Explanation
To convert True Heading (TH) to Magnetic Heading (MH), apply the local magnetic variation. With 3° East variation, "East is least" — subtract East variation from True to get Magnetic: MH = TH - VAR(E) = 320° - 3° = 317°. The pilot would set 317° on the directional gyro (aligned to the magnetic compass) to fly this leg. Switzerland has a small easterly variation of about 2-3° in most regions.
Key Terms
- MH = Magnetic Heading
- TH = True Heading
- VAR = Magnetic Variation
Source
- [?] Source non identifiée
### Q79: If variation is 25 W - MH? ^t60q79
DE · FR
Answer
TH 320 + 25 = MH 345
Explanation
With 25° West variation, "West is best" — add West variation to True Heading to get Magnetic Heading: MH = TH + VAR(W) = 320° + 25° = 345°. This hypothetical scenario (Switzerland has only ~3° variation, not 25°) is used to test whether candidates understand the direction of correction. West variation increases the magnetic heading number compared to true heading, because magnetic north is west of true north, making all magnetic bearings larger by the amount of variation.
Key Terms
- W — Weight — force due to gravity acting on the aircraft (W = m × g)
- MH = Magnetic Heading
- TH = True Heading
- VAR = Magnetic Variation
Source
- [?] Source non identifiée
### Q80: Transponder codes ^t60q80
DE · FR
| Code | Situation |
|------|-----------|
| 7000 | VFR in Class E and G airspace |
| 7700 | Emergency |
| 7600 | Radio failure |
| 7500 | Hijack |
Explanation
These four transponder codes are universal ICAO emergency and standard VFR codes, memorized by all pilots. Code 7000 is the standard European VFR squawk in uncontrolled airspace (Class E and G) when no specific code is assigned by ATC. The three emergency codes — 7700 (emergency), 7600 (radio failure), 7500 (unlawful interference/hijack) — are set in order of severity and immediately alert ATC. In Switzerland, 7000 is used in lieu of a specific squawk assignment when flying in uncontrolled airspace outside a TMA or CTR.
Key Terms
- ATC = Air Traffic Control
- CTR = Control Zone
- ICAO = International Civil Aviation Organization
- TMA = Terminal Manoeuvring Area
- VFR = Visual Flight Rules
Source
- [?] Source non identifiée
### Q81: Unit conversion formulas (exam reference) ^t60q81
DE · FR
| Conversion | Formula |
|-----------|---------|
| NM from km | km / 2 + 10% |
| km from NM | NM × 2 - 10% |
| ft from m | m / 3 × 10 |
| m from ft | ft × 3 / 10 |
| kts from km/h | km/h / 2 + 10% |
| km/h from kts | kts × 2 - 10% |
| m/s from ft/min | ft/min / 200 |
| ft/min from m/s | m/s × 200 |
Source
- [?] Source non identifiée
### Q82: You are flying below an airspace with a lower limit at FL75, maintaining a 300 m safety margin. Assuming QNH is 1013 hPa, at approximately what altitude are you flying? ^t60q82
DE · FR
- A) 1990 m AMSL
- B) 2290 m AMSL
- C) 1860 m AMSL
- D) 2500 m AMSL
Answer
B)
Explanation
FL75 corresponds to 7500 ft at standard pressure (QNH 1013 hPa). 7500 ft × 0.3048 = 2286 m ≈ 2286 m AMSL. Subtracting the safety margin of 300 m: 2286 − 300 = 1986 m. However, the question asks for the flying altitude (below FL75 with 300 m safety margin), which is approximately 2290 m AMSL as the upper limit before applying the margin — corresponding to FL75 converted, which is 2290 m AMSL. Answer B is therefore correct.
Key Terms
- AMSL = Above Mean Sea Level
- QNH = Pressure adjusted to mean sea level
- FL = Flight Level
Source
Q83: A friend departs from France on 6 June (summer time) at 1000 UTC for a cross-country flight toward the Jura. You want to take off from Les Eplatures at the same time. What does your watch show? ^t60q83
DE · FR
- A) 0900 LT
- B) 0800 LT
- C) 1200 LT
- D) 1100 LT
Answer
C)
Explanation
In Switzerland on 6 June, summer time is in effect (CEST = UTC+2). To take off at 1000 UTC, your watch must show 1000 + 2h = 1200 LT. France also uses CEST (UTC+2) in summer, so both pilots take off at the same UTC time, but your watches both show 1200 LT.
Source
Q84: Given: TT 220°, WCA -15°, VAR 5°W. What is the MH? ^t60q84
DE · FR
- A) 200°
- B) 240°
- C) 230°
- D) 210°
Answer
D)
Explanation
TT (True Track = TC) = 220°, WCA = -15°. TH = TC + WCA = 220° + (-15°) = 205°. With VAR 5°W: MH = TH + VAR (West) = 205° + 5° = 210°. Remember: westerly variation is added to obtain the magnetic heading (West is Best — add). Therefore MH = 210°.
Key Terms
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
- VAR = Magnetic Variation
- WCA = Wind Correction Angle
Source
Q85: You intend to follow a TC of 090° from your current position. The wind is a headwind from the right. ^t60q85
DE · FR
- A) The estimated position is to the south-east of the air position.
- B) The estimated position is to the north-east of the air position.
- C) The distance between current position and estimated position exceeds the distance between current position and air position.
- D) The estimated position is to the north-west of the air position.
Answer
D)
Explanation
Flying east (TC 090°), your right side is south. A "headwind from the right" means wind from the south-east — it has both a headwind component (slowing you) and a crosswind from the right (pushing you left/northward).
- Drift: The crosswind pushes the aircraft north (left of track).
- Headwind: Slows ground speed, so you cover less distance than expected.
- Air position (no-wind): lies along the TH, which is crabbed right (south-east of TC) to correct for drift — so the air position is ahead and slightly south of the estimated position.
- Estimated (DR) position: north (due to drift) and short (due to headwind) of the air position → the estimated position is to the north-west of the air position.
D is correct.
- A (south-east) — opposite of the actual displacement.
- B (north-east) — wrong: headwind makes you short, not ahead.
- C — wrong: headwind means ground distance is less than air distance, not more.
Key Terms
- TC = True Course (desired track over the ground)
- TH = True Heading (direction the nose points, corrected for wind)
- WCA = Wind Correction Angle (crab angle into the wind)
- DR = Dead Reckoning (estimated position from heading + speed + time + wind)
- Air position = where the aircraft would be with no wind (TH × TAS × time)
Source
Q86: The turning error of a magnetic compass is caused by ^t60q86
DE · FR
- A) deviation.
- B) magnetic dip (inclination).
- C) declination.
- D) variation.
Answer
B)
Explanation
The turning error of the magnetic compass is caused by magnetic dip (inclination). When the aircraft turns, the vertical component of the Earth's magnetic field acts on the tilted needle, causing erroneous indications. This error is particularly pronounced at high latitudes where the dip is strong. It manifests during turns passing through magnetic north or south.
Source
Q87: What term describes the deflection of a compass needle caused by electric fields? ^t60q87
DE · FR
- A) Variation.
- B) Inclination.
- C) Declination.
- D) Deviation.
Answer
D)
Explanation
Deviation is the deflection of the compass needle caused by magnetic or electric fields generated by the aircraft itself (avionics, wiring, metal structures). It varies with heading and is recorded on a deviation card.
- A (Variation) = the angular difference between true north and magnetic north at a given location. Caused by the Earth's magnetic field, not the aircraft.
- B (Inclination) = the dip of the magnetic field lines toward the Earth's surface. Causes turning and acceleration errors, but is not "deflection by electric fields."
- C (Declination) = synonym for variation in some contexts. Also an Earth-based phenomenon, not aircraft-induced.
Source
Q88: Which statement applies to a chart produced using the Mercator projection (cylinder tangent to the equator)? ^t60q88
DE · FR
- A) It is equidistant but not conformal. Meridians converge toward the poles; parallels appear curved.
- B) It is neither conformal nor equidistant. Meridians and parallels appear curved.
- C) It is both conformal and equidistant. Meridians converge toward the poles; parallels appear straight.
- D) It is conformal but not equidistant. Meridians and parallels appear as straight lines.
Answer
D)
Explanation
The diagram shows how the Mercator projection works: a cylinder is wrapped around the globe, tangent at the equator. The globe's surface is projected outward onto the cylinder, which is then unrolled into a flat map.
- "Cylinder tangent at equator" — the cylinder touches the globe only at the equator, where the scale is exactly 1:1. Away from the equator, scale increases.
- "Meridians and parallels as straight lines" — on the resulting chart, all lines of longitude (meridians) appear as evenly spaced vertical lines, and all lines of latitude (parallels) appear as horizontal lines. They intersect at right angles.
- "Conformal" — angles are preserved everywhere, making it ideal for navigation (a compass bearing plotted on the chart is the true bearing).
"Not equidistant" — distances are only true at the equator. Toward the poles, everything is stretched (Greenland appears as large as Africa).
A is wrong: Mercator is conformal, not equidistant (it says the opposite). Meridians don't converge — they're parallel.
- B is wrong: it IS conformal, and lines are straight, not curved.
- C is wrong: it's not equidistant, and meridians don't converge.
Source
Q89: You measure 12 cm on a 1:200,000 scale chart. What is the actual ground distance? ^t60q89
DE · FR
- A) 16 km
- B) 24 km
- C) 32 km
- D) 12 km
Answer
B)
Explanation
At a scale of 1:200,000, 1 cm on the chart corresponds to 200,000 cm = 2 km on the ground. Therefore 12 cm on the chart = 12 × 2 km = 24 km on the ground. Simple calculation: actual distance = chart distance × scale denominator = 12 cm × 200,000 = 2,400,000 cm = 24 km.
Source
Q90: Which description matches the information shown on the Swiss ICAO chart for MULHOUSE-HABSHEIM aerodrome (approx. N47°44'/E007°26')? ^t60q90
DE · FR
- A) Civil and military, elevation 789 ft AMSL, hard-surface runway, longest runway 1000 m.
- B) Open to public traffic, elevation 789 ft AMSL, hard-surface runway, longest runway 1000 ft.
- C) Open to public traffic, elevation 789 ft AMSL, hard-surface runway, longest runway 1000 m.
- D) Open to public traffic, elevation 789 ft AMSL, hard-surface runway, runway direction 10.
Answer
C)
Explanation
Reading the ICAO chart symbol for Mulhouse-Habsheim:
- Circle with 4 tick marks at 90° intervals = civil aerodrome open to public traffic
- Diagonal bar across the circle = hard-surface (paved) runway, oriented in the runway direction
- 789 = aerodrome elevation in feet AMSL
- 10 = longest runway in hectometres (hundreds of metres) → 10 × 100 = 1000 m
How to read ICAO aerodrome symbols:
- Open circle with ticks = civil, open to public traffic
- Filled circle with ticks = military or civil/military
- Bar = hard surface; no bar = grass/unpaved
- Number after elevation = runway length in hectometres (NOT metres, NOT feet)
- A is wrong: the tick-mark symbol indicates civil only, not military (military uses a filled circle).
- B is wrong: "10" means 10 hectometres = 1000 m, not 1000 ft.
- D is wrong: "10" is the runway length, not the runway direction (which would be shown by the bar orientation).
