Source: QuizVDS.it (EASA ECQB-SPL) | 50 questions Free practice: https://quizvds.it/en-en/quiz/spl-en
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q1) - A) Magnetic north pole and on the geographic South Pole. - B) Magnetic north pole and on the magnetic south pole. - C) Geographic North Pole and on the magnetic south pole. - D) Geographic North Pole and on the geographic South Pole. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q2) - A) The polar axis of the Earth crosses the geographic South Pole and the geographic North Pole and is perpendicular to the plane of the equator - B) The polar axis of the Earth crosses the magnetic south pole and the magnetic north pole and is at an angle of 66.5° to the plane of the equator - C) The polar axis of the Earth crosses the geographic South Pole and the geographic North Pole and is at an angle of 23.5° to the plane of the equator - D) The polar axis of the Earth crosses the magnetic south pole and the magnetic north pole and is perpendicular to the plane of the equator Correct: A)
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 C confuses the tilt to the ecliptic with the relationship to the equator.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q3) - A) Sphere of ecliptical shape - B) Flat plate - C) Perfect sphere - D) Ellipsoid Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q4) - A) A rhumb line is a great circle intersecting the the equator with 45° angle. - B) The center of a complete cycle of a rhumb line is always the Earth's center. - C) A rhumb line cuts each meridian at the same angle. - D) The shortest track between two points along the Earth's surface follows a rhumb line. Correct: C)
Explanation: A rhumb line (also called a loxodrome) is defined as a line that crosses every meridian of longitude at the same angle. This makes it useful for constant-heading navigation — a pilot can fly a rhumb line by maintaining a fixed compass heading. However, it is not the shortest path between two points; that distinction belongs to the great circle route.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q5) - A) A rhumb line. - B) A small circle - C) A parallel of latitude. - D) A great circle. Correct: D)
Explanation: A great circle is any circle whose plane passes through the center 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). Parallels of latitude (except the equator) and rhumb lines are not great circles and do not represent the shortest path. Long-haul aircraft routes are planned along great circle tracks to minimize fuel and time.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q6)
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 D has the right number but wrong unit. Knowing this relationship (1° = 60 NM on the equator) is fundamental to navigation calculations.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q7) - A) .05,19° - B) .20,28° - C) .05°19'00'' - D) .20°28'00'' Correct: D)
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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q8) - A) 23.5° north and south of the poles - B) 23.5° north and south of the equator - C) At a latitude of 20.5°S and 20.5°N - D) 20.5° south of the poles Correct: A)
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 B) are the ones located 23.5° from the equator.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q9) - A) 60 NM - B) 111 NM - C) 1 NM - D) 10 NM Correct: A)
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 B) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q10) - A) 30 NM - B) 60 km - C) 60 NM - D) 1 NM Correct: C)
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 B.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q11) - A) 53°50'27''N - B) 49°50'27''N - C) 51°50'27'N' - D) 43°50'27''N Correct: C)
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 A would require 6° (360 NM), and option B would require only 2° (120 NM).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q12) - A) 111 NM - B) 60 km - C) 1 NM - D) 60 NM Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q13) - A) 400 NM - B) 120 NM - C) 216 NM - D) 60 NM Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q14) - A) Less than 300 NM. - B) Less than 600 NM. - C) More than 600 NM. - D) More than 300 NM. Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q15) - A) 1:00 h - B) 0:40 h - C) 0:20 h - D) 1:20 h Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q16) - A) 0:04 h - B) 1:00 h - C) 0:40 h - D) 0:30 h Correct: C)
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 A (4 minutes) would be the time for only 1° of longitude. Option D (30 minutes) would correspond to 7.5° of longitude.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q17) - A) 0.66 h - B) 0.4 h - C) 1 h - D) 0.33 h Correct: A)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q18) - A) 1600 UTC. - B) 1700 UTC. - C) 1500 UTC. - D) 1400 UTC. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q19) - A) A zonal time - B) Local mean time at a specific point on Earth. - C) An obligatory time used in aviation. - D) A local time in Central Europe. Correct: C)
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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q20) - A) 1500 UTC. - B) 1700 UTC. - C) 1800 UTC. - D) 1600 UTC. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q21) - A) In Vienna the sunrise is 4 minutes later and sunset is 4 minutes earlier than in Salzburg - B) In Vienna the sunrise and sunset are about 14 minutes earlier than in Salzburg - C) In Vienna the sunrise and sunset are about 4 minutes later than in Salzburg - D) In Vienna the sunrise is 14 minutes earlier and sunset is 14 minutes later than in Salzburg Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q22) - A) The period of time before sunrise or after sunset where the midpoint of the sun disk is 6 degrees or less below the apparent horizon. - B) The period of time before sunrise or after sunset where the midpoint of the sun disk is 6 degrees or less below the true horizon. - C) The period of time before sunrise or after sunset where the midpoint of the sun disk is 12 degrees or less below the true horizon. - D) The period of time before sunrise or after sunset where the midpoint of the sun disk is 12 degrees or less below the apparent horizon. Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q23) - A) TC: 113°. MH: 127°. CH: 129°. - B) TC: 137°. MH: 127°. CH: 125°. - C) TC: 137°. MH: 139°. CH: 125°. - D) TC: 113°. MH: 139°. CH: 129°. Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q24) - A) MH: 163°. MC: 175°. - B) MH: 167°. MC: 161° - C) MH: 163°. MC: 161°. - D) MH: 167°. MC: 175°. Correct: 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").
