Source: exam.quizvds.it (EASA ECQB-SPL) | 78 questions
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.
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.
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.
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.
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).
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.
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.
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.
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.
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: B)
Explanation: Magnetic Course (MC) is defined as the angle measured at the aircraft's position between magnetic north and the intended course line, measured clockwise from 0° to 360°. It differs from True Course, which is measured from geographic (true) north. Option A describes a magnetic bearing to the pole, not a course angle. Option C is the definition of True Course. Option D describes the direction to the geographic North Pole (true north reference).
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.
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: 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: 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.
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.
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.
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).
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.
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.
Correct: A)
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: D)
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: A)
Explanation: With a true course of 255° and wind from 200° at 10 kt, the wind has a component from the left-front (southerly wind pushing the aircraft to the right of track). To maintain the 255° course, the pilot must crab slightly into the wind — heading to the left, i.e., a smaller heading number. Applying the WCA formula (WCA ≈ sin⁻¹(wind speed × sin(wind angle off nose) / TAS) ≈ sin⁻¹(10 × sin55° / 100) ≈ sin⁻¹(0.082) ≈ 5°), the true heading is approximately 255° − 5° = 250°.
Correct: B)
Explanation: With a true course of 165° and wind from 130° at 20 kt, the wind comes from ahead-left (approximately 35° off the left nose). The crosswind component pushes the aircraft to the right of the intended track, so the pilot must crab left — heading to a smaller bearing. WCA ≈ sin⁻¹(20 × sin35° / 90) ≈ sin⁻¹(0.128) ≈ 7°. True heading = 165° − 7° = 158°. Options A, C, and D are inconsistent with this vector calculation.
Correct: D)
Explanation: With a true course of 040° and wind from 350° at 30 kt, the wind is from ahead-left (50° off the left of the course). The wind has a headwind component: 30 × cos50° ≈ 19 kt headwind, reducing groundspeed. The crosswind component: 30 × sin50° ≈ 23 kt causes a WCA of about 7° right. GS = TAS × cos(WCA) − headwind component ≈ 180 × cos7° − 19 ≈ 179 − 19 ≈ 160 kt… More precisely using vector arithmetic, GS ≈ 159 kt, matching option D.
Correct: A)
Explanation: With a true course of 120° and wind from 150° at 12 kt, the wind is from approximately 30° to the right of the course line (from behind-right). This pushes the aircraft to the left of track, requiring the pilot to crab right — applying a positive WCA. WCA ≈ sin⁻¹(12 × sin30° / 120) = sin⁻¹(0.05) ≈ 3° to the right. Options B and C are too large; option D is in the wrong direction.
Correct: B)
Explanation: Using the closing angle method: the track error is 7 NM in 55 NM flown, giving an opening angle of 7/55 × 60 ≈ 7.6° ≈ 8° off track. The remaining distance to B is 120 − 55 = 65 NM. The closing angle needed to reach B = 7/65 × 60 ≈ 6.5° ≈ 7°. Total course change = opening angle + closing angle ≈ 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left). This matches option B.
Correct: D)
Explanation: GPS requires signals from at least four satellites for a precise three-dimensional position fix with integrity verification. Three satellites provide a 2D fix (latitude and longitude only); the fourth satellite provides the altitude dimension and, critically, allows the receiver to solve for clock error and verify the solution. A fifth satellite enables Receiver Autonomous Integrity Monitoring (RAIM). Two satellites are insufficient for any reliable position fix.
Correct: D)
Explanation: During visual navigation, large linear features — rivers, railways, and highways — are the most reliable ground references because they are prominent, unambiguous, correctly depicted on aeronautical charts, and visible from distance. Power lines (option A) are difficult to spot and hazardous to fly near. Farm tracks and creeks (option B) are too numerous and similar to distinguish reliably. Border lines (option C) are invisible from the air.
Correct: C)
Explanation: The circumference of the Earth at the equator is approximately 21,600 nautical miles (NM), which corresponds to 360° × 60 NM/° = 21,600 NM. This is a fundamental navigation fact: one degree of arc on the Earth's surface equals 60 NM, and one minute of arc equals 1 NM. The other values in km are incorrect: the actual circumference is about 40,075 km, not 10,800 or 12,800 km; 40,000 NM is also far too large.
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.
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.
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.
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.
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).
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.
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.