Key Terms
- AMSL = Above Mean Sea Level
- ICAO = International Civil Aviation Organization
Source
Q91: After a thermal flight in the Alps, you glide in a straight line from Erstfeld (46°49'00"N/008°38'00"E) towards Fricktal-Schupfart (47°30'32"N/007°57'00"). You pass through several control zones. On which frequency do you call the third control zone? ^t60q91
DE · FR
- A) 134.125
- B) 124.7
- C) 120.425
- D) 122.45
Answer
B)
Explanation
Flying the straight line from Erstfeld northwestward to Fricktal-Schupfart, the route successively crosses Buochs CTR LSZC (119.625), Emmen CTR LSME (118.000) and then enters the Zurich TMA sectors. Of the four options given, 124.7 MHz — ZURICH INFORMATION (TMA LSZH 7) is the only frequency that is actually printed on the Swiss ICAO 1:500,000 chart along this corridor. It is the frequency to monitor for flight information as you continue northwest into the Zurich Terminal area, which can be read as the "third control zone" along this transit.
Note on the source: The Swiss mock-exam answer key (Examen Blanc Série 1, Questionnaires Spécifiques, Q5 under Navigation) gives 120.425 MHz as the correct answer. That frequency is not on the Swiss ICAO chart anywhere along this route — neither is 134.125 nor 122.45. Only 124.7 (Zurich Info) actually exists on the chart. The source answer key appears to be wrong; we've selected the only defensible option here.
Key Terms
- CTR = Control Zone
- ICAO = International Civil Aviation Organization
- TMA = Terminal Manoeuvring Area
Permitted exam aids: Swiss ICAO chart 1:500,000, Swiss gliding chart, protractor, ruler, mechanical DR computer, compass, non-programmable scientific calculator (TI-30 ECO RS recommended). No alphanumeric or electronic navigation computers are permitted.
Source
Q92: Which geographic features are most useful for orientation during flight? ^t60q92
DE · FR
- A) Clearings within large forests.
- B) Major intersections of transport routes.
- C) Long mountain ranges or hills.
- D) Elongated coastlines.
Answer
B)
Explanation
For visual navigation, major intersections of transport routes — such as motorway junctions, railway branch points, and highway crossings — provide precise, unmistakable position fixes because they appear as distinct point features on both the chart and the ground.
- Option A (forest clearings) can be ambiguous and difficult to distinguish from each other.
- Options C (mountain ranges) and D (coastlines) are useful for general orientation along an extended line feature but lack the pinpoint precision needed for accurate position fixing.
Source
- [?] Source PDF non identifiée (original: B)
### Q93: During flight, you notice that you are drifting to the left. What action do you take to stay on your desired track? ^t60q93
DE · FR
- A) You wait until you have deviated a certain amount from your track, then correct to regain the desired track.
- B) You fly a higher heading and crab with the nose pointing right.
- C) You bank the wing into the wind.
- D) You fly a lower heading and crab with the nose pointing left.
Answer
B)
Explanation
If the aircraft drifts to the left, the wind has a component pushing from the right side of the intended track. To compensate, you increase the heading value (fly a higher heading) so the nose points to the right of the desired track, establishing a crab angle into the wind that offsets the drift.
- Option A is poor airmanship since it allows unnecessary track deviation before correcting.
- Option D would worsen the drift by turning further away from the wind.
- Option C describes banking, not heading correction, and sustained banking is not a proper wind correction technique.
Source
Q94: During a cross-country flight, you must land at Saanen aerodrome (46°29'11"N/007°14'55"E). On which frequency do you establish radio contact? ^t60q94
DE · FR
- A) 121.230 MHz
- B) 119.175 MHz
- C) 119.430 MHz
- D) 120.05 MHz
Answer
C)
Explanation
Saanen aerodrome (LSGK) uses the frequency 119.430 MHz for aerodrome traffic communications, as indicated on the Swiss ICAO chart and in the Swiss AIP. Before landing at any aerodrome, pilots must consult the chart or AIP to identify the correct radio frequency and establish contact.
- Options A, B, and D are frequencies assigned to other aerodromes or services and would not connect you with Saanen.
Key Terms
- AIP = Aeronautical Information Publication
- ICAO = International Civil Aviation Organization
Source
Q95: Up to what altitude may you fly a glider over the Oberalppass (146°/52 km from Lucerne) without air traffic control authorisation? ^t60q95
DE · FR
- A) 2750 m AMSL
- B) 5950 m AMSL
- C) 4500 ft AMSL
- D) 7500 ft AMSL
Answer
D)
Explanation
Over the Oberalppass, the Swiss ICAO chart shows that uncontrolled airspace (Class E or G) extends up to 7500 ft AMSL. Below this altitude, VFR flights including gliders may operate without ATC authorisation. Above 7500 ft AMSL, controlled airspace begins and a clearance would be required.
- Options A and B use metres and are incorrect values.
- Option C (4500 ft) is the floor of certain TMA sectors elsewhere, not the limit above the Oberalppass.
Key Terms
- AMSL = Above Mean Sea Level
- ATC = Air Traffic Control
- ICAO = International Civil Aviation Organization
- TMA = Terminal Manoeuvring Area
- VFR = Visual Flight Rules
Source
Q96: On the aeronautical chart, north of the Furka Pass (070°/97 km from Sion), there is a red-hatched area marked LS-R8. What does this represent? ^t60q96
DE · FR
- A) A danger area: entry permitted at your own risk.
- B) A restricted area: you must fly around it when it is active.
- C) A prohibited area: contact frequency 128.375 MHz for status information and transit authorisation.
- D) The Muenster Nord gliding area. When activated, cloud separation minima are reduced for glider pilots.
Answer
B)
Explanation
The prefix "R" in LS-R8 designates a Restricted area under the Swiss airspace classification system. When a restricted area is active, entry is prohibited unless specific authorisation has been obtained, and pilots must circumnavigate it. Activation status is published via DABS (Daily Airspace Bulletin Switzerland) or available from ATC.
- Option A describes a danger area (LS-D), where transit is permitted at the pilot's own risk.
- Option C describes a prohibited area (LS-P), which is a different and more restrictive category.
- Option D describes a gliding sector with reduced cloud separation, which is unrelated to the R designation.
Key Terms
ATC = Air Traffic Control
Source
Q97: The coordinates 46°45'43" N / 006°36'48'' correspond to which aerodrome? ^t60q97
DE · FR
- A) Lausanne
- B) Yverdon
- C) Motiers
- D) Montricher
Answer
C)
Explanation
Plotting the coordinates 46 degrees 45 minutes 43 seconds N / 006 degrees 36 minutes 48 seconds E on the Swiss ICAO chart places the position at Motiers aerodrome (LSGM), located in the Val de Travers in the canton of Neuchatel.
- Option A (Lausanne) is situated further south and west along Lake Geneva.
- Option B (Yverdon) lies to the southwest near the southern end of Lake Neuchatel.
- Option D (Montricher) is located in the Jura foothills west of Lausanne.
- Accurate coordinate plotting on the chart confirms option C.
Key Terms
ICAO = International Civil Aviation Organization
Source
Q98: After a thermal flight in the Alps, you plan to fly in a straight line from the Gemmi Pass (171°/58 km from Bern Belp) to Grenchen aerodrome. Which magnetic course (MC) do you select? ^t60q98
DE · FR
- A) 172°
- B) 168°
- C) 352°
- D) 348°
Answer
D)
Explanation
The Gemmi Pass lies south-southeast of Grenchen, so the true course from Gemmi to Grenchen is roughly north-northwest (approximately 345-350 degrees true). Applying the Swiss magnetic variation of approximately 2-3 degrees East (MC = TC minus easterly variation) yields a magnetic course close to 348 degrees.
- Options A and B point roughly southward, which would be the reverse direction.
- Option C (352 degrees) does not account for the magnetic variation correction.
Key Terms
- MC = Magnetic Course
- TC = True Course
Source
Q99: On a cross-country flight from Birrfeld aerodrome (47°26'N, 008°13'E) you turn at Courtelary aerodrome (47°10'N, 007°05'E). On the return leg you land at Grenchen aerodrome (47°10'N, 007°25'E). According to the Swiss gliding chart, the distance flown is ^t60q99
DE · FR
- A) 58 km
- B) 232 km
- C) 115 km
- D) 156 km
Answer
C)
Explanation
The flight consists of two legs measured on the Swiss gliding chart: Birrfeld to Courtelary (approximately 58 km southwest) and Courtelary to Grenchen (approximately 57 km returning northeast but landing short of Birrfeld). The total distance of both legs is approximately 115 km.
- Option A (58 km) accounts for only the first leg.
- Option B (232 km) is roughly double the correct total.
- Option D (156 km) likely adds a third leg back to Birrfeld, but the pilot landed at Grenchen.
Source
Q100: What onboard equipment does your aircraft need for you to determine your position using a VDF bearing? ^t60q100
DE · FR
- A) Transponder.
- B) GPS.
- C) Onboard VOR equipment.
- D) Onboard radio.
Answer
C)
Explanation
VDF (VHF Direction Finding) is a ground-based service in which the station determines the bearing of the aircraft's radio transmission. To use a VDF bearing for position determination, the aircraft needs onboard VOR equipment (VHF omnidirectional range receiver) to interpret and display the bearing information provided by the ground station.
- Option A (transponder) is used for radar identification, not VDF bearings.
- Option B (GPS) is a satellite-based system unrelated to VDF.
- Option D (onboard radio) allows communication but alone does not provide the means to interpret bearing data.
Key Terms
VHF = Very High Frequency
Source
Q101: Which phenomenon is most likely to degrade GPS indications? ^t60q101
DE · FR
- A) High, dense cloud layers.
- B) Thunderstorm areas.
- C) Frequent heading changes.
- D) Flying low in mountainous terrain.
Answer
D)
Explanation
GPS signals are microwave transmissions from orbiting satellites that require a clear line of sight between the satellite and the receiver. When flying low in mountainous terrain, surrounding peaks and ridgelines mask portions of the sky, reducing the number of visible satellites and degrading the geometric dilution of precision (GDOP). This can lead to inaccurate position fixes or complete signal loss.
- Option A (cloud layers) does not affect microwave GPS signals.
- Option B (thunderstorms) do not block GPS signals.
- Option C (heading changes) have no effect on satellite signal reception.
Source
- [?] Source PDF non identifiée (original: D)
### Q102: Given: MC 225 degrees, magnetic declination (variation) 5 degrees E. What is the TC? ^t60q102
DE · FR
- A) 225 degrees
- B) Parameters are insufficient to answer this question.
- C) 230 degrees
- D) 220 degrees
Answer
D)
Explanation
True Course (TC) is calculated from Magnetic Course (MC) by accounting for magnetic declination. With easterly variation, magnetic north lies east of true north, so MC is larger than TC. The formula is TC = MC minus East variation: 225 degrees minus 5 degrees = 220 degrees.
Key Terms
- MC = Magnetic Course
- TC = True Course
Source
- [?] Source PDF non identifiée (original: D)
### Q103: In poor visibility, you fly from Gruyeres (222°/46 km from Bern) towards Lausanne (051°/52 km from Geneva). Which true course (TC) do you select? ^t60q103
DE · FR
- A) 282 degrees
- B) 268 degrees
- C) 082 degrees
- D) 261 degrees
Answer
D)
Explanation
Using the radial and distance references to plot both positions on the Swiss ICAO chart — Gruyeres at 222 degrees/46 km from Bern and Lausanne at 051 degrees/52 km from Geneva — and measuring the true course between them with a protractor yields approximately 261 degrees (roughly west-southwest).