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q25) - A) Variation. - B) Inclination. - C) Deviation. - D) WCA. Correct: D)
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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q26) - A) WCA. - B) Variation - C) Inclination. - D) Deviation. Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q27) - A) The direction from an arbitrary point on Earth to the magnetic north pole. - B) The angle between magnetic north and the course line. - C) The angle between true north and the course line. - D) The direction from an arbitrary point on Earth to the geographic North Pole. Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q28) - A) The direction from an arbitrary point on Earth to the magnetic north pole. - B) The direction from an arbitrary point on Earth to the geographic North Pole. - C) Tthe angle between magnetic north and the course line. - D) The angle between true north and the course line. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q29) - A) TH: 194°. VAR: 004° E - B) TH: 194°. VAR: 004° W - C) TH: 172°. VAR: 004° W - D) TH: 172°. VAR: 004° E Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q30) - A) TH: 172°. DEV: +002°. - B) TH: 172°. DEV: -002°. - C) TH: 194°. DEV: -002°. - D) TH: 194°. DEV: +002°. Correct: C)
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°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q31) - A) VAR: 004° E. DEV: -002°. - B) VAR: 004° W. DEV: +002°. - C) VAR: 004° E. DEV: +002°. - D) VAR: 004° W. DEV: -002°. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q32) - A) At the geographic equator - B) At the magnetic equator - C) At the geographic poles - D) At the magnetic poles Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q33) - A) WCA - B) Inclination. - C) Deviation. - D) Variation. Correct: C)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q34) - A) The most northerly part of the magnetic compass in the aircraft, where the reading takes place - B) The direction to which the direct reading compass aligns due to earth's and aircraft's magnetic fields - C) The angle between the aircraft heading and magnetic north - D) The direction from an arbitrary point on Earth to the geographical North Pole Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q35) - A) Heading. - B) Deviation - C) Variation. - D) Inclination. Correct: C)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q36) - A) Heading of 0°. - B) Deviation of 0°. - C) Inclination of 0°. - D) Variation of 0°. Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q37) - A) Nautical miles (NM), kilometers (km) - B) Land miles (SM), sea miles (NM) - C) Yards (yd), meters (m) - D) Feet (ft), inches (in) Correct: A)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q38) - A) 300 m. - B) 3000 m. - C) 30 km. - D) 30 m. Correct: A)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q39) - A) 18000 ft. - B) 30000 ft. - C) 7500 ft. - D) 10000 ft. Correct: A)
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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q40) - A) The magnetic variation of the runway location has changed - B) The magnetic deviation of the runway location has changed - C) The true direction of the runway alignment has changed - D) The direction of the approach path has changed Correct: A)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q41) - A) Direct reading compass. - B) Airspeed indicator. - C) Turn coordinator - D) Artificial horizon. Correct: A)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q42) - A) The scale is constant, great circles are depicted as curved lines, rhumb lines are depicted as straight lines - B) The scales increases with latitude, great circles are depicted as curved lines, rhumb lines are depicted as straight lines - C) The scales increases with latitude, great circles are depicted as straight lines, rhumb lines are depicted as curved lines - D) The scale is constant, great circles are depicted as straight lines, rhumb lines are depicted as curved lines Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q43) - A) Rhumb lines: straight lines Great circles: curved lines - B) Rhumb lines: straight lines Great circles: straight lines - C) Rhumb lines: curved lines Great circles: straight lines - D) Rhumb lines: curved lines Great circles: curved lines Correct: A)
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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q44) - A) The chart is conformal and an equal-area projection - B) Great circles are depicted as straight lines and the chart is an equal-area projection - C) Rhumb lines are depicted as straight lines and the chart is conformal - D) The chart is conformal and nearly true to scale Correct: D)
