### Q1: Through which points does the Earth's rotational axis pass? ^t60q1 - A) The geographic North Pole and the magnetic south pole. - B) The magnetic north pole and the geographic South Pole. - C) The geographic North Pole and the geographic South Pole. - D) The magnetic north pole and the magnetic south pole. **Correct: C)** > **Explanation:** The Earth's rotational axis is the physical axis around which the planet spins, and it passes through the geographic (true) poles — not the magnetic poles. The geographic poles are fixed points defined by the rotational axis, while the magnetic poles are offset from them and drift over time due to changes in the Earth's molten core. ### Q2: Which statement correctly describes the polar axis of the Earth? ^t60q2 - A) It passes through the geographic South Pole and the geographic North Pole and is tilted 23.5° relative to the equatorial plane. - B) It passes through the magnetic south pole and the magnetic north pole and is tilted 66.5° relative to the equatorial plane. - C) It passes through the magnetic south pole and the magnetic north pole and is perpendicular to the equatorial plane. - D) It passes through the geographic South Pole and the geographic North Pole and is perpendicular to the equatorial plane. **Correct: D)** > **Explanation:** The polar axis passes through the geographic poles and is perpendicular (90°) to the plane of the equator by definition. The Earth's axis is indeed tilted 23.5° relative to the plane of its orbit around the sun (the ecliptic), but it is perpendicular to the equatorial plane — those two facts are consistent and not contradictory. Option A confuses the tilt to the ecliptic with the relationship to the equator. ### Q3: For navigation systems, which approximate geometrical shape best represents the Earth? ^t60q3 - A) A flat plate. - B) An ellipsoid. - C) A sphere of ecliptical shape. - D) A perfect sphere. **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. ### Q4: Which of the following statements about a rhumb line is correct? ^t60q4 - A) The shortest path between two points on the Earth follows a rhumb line. - B) A rhumb line crosses each meridian at an identical angle. - C) The centre of a complete rhumb line circuit is always the centre of the Earth. - D) A rhumb line is a great circle that meets the equator at 45°. **Correct: B)** > **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. ### Q5: The shortest route between two points on the Earth's surface follows a segment of... ^t60q5 - A) A small circle - B) A great circle. - C) A rhumb line. - D) A parallel of latitude. **Correct: B)** > **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. ### Q6: What is the approximate circumference of the Earth measured along the equator? See figure (NAV-002) ^t60q6 ![Earth Globe](figures/t60_q6.svg) - A) 40000 NM. - B) 21600 NM. - C) 10800 km. - D) 12800 km. **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. ### Q7: What is the latitude difference between point A (12°53'30''N) and point B (07°34'30''S)? ^t60q7 - A) .20°28'00'' - B) .05°19'00'' - C) .20,28° - D) .05,19° **Correct: A)** > **Explanation:** When two points are on opposite sides of the equator, the difference in latitude is the sum of their respective latitudes. Here: 12°53'30''N + 07°34'30''S = 20°28'00''. Converting minutes: 53'30'' + 34'30'' = 88'00'' = 1°28'00'', so 12° + 7° + 1°28' = 20°28'00''. Always add latitudes when they are in opposite hemispheres (N and S). ### Q8: At what positions are the two polar circles located? ^t60q8 - A) 23.5° north and south of the equator - B) At a latitude of 20.5°S and 20.5°N - C) 20.5° south of the poles - D) 23.5° north and south of the poles **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 A) are the ones located 23.5° from the equator. ### Q9: Along a meridian, what is the distance between the 48°N and 49°N parallels of latitude? ^t60q9 - A) 111 NM - B) 10 NM - C) 60 NM - D) 1 NM **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 A) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations. ### Q10: Along any line of