- Options A and B give headings too far to the northwest.
- Option C points east-northeast, which would be the reverse direction entirely.
Key Terms
- ICAO = International Civil Aviation Organization
- TC = True Course
Source
- [?] Source PDF non identifiée (original: C)
### Q104: You want to determine your position using a VDF bearing, but the controller reports the signals are too weak for assessment. What is the likely reason? ^t60q104
DE · FR
- A) Your transponder has too low a transmitting power.
- B) Atmospheric interference weakens the signals.
- C) You are flying too low, and the theoretical line-of-sight (quasi-optical) link is insufficient.
- D) The onboard radio communication system is defective.
Answer
C)
Explanation
VDF operates on VHF frequencies, which propagate in a quasi-optical (line-of-sight) manner. If the aircraft is flying too low, the curvature of the Earth or intervening terrain blocks the signal path between the aircraft and the ground station, resulting in weak or undetectable signals.
- Option A is irrelevant because transponders are not used for VDF bearings.
- Option B overstates atmospheric effects, which are negligible for VHF under normal conditions.
- Option D (defective radio) is possible but less likely than the geometric limitation described in option C.
Key Terms
- D — Drag
- VHF = Very High Frequency
Source
- [?] Source PDF non identifiée (original: A)
### Q105: What does the term "agonic line" mean? ^t60q105
DE · FR
- A) A line along which the magnetic declination is 0 degrees.
- B) All regions where the magnetic declination is greater than 0 degrees.
- C) Any line connecting regions with the same magnetic declination.
- D) Disturbance zones where the Earth's magnetic field lines are strongly deflected (e.g. by ferrous rock), causing large declination variations over a small area.
Answer
A)
Explanation
The agonic line is a specific isogonic line along which the magnetic declination (variation) is exactly zero degrees — meaning true north and magnetic north are aligned. Along this line, a magnetic compass points directly to geographic north without any correction needed.
- Option B describes a region, not a line, and is not a recognized navigational term.
- Option C defines the broader category of isogonic lines, of which the agonic line is a special case.
- Option D describes local magnetic anomalies, not the agonic line.
Source
- [?] Source PDF non identifiée (original: B)
### Q106: What is 4572 m expressed in feet? ^t60q106
DE · FR
- A) 1500 ft
- B) 15000 ft
- C) 13935 ft
- D) 1393 ft
Answer
B)
Explanation
To convert metres to feet, multiply by the conversion factor 3.2808 (since 1 metre = 3.2808 feet). Calculating: 4572 m multiplied by 3.2808 = 15,000 ft. This is a standard altitude conversion that aviation pilots should be able to perform quickly.
- Option A (1500 ft) and option D (1393 ft) are an order of magnitude too small.
- Option C (13,935 ft) results from an incorrect conversion factor.
Source
- [?] Source PDF non identifiée (original: D)
### Q107: Which of the following statements is correct? ^t60q107
DE · FR
- A) The distance between two degrees of longitude or latitude is always equal to 60 NM (111 km).
- B) The distance between two degrees of latitude equals 60 NM (111 km) at the equator and decreases steadily towards the poles.
- C) The distance between two degrees of longitude is always equal to 60 NM (111 km).
- D) The distance between two degrees of longitude equals 60 NM (111 km) only at the equator.
Answer
D)
Explanation
Lines of longitude (meridians) converge toward the poles, so the distance between two degrees of longitude is greatest at the equator (60 NM or 111 km) and decreases to zero at the poles, following the cosine of the latitude. This is a fundamental property of the spherical coordinate system.
- Option A is wrong because longitude spacing varies with latitude.
- Option B incorrectly describes latitude: the distance between two degrees of latitude is approximately constant at 60 NM everywhere, not decreasing toward the poles.
- Option C makes the same error as A for longitude alone.
Key Terms
NM = Nautical Mile(s)
Source
Q108: Which value must you mark on the navigation chart before a cross-country flight? ^t60q108
DE · FR
- A) True heading (TH)
- B) Magnetic heading (MH)
- C) True course (TC)
- D) Compass heading (CH)
Answer
C)
Explanation
On a navigation chart, the course line is drawn relative to the chart's grid, which is oriented to geographic (true) north. Therefore, the value measured and marked on the chart is the True Course (TC) — the angle between true north and the intended track line. Magnetic heading (option B), true heading (option A), and compass heading (option D) all incorporate corrections for wind, magnetic variation, or compass deviation that are calculated separately during flight planning, not drawn on the chart itself.
Key Terms
- CH = Compass Heading
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
Source
Q109: In flight, you notice a drift to the right. How do you correct? ^t60q109
DE · FR
- A) By correcting the heading to the right
- B) By flying more slowly
- C) By increasing the heading value
- D) By decreasing the heading value
Answer
C)
Explanation
If the aircraft drifts to the right, the wind has a component pushing from the left side. To counteract this drift and maintain the desired track, you must turn into the wind by increasing the heading value (turning the nose further to the right to establish a crab angle into the wind component).
- Option A is vague but could be interpreted as correct — however, option C is more precise in specifying the heading adjustment.
- Option B (flying more slowly) would actually increase the drift angle.
- Option D (decreasing the heading) would turn away from the wind and worsen the drift.
Source
Q110: Up to what maximum altitude may you fly a glider over Lenzburg (255°/28 km from Zurich) without notification or authorisation? ^t60q110
DE · FR
- A) 5950 m AMSL
- B) 2000 m AMSL
- C) 4500 ft AMSL
- D) 1700 m AMSL
Answer
D)
Explanation
Lenzburg lies beneath the Zurich TMA structure. According to the Swiss ICAO chart, the lowest TMA sector in this area has its floor at 1700 m AMSL. Below this altitude, the airspace is uncontrolled (Class E or G), and gliders may fly without ATC notification or authorisation. Above 1700 m AMSL, you enter controlled airspace requiring a clearance.
- Options A and B are incorrect altitude values.
- Option C (4500 ft, approximately 1370 m) is below the actual limit and would unnecessarily restrict your flight.
Key Terms
- AMSL = Above Mean Sea Level
- ATC = Air Traffic Control
- ICAO = International Civil Aviation Organization
- TMA = Terminal Manoeuvring Area
Source
Q111: How does the map grid appear in a Lambert (normal conic) projection? ^t60q111
DE · FR
- A) Meridians and parallels form parallel straight lines.
- B) Meridians are parallel to each other, parallels form converging straight lines.
- C) Meridians form converging straight lines, parallels form parallel curves.
- D) Meridians and parallels form equidistant curves.
Answer
C)
Explanation
In a Lambert conformal conic projection, the cone is placed over the globe so that meridians project as straight lines converging toward the apex (the pole), while parallels of latitude appear as concentric arcs (parallel curves) centered on the pole. This projection preserves angles (conformality), making it ideal for aeronautical charts.
- Option A describes a cylindrical projection like Mercator.
- Option B reverses the characteristics of meridians and parallels.
- Option D does not describe any standard cartographic projection.
Source
Q112: You depart from Bern on 10 June (summer time) at 1030 LT. The flight duration is 80 minutes. At what UTC time do you land? ^t60q112
DE · FR
- A) 1050 UTC.
- B) 1350 UTC.
- C) 1250 UTC.
- D) 0950 UTC.
Answer
D)
Explanation
On 10 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure at 1030 LT (CEST) equals 0830 UTC. Adding 80 minutes of flight time: 0830 + 0080 = 0950 UTC.
- Option A (1050 UTC) appears to use UTC+1 instead of UTC+2.
- Option B (1350 UTC) adds the time difference instead of subtracting it.
- Option C (1250 UTC) likely applies only a one-hour offset and rounds incorrectly.
Source
Q113: What are the coordinates of Bellechasse aerodrome (285°/28 km from Bern)? ^t60q113
DE · FR
- A) 47 degrees 22' N / 008 degrees 14' E
- B) 47 degrees 11' S / 008 degrees 13' W
- C) 46 degrees 59' S / 007 degrees 08' W
- D) 46 degrees 59' N / 007 degrees 08' E
Answer
D)
Explanation
Bellechasse aerodrome (LSGE) is located west-northwest of Bern, near the town of Bellechasse in the canton of Fribourg. Plotting the position at 285 degrees/28 km from Bern on the Swiss ICAO chart yields coordinates of approximately 46 degrees 59 minutes N / 007 degrees 08 minutes E.
- Options B and C use South and West designations, which are impossible for locations in Switzerland (Northern Hemisphere, east of the Greenwich meridian).
- Option A places the aerodrome too far north and east.
Key Terms
ICAO = International Civil Aviation Organization
Source
Q114: During a cross-country flight, "POOR GPS COVERAGE" appears on the screen. What could be the cause? ^t60q114
DE · FR
- A) Poor GPS coverage is a consequence of the twilight effect.
- B) The position of a satellite has changed significantly and requires a readjustment procedure.
- C) Your device is receiving an insufficient number of satellite signals, possibly due to terrain configuration blocking them.
- D) The indication may be the result of severe nearby thunderstorms.
Answer
C)
Explanation
The "POOR GPS COVERAGE" message indicates that the receiver cannot track enough satellites with adequate geometry for a reliable position fix. The most common cause during cross-country glider flights is terrain masking — flying in deep valleys or near steep mountain faces that block satellite signals from view.
- Option A (twilight effect) is not a recognized GPS phenomenon.
- Option B overstates how satellite repositioning works, as GPS receivers continuously update orbital data without manual intervention.
- Option D (thunderstorms) does not affect GPS microwave signals.
Source
Q115: The magnetic compass of an aircraft is affected by metallic parts and electrical equipment. What is this influence called? ^t60q115
DE · FR
- A) Variation
- B) Declination
- C) Deviation
- D) Inclination
Answer
C)
Explanation
Deviation is the error in a magnetic compass caused by local magnetic fields from the aircraft's own metallic structure, electrical wiring, and electronic equipment. It varies with heading and is recorded on a deviation card in the cockpit.
- Option A (variation) and option B (declination) both refer to the angular difference between true north and magnetic north, which is a property of the Earth's magnetic field, not the aircraft.
- Option D (inclination or dip) is the angle at which the Earth's magnetic field lines intersect the surface, which affects compass behavior but is not the same as the aircraft-induced error.
Source
Q116: You plan a cross-country flight Courtelary (315°/43 km from Bern-Belp) - Dittingen (192°/18 km from Basel-Mulhouse) - Birrfeld (265°/24 km from Zurich) - Courtelary. What is the total distance? ^t60q116
DE · FR
- A) 315 km
- B) 97 km
- C) 210 km
- D) 189 km
Answer
D)
Explanation
This is a closed triangular cross-country route with three legs: Courtelary to Dittingen, Dittingen to Birrfeld, and Birrfeld back to Courtelary. Each position is plotted on the Swiss ICAO 1:500,000 chart using the given radial/distance references, and the leg distances are measured with a ruler. The sum of all three legs yields approximately 189 km.
- Option A (315 km) is far too long.
- Option B (97 km) accounts for only about half the route.
- Option C (210 km) overestimates by roughly 20 km.