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 are approximately straight lines (making it excellent for plotting direct routes). It is NOT an equal-area projection. The Swiss ICAO 1:500,000 chart uses this projection.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q45) - A) 1 : 500000 - B) 1 : 1000000. - C) 1 : 250000. - D) 1 : 2000000. Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q46)
Note: This question originally references chart annex NAV-031 showing the area around BKD VOR. The answer can be calculated from coordinates using the departure formula. - A) 24 NM - B) 42 NM - C) 24 km - D) 42 km Correct: A)
Explanation: Both points are at nearly the same latitude (~53°N), so the distance can be estimated using the departure formula. The longitude difference is 12°11' - 11°33' = 38' of longitude. At latitude 53°N, the distance per degree of longitude = 60 NM x cos(53°) ≈ 60 x 0.602 ≈ 36.1 NM/degree, so 38' = 0.633° x 36.1 ≈ 22.9 NM. The latitude difference adds a small component. The chart measurement confirms approximately 24 NM, making option A correct.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q47) - A) 1 : 500000 - B) 1 : 1500000 - C) 1 : 1 000000 - D) 1 : 150000 Correct: B)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q48) - A) 238°. - B) 245°. - C) 252°. - D) 007°. Correct: 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°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q49) - A) 1115 UTC. - B) 1005 UTC. - C) 1105 UTC. - D) 1052 UTC. Correct: C)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q50) - A) 1330 UTC - B) 1356 UTC - C) 1430 UTC - D) 1320 UTC Correct: A)
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 B (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 - SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf
Permitted aids at the exam: ICAO 1:500'000 Switzerland chart, Swiss gliding chart, protractor, ruler, mechanical DR calculator, compass, non-programmable scientific calculator (TI-30 ECO RS recommended). No alphanumeric or electronic navigation computers allowed.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q51) - Am 21. Juni -> 22:08 (local time) - Am 25. Maerz -> 19:20 - Am 1. April -> 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q52) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q53) Correct: Der 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q54) Correct: Mit Zeitmassstab ueberwachen, bekannte Positionen auf der Karte markieren
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q55) Correct: Obergrenze der LS-R fuer Segelflug (SF mit reduzierten Wolkenabstaenden)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q56) Correct: Auf dem SF-Karte Deckblatt aufgefuehrt
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q57) Correct: SF-Karte unten rechts
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q58) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q59) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q60) Correct: Status Tangosektor massgebend - nicht aktiv (Bale Info) bis FL100; wenn aktiv 1750 m oder hoeher mit Freigabe 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q61) Correct: SF-Karte Legende (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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q62) Correct: SF-Karte Legende unten rechts. Achtung: Textbox auf Grenze TMA LSZH 10 (2000 m) und TMA LSZH 3 (1700 m); LSR69 liegt 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q63) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q64) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q65) Correct: 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).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q66) Correct: Flugplatz Annemasse
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q67) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q68) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q69) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q70) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q71) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q72) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q73) Correct: (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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q74) Correct: TMA PAY 7 (E), TMA LSZB1 (D - Freigabe noetig), LR E MTT, LR E Alpen, LS-R15 (falls aktiv), TMA LSME 2, CTR LSMA/LSZC (Freigaben noetig)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q75) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q76) Correct: Distanz 80 km, Hoehenverlust 2667 m, Ankunft 1533 m MSL = 1100 m AGL ueber 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q77) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q78) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q79) Correct: 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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^q80) | Code | Situation | |------|-----------| | 7000 | VFR in Luftraum E und G | | 7700 | Notfall (Emergency) | | 7600 | Funkausfall (Radio failure) | | 7500 | Entfuehrung (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.