longitude, what distance corresponds to one degree of latitude? ^t60q10 - A) 30 NM - B) 1 NM - C) 60 km - D) 60 NM **Correct: D)** > **Explanation:** One degree of latitude = 60 arcminutes, and since 1 NM equals exactly 1 arcminute of latitude along a meridian, 1° of latitude = 60 NM. This relationship holds along any meridian because all meridians are great circles. In SI units, 1° of latitude ≈ 111 km, not 60 km as stated in option C. ### Q11: Point A lies at exactly 47°50'27''N latitude. Which point is precisely 240 NM north of A? ^t60q11 - A) 49°50'27''N - B) 43°50'27''N - C) 53°50'27''N - D) 51°50'27'N' **Correct: D)** > **Explanation:** Converting 240 NM to degrees of latitude: 240 NM / 60 NM per degree = 4°. Adding 4° to 47°50'27''N gives 51°50'27''N. Moving north increases the latitude value. Option C would require 6° (360 NM), and option A would require only 2° (120 NM). ### Q12: Along the equator, what is the distance between the 150°E and 151°E meridians? ^t60q12 - A) 1 NM - B) 60 NM - C) 60 km - D) 111 NM **Correct: B)** > **Explanation:** On the equator, meridians of longitude are separated by great circle arcs, and 1° of longitude along the equator equals 60 NM — the same as 1° of latitude along any meridian, because the equator is also a great circle. At higher latitudes, the distance between meridians decreases (multiplied by cos(latitude)), but at the equator it is exactly 60 NM per degree. ### Q13: When two points A and B on the equator are separated by exactly one degree of longitude, what is the great circle distance between them? ^t60q13 - A) 216 NM - B) 120 NM - C) 60 NM - D) 400 NM **Correct: C)** > **Explanation:** The equator itself is a great circle, so the great circle distance between two points on the equator separated by 1° of longitude is simply 60 NM (1° x 60 NM/degree). This is the same principle as measuring along a meridian. Any confusion arises if one tries to calculate using km instead — 1° ≈ 111 km on the equator, but the question asks for NM. ### Q14: Consider two points A and B on the same parallel of latitude (not the equator). A is at 010°E and B at 020°E. The rhumb line distance between them is always... ^t60q14 - A) More than 600 NM. - B) More than 300 NM. - C) Less than 300 NM. - D) Less than 600 NM. **Correct: D)** > **Explanation:** The rhumb line distance between points on the same parallel of latitude is: 10° x 60 NM x cos(latitude). Since cos(latitude) is always less than 1 for any latitude other than the equator (where it equals exactly 60 NM x 10 = 600 NM), the rhumb line distance is always strictly less than 600 NM. At the equator it would equal 600 NM, but since they are specifically "not on the equator," the distance is always less than 600 NM. ### Q15: How much time elapses as the sun traverses 20° of longitude? ^t60q15 - A) 0:20 h - B) 1:20 h - C) 0:40 h - D) 1:00 h **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. ### Q16: How much time passes as the sun crosses 10° of longitude? ^t60q16 - A) 0:30 h - B) 0:40 h - C) 1:00 h - D) 0:04 h **Correct: B)** > **Explanation:** Using the same principle as Q15: the Earth rotates 15° per hour, so 10° corresponds to 10/15 hours = 2/3 hour = 40 minutes = 0:40 h. Option D (4 minutes) would be the time for only 1° of longitude. Option A (30 minutes) would correspond to 7.5° of longitude. ### Q17: The sun traverses 10° of longitude. What is the corresponding time difference? ^t60q17 - A) 0.33 h - B) 1 h - C) 0.4 h - D) 0.66 h **Correct: D)** > **Explanation:** This is the same calculation as Q16 but expressed as a decimal fraction of an hour: 10° / 15°/h = 0.6667 h ≈ 0.66 h (40 minutes in decimal hours). Note that Q16 and Q17 appear to ask the same question but expect different answer formats — Q16 expects 0:40 h (40 minutes) while Q17 expects 0.66 h (the decimal equivalent). Both represent the same 40-minute time difference. ### Q18: If Central European Summer Time (CEST) is UTC+2, what is the UTC equivalent of 1600 CEST? ^t60q18 - A) 1400 UTC. - B) 1600 UTC. - C) 1500 UTC. - D) 1700 UTC. **Correct: A)** > **Explanation:** UTC+2 means CEST