Key Terms
ICAO = International Civil Aviation Organization
Source
Q117: Your GPS displays heights in metres, but you need feet. Can you change this? ^t60q117
DE · FR
- A) No, only the electronics workshop of a maintenance company can change the unit settings.
- B) Yes, you change the distance units of measurement in the settings options (SETTING MODE).
- C) Yes, you change the units of measurement in the aeronautical database (DATA BASE).
- D) No, your device is certified M (metric) and cannot be changed.
Answer
B)
Explanation
Modern aviation GPS units allow pilots to change the display units (metres, feet, kilometres, nautical miles, etc.) through the device's settings menu (SETTING MODE). This is a simple user-accessible configuration change that does not require any maintenance intervention.
- Option A incorrectly suggests that a workshop visit is needed.
- Option C confuses the aeronautical database (which contains waypoints and airspace data) with display settings.
- Option D invents a certification restriction that does not exist for GPS unit settings.
Source
Q118: On a map, 5 cm correspond to a distance of 10 km. What is the scale? ^t60q118
DE · FR
- A) 1:100,000
- B) 1:20,000
- C) 1:500,000
- D) 1:200,000
Answer
D)
Explanation
To determine map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio of map distance to real distance is 5 cm to 1,000,000 cm, which simplifies to 1 cm representing 200,000 cm, giving a scale of 1:200,000.
- Option A (1:100,000) would mean 5 cm = 5 km.
- Option B (1:20,000) would mean 5 cm = 1 km.
- Option C (1:500,000) would mean 5 cm = 25 km.
- Only 1:200,000 produces the correct 5 cm = 10 km relationship.
Source
Q119: During a long approach over a difficult navigation area, which method is most effective? ^t60q119
DE · FR
- A) Orient the map to the north.
- B) Constantly monitor the compass.
- C) Monitor time with the time ruler; mark known positions on the map.
- D) Track your position on the map with your thumb.
Answer
C)
Explanation
Over a difficult navigation area during a long approach, the most effective technique is to use time-based dead reckoning: monitor elapsed time with a time ruler (marking planned time checkpoints along the route) and confirm your position by identifying ground features as they appear, marking each verified position on the map. This combines time estimation with visual confirmation for maximum accuracy.
- Option A (orienting to north) is a basic step but alone does not solve navigation difficulties.
- Option B (monitoring the compass) maintains heading but provides no position information.
- Option D (thumb tracking) works well for shorter legs but is less systematic for long approaches.
Source
Q120: If you are south of the Montreux - Thun - Lucerne - Rapperswil line, on which frequency do you communicate with other glider pilots? ^t60q120
DE · FR
- A) 123.450 MHz
- B) 125.025 MHz
- C) 122.475 MHz
- D) 123.675 MHz
Answer
C)
Explanation
In Switzerland, glider-to-glider communication frequencies are divided geographically. South of the Montreux-Thun-Lucerne-Rapperswil line, the designated common glider frequency is 122.475 MHz. This frequency is used for traffic awareness, thermal information sharing, and safety communication among glider pilots operating in the southern Swiss Alps and surrounding areas. The other listed frequencies are either assigned to the northern sector or serve different aviation purposes.
Source
Q121: What does the designation LS-R6, shown as a red hatched area north of Grindelwald (127°/52 km from Bern), mean? ^t60q121
DE · FR
- A) Restricted zone for gliders. Once activated, minimum cloud separation distances are reduced for gliders.
- B) Danger zone, transit prohibited (helicopter EMS and special flights exempted).
- C) Prohibited zone; activity information and authorization for transit on frequency 135.475 MHz.
- D) Restricted zone; entry prohibited when active (helicopter EMS flights exempted).
Answer
D)
Explanation
LS-R6 is a restricted area (the "R" stands for Restricted in Swiss airspace classification). When active, entry is prohibited for all aircraft except helicopter emergency medical service (EMS) flights, which are exempted due to their life-saving mission.
- Option A incorrectly describes it as merely reducing cloud separation distances.
- Option B misclassifies it as a danger zone (that would be LS-D).
- Option C describes a prohibited zone (LS-P), which is a different category entirely.
Source
- [ ] ~ VV Q27 p.151 (clé: A, original: D)
### Q122: How do you find the magnetic declination (variation) values for a given location? ^t60q122
DE · FR
- A) By calculating the difference between the course measured on the chart and the compass heading.
- B) Using the declination table found in the balloon flight manual (AFM).
- C) By calculating the angle between the local meridian and the Greenwich meridian.
- D) Using the isogonic lines shown on the aeronautical chart.
Answer
D)
Explanation
Magnetic declination (variation) is found by reading the isogonic lines printed on aeronautical charts such as the Swiss ICAO 1:500,000 chart. Isogonic lines connect points of equal magnetic declination and are updated periodically to reflect the slow drift of Earth's magnetic field.
- Option A describes a method for finding deviation, not declination.
- Option B references a balloon flight manual, which is irrelevant for glider operations.
- Option C describes the definition of longitude, not magnetic declination.
Key Terms
ICAO = International Civil Aviation Organization
Source
Q123: In flight, you notice a drift to the left. How do you correct? ^t60q123
DE · FR
- A) By modifying the heading to the left
- B) By increasing the heading value
- C) By decreasing the heading value
- D) By flying more quickly
Answer
B)
Explanation
If the aircraft drifts to the left, the wind is pushing it from the right side of the flight path. To correct, the pilot must turn into the wind by increasing the heading value (turning right). This applies a wind correction angle that offsets the crosswind component. Turning left (option A) or decreasing the heading (option C) would worsen the drift. Flying faster (option D) reduces drift angle slightly but does not correct it — proper heading adjustment is the correct technique.
Source
Q124: What does the indication GND on the cover of the gliding chart (top left, approximately 15 NM west of St Gallen-Altenrhein, 088°/75 km from Zurich-Kloten) mean? ^t60q124
DE · FR
- A) Normal cloud separation distances always apply inside the zones designated GND.
- B) Does not apply to gliding.
- C) Reduced cloud separation distances apply inside the zones designated GND during MIL flying service hours.
- D) Reduced cloud separation distances apply inside the zones designated GND outside MIL flying service hours.
Answer
D)
Explanation
The GND designation on the Swiss gliding chart indicates that reduced cloud separation distances are permitted inside the designated zones outside military flying service hours. When the military is not active, glider pilots benefit from relaxed minima in these areas.
- Option A is incorrect because the whole point of the designation is to allow reduced, not normal, distances.
- Option B is wrong because it specifically applies to gliding operations.
- Option C reverses the timing — the reduced distances apply outside, not during, military hours.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: C)
### Q125: Given: TC 180 degrees, MC 200 degrees. What is the magnetic declination (variation)? ^t60q125
DE · FR
- A) 20 degrees E.
- B) 10 degrees on average.
- C) 20 degrees W.
- D) Additional parameters are missing to answer this question.
Answer
C)
Explanation
Magnetic declination (variation) is the difference between True Course (TC) and Magnetic Course (MC), calculated as: Variation = TC - MC = 180° - 200° = -20°. A negative value indicates West declination, so the answer is 20°W. The mnemonic "variation west, magnetic best" (magnetic heading is greater) confirms this: when MC is greater than TC, variation is West.
- Option A gives the wrong direction (East).
- Option B is an arbitrary average.
- Option D is incorrect because TC and MC are sufficient to determine variation.
Key Terms
- MC = Magnetic Course
- TC = True Course
Source
- [?] Source PDF non identifiée (original: D)
### Q126: During a triangle flight Grenchen (350°/31 km from Bern-Belp) - Kagiswil (090°/57 km from Bern-Belp) - Buttwil (221°/28 km from Zurich-Kloten) - Grenchen, on the return from Buttwil you must land at Langenthal (032°/35 km from Bern-Belp). What is the straight-line distance flown? ^t60q126
DE · FR
- A) 257 km
- B) 154 km
- C) 145 km
- D) 178 km
Answer
D)
Explanation
The total distance is the sum of the individual legs: Grenchen to Kagiswil, Kagiswil to Buttwil, and Buttwil to Langenthal (since the pilot diverted instead of returning to Grenchen). Measuring these legs on the 1:500,000 ICAO chart using the given radial/distance references from Bern-Belp and Zurich-Kloten yields a total of approximately 178 km.
- Option A (257 km) is too long and likely adds an extra leg.
- Option B (154 km) and option C (145 km) are too short, probably omitting one leg of the route.
Key Terms
ICAO = International Civil Aviation Organization
Source
- [?] Source PDF non identifiée (original: A)
### Q127: South of Gruyeres aerodrome there is a zone designated LS-D7. What is this? ^t60q127
DE · FR
- A) A danger zone with an upper limit of 9000 ft above mean sea level.
- B) A prohibited zone with an upper limit of 9000 ft above mean sea level.
- C) A prohibited zone with a lower limit of 9000 ft above ground level.
- D) A danger zone with a lower limit of 9000 ft above ground level.
Answer
A)
Explanation
The prefix "D" in LS-D7 designates a Danger zone under the Swiss airspace classification system. The upper limit of this zone is 9000 ft AMSL (above mean sea level).
- Option B incorrectly calls it a prohibited zone (that would be LS-P).
- Options C and D refer to a "lower limit" of 9000 ft, which would mean the zone starts at 9000 ft rather than ending there — and both also either misclassify the zone type or use the wrong altitude reference (AGL vs.
- AMSL).
Key Terms
- AMSL = Above Mean Sea Level
- AGL = Above Ground Level
Source
Q128: On a map, 4 cm correspond to 10 km. What is the scale? ^t60q128
DE · FR
- A) 1:25,000
- B) 1:100,000
- C) 1:400,000
- D) 1:250,000
Answer
D)
Explanation
To find the map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio is 4 cm on the map to 1,000,000 cm in reality, so 1 cm represents 250,000 cm, giving a scale of 1:250,000.
- Option A (1:25,000) would mean 4 cm = 1 km.
- Option B (1:100,000) would mean 4 cm = 4 km.
- Option C (1:400,000) would mean 4 cm = 16 km.
- Only 1:250,000 yields the correct 4 cm = 10 km relationship.
Source
Q129: Up to what altitude does the Locarno CTR (352°/18 km from Lugano-Agno) extend? ^t60q129
DE · FR
- A) 3950 m AMSL.
- B) 3950 ft AGL.
- C) FL 125.
- D) 3950 ft AMSL.
Answer
D)
Explanation
The Locarno CTR (Control Zone) extends from the surface up to 3,950 ft AMSL (above mean sea level), as published on the Swiss aeronautical charts.
- Option A confuses feet with metres — 3,950 m would be approximately 12,960 ft, far too high for a CTR.
- Option B uses AGL (above ground level), which is not how this CTR's upper limit is defined.
- Option C (FL 125) refers to a flight level reference that is unrelated to this particular CTR boundary.
Key Terms
- AMSL = Above Mean Sea Level
- AGL = Above Ground Level
- FL = Flight Level
- CTR = Control Zone
Source
Q130: You are above Fraubrunnen (north of Bern-Belp airport), N47°05'/E007°32', at 4500 ft AMSL. Your height above the ground is approximately 3000 ft. In which airspace are you? ^t60q130
DE · FR
- A) Airspace class D, TMA BERN 2.
- B) Airspace class G.
- C) Airspace class E.
- D) Airspace class D, CTR BERN.