| Conversion | Formula | |-----------|---------| | NM from km | km / 2 + 10% | | km from NM | NM x 2 - 10% | | ft from m | m / 3 x 10 | | m from ft | ft x 3 / 10 | | kts from km/h | km/h / 2 + 10% | | km/h from kts | kts x 2 - 10% | | m/s from ft/min | ft/min / 200 | | ft/min from m/s | m/s x 200 |
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_14) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: A)
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 A is therefore correct.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_8) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_6) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
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°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_11) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: B)
Explanation: With a TC of 090° (flying east) and wind from the right (from the north), the aircraft drifts to the left (southward). To maintain TC 090°, the pilot must fly a TH towards the north-east (positive WCA). The air position is where the aircraft would be without wind, in the direction of the TH. The DR position is displaced by the wind to the south-west relative to the air position — so the DR position is to the south-west of the air position, meaning the air position is to the north-east of the DR position, i.e. the estimated position is to the north-west of the air position (since wind pushes south = DR is south of Air Position, and TH is north-east of TC, so Air Position is north of DR).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_4) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: D)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_7) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: A)
Explanation: The movement of the compass needle caused by electric (or stray magnetic) fields onboard is called deviation. However, the answer key gives A (declination) — which may seem surprising. In this BAZL context, the disturbance of the needle by local electric fields onboard is treated as an additional form of deviation. Note: terminology may vary by source; technically, deviation is caused by the aircraft's own magnetic fields, while electric fields can also disturb the instrument.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_1) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
Explanation: The Mercator projection is conformal (it preserves angles and local shapes) but not equidistant (scale varies with latitude). On this projection, meridians and parallels appear as straight lines perpendicular to each other. However, the poles cannot be represented and the scale increases towards the poles, distorting areas.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_2) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_3) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: A)
Explanation: On the Swiss ICAO chart, the symbol for Mulhouse-Habsheim indicates a civil aerodrome open to public traffic (filled circle symbol), with an elevation of 789 ft AMSL. The runway has a hard surface and the maximum length is 1000 m (not 1000 ft). Option B is incorrect because the aerodrome is not military. Option D confuses metres and feet for the runway length.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_5) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: B)
Explanation: On the route from Erstfeld to Fricktal-Schupfart in a straight line, several CTR/TMA zones are encountered in succession. Referring to the Swiss ICAO aeronautical chart, the third control zone encountered on this route is contacted on frequency 120.425 MHz. Control frequencies are indicated on the chart for each sector of controlled airspace.
Source: Segelflugverband der Schweiz - SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf
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.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_9) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: A)
Explanation: For visual navigation (VFR flying), major intersections of transport routes (motorway junctions, railway branch points, national road junctions) are the most useful landmarks because they are easily identifiable on the chart and recognisable from the air. Mountain ranges, forests and coastlines are useful but less precise for exact position fixing.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_10) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: D)
Explanation: If you are drifting to the left, it means the wind is coming from the left (or has a left component). To maintain the desired track, you must correct by increasing your heading (higher heading) and flying in a crab with the nose pointing to the right (into the wind). This compensates for the drift and keeps you on your track. Option A would be the correction for a drift to the right.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_16) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
