111 questions
Correct: A)
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.
Correct: C)
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 B confuses the tilt to the ecliptic with the relationship to the equator.
Correct: 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.
Correct: A)
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.
Correct: C)
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.
Correct: 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.
Correct: B)
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).
Correct: 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 B) are the ones located 23.5° from the equator.
Correct: 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 D) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations.
Correct: B)
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 D.
Correct: B)
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 D would require only 2° (120 NM).
Correct: A)
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.
Correct: B)
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.
Correct: C)
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.
Correct: 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.
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 B (4 minutes) would be the time for only 1° of longitude. Option D (30 minutes) would correspond to 7.5° of longitude.
Correct: C)
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.
Correct: B)
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.
Correct: 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).
Correct: B)
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.
Correct: A)
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.
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.
Correct: C)
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.
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°. East variation is subtracted when converting from True to Magnetic ("East is least").
Correct: 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).
Correct: D)
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.
Correct: C)
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.
Correct: 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.
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.
Correct: 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°.
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.
Correct: D)
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.
Correct: 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.
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.
Correct: 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.
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.
Correct: C)
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.
Correct: C)
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.
Correct: C)
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).
Correct: D)
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.
Correct: C)
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.
Correct: 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.
Correct: C)
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).
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.
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.
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.
Correct: D)
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 D correct.
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.
Correct: B)
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°.
Correct: 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.
Correct: 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.
Correct: C)
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 C is therefore correct.
Correct: B)
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.
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°.
Correct: C)
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).
Correct: 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.
Correct: B)
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.
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.
Correct: B)
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 A is incorrect because the aerodrome is not military. Option D confuses metres and feet for the runway length.
Correct: C)
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.
Correct: B)
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.
Correct: C)
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.
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.
Correct: B)
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.
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.
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.
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.
Correct: D)
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).
Correct: B)
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.
Correct: C)
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.
Correct: C)
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).
Correct: C)
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°.
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.
Correct: A)
Explanation: The agonic line is the line along which magnetic declination is zero (0°). Isogonic lines connect points of equal magnetic declination.
Correct: B)
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.
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.
Correct: D)
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.
Correct: D)
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.
Correct: C)
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.
Correct: B)
Explanation: In Lambert projection (normal conic), meridians form converging straight lines toward the pole and parallels form parallel curved arcs.
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.
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.
Correct: C)
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.
Correct: C)
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.
Correct: D)
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).
Correct: D)
Explanation: Yes, you can change measurement units to meters in the GPS SETTING MODE. This modification does not require electronic workshop intervention.
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.
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.
Correct: B)
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).
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.
Correct: C)
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.
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.
Correct: A)
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.
Correct: D)
Explanation: Triangle: Grenchen - Kägiswil (90°/57 km from Bern) - Buttwil (221°/28 km from Zürich) - return + Langenthal detour. Estimated distance ≈ 178 km per chart.
Correct: B)
Explanation: LS-D7 is a danger zone (D = Danger). The upper limit of 9000 ft is above mean sea level (AMSL).
Correct: D)
Explanation: Scale: 4 cm = 10 km = 1,000,000 cm. So 1 cm = 250,000 cm → scale 1:250,000.
Correct: C)
Explanation: The Locarno CTR extends up to 3,950 ft AMSL. Not AGL and not FL.
Correct: C)
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.
Correct: D)
Explanation: Yes, you can change distance units (NM to KM) in the GPS SETTING MODE. No technical intervention is required.
Correct: B)
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).
Correct: C)
Explanation: 54 NM × 1.852 km/NM = 100.00 km. (1 NM = 1.852 km exactly).
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.
Correct: C)
Explanation: An isogonic line connects regions of equal magnetic declination. The agonic line is the special case where declination is 0°.
Correct: B)
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°.
Correct: C)
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.
Correct: A)
Explanation: Coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome (Gstaad). Sion is further east (007°20'E and 46°13'N).
Correct: C)
Explanation: Geographic longitude is the distance from the 0° meridian (Greenwich), expressed in degrees East or West. It is an angular coordinate.
Correct: B)
Explanation: True altitude is calculated from QNH altitude by correcting for non-standard temperature. The ISA temperature at 6500 ft QNH altitude is approximately +3°C (ISA = 15°C − 2°C/1000 ft × 6.5 ≈ +2°C). The OAT is −9°C, meaning the air is colder than ISA. Cold air is denser, so the aircraft is actually lower than the pressure altitude indicates — true altitude is less than QNH altitude. Using the ICAO correction formula (approx. 4 ft per 1°C per 1000 ft), the temperature deviation is about −11°C at ~6500 ft, giving a correction of roughly −250 ft, yielding approximately 6250 ft true altitude.
Correct: C)
Explanation: At a pressure altitude of 7000 ft and QNH altitude of 6500 ft, the aircraft is 500 ft above QNH. OAT is +11°C. ISA temperature at ~7000 ft is approximately +1°C (15 − 2×7 = +1°C). OAT of +11°C is +10°C above ISA — warmer air is less dense, so the aircraft is higher than indicated. Applying the standard correction of ~4 ft per 1°C per 1000 ft: +10°C × ~4 ft/°C/1000 ft × 6.5 ≈ +250 ft above QNH altitude. 6500 + 250 = 6750 ft true altitude.
Correct: C)
Explanation: At pressure altitude 7000 ft, QNH altitude 6500 ft, and OAT +21°C: ISA temperature at ~7000 ft is approximately +1°C. OAT of +21°C is +20°C above ISA — significantly warmer, meaning less dense air and the aircraft is higher than QNH. The temperature correction (≈ 4 ft/°C/1000 ft × +20°C × 6.5) yields approximately +500 ft, so true altitude ≈ 6500 + 500 = 7000 ft. When OAT closely matches the temperature that would produce standard pressure at that altitude, true and pressure altitudes converge near 7000 ft.
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.
Correct: C)
Explanation: Terrestrial navigation means the pilot navigates visually by identifying and matching actual ground features — roads, rivers, towns, railways — to the aeronautical chart. This technique does not rely on instruments (option A), GPS (option B), or celestial bodies (option D). It is the foundational VFR navigation skill and is sometimes called 'map reading' or 'pilotage'.