is 2 hours ahead of UTC. To convert from local time to UTC, subtract the offset: 1600 CEST - 2 hours = 1400 UTC. A simple mnemonic: "to get UTC, subtract the positive offset." This is critical in aviation as all flight plans, ATC communications, and NOTAMs use UTC regardless of local time zone. ### Q19: What is UTC? ^t60q19 - A) A local time in Central Europe. - B) Local mean time at a specific point on Earth. - C) A zonal time - D) The mandatory time reference used in aviation. **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). ### Q20: If Central European Time (CET) is UTC+1, what is the UTC equivalent of 1700 CET? ^t60q20 - A) 1800 UTC. - B) 1500 UTC. - C) 1600 UTC. - D) 1700 UTC. **Correct: C)** > **Explanation:** CET is UTC+1, meaning it is 1 hour ahead of UTC. To convert to UTC, subtract the offset: 1700 CET - 1 hour = 1600 UTC. Switzerland uses CET (UTC+1) in winter and CEST (UTC+2) in summer — knowing the current offset is essential when filing flight plans or reading NOTAMs. ### Q21: Vienna (LOWW) is at 016°34'E and Salzburg (LOWS) at 013°00'E, both at approximately the same latitude. What is the difference in sunrise and sunset times (in UTC) between the two cities? (2,00 P.) ^t60q21 - A) In Vienna sunrise is 14 minutes earlier and sunset is 14 minutes later than in Salzburg - B) In Vienna sunrise and sunset are about 14 minutes earlier than in Salzburg - C) In Vienna sunrise is 4 minutes later and sunset is 4 minutes earlier than in Salzburg - D) In Vienna sunrise and sunset are about 4 minutes later than in Salzburg **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. ### Q22: How is "civil twilight" defined? ^t60q22 - A) The interval before sunrise or after sunset when the sun's centre is no more than 6° below the true horizon. - B) The interval before sunrise or after sunset when the sun's centre is no more than 12° below the apparent horizon. - C) The interval before sunrise or after sunset when the sun's centre is no more than 6° below the apparent horizon. - D) The interval before sunrise or after sunset when the sun's centre is no more than 12° below the true horizon. **Correct: A)** > **Explanation:** Civil twilight is the period when the sun's center is between 0° and 6° below the true (geometric) horizon — there is still sufficient natural light for most outdoor activities without artificial lighting. The true horizon (geometric) is used in the formal definition, not the apparent horizon (which is affected by refraction). Nautical twilight uses 12°, and astronomical twilight uses 18° below the true horizon. In aviation regulations, civil twilight often defines the boundary for day/night VFR operations. ### Q23: Given: WCA: -012°; TH: 125°; MC: 139°; DEV: 002°E. Determine TC, MH, and CH. (2,00 P.) ^t60q23 - A) TC: 113°. MH: 139°. CH: 125°. - B) TC: 137°. MH: 127°. CH: 125°. - C) TC: 137°. MH: 139°. CH: 125°. - D) TC: 113°. MH: 127°. CH: 129°. **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. ### Q24: Given: TC: 179°; WCA: -12°; VAR: 004° E; DEV: +002°. What are MH and MC? ^t60q24 - A) MH: 163°. MC: 175°. - B) MH: 167°. MC: 175°. - C) MH: 167°. MC: 161° - D) MH: 163°. MC: 161°. **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"). ### Q25: The angular difference between the true course and the true heading is known as the... ^t60q25 - A) Variation. - B) WCA. - C) Deviation. - D) Inclination. **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). ### Q26: The angular difference between the magnetic course and the true course is called... ^t60q26 - A) Deviation. - B) WCA. - C) Variation - D) Inclination. **Correct: C)** > **Explanation:** Magnetic variation (also called declination) is the angle between true north (geographic) and magnetic north at any given location, which creates a difference between the true course and the magnetic course. Variation changes with location and over time as the magnetic poles shift. Deviation is the error introduced by the aircraft's own magnetic field on the compass, affecting the difference between magnetic north and compass north. ### Q27: How is "magnetic