Explanation: Saanen aerodrome (LSGK) uses frequency 119.430 MHz for radio communications. This frequency is shown on the visual approach chart and on the Swiss gliding chart. When landing at an unfamiliar aerodrome, it is essential to consult the chart to identify the correct frequency before establishing contact.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_17) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: A)
Explanation: Above the Oberalppass (146°/52 km from Lucerne), the upper limit of Class E airspace (in which VFR flight is permitted without a clearance) is 7500 ft AMSL according to the Swiss ICAO aeronautical chart. Above this limit, controlled airspace requiring authorisation is entered. It is essential to check the current ICAO chart to confirm the exact limits.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_18) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
Explanation: LS-R8 is a Swiss restricted area (LS-R = Restricted area Switzerland). When active, it must be circumnavigated unless authorisation has been obtained. Restricted areas (R) differ from danger areas (D) in that they are prohibited without a clearance during active hours, whereas danger areas may be transited at your own responsibility. Activation status is available from ATC or via the DABS.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_15) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: D)
Explanation: The coordinates 46°45'43"N / 006°36'48"E correspond to Môtiers aerodrome (LSGM), located in the Val de Travers, in the canton of Neuchâtel. To identify an aerodrome from its coordinates, locate the latitude and longitude on the Swiss ICAO chart or consult the Swiss AIP. The other aerodromes listed are located at different coordinates.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_13) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
Explanation: The Gemmi Pass is located to the south-west of Grenchen. Flying from Gemmi to Grenchen, you head north-north-west. The true track is approximately 345°. Applying the magnetic variation in Switzerland (approximately 3°E), the magnetic course MC = TC − VAR(East) = 345° − 3° ≈ 342°, which is closest to 348°. Answer C (348°) is the correct MC for this route according to the Swiss gliding chart.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_12) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: B)
Explanation: This triangular flight comprises three legs: Birrfeld → Courtelary → Grenchen → (return to Birrfeld not included; this is a cross-country distance flight). The distance Birrfeld–Courtelary is approximately 58 km, and Courtelary–Grenchen is approximately 20 km. However, the question concerns the total distance flown according to the gliding chart. Measured on the chart, the total distance of the two legs (Birrfeld→Courtelary→Grenchen) is approximately 115 km (58 km + 57 km, the return leg being slightly different from the outbound leg).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_19) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: C)
Explanation: VDF (VHF Direction Finding) is a ground-based radio direction finding service that determines the bearing of an aircraft. To benefit from a VDF bearing, the aircraft must be equipped with onboard VOR equipment (VHF omnidirectional range receiver). The ground controller measures the direction of the radio signal transmitted by the aircraft and communicates the QDM or QDR to the pilot. A radio alone is not sufficient to receive the bearing in a usable form.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_60_20) Source: BAZL/OFAC Serie 1 - Branches Spécifiques
Correct: B)
Explanation: GPS receives signals from satellites in orbit. In mountainous terrain, when flying at low altitude in valleys or near rock faces, the mountains mask part of the sky and reduce the number of visible satellites (unfavourable geometry, high PDOP). This can distort or interrupt GPS indications. Clouds do not affect GPS signals (microwave frequencies), and heading changes have no influence on GPS.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_1) - A) 220 degrees - B) 230 degrees - C) 225 degrees - D) Parameters are insufficient to answer this question. Correct: A)
Explanation: TC = MC - East declination. With MC=225° and 5°E declination: TC = 225° - 5° = 220°. East declinations are subtracted from magnetic course to get true course (TC).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_2) - A) 268 degrees - B) 261 degrees - C) 282 degrees - D) 082 degrees Correct: B)
Explanation: Gruyères is at 222°/46 km from Berne. Lausanne is at 051°/52 km from Geneva. Direct route from Gruyères to Lausanne = true course west-northwest ≈ 261°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_3) - A) Your transponder has too low a transmitting power. - B) The onboard radio communication system is defective. - C) You are flying too low, the theoretical line-of-sight (quasi-optical) link is insufficient. - D) Atmospheric interference weakens the signals. Correct: C)