Answer
C)
Explanation
At Fraubrunnen (north of Bern-Belp) at 4500 ft AMSL, the aircraft is below the BERN 2 TMA, which begins at 5500 ft AMSL in this area, and above the Bern CTR, which only extends to a lower altitude. This places the aircraft in Class E airspace.
- Option A is wrong because the TMA floor is above the aircraft.
- Option D is incorrect because the Bern CTR does not extend this far north or this high.
- Option B (Class G) applies to uncontrolled airspace below the Class E floor, which the aircraft is above.
Key Terms
- AMSL = Above Mean Sea Level
- CTR = Control Zone
- TMA = Terminal Manoeuvring Area
Source
Q131: Your GPS displays distances in NM, but you need km for your calculations. Can you change this? ^t60q131
DE · FR
- A) No, only the electronics workshop of a maintenance company can change the unit settings.
- B) No, your device is not certified M (metric).
- C) Yes, you change the distance units of measurement in the setting mode (SETTING MODE).
- D) Yes, you change the units of measurement in the database (AVIATION DATA BASE).
Answer
C)
Explanation
Modern aviation GPS units allow the pilot to change distance display units (NM to km or vice versa) through the device's SETTING MODE menu. This is a simple user preference and requires no technical workshop intervention.
- Option A is incorrect because unit changes are user-accessible.
- Option B incorrectly suggests certification locks prevent the change.
- Option D confuses the aviation database (which contains waypoints and airspace data) with the display settings menu.
Key Terms
NM = Nautical Mile(s)
Source
Q132: You depart from Bern on 5 June (summer time) at 0945 UTC for a glider flight lasting 45 minutes. At what local time do you land? ^t60q132
DE · FR
- A) 0930 LT.
- B) 1130 LT.
- C) 0830 LT.
- D) 1230 LT.
Answer
B)
Explanation
On 5 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure is at 0945 UTC, and the flight lasts 45 minutes, so landing occurs at 0945 + 0045 = 1030 UTC. Converting to local time: 1030 UTC + 2 hours = 1230 CEST. However, the correct answer given is B (1130 LT), which would correspond to UTC+1 conversion. This suggests the question intends standard CET (UTC+1) or uses a different convention.
- Options A and C yield times before departure, which are impossible, and option D overshoots.
Source
Q133: 54 NM correspond to: ^t60q133
DE · FR
- A) 27.00 km.
- B) 29.16 km.
- C) 100.00 km.
- D) 92.60 km.
Answer
C)
Explanation
The conversion factor is 1 NM = 1.852 km. Therefore 54 NM x 1.852 km/NM = 100.008 km, which rounds to 100.00 km.
- Option A (27 km) appears to divide by 2 instead of multiplying by 1.852.
- Option B (29.16 km) uses an incorrect conversion factor.
- Option D (92.60 km) is close to the correct value but uses an inaccurate conversion ratio.
- Knowing the NM-to-km conversion factor of 1.852 is essential for cross-country flight planning.
Key Terms
- D — Drag
- NM = Nautical Mile(s)
Source
Q134: Which statement about GPS is correct? ^t60q134
DE · FR
- A) GPS has the advantage of always providing accurate indications, as it is not affected by interference.
- B) GPS is a very accurate means of determining position, but satellite signal disruptions must be expected. The current position must therefore always be verified against significant ground references.
- C) Thanks to its accuracy, GPS replaces terrestrial navigation and warns against inadvertent entry into controlled airspace.
- D) Once switched on, GPS automatically receives current information about airspace structure, frequencies, etc.; an up-to-date aeronautical database is therefore always available.
Answer
B)
Explanation
GPS is highly accurate for position determination, but satellite signals can be disrupted by terrain shading, atmospheric conditions, or intentional interference. Pilots must always cross-check GPS position against visual ground references.
- Option A is wrong because GPS is susceptible to interference and signal loss.
- Option C overstates GPS capability — it does not replace basic pilotage skills, and airspace warnings depend on database currency.
- Option D is incorrect because GPS does not automatically update its aviation database; this requires manual updates by the user.
Source
- [?] Source PDF non identifiée (original: C)
### Q135: What is meant by an "isogonic line"? ^t60q135
DE · FR
- A) Any line connecting regions with the same temperature.
- B) Any line connecting regions where the magnetic declination is 0 degrees.
- C) Any line connecting regions with the same magnetic declination.
- D) Any line connecting regions with the same atmospheric pressure.
Answer
C)
Explanation
An isogonic line connects all points on a chart that have the same magnetic declination (variation). These lines are printed on aeronautical charts to help pilots convert between true and magnetic bearings.
- Option A describes an isotherm (equal temperature).
- Option B describes the agonic line, which is the special case where declination equals zero — a subset, not the general definition.
- Option D describes an isobar (equal pressure).
Source
- [?] Source PDF non identifiée (original: C)
### Q136: In poor visibility, you fly from the Saentis (110°/65 km from Zurich-Kloten) towards Amlikon (075°/40 km from Zurich-Kloten). Which true course (TC) do you select? ^t60q136
DE · FR
- A) 147 degrees
- B) 227 degrees
- C) 328 degrees
- D) 318 degrees
Answer
C)
Explanation
Plotting both positions relative to Zurich-Kloten on the chart, the Saentis lies to the southeast (110°/65 km) and Amlikon to the east-northeast (075°/40 km). The route from Saentis to Amlikon heads northwest, yielding a true course of approximately 328°.
- Option D (318°) is close but inaccurate based on the chart plot.
- Options A (147°) and B (227°) point in roughly the opposite direction — southeast and southwest respectively — which would take the pilot away from the destination.
Key Terms
TC = True Course
Source
- [?] Source PDF non identifiée (original: C)
### Q137: What onboard equipment must your glider have for you to determine your position using a VDF bearing? ^t60q137
DE · FR
- A) An emergency transmitter (ELT).
- B) A transponder.
- C) An onboard radio communication system.
- D) A GPS.
Answer
C)
Explanation
VDF (VHF Direction Finding) works by having a ground station take a bearing on the pilot's radio transmission. The only equipment the aircraft needs is a standard VHF radio communication system — the pilot transmits, and the ground station determines the direction.
- Option A (ELT) is for emergency location, not routine position finding.
- Option B (transponder) is for radar identification, not VDF.
- Option D (GPS) determines position independently and is not related to VDF bearings.
Key Terms
- ELT = Emergency Locator Transmitter
- VHF = Very High Frequency
Source
Q138: How does the map grid appear in a normal cylindrical projection (Mercator projection)? ^t60q138
DE · FR
- A) Meridians form converging straight lines, parallels form parallel curves.
- B) Meridians and parallels form equidistant curves.
- C) Meridians and parallels form parallel straight lines.
- D) Meridians are parallel to each other, parallels form converging straight lines.
Answer
C)
Explanation
In a Mercator (normal cylindrical) projection, both meridians and parallels appear as straight lines that intersect at right angles, forming a rectangular grid. Meridians are evenly spaced vertical lines and parallels are horizontal lines (though their spacing increases toward the poles).
- Option A describes a conic projection where meridians converge.
- Option B incorrectly calls them curves.
- Option D reverses the convergence — in a Mercator projection, neither meridians nor parallels converge.
Source
Q139: Up to what maximum altitude may you fly a glider over Burgdorf (035°/19 km from Bern-Belp) without notification or authorisation? ^t60q139
DE · FR
- A) 3050 m AMSL.
- B) 5500 ft AGL.
- C) 1700 m AGL.
- D) 1700 m AMSL.
Answer
D)
Explanation
Above Burgdorf, the lower boundary of the Bern TMA is at 1700 m AMSL. Below this altitude, a glider may fly freely without notification or authorization in Class E or G airspace.
- Option A (3050 m AMSL) represents a higher TMA boundary that applies in a different area.
- Option B (5500 ft AGL) uses an AGL reference which is incorrect for this airspace boundary.
- Option C (1700 m AGL) confuses the reference — the limit is AMSL, not above ground level.
Key Terms
- AMSL = Above Mean Sea Level
- AGL = Above Ground Level
- TMA = Terminal Manoeuvring Area
Source
Q140: What is the name of the location at coordinates 46°29' N / 007°15' E? ^t60q140
DE · FR
- A) The Sanetsch Pass
- B) Sion airport
- C) Saanen aerodrome
- D) The Gstaad/Grund heliport
Answer
C)
Explanation
The coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome, which serves the Gstaad area in the Bernese Oberland.
- Option B (Sion airport) is located further south and slightly east, at approximately 46°13'N / 007°20'E.
- Option A (Sanetsch Pass) is a mountain pass between Sion and the Bernese Oberland at a different position.
- Option D (Gstaad/Grund heliport) is nearby but has different precise coordinates.
Source
- [?] Source PDF non identifiée (original: D)
### Q141: What is meant by the "geographic longitude" of a location? ^t60q141
DE · FR
- A) The distance from the equator, expressed in kilometres.
- B) The distance from the equator, expressed in degrees of longitude.
- C) The distance from the north pole, expressed in degrees of latitude.
- D) The distance from the 0 degree meridian, expressed in degrees of longitude.
Answer
D)
Explanation
Geographic longitude is the angular distance measured east or west from the Prime Meridian (0° at Greenwich) to the local meridian passing through the given location, expressed in degrees (0° to 180°E or W).
- Options A and B incorrectly reference the equator — distance from the equator is latitude, not longitude.
- Option C describes a co-latitude measurement from the north pole, which is also a form of latitude.
- Only option D correctly identifies longitude as the angular measure from the Greenwich meridian.
Source
Q142: The term 'magnetic course' (MC) is defined as ^t60q142
DE · FR
- A) The direction from an arbitrary point on Earth to the geographic North Pole.
- B) The direction from an arbitrary point on Earth to the magnetic north pole.
- C) The angle between true north and the course line.
- D) The angle between magnetic north and the course line.
Answer
D)
Explanation
Magnetic Course (MC) is defined as the angle measured clockwise from magnetic north to the intended course line over the ground. It is the course referenced to the Earth's magnetic field rather than to true (geographic) north.
- Option A describes the direction of true north.
- Option B describes the direction to the magnetic north pole, not a course angle.
- Option C defines True Course (TC), which is referenced to geographic north rather than magnetic north.
Key Terms
- MC = Magnetic Course
- TC = True Course
Source
Q143: An aircraft is flying at FL 75 with an outside air temperature (OAT) of -9°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q143
DE · FR
- A) 6500 ft.
- B) 7000 ft.
- C) 6250 ft.
- D) 6750 ft
Answer
C)
Explanation
True altitude accounts for non-standard temperature effects on pressure altitude. ISA temperature at approximately 6500 ft is about +2°C (15° - 2°/1000 ft x 6.5). With OAT of -9°C, the air is approximately 11°C colder than ISA. Cold air is denser, meaning pressure levels are compressed closer to the ground, so the aircraft is actually lower than the altimeter indicates. Using the correction of roughly 4 ft per 1°C per 1000 ft: 11°C x 4 x 6.5 = approximately 286 ft below QNH altitude, yielding about 6250 ft true altitude.
- Options A, B, and D all overestimate the true altitude.
Key Terms
- QNH = Pressure adjusted to mean sea level
- ISA = International Standard Atmosphere
- FL = Flight Level
Source
- [?] Source PDF non identifiée (original: A)
### Q144: An aircraft flies at a pressure altitude of 7000 ft with OAT +11°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q144
DE · FR
- A) 6750 ft.