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: 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.
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: Ground speed is 107 kt and distance is 100 NM. Flight time = 100/107 hours = 0.935 h = 56 minutes. ETD is 0933 UTC; ETA = 0933 + 0056 = 1029 UTC. Options A, C, and D all differ from this calculation and are incorrect.
Correct: B)
Explanation: Ground speed = distance / time = 100 km / (56/60 h) = 100 × 60/56 ≈ 107 km/h. The result is in km/h since the distance was given in km and time in minutes. Option A (93 kt) confuses units; option C (198 kt) is far too high; option D (58 km/h) would be the result of an arithmetic error. 107 km/h correctly answers the question.
Correct: C)
Explanation: Groundspeed = TAS − headwind = 180 − 25 = 155 kt. Flight time = 2 h 25 min = 2.417 h. Distance = 155 × 2.417 ≈ 375 NM. Option A (693 NM) uses TAS without subtracting the headwind. Option B (202 NM) appears to use the headwind component only. Option D (435 NM) uses TAS without headwind correction.
Correct: B)
Explanation: The wind is from 140° at 20 kt and the true course is 177°. The wind is approximately 37° to the left of the course, so it pushes the aircraft to the right of track — the pilot must crab left (reduce heading). WCA ≈ sin⁻¹(20 × sin37° / GS). Given GS = 160 kt, WCA ≈ sin⁻¹(12.0/160) ≈ sin⁻¹(0.075) ≈ 4.3°. True heading = 177° − 4° = 173°. Options A, C, and D yield incorrect headings for this wind scenario.
Correct: C)
Explanation: With a true course of 040° and wind from 350° at 30 kt, the wind angle relative to the course is 50° from the left. The crosswind component = 30 × sin50° ≈ 23 kt pushes the aircraft right of track; to maintain the 040° course the aircraft must crab left (negative WCA). WCA ≈ −sin⁻¹(23/180) ≈ −7°. The negative sign confirms a left correction (option C: −7°). Options A and D show right corrections, which would be wrong for this wind direction.
Correct: D)
Explanation: With a direct headwind of 25 kt (wind from 090° on a 270° course), groundspeed = TAS + tailwind = 100 + 25 = 125 kt. Distance is 100 NM, so flight time = 100/125 = 0.8 h = 48 min. However, since the aircraft flies toward the wind source (west), the wind from the east is actually a tailwind. GS = 100 + 25 = 125 kt. Option A (120 kt) is close but reflects only partial wind addition; option B (131 kt) and option C (117 kt) are also incorrect by varying amounts.
Correct: B)
Explanation: The GPS CDI (Course Deviation Indicator) bar shows lateral track error as an absolute distance in nautical miles, not as an angular deviation in degrees. The full-scale deflection of the bar depends on the operating mode: in terminal mode it is typically ±1 NM, in en-route mode ±5 NM, and in approach mode ±0.3 NM. Options A and C incorrectly state that the deviation is angular (in degrees). Option D incorrectly states the fixed scale as ±10 NM.
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).
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).
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.
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").
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°.
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: 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.
Correct: A)
Explanation: Using the chart coordinates: BKD is at 53°02'N, 011°33'E and EDBU is at 53°11'N, 012°11'E. The latitude difference is 9' (= 9 NM north-south component). The longitude difference is 38'; at 53°N, 1' of longitude ≈ cos53° NM ≈ 0.60 NM, so 38' × 0.60 ≈ 22.8 NM east-west component. Total distance ≈ √(9² + 23²) ≈ √(81 + 529) ≈ √610 ≈ 24.7 NM ≈ 24 NM. The km options (options C and D) are incorrect units for this aeronautical distance.
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°.
Correct: A)
Explanation: Groundspeed = TAS + tailwind = 120 + 35 = 155 kt. Flight time = 185 NM / 155 kt = 1.194 h ≈ 1 h 12 min. Option B (2 h 11 min) is far too long and appears to use only TAS. Option C (50 min) would require a much higher groundspeed. Option D (1 h 32 min) would correspond to a groundspeed of about 120 kt, ignoring the tailwind.
Correct: A)
Explanation: With wind from 090° at 25 kt on a 270° course, the wind is a direct tailwind, giving GS = TAS + wind = 100 + 25 = 125 kt. Flight time = 100 NM / 125 kt = 0.8 h = 48 min. Option B (37 min) would require a GS of about 162 kt. Option C (84 min) would be the result if the wind were treated as a headwind. Option D (62 min) reflects an incorrect intermediate GS.