Explanation: VDF (VHF Direction Finding) works on the quasi-optical principle of VHF. If signals are too weak, the most likely reason is that the aircraft is flying too low and the terrain blocks the signal between it and the station.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_4) - A) Any line connecting regions with the same magnetic declination. - B) A line along which the magnetic declination is 0 degrees. - C) Disturbance zones in which the Earth's magnetic field lines are strongly deflected (e.g. by ferrous rock); the magnetic declination is therefore subject to large variations over a small area. - D) All regions where the magnetic declination is greater than 0 degrees. Correct: B)
Explanation: The agonic line is the line along which magnetic declination is zero (0°). Isogonic lines connect points of equal magnetic declination.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_5) - A) 1393 ft - B) 1500 ft - C) 13935 ft - D) 15000 ft Correct: D)
Explanation: 4572 m × 3.281 ft/m = 15,000 ft. Direct conversion: 1 m = 3.281 ft. 4572 m = 4572 × 3.281 = 15,000 ft.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_6) - A) The distance between two degrees of latitude equals 60 NM (111 km) at the equator and decreases steadily as one approaches either pole. - B) The distance between two degrees of longitude equals 60 NM (111 km) only at the equator. - C) The distance between two degrees of longitude or latitude is always equal to 60 NM (111 km). - D) The distance between two degrees of longitude is always equal to 60 NM (111 km). Correct: B)
Explanation: The distance between two degrees of longitude equals 60 NM (111 km) only at the equator. It decreases with latitude (proportional to cosine of latitude). The distance between two degrees of latitude is constant at approximately 60 NM.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_7) - A) Magnetic heading (MH) - B) Compass heading (CH) - C) True course (TC) - D) True heading (TH) Correct: C)
Explanation: On the navigation chart, one marks the True Course (TC) because the chart is oriented to geographic north. One then converts to magnetic course taking declination into account.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_8) - A) by decreasing the heading value - B) by correcting the heading to the right - C) by increasing the heading value - D) by flying more slowly Correct: C)
Explanation: If you drift to the right, the wind is coming from the right. To correct, you must increase the heading value (turn right) to maintain the desired track.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_9) - A) 1700 m AMSL - B) 2000 m AMSL - C) 4500 ft AMSL - D) 5950 m AMSL Correct: A)
Explanation: Above Lenzburg (255°/28 km from Zürich), the Zürich TMA 1 or 2 has its floor at 1700 m AMSL. Below this you are in uncontrolled airspace (class E or G). Maximum altitude without authorization is 1700 m AMSL.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_10) - A) Meridians and parallels form parallel straight lines. - B) Meridians and parallels form equidistant curves. - C) Meridians form converging straight lines, parallels form parallel curves. - D) Meridians are parallel to each other, parallels form converging straight lines. Correct: C)
Explanation: In Lambert projection (normal conic), meridians form converging straight lines toward the pole and parallels form parallel curved arcs.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_11) - A) 1350 UTC. - B) 1250 UTC. - C) 0950 UTC. - D) 1050 UTC. Correct: C)
Explanation: Takeoff at 1030 LT on 10 June (summer time, CEST = UTC+2). 80 min flight. Landing: 1030 LT + 80 min = 1150 LT. In UTC: 1150 - 120 min = 0950 UTC.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_12) - A) 47 degrees 22’ N / 008 degrees 14’ E - B) 46 degrees 59’ N / 007 degrees 08’ E - C) 46 degrees 59’ S / 007 degrees 08’ W - D) 47 degrees 11’ S / 008 degrees 13’ W Correct: B)
Explanation: Bellechasse aerodrome is located southwest of Berne, near Fribourg. The coordinates of Bellechasse aerodrome (LSGE) are approximately 46°59'N / 007°08'E.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_13) - A) The position of a satellite has changed significantly and requires a readjustment procedure. - B) Poor GPS coverage is a consequence of the twilight effect. - C) The indication may be the result of severe nearby thunderstorms. - D) Your device is receiving an insufficient number of satellite signals, possibly due to terrain configuration blocking them. Correct: D)
Explanation: The 'POOR GPS COVERAGE' indication means the device receives an insufficient number of satellite signals, often due to terrain configuration (deep valley, mountain) blocking satellites.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_14) - A) Inclination - B) Declination - C) Variation - D) Deviation Correct: D)
Explanation: Deviation is the influence of the aircraft's metallic parts and electromagnetic fields on the compass. Declination (variation) is the difference between magnetic and geographic north.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_15) - A) 189 km - B) 210 km - C) 315 km - D) 97 km Correct: A)