- B) 6500 ft.
- C) 7000 ft
- D) 6250 ft.
Answer
A)
Explanation
At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +11°C is about 9-10°C warmer than ISA. In warmer-than-standard air, the atmosphere is expanded, so the aircraft sits higher than the altimeter indicates. Applying the temperature correction (approximately +10°C x 4 ft/°C/1000 ft x 6.5 = +260 ft) to the QNH altitude gives approximately 6500 + 250 = 6750 ft true altitude.
- Option B ignores the temperature correction entirely.
- Options C and D either overcorrect or correct in the wrong direction.
Key Terms
- QNH = Pressure adjusted to mean sea level
- ISA = International Standard Atmosphere
- OAT — Outside Air Temperature
- D — Drag
Source
- [?] Source PDF non identifiée (original: D)
### Q145: An aircraft flies at a pressure altitude of 7000 ft with OAT +21°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q145
DE · FR
- A) 7000 ft.
- B) 6250 ft.
- C) 6750 ft.
- D) 6500 ft
Answer
A)
Explanation
At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +21°C means the air is about 19-20°C warmer than standard. Warm air expands, placing the aircraft significantly higher than indicated. The correction is approximately +20°C x 4 ft/°C/1000 ft x 6.5 = +520 ft, yielding about 6500 + 500 = 7000 ft true altitude. This large warm correction brings the true altitude up to match the pressure altitude.
- Options B, C, and D underestimate the warm-air correction effect.
Key Terms
- QNH = Pressure adjusted to mean sea level
- ISA = International Standard Atmosphere
Source
- [?] Source PDF non identifiée (original: D)
### Q146: Given: True course: 255°. TAS: 100 kt. Wind: 200°/10 kt. The true heading equals ^t60q146
DE · FR
- A) 275°.
- B) 265°.
- C) 245°.
- D) 250°.
Answer
D)
Explanation
With TC 255° and wind from 200°, the wind comes from approximately 55° to the left of the course line. This crosswind pushes the aircraft to the right of track. To compensate, the pilot must crab into the wind (turn left), reducing the heading below the course value. The wind correction angle is approximately sin^-1(10 x sin55° / 100) = sin^-1(0.082) = about 5°. True heading = 255° - 5° = 250°.
- Option A (275°) and B (265°) incorrectly add to the heading.
- Option C (245°) overcorrects by 10°.
Key Terms
- TAS = True Airspeed
- TC = True Course
Source
- [?] Source PDF non identifiée (original: D)
### Q147: Given: True course: 165°. TAS: 90 kt. Wind: 130°/20 kt. Distance: 153 NM. The true heading equals ^t60q147
DE · FR
- A) 165°.
- B) 126°.
- C) 152°.
- D) 158°.
Answer
D)
Explanation
The wind from 130° on a 165° course comes from approximately 35° to the left of the nose, pushing the aircraft right of track. The pilot must crab left to compensate. WCA = sin^-1(20 x sin35° / 90) = sin^-1(0.127) = approximately 7°. True heading = 165° - 7° = 158°.
- Option A (165°) applies no wind correction.
- Option B (126°) overcorrects massively.
- Option C (152°) applies too large a correction of 13°.
- Only 158° properly accounts for the crosswind component.
Key Terms
- NM = Nautical Mile(s)
- TAS = True Airspeed
- WCA = Wind Correction Angle
Source
- [?] Source PDF non identifiée (original: C)
### Q148: An aircraft follows a true course (TC) of 040° at a constant TAS of 180 kt. The wind vector is 350°/30 kt. The groundspeed (GS) equals ^t60q148
DE · FR
- A) 172 kt.
- B) 155 kt.
- C) 168 kt.
- D) 159 kt.
Answer
D)
Explanation
With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left-front. The headwind component is 30 x cos50° = approximately 19 kt, and the crosswind component is 30 x sin50° = approximately 23 kt. The wind correction angle is about 7°, and the groundspeed is calculated from the navigation triangle as TAS minus the effective headwind component, approximately 180 - 21 = 159 kt.
- Options A (172 kt) and C (168 kt) underestimate the headwind effect.
- Option B (155 kt) overestimates it.
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- TC = True Course
Source
- [?] Source PDF non identifiée (original: D)
### Q149: Given: True course: 120°. TAS: 120 kt. Wind: 150°/12 kt. The WCA equals ^t60q149
DE · FR
- A) 6° to the left.
- B) 3° to the left.
- C) 3° to the right.
- D) 6° to the right.
Answer
C)
Explanation
With TC 120° and wind from 150°, the wind comes from 30° to the right of and behind the course line. This pushes the aircraft to the left of track, requiring the pilot to crab to the right. WCA = sin^-1(12 x sin30° / 120) = sin^-1(6/120) = sin^-1(0.05) = approximately 3° to the right.
- Options A and B indicate left corrections, which would worsen the drift.
- Option D (6° right) doubles the actual correction angle needed.
Key Terms
- TAS = True Airspeed
- TC = True Course
- WCA = Wind Correction Angle
Source
Q150: The distance from 'A' to 'B' is 120 NM. At 55 NM from 'A' the pilot finds a deviation of 7 NM to the right. What approximate course change is needed to reach 'B' directly? ^t60q150
DE · FR
- A) 8° left
- B) 6° left
- C) 15° left
- D) 14° left
Answer
D)
Explanation
Using the 1:60 rule, the opening angle (track error from A) is (7/55) x 60 = approximately 7.6° or about 8°. The remaining distance to B is 120 - 55 = 65 NM, so the closing angle to reach B is (7/65) x 60 = approximately 6.5° or about 6°. The total course correction needed is the sum of both angles: 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left).
- Option C (15°) slightly overestimates.
- Option A (8°) only accounts for the opening angle.
- Option B (6°) only accounts for the closing angle.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: D)
### Q151: How many satellites are required for a precise and verified three-dimensional position fix? ^t60q151
DE · FR
- A) Five
- B) Two
- C) Three
- D) Four
Answer
D)
Explanation
A GPS receiver needs signals from at least four satellites for a three-dimensional position fix (latitude, longitude, and altitude). Three satellites would provide only a two-dimensional fix, and the fourth is needed to solve for the receiver's clock error in addition to three spatial coordinates.
- Option A (five) describes what is needed for RAIM (Receiver Autonomous Integrity Monitoring), not a basic 3D fix.
- Option B (two) and option C (three) are insufficient for a full 3D position with clock correction.
Source
- [ ] ✓ VV Q2 p.146 (clé: B, original: C)
### Q152: Which ground features should be preferred for orientation during visual flight? ^t60q152
DE · FR
- A) Farm tracks and creeks
- B) Border lines
- C) Power lines
- D) Rivers, railroads, highways
Answer
D)
Explanation
Rivers, railroads, and highways are the preferred visual navigation references because they are large, prominent linear features that are easily identifiable from altitude and accurately depicted on aeronautical charts.
- Option A (farm tracks and creeks) are too small and numerous to reliably distinguish from the air.
- Option B (border lines) are invisible — there are no physical markings on the ground.
- Option C (power lines) are extremely difficult to see from altitude and pose a collision hazard when flying low.
Source
- [?] Source PDF non identifiée (original: B)
### Q153: What is the approximate circumference of the Earth at the equator? ^t60q153
DE · FR
- A) 40000 NM.
- B) 12800 km.
- C) 21600 NM.
- D) 10800 km.
Answer
C)
Explanation
The Earth's equatorial circumference is approximately 21,600 NM. This derives from the fundamental navigation relationship: 360° of longitude x 60 NM per degree = 21,600 NM, since one nautical mile equals one minute of arc on a great circle. In metric terms, the circumference is about 40,075 km, but that does not match any of the other options correctly.
- Option A (40,000 NM) is nearly double the correct NM value.
- Options B (12,800 km) and D (10,800 km) are both far below the actual metric circumference.
Key Terms
NM = Nautical Mile(s)
Source
Q154: Given: True course from A to B: 352°. Ground distance: 100 NM. GS: 107 kt. ETD: 0933 UTC. The ETA is ^t60q154
DE · FR
- A) 1146 UTC.
- B) 1029 UTC.
- C) 1045 UTC.
- D) 1129 UTC.
Answer
B)
Explanation
Flight time equals distance divided by groundspeed: 100 NM / 107 kt = 0.935 hours = 56 minutes. Adding 56 minutes to the ETD of 0933 UTC gives 0933 + 0056 = 1029 UTC.
- Option A (1146 UTC) would imply a flight time of over 2 hours.
- Option C (1045 UTC) implies 72 minutes, suggesting a groundspeed of about 83 kt.
- Option D (1129 UTC) implies nearly 2 hours of flight time.
- Only 1029 UTC matches the 56-minute calculation.
Key Terms
- ETA = Estimated Time of Arrival
- ETD = Estimated Time of Departure
- GS = Ground Speed
- NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: C)
### Q155: An aircraft travels 100 km in 56 minutes. The ground speed (GS) equals ^t60q155
DE · FR
- A) 198 kt.
- B) 93 kt
- C) 58 km/h
- D) 107 km/h.
Answer
D)
Explanation
Groundspeed = distance / time = 100 km / (56/60 hours) = 100 x (60/56) = 107.1 km/h. Since the distance is given in kilometres, the result is naturally in km/h.
- Option A (198 kt) is far too high and appears to be a unit conversion error.
- Option B (93 kt) would be correct if the distance were in NM, not km.
- Option C (58 km/h) results from dividing 56 by something incorrectly.
- Only 107 km/h correctly applies the speed formula.
Key Terms
- GS = Ground Speed
- NM = Nautical Mile(s)
Source
Q156: An aircraft flies with TAS 180 kt and a headwind component of 25 kt for 2 hours and 25 minutes. The distance flown equals ^t60q156
DE · FR
- A) 435 NM.
- B) 693 NM.
- C) 375 NM.
- D) 202 NM.
Answer
C)
Explanation
Groundspeed = TAS minus headwind = 180 - 25 = 155 kt. Flight time = 2 hours 25 minutes = 2.417 hours. Distance = GS x time = 155 x 2.417 = 374.6 NM, approximately 375 NM.
- Option A (435 NM) incorrectly uses TAS (180 x 2.417 = 435) without subtracting the headwind.
- Option B (693 NM) appears to add the headwind instead of subtracting it.
- Option D (202 NM) likely uses only the headwind component for the calculation.
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: D)
### Q157: Given: GS 160 kt, TC 177°, wind vector 140°/20 kt. The true heading (TH) equals ^t60q157
DE · FR
- A) 184°.
- B) 173°.
- C) 180°
- D) 169°.
Answer
B)
Explanation
The wind from 140° on a 177° true course comes from approximately 37° to the left of the course, pushing the aircraft to the right. The pilot must crab left to compensate. WCA = sin^-1(20 x sin37° / 160) = sin^-1(12/160) = sin^-1(0.075) = approximately 4°. True heading = 177° - 4° = 173°.
- Option A (184°) incorrectly turns right into the drift.
- Option C (180°) applies only a 3° correction in the wrong direction.
- Option D (169°) overcorrects by 8°.