course" (MC) defined? ^t60q27 - A) The angle between true north and the course line. - B) The direction from any point on Earth toward the geographic North Pole. - C) The direction from any point on Earth toward the magnetic north pole. - D) The angle between magnetic north and the course line. **Correct: D)** > **Explanation:** The magnetic course is the direction of the intended flight path (course line) measured clockwise from magnetic north. It differs from the true course by the local magnetic variation. Pilots use magnetic course because aircraft compasses point to magnetic north, making magnetic references more directly usable for navigation without additional corrections. ### Q28: How is "True Course" (TC) defined? ^t60q28 - A) The angle between true north and the course line. - B) The direction from any point on Earth toward the magnetic north pole. - C) The angle between magnetic north and the course line. - D) The direction from any point on Earth toward the geographic North Pole. **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. ### Q29: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and VAR? (2,00 P.) ^t60q29 - A) TH: 172°. VAR: 004° W - B) TH: 194°. VAR: 004° W - C) TH: 194°. VAR: 004° E - 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. ### Q30: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and DEV? (2,00 P.) ^t60q30 - A) TH: 172°. DEV: -002°. - B) TH: 194°. DEV: +002°. - C) TH: 172°. DEV: +002°. - D) TH: 194°. DEV: -002°. **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°. ### Q31: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. Determine VAR and DEV. (2,00 P.) ^t60q31 - A) VAR: 004° E. DEV: +002°. - B) VAR: 004° W. DEV: -002°. - C) VAR: 004° W. DEV: +002°. - D) VAR: 004° E. DEV: -002°. **Correct: B)** > **Explanation:** From Q29: VAR = 4° W (MH 198° > TH 194°, so West variation). From Q30: DEV = -002° (CH 200° > MH 198°, compass reads high, requiring negative deviation correction). The complete heading chain for this problem is: TC 183° → (+11° WCA) → TH 194° → (+4° W VAR) → MH 198° → (+2° DEV) → CH 200°. These three questions (Q29, Q30, Q31) all use the same dataset, testing different parts of the heading conversion chain. ### Q32: At what location does magnetic inclination reach its minimum value? ^t60q32 - A) At the geographic poles - B) At the geographic equator - C) At the magnetic equator - D) At the magnetic poles **Correct: C)** > **Explanation:** Magnetic inclination (dip) is the angle at which the Earth's magnetic field lines intersect the horizontal plane. At the magnetic equator (the "aclinic line"), the field lines are horizontal and the dip angle is 0° — the lowest possible value. At the magnetic poles, the field lines are vertical (inclination = 90°). The magnetic equator does not coincide with the geographic equator. ### Q33: The angular difference between compass north and magnetic north is referred to as... ^t60q33 - A) Variation. - B) Deviation. - C) Inclination. - D) WCA **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. ### Q34: What does "compass north" (CN) refer to? ^t60q34 - A) The angle between the aircraft heading and magnetic north - B) The direction to which the direct reading compass aligns under the combined influence of the Earth's and the aircraft's magnetic fields - C) The direction from any point on Earth toward the geographic North Pole - D) The most northerly reading point on the magnetic compass in the aircraft **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. ### Q35: An "isogonal" or "isogonic line" on an aeronautical chart connects all points sharing the same value of... ^t60q35 - A) Deviation - B) Inclination. - C) Heading. - D) Variation. **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. ### Q36: An "agonic line" on the Earth or on an aeronautical chart connects all points where the... ^t60q36 - A) Heading is 0°. - B) Inclination is 0°. - C) Variation is 0°. - D) Deviation is 0°. **Correct: C)** > **Explanation:** The agonic line is a special isogonic line where magnetic variation equals zero — meaning true north and