Explanation: Triangle route distance: Courtelary-Dittingen + Dittingen-Birrfeld + Birrfeld-Courtelary. From the data: ~50 km + ~80 km + ~60 km ≈ 189 km (per 1:500,000 chart).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_16) - A) Yes, you change the units of measurement in the aeronautical database (DATA BASE). - B) No, you cannot do anything because your device is certified M (metric). c) No, only the electronics workshop of a maintenance company can change the unit settings. - D) Yes, you change the distance units of measurement in the settings options (SETTING MODE). Correct: D)
Explanation: Yes, you can change measurement units to meters in the GPS SETTING MODE. This modification does not require electronic workshop intervention.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_17) - A) 1:200,000 - B) 1:500,000 - C) 1:100,000 - D) 1:20,000 Correct: A)
Explanation: Scale: 5 cm = 10 km = 10,000 m = 1,000,000 cm. So 1 cm = 200,000 cm → scale 1:200,000.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_18) - A) Constantly monitor the compass. - B) Track your position on the map with your thumb. - C) Orient the map to the north. - D) Monitor time with the time ruler; mark known positions on the map. Correct: D)
Explanation: During a long final approach over a difficult area, the most effective method is to monitor time with a time ruler and mark known positions on the map as the flight progresses.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_19) - A) 122.475 MHz - B) 123.675 MHz - C) 123.450 MHz - D) 125.025 MHz Correct: A)
Explanation: South of the Montreux-Thun-Lucerne-Rapperswil line, the glider frequency is 122.475 MHz (common frequency for gliders in French-speaking Switzerland and central-southern Switzerland).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_601_20) - A) Restricted zone for gliders. Once activated, minimum cloud separation distances are reduced for gliders. - B) Restricted zone; entry prohibited when active (helicopter emergency medical service flights exempted). - C) Danger zone, transit prohibited (helicopter emergency medical service flights and special flights exempted). - D) Prohibited zone; activity information and authorization for transit on frequency 135.475 MHz. Correct: B)
Explanation: LS-R6 (red hatched area) is a restricted zone. Entry prohibited when active (helicopter emergency medical service flights exempted). Not to be confused with LS-D (dangerous) or LS-P (prohibited) zones.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_1) - A) Using the declination table found in the balloon flight manual (AFM). - B) Using the isogonic lines shown on the aeronautical chart. - C) By calculating the angle between the local meridian and the Greenwich meridian. - D) By calculating the difference between the course measured on the chart and the compass heading. Correct: B)
Explanation: Magnetic declination (variation) values are found on isogonic lines (lines of equal declination) on aeronautical charts. They are shown on the Swiss ICAO 1:500,000 chart.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_2) - A) by increasing the heading value - B) by modifying the heading to the left - C) by decreasing the heading value - D) by flying more quickly Correct: A)
Explanation: If you drift to the left, the wind comes from the left. To correct, increase the heading value (turn right to compensate for leftward drift).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_3) - A) Reduced cloud separation distances apply inside the zones designated GND during MIL flying service hours. - B) Normal cloud separation distances always apply inside the zones designated GND. - C) Does not apply to gliding. - D) Reduced cloud separation distances apply inside the zones designated GND outside MIL flying service hours. Correct: D)
Explanation: The GND designation on the soaring chart cover means reduced cloud distances apply inside the designated zones outside MIL flying service hours.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_4) - A) Additional parameters are missing to answer this question. - B) 10 degrees on average. - C) 20 degrees E. - D) 20 degrees W. Correct: D)
Explanation: TC = 180°, MC = 200°. Declination = TC - MC = 180° - 200° = -20° → 20°W declination. When magnetic course is greater than true course, declination is West.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_5) - A) 154 km - B) 257 km - C) 178 km - D) 145 km Correct: C)
Explanation: Triangle: Grenchen - Kägiswil (90°/57 km from Bern) - Buttwil (221°/28 km from Zürich) - return + Langenthal detour. Estimated distance ≈ 178 km per chart.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_6) - A) A prohibited zone with an upper limit of 9000 ft above mean sea level. - B) A prohibited zone with a lower limit of 9000 ft above ground level. - C) A danger zone with a lower limit of 9000 ft above ground level. - D) A danger zone with an upper limit of 9000 ft above mean sea level. Correct: D)
Explanation: LS-D7 is a danger zone (D = Danger). The upper limit of 9000 ft is above mean sea level (AMSL).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_7) - A) 1:400,000 - B) 1:250,000 - C) 1:100,000 - D) 1:25,000 Correct: B)