Key Terms
- GS = Ground Speed
- TC = True Course
- TH = True Heading
- WCA = Wind Correction Angle
Source
- [?] Source PDF non identifiée (original: C)
### Q158: An aircraft follows TC 040° at a constant TAS of 180 kt. The wind vector is 350°/30 kt. The wind correction angle (WCA) equals ^t60q158
DE · FR
- A) .+ 5°
- B) . - 9°
- C) .+ 11°
- D) .- 7°
Answer
D)
Explanation
With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left side. The crosswind component = 30 x sin50° = approximately 23 kt pushes the aircraft to the right of track. To maintain course, the pilot crabs left (negative WCA). WCA = -sin^-1(23/180) = -sin^-1(0.128) = approximately -7°.
- Option A (+5°) and C (+11°) are in the wrong direction (right instead of left).
- Option B (-9°) overcorrects the wind effect.
Key Terms
- TAS = True Airspeed
- TC = True Course
- WCA = Wind Correction Angle
Source
- [?] Source PDF non identifiée (original: B)
### Q159: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The ground speed (GS) equals ^t60q159
DE · FR
- A) 117 kt.
- B) 131 kt.
- C) 125 kt.
- D) 120 kt.
Answer
C)
Explanation
The aircraft flies on TC 270° (westbound) and the wind blows from 090° (east). Since the wind comes from directly behind the aircraft, it is a pure tailwind. Groundspeed = TAS + tailwind = 100 + 25 = 125 kt. There is no crosswind component, so no wind correction angle is needed.
- Option A (117 kt) and D (120 kt) underestimate the tailwind effect.
- Option B (131 kt) overestimates it.
- The direct tailwind simply adds to TAS.
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- NM = Nautical Mile(s)
- TC = True Course
Source
- [ ] ~ VV Q81 p.163 (clé: C, original: D)
### Q160: When using GPS for tracking to the next waypoint, a deviation bar with dots is displayed. Which interpretation is correct? ^t60q160
DE · FR
- A) The bar deviation from centre shows track error as angular distance in degrees; full-scale deflection is +-10°.
- B) The bar deviation from centre shows track error as absolute distance in NM; full-scale deflection depends on the GPS operating mode.
- C) The bar deviation from centre shows track error as angular distance in degrees; full-scale deflection depends on the GPS operating mode.
- D) The bar deviation from centre shows track error as absolute distance in NM; full-scale deflection is +-10 NM.
Answer
B)
Explanation
The GPS CDI (Course Deviation Indicator) displays lateral track error as an absolute distance in nautical miles, not as angular degrees like a VOR CDI. The full-scale deflection varies by operating mode: typically +/-5 NM in en-route mode, +/-1 NM in terminal mode, and +/-0.3 NM in approach mode.
- Options A and C incorrectly state the deviation is angular.
- Option D incorrectly states a fixed +/-10 NM scale regardless of mode.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: C)
### Q161: What is the approximate distance from Schänis (LSZX) to Sion (LSGS)? ^t60q161
DE · FR
- Schänis (LSZX): 47°10′30″N 009°02′24″E
- Sion (LSGS): 46°13′09″N 007°20′07″E
- A) 60 NM
- B) 75 NM
- C) 90 NM
- D) 110 NM
Answer
C)
Explanation
This is a classic long alpine glider cross-country. Apply the equirectangular formula:
- Δlat = 47°10′30″ − 46°13′09″ = 57′21″ ≈ 57.4 NM south.
- Δlon = 009°02′24″ − 007°20′07″ = 1°42′17″ ≈ 102.3′ west.
- Mean latitude ≈ 46.7°, so 1′ of longitude ≈ cos(46.7°) ≈ 0.686 NM, giving the east component ≈ 102.3 × 0.686 ≈ 70.2 NM.
- Total distance = √(57.4² + 70.2²) ≈ √(3295 + 4928) ≈ 90.7 NM.
Option C (90 NM) matches.
- A (60 NM) only accounts for the latitude component.
- B (75 NM) underestimates the longitude contribution.
- D (110 NM) overshoots — would require a lon factor of ≈1 instead of cos(φ).
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: D)
### Q162: An aircraft flies with TAS 120 kt and experiences 35 kt tailwind. How much time is needed for a distance of 185 NM? ^t60q162
DE · FR
- A) 2 h 11 min
- B) 0 h 50 min
- C) 1 h 12 min
- D) 1 h 32 min
Answer
C)
Explanation
Groundspeed = TAS + tailwind = 120 + 35 = 155 kt. Flight time = distance / GS = 185 / 155 = 1.194 hours = 1 hour 12 minutes.
- Option A (2 h 11 min) appears to use TAS alone without the tailwind (185/85 does not work either — likely a calculation error).
- Option B (50 min) would require a GS of about 222 kt.
- Option D (1 h 32 min) corresponds to using TAS of 120 kt without adding the tailwind (185/120 = 1.54 h = 1 h 32 min).
Key Terms
- TAS = True Airspeed
- GS = Ground Speed
- D — Drag
- NM = Nautical Mile(s)
Source
- [?] Source non identifiée
### Q163: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The flight time equals ^t60q163
DE · FR
- A) 62 Min.
- B) 37 Min.
- C) 48 Min.
- D) 84 Min.
Answer
C)
Explanation
Flying on TC 270° with wind from 090° means the wind is a direct tailwind (blowing from directly behind). GS = TAS + tailwind = 100 + 25 = 125 kt. Flight time = 100 NM / 125 kt = 0.80 hours = 48 minutes.
- Option D (84 min) would result from treating the 25 kt wind as a headwind (GS = 75 kt).
- Option A (62 min) corresponds to a GS of about 97 kt.
- Option B (37 min) would require an unrealistically high GS of about 162 kt.
Key Terms
- GS = Ground Speed
- TAS = True Airspeed
- NM = Nautical Mile(s)
- TC = True Course
Source
- [?] Source PDF non identifiée (original: B)
### Q164: Which answer completes the flight plan (marked cells)? ^t60q164
DE · FR
- A) TH: 185°. MH: 185°. MC: 180°.
- B) TH: 173°. MH: 174°. MC: 178°.
- C) TH: 173°. MH: 184°. MC: 178°.
- D) TH: 185°. MH: 184°. MC: 178°.
Answer
D)
Explanation
The flight plan conversion chain proceeds from True Course through wind correction to True Heading (TH), then applying magnetic variation to get Magnetic Heading (MH), and finally accounting for compass deviation for Magnetic Course (MC). The values TH 185°, MH 184°, and MC 178° are consistent with the sequential application of a small wind correction angle, a 1° easterly variation, and compass deviation.
- Options A, B, and C contain inconsistencies in the TC-to-TH-to-MH-to-MC conversion chain that do not satisfy the given flight plan parameters.
Key Terms
- MC = Magnetic Course
- MH = Magnetic Heading
- TC = True Course
- TH = True Heading
Source
- [?] Source PDF non identifiée (original: D)
### Q165: What is meant by the term "terrestrial navigation"? ^t60q165
DE · FR
- A) Orientation by instrument readings during visual flight
- B) Orientation by ground features during visual flight
- C) Orientation by GPS during visual flight
- D) Orientation by ground celestial objects during visual flight
Answer
B)
Explanation
Terrestrial navigation (also known as pilotage or map reading) is the technique of orienting the aircraft by visually identifying ground features — towns, rivers, roads, railways, lakes — and matching them to the aeronautical chart.
- Option A describes instrument navigation, which relies on cockpit instruments rather than visual ground references.
- Option C describes GPS navigation, a satellite-based method.
- Option D confuses terrestrial with celestial navigation, which uses stars and other astronomical bodies for position determination.
Source
- [?] Source non identifiée
### Q166: What flight time is required for a distance of 236 NM at a ground speed of 134 kt? ^t60q166
DE · FR
- A) 0:46 h
- B) 0:34 h
- C) 1:46 h
- D) 1:34 h
Answer
C)
Explanation
Flight time = distance / groundspeed = 236 NM / 134 kt = 1.761 hours. Converting the decimal fraction: 0.761 x 60 = 45.7 minutes, approximately 46 minutes, giving a total of 1 hour 46 minutes.
- Option A (0:46 h) has the correct minutes but is missing the full hour.
- Option D (1:34 h) would correspond to a GS of about 144 kt.
- Option B (0:34 h) is far too short for this distance at this speed.
Key Terms
- GS = Ground Speed
- NM = Nautical Mile(s)
Source
Q167: What is the true course (TC) from Birrfeld (LSZF) to Grenchen (LSZG)? ^t60q167
DE · FR
- Birrfeld (LSZF): 47°26′35″N 008°14′00″E
- Grenchen (LSZG): 47°10′54″N 007°25′02″E
- A) 215°
- B) 230°
- C) 245°
- D) 060°
Answer
C)
Explanation
Grenchen lies south-west of Birrfeld (lower latitude and further west), so the course must be in the SW quadrant (180°–270°).
- Δlat = 47°10′54″ − 47°26′35″ = −15′41″ ≈ −15.7 NM (south component, dN).
- Δlon = 007°25′02″ − 008°14′00″ = −48′58″ ≈ −49.0′ (west).
- Mean latitude ≈ 47.3°, so 1′ of longitude ≈ cos(47.3°) ≈ 0.678 NM, giving the east component ≈ −49.0 × 0.678 ≈ −33.2 NM (dE).
- TC = atan2(dE, dN) = atan2(−33.2, −15.7) → third-quadrant bearing ≈ 245°.
Option C (245°) is correct.
- A (215°) is SW but too far south of the true track.
- B (230°) is in the right quadrant but underestimates the westerly component.
- D (060°) is the reciprocal — the course from Grenchen back to Birrfeld, not the forward direction asked.
Key Terms
TC = True Course
Source
- [?] Source PDF non identifiée (original: C)
### Q168: What does the 1:60 rule mean? ^t60q168
DE · FR
- A) 10 NM lateral offset at 1° drift after 60 NM
- B) 60 NM lateral offset at 1° drift after 1 NM
- C) 1 NM lateral offset at 1° drift after 60 NM
- D) 6 NM lateral offset at 1° drift after 10 NM
Answer
C)
Explanation
The 1:60 rule is a mental math shortcut stating that at a distance of 60 NM, a 1° track error produces approximately 1 NM of lateral offset. Mathematically, this works because the arc length of 1° on a 60 NM radius circle is 2 x pi x 60 / 360 = approximately 1.047 NM, close enough to 1 NM for practical navigation.
- Option A (10 NM offset) is ten times too large.
- Option B reverses the distance and offset.
- Option D (6 NM at 10 NM) is geometrically inconsistent with the rule.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: C)
### Q169: An aircraft follows TC 220° at a constant TAS of 220 kt. The wind vector is 270°/50 kt. The ground speed (GS) equals ^t60q169
DE · FR
- A) 135 kt.
- B) 170 kt.
- C) 185 kt.
- D) 255 kt.
Answer
C)
Explanation
With TC 220° and wind from 270°, the wind angle is 50° from the right-front of the aircraft. The headwind component = 50 x cos50° = approximately 32 kt, and the crosswind component = 50 x sin50° = approximately 38 kt. Using the navigation wind triangle, the groundspeed works out to approximately 185 kt after accounting for both the headwind reduction and the crab angle.
- Option D (255 kt) would require a tailwind.
- Option A (135 kt) subtracts the full wind speed.
- Option B (170 kt) overcorrects for the headwind component.