Correct: A)
Explanation: This flight plan question involves converting from True Course to Magnetic Heading using variation and wind correction. The correct answer TH: 185°, MH: 184°, MC: 178° reflects the sequential application of wind correction angle (WCA) to obtain true heading, then magnetic variation to convert to magnetic heading, and finally compass deviation to obtain compass heading (or vice versa). The other options contain inconsistencies in the conversion chain that do not satisfy the navigation triangle for the given parameters.
Correct: C)
Explanation: Terrestrial navigation (also called visual navigation or pilotage) means the pilot orients the aircraft by visually identifying ground features and matching them to a topographic or aeronautical chart. This is distinct from instrument navigation (option B), GPS navigation (option D), and celestial navigation. Option A ('celestial object') incorrectly conflates terrestrial with astronomical navigation.
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.
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.
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).
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.
Correct: D)
Explanation: Flight time = distance / groundspeed = 236 NM / 134 kt = 1.761 h. To convert to hours and minutes: 0.761 × 60 ≈ 46 min, giving 1:46 h. Option A (1:34 h) would correspond to about 150 kt groundspeed. Options B (0:34 h) and C (0:46 h) are well under an hour and far too short for 236 NM at 134 kt.
Correct: D)
Explanation: On the aeronautical chart, Uelzen (EDVU) lies to the south-west of Neustadt (EDAN) — Neustadt is further north and further east. The true course from Neustadt to Uelzen is therefore in a south-westerly direction (~241°), while the reciprocal course from Uelzen to Neustadt is north-easterly (~061°). The question asks for the course FROM Uelzen TO Neustadt, which is approximately 061°. Option A (241°) is the reciprocal. Options B (055°) and C (235°) are close but do not match the plotted bearing accurately.
Correct: B)
Explanation: The 1:60 rule states that at 60 NM from a reference point, 1° of angular track error produces a lateral offset of exactly 1 NM. This is because the arc length of 1° on a circle of radius 60 NM equals approximately 1 NM (since 2π × 60 / 360 ≈ 1.047 NM ≈ 1 NM). This rule is used to quickly estimate track corrections without a computer. Options A, C, and D misstate either the angle, the distance, or the offset relationship.
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.
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.
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.
Correct: A)
Explanation: With a true course of 220° and wind from 270° at 50 kt, the wind angle relative to the course is 50° from the right (270° − 220° = 50°). The headwind component = 50 × cos50° ≈ 32 kt and the crosswind component = 50 × sin50° ≈ 38 kt. GS ≈ √((TAS − headwind)² + crosswind²)... more precisely using the navigation triangle: GS ≈ TAS − (headwind component) corrected for crab angle. Vector calculation yields approximately 185 kt. Options B (255 kt) and D (135 kt) are too high and too low respectively; option C (170 kt) is slightly too low.
Correct: D)
Explanation: Using the 1:60 rule: the track error is 4.5 NM in 45 NM flown, giving an opening angle of 4.5/45 × 60 = 6° (the aircraft is north of track, heading 090°). The remaining distance = 90 − 45 = 45 NM. The closing angle = 4.5/45 × 60 = 6°. Total correction = 6° + 6° = 12° to the right (south, since the aircraft is north of track). Option A (18°) and option B (9°) are arithmetically incorrect; option C (6°) only accounts for the closing angle.
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.
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.
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: The distance from Neustadt (EDAN) to Uelzen (EDVU) can be calculated from the coordinates: latitude difference = 53°22'N − 52°59'N = 23' ≈ 23 NM north-south. Longitude difference = 011°37'E − 10°28'E = 69'; at ~53°N, 1' longitude ≈ 0.60 NM, so 69' × 0.60 ≈ 41.4 NM east-west. Total ≈ √(23² + 41.4²) ≈ √(529 + 1714) ≈ √2243 ≈ 47 NM ≈ 46 NM. Options C and D in km (78 km) would equal ~42 NM, which is too low; option A (46 km ≈ 25 NM) is far too short.
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 B), GPS (option D), or celestial bodies (option A). It is the foundational VFR navigation skill and is sometimes called 'map reading' or 'pilotage'.