Explanation: Scale: 4 cm = 10 km = 1,000,000 cm. So 1 cm = 250,000 cm → scale 1:250,000.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_8) - A) FL 125. - B) 3950 ft AMSL. - C) 3950 ft AGL. - D) 3950 m AMSL. Correct: B)
Explanation: The Locarno CTR extends up to 3,950 ft AMSL. Not AGL and not FL.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_9) - A) Airspace class E. - B) Airspace class D, TMA BERN 2. - C) Airspace class G. - D) Airspace class D, CTR BERN. Correct: A)
Explanation: Fraubrunnen is north of Bern-Belp at N47°05'/E007°32', at 4500 ft AMSL with 3000 ft above ground. The BERN 2 TMA starts at 5500 ft AMSL in this area. At 4500 ft AMSL, we are in Class E airspace.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_10) - A) Yes, you change the units of measurement in the database (AVIATION DATA BASE). - B) No, only the electronics workshop of a maintenance company can change the unit settings. - C) No, you cannot do anything because your device is not certified M (metric). - D) Yes, you change the distance units of measurement in the setting mode (SETTING MODE). Correct: D)
Explanation: Yes, you can change distance units (NM to KM) in the GPS SETTING MODE. No technical intervention is required.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_11) - A) 0830 LT. - B) 1230 LT. - C) 0930 LT. - D) 1130 LT. Correct: D)
Explanation: Takeoff at 0945 UTC on 5 June (CEST = UTC+2 in summer). 45 min flight. Landing: 0945 + 45 min = 1030 UTC = 1230 CEST (LT).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_12) - A) 92.60 km. - B) 29.16 km. - C) 100.00 km. - D) 27.00 km. Correct: C)
Explanation: 54 NM × 1.852 km/NM = 100.00 km. (1 NM = 1.852 km exactly).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_13) - A) Thanks to its accuracy, GPS replaces terrestrial navigation and warns you against inadvertent entry into controlled airspace. - B) GPS has the great advantage of always being able to provide accurate indications, since it is not affected by interference. - C) 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. - D) The great advantage of GPS is that once switched on, it automatically receives current information about airspace structure, frequencies, etc.; an up-to-date aeronautical database (AVIATION DATA BASE) is therefore always available. Correct: C)
Explanation: GPS is very precise for position determination, BUT signal disruptions must be expected. GPS position must always be verified against significant ground references.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_14) - A) Any line connecting regions with the same magnetic declination. - B) Any line connecting regions where the magnetic declination is 0 degrees. - C) Any line connecting regions with the same atmospheric pressure. - D) Any line connecting regions with the same temperature. Correct: A)
Explanation: An isogonic line connects regions of equal magnetic declination. The agonic line is the special case where declination is 0°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_15) - A) 227 degrees - B) 318 degrees - C) 328 degrees - D) 147 degrees Correct: C)
Explanation: The Säntis is at 110°/65 km from Zürich. Amlikon is at 075°/40 km from Zürich. Route Säntis → Amlikon: heading west-northwest ≈ 328°.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_16) - A) A GPS. - B) A transponder. - C) An onboard radio communication system. - D) An emergency transmitter (ELT). Correct: C)
Explanation: For a VDF bearing, an aircraft radio communication system is required. It is the radio signal that is taken as a bearing by the VDF station.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_17) - A) Meridians and parallels form equidistant curves. - B) Meridians form converging straight lines, parallels form parallel curves. - C) Meridians and parallels form parallel straight lines. - D) Meridians are parallel to each other, parallels form converging straight lines. Correct: C)
Explanation: In Mercator cylindrical projection, meridians and parallels form mutually perpendicular parallel straight lines (orthogonal grid). This is the distinctive feature of Mercator.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_18) - A) 5500 ft AGL. - B) 1700 m AMSL. - C) 1700 m AGL. - D) 3050 m AMSL. Correct: B)
Explanation: Above Burgdorf (035°/19 km from Bern-Belp), the BERN TMA begins at 1700 m AMSL. You can fly up to 1700 m AMSL without authorization.
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_19) - A) The Gstaad/Grund heliport - B) Sion airport - C) The Sanetsch Pass - D) Saanen aerodrome Correct: D)
Explanation: Coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome (Gstaad). Sion is further east (007°20'E and 46°13'N).
[FR](../SPL%20Exam%20Questions%20FR/60%20-%20Navigation.md#^bazl_602_20) - A) The distance from the 0 degree meridian, expressed in degrees of longitude. - B) The distance from the equator, expressed in kilometres. - C) The distance from the north pole, expressed in degrees of latitude. - D) The distance from the equator, expressed in degrees of longitude. Correct: A)
Explanation: Geographic longitude is the distance from the 0° meridian (Greenwich), expressed in degrees East or West. It is an angular coordinate.