Key Terms
- GS = Ground Speed
- TAS = True Airspeed
- TC = True Course
Source
- [ ] ~ VV Q81 p.163 (clé: C, original: A)
### Q170: An aeroplane has a heading of 090°. The distance to fly is 90 NM. After 45 NM the aeroplane is 4.5 NM north of the planned flight path. What corrected heading is needed to reach the destination directly? ^t60q170
DE · FR
- A) 9° to the right
- B) 6° to the right
- C) 12° to the right
- D) 18° to the right
Answer
C)
Explanation
Applying the 1:60 rule: the opening angle (track error) = (4.5 / 45) x 60 = 6° off track to the north. The remaining distance is 90 - 45 = 45 NM. The closing angle to reach the destination = (4.5 / 45) x 60 = 6°. Total correction = opening angle + closing angle = 6° + 6° = 12° to the right (south), since the aircraft has drifted north of track.
- Option A (9°) is too small.
- Option B (6°) accounts for only the closing angle.
- Option D (18°) is too aggressive and would overshoot the correction.
Key Terms
NM = Nautical Mile(s)
Source
- [?] Source PDF non identifiée (original: B)
### Q171: What is the distance from Samedan (LSZS) to Lugano (LSZA)? ^t60q171
DE · FR
- Samedan (LSZS): 46°32′04″N 009°53′02″E
- Lugano (LSZA): 46°00′15″N 008°54′38″E
- A) 35 NM
- B) 45 NM
- C) 51 NM
- D) 65 NM
Answer
C)
Explanation
Equirectangular approximation across the southern Swiss Alps:
- Δlat = 46°32′04″ − 46°00′15″ = 31′49″ ≈ 31.8 NM south.
- Δlon = 009°53′02″ − 008°54′38″ = 58′24″ ≈ 58.4′ west.
- Mean latitude ≈ 46.3°, so 1′ of longitude ≈ cos(46.3°) ≈ 0.691 NM, giving the west component ≈ 58.4 × 0.691 ≈ 40.4 NM.
- Total distance = √(31.8² + 40.4²) ≈ √(1012 + 1630) ≈ 51.4 NM.
Option C (51 NM) is the best match.
- A (35 NM) only accounts for the latitude component.
- B (45 NM) underestimates the longitude contribution.
- D (65 NM) overshoots; it would be about right if the longitude were not shortened by cos(φ).
Key Terms
- NM = Nautical mile (1′ of latitude ≈ 1 NM ≈ 1.852 km).
- LSZS = Samedan aerodrome (Engadin).
- LSZA = Lugano aerodrome (Ticino).
Source
- [?] Source non identifiée
### Q172: What does the term terrestrial navigation mean? ^t60q172
DE · FR
- A) Orientation by GPS during visual flight
- B) Orientation by ground features during visual flight
- C) Orientation by instrument readings during visual flight
- D) Orientation by ground celestial objects during visual flight
Answer
B)
Explanation
Terrestrial navigation is the method of navigating by visually identifying ground features such as roads, rivers, railways, towns, and lakes, and matching them to an aeronautical chart. It is the primary VFR navigation technique and sometimes called pilotage or map reading.
- Option A (GPS) is satellite-based navigation.
- Option C (instruments) describes instrument navigation or dead reckoning.
- Option D confuses terrestrial (ground-based) with celestial (star-based) navigation methods.
Key Terms
VFR = Visual Flight Rules
Source
- [?] Source non identifiée
### Q173: What does QNH mean? ^t60q173
DE · FR
- A) The altimeter reads zero when set to QNH on the ground.
- B) The altimeter reads the pressure altitude when set to QNH.
- C) The altimeter reads zero when airborne after setting QNH.
- D) The altimeter reads the aerodrome elevation when set on the ground.
Answer
D)
Explanation
QNH is the altimeter setting that causes the altimeter to indicate the field elevation (above mean sea level) when the aircraft is on the ground. In other words, setting QNH on the Kollsman window makes the altimeter read the actual altitude above sea level of the aerodrome. In flight, the altimeter then shows the aircraft's altitude AMSL.
- Option A describes QFE, not QNH (QFE causes the altimeter to read zero on the ground).
- Options B and C are incorrect; QNH relates altitude to sea level, not pressure altitude.
Key Terms
- QNH = Altimeter setting for altitude above mean sea level (reads field elevation on ground)
- QFE = Altimeter setting that reads zero on the ground at a specific aerodrome
- AMSL = Above Mean Sea Level
Source
Q174: You forgot to set the QNH before take-off and are now airborne. What should you do? ^t60q174
DE · FR
- A) Land immediately, as further flight is not permitted.
- B) Use the standard pressure setting (1013.25 hPa) for the remainder of the flight.
- C) Continue without adjustment; the error is negligible.
- D) Request the current QNH by radio and set it on the altimeter.
Answer
D)
Explanation
If QNH was not set before departure, the correct action is to request the current QNH via radio (from a ground station, ATIS, or ATC) and set it on the altimeter as soon as possible. Flying with an incorrectly set altimeter poses a safety risk, particularly in mountainous terrain or controlled airspace.
- Option A is too extreme; the situation can be corrected in flight.
- Option B (standard setting 1013.25 hPa) is used above the transition altitude, not as a substitute for QNH at low altitudes.
- Option C is wrong; the error can be significant depending on local pressure.
Key Terms
- QNH = Local altimeter setting for sea-level altitude reference
- ATIS = Automatic Terminal Information Service (broadcasts QNH and weather)
- Transition altitude = Altitude below which QNH is used; above which standard setting (1013) applies
Source
Q175: On the Swiss soaring chart, the text "NIL" appears in a soaring zone near Langenthal. What does this mean for cloud separation? ^t60q175
DE · FR
- A) Normal VFR cloud separation distances apply.
- B) Flight in cloud is permitted in this zone.
- C) Cloud separation is reduced to 150 m horizontal and 100 ft vertical.
- D) No flight is permitted when clouds are present.
Answer
A)
Explanation
The designation "NIL" in the cloud separation column of a soaring zone on the Swiss soaring chart means that no special (reduced) cloud separation applies - the standard VFR cloud separation distances are required. This contrasts with zones that specify reduced minima (e.g., 150 m / 300 ft). Pilots must apply the full standard VFR cloud clearances in NIL zones.
- Options B and C describe reduced separation conditions, which do not apply when NIL is stated.
- Option D is incorrect; NIL does not prohibit flight near clouds, it simply requires standard separation.
Key Terms
- NIL = No reduced cloud separation; standard VFR minima apply
- Soaring zone = Designated area on the Swiss soaring chart with specific soaring conditions
- VFR cloud separation = Standard visibility and cloud distance requirements for visual flight
Source
Q176: During which period of the year are Class E airspace soaring periods active in Switzerland? ^t60q176
DE · FR
- A) Year-round, daily from sunrise to sunset.
- B) June 1 to September 30.
- C) April 1 to October 31.
- D) May 1 to August 31.
Answer
C)
Explanation
The soaring periods within Class E airspace in Switzerland are active from April 1 to October 31. During this period, designated soaring zones within Class E may be in use by gliders under the conditions published on the Swiss soaring chart. Outside this period, the special soaring provisions do not apply.
- Option A is incorrect; the soaring periods are seasonal, not year-round.
- Options B and D give the wrong date ranges.
Key Terms
- Class E airspace = Controlled airspace where IFR flights receive ATC separation; VFR flights may operate without clearance
- Soaring period = Season during which special glider soaring conditions apply in Swiss airspace
Source
Q177: When reading the military activity notes on the Swiss soaring chart, what should glider pilots pay particular attention to? ^t60q177
DE · FR
- A) Weekday morning activity hours only.
- B) Night flights, especially during winter months.
- C) Weekend activity during summer holidays.
- D) Instrument approach procedures at military aerodromes.
Answer
B)
Explanation
Military activity notes on the Swiss soaring chart include information about night operations, which are particularly relevant in winter when darkness falls earlier. Glider pilots should be aware of military night training flights in certain areas, as these may affect airspace availability or safety. The notes indicate periods and types of military activity that are not standard daytime operations.
- Option A is incomplete; military activity can occur outside weekday mornings.
- Options C and D are not the primary focus of these notes for glider pilots.
Key Terms
- Military activity notes = Annotations on the Swiss soaring chart describing military operations in specific areas
- Night operations = Military flights conducted during darkness, relevant for airspace management
Source
Q178: Who is responsible for activating the Dittingen-Nord soaring sector? ^t60q178
DE · FR
- A) The nearest ATC centre.
- B) The Swiss NOTAM office (AIS).
- C) The airfield duty officer (chef de place).
- D) The regional gliding federation representative.
Answer
C)
Explanation
The Dittingen-Nord soaring sector is activated by the airfield duty officer (chef de place / Platzchef) at Dittingen aerodrome. This local activation process is typical for Swiss soaring sectors that are tied to specific aerodromes - the on-site responsible person coordinates the activation of the sector based on actual soaring activity. This information is published on the Swiss soaring chart.
- Options A and B involve ATC/AIS, which manage instrument and controlled airspace, not local soaring sector activations.
- Option D is not the defined responsible authority for sector activation.
Key Terms
- Chef de place = Airfield duty officer responsible for local aerodrome operations
- Dittingen-Nord = Soaring sector near Dittingen aerodrome in northwestern Switzerland
- Sector activation = Process by which a soaring sector is put into use for the day
Source
Q179: What is the radio frequency used by retrieve teams operating in the Alps? ^t60q179
DE · FR
- A) 121.500 MHz (emergency frequency).
- B) 123.500 MHz (air-to-air).
- C) 122.475 MHz.
- D) 126.700 MHz (Swiss gliding common frequency).
Answer
C)
Explanation
The frequency 122.475 MHz is designated for retrieve teams (ground crews picking up landed-out gliders) operating in the Alps region. This frequency allows the pilot and retrieve crew to communicate when outside normal gliding club radio range. It is published on the Swiss soaring chart and in relevant Swiss aeronautical information.
- Option A (121.5 MHz) is the international emergency/distress frequency and must not be used for routine communication.
- Option B (123.5 MHz) is used for general air-to-air communication, not specifically retrieve operations.
- Option D is not the designated retrieve frequency.
Key Terms
- Retrieve team = Ground crew responsible for collecting a glider and pilot after an off-field landing
- 122.475 MHz = Retrieve team frequency for Alpine operations (published on Swiss soaring chart)
Source
Q180: Where can a glider pilot find information about soaring conditions and procedures in Class D and Class C airspace in Switzerland? ^t60q180
DE · FR
- A) In the soaring weather bulletin.
- B) In the GAFOR forecast.
- C) On the KOSIF system.
- D) On the Swiss soaring chart.
Answer
D)
Explanation
Information about soaring conditions and procedures in Class D and Class C airspace in Switzerland is published on the Swiss soaring chart (1:300,000). This chart contains the soaring sectors, altitude limits, frequencies, activation periods, and special conditions applicable to gliders in controlled airspace.
- Option A (soaring weather bulletin) provides general or meteorological information, not airspace-specific details.
- Option B (GAFOR) is a meteorological forecast for general aviation routing, not airspace information.
- Option C (KOSIF) is a military information system, not intended for this type of soaring information.
Key Terms
- Class D = Controlled airspace (ATC clearance required, full services provided)
- Class C = Controlled airspace (separation for all flights, clearance required)
- Swiss soaring chart = 1:300,000 aeronautical chart specifically for glider pilots in Switzerland
- GAFOR = General Aviation FORecast (route weather forecast)
Source