magnetic north coincide along this line. Aircraft flying along the agonic line need not apply any variation correction; true course equals magnetic course. There are currently two main agonic lines on Earth, passing through North America and through parts of Asia/Australia. ### Q37: Which are the official standard units for horizontal distances in aeronautical navigation? ^t60q37 - A) Land miles (SM), sea miles (NM) - B) Feet (ft), inches (in) - C) Yards (yd), meters (m) - D) Nautical miles (NM), kilometers (km) **Correct: D)** > **Explanation:** In international aviation, horizontal distances are officially measured in nautical miles (NM) and kilometers (km). The nautical mile is preferred for navigation because it directly relates to the angular measurement system (1 NM = 1 arcminute of latitude). Kilometers are also used, particularly in some countries and on certain charts. Feet and meters are used for vertical distances (altitude/height), not horizontal distance. ### Q38: How many metres are equivalent to 1000 ft? ^t60q38 - A) 30 m. - B) 3000 m. - C) 30 km. - D) 300 m. **Correct: D)** > **Explanation:** 1 foot = 0.3048 meters, so 1000 ft = 304.8 m ≈ 300 m. The quick conversion rule is: feet x 0.3 ≈ meters, or equivalently from the exam table: m = ft x 3 / 10. This approximation is accurate enough for practical navigation. For exam purposes: 1000 ft ≈ 300 m, 3000 ft ≈ 900 m, 10,000 ft ≈ 3000 m. ### Q39: How many feet correspond to 5500 m? ^t60q39 - A) 10000 ft. - B) 7500 ft. - C) 30000 ft. - D) 18000 ft. **Correct: D)** > **Explanation:** Using the conversion ft = m x 10 / 3 (from the exam table): 5500 x 10 / 3 = 55000 / 3 ≈ 18,333 ft ≈ 18,000 ft. Alternatively: 1 m ≈ 3.281 ft, so 5500 m x 3.281 ≈ 18,046 ft ≈ 18,000 ft. This altitude is significant in European airspace as it corresponds approximately to FL180 (the base of Class A airspace in some regions). ### Q40: What might cause the runway designation at an aerodrome to change (e.g. from runway 06 to runway 07)? ^t60q40 - A) The direction of the approach path has changed - B) The magnetic variation at the runway location has changed - C) The magnetic deviation at the runway location has changed - D) The true direction of the runway alignment has changed **Correct: B)** > **Explanation:** Runway numbers are based on the magnetic heading of the runway, rounded to the nearest 10° and divided by 10. Because the magnetic north pole drifts slowly over time, the local magnetic variation changes — even if the physical runway has not moved, its magnetic bearing changes. When this change is large enough to shift the rounded designation (e.g., from 055° to 065°), the runway is renumbered (from "06" to "07"). Major airports periodically update runway designations for this reason. ### Q41: Which flight instrument is affected by electronic devices operated on board the aircraft? ^t60q41 - A) Airspeed indicator. - B) Turn coordinator - C) Artificial horizon. - D) Direct reading compass. **Correct: D)** > **Explanation:** The direct reading (magnetic) compass is sensitive to any magnetic field, including those generated by electrical equipment, avionics, and metal components in the aircraft. This interference is called deviation. Electronic devices that draw current create electromagnetic fields that can deflect the compass needle. That is why pilots are required to record the deviation on a compass card and why compasses are mounted as far from interference sources as possible. ### Q42: What are the key characteristics of a Mercator chart? ^t60q42 - A) Scale increases with latitude, great circles appear curved, rhumb lines appear straight - B) Constant scale, great circles appear straight, rhumb lines appear curved - C) Scale increases with latitude, great circles appear straight, rhumb lines appear curved - D) Constant scale, great circles appear curved, rhumb lines appear straight **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. ### Q43: On a direct Mercator chart, how do rhumb lines and great circles appear? ^t60q43 - A) Rhumb lines: curved lines; Great circles: curved lines - B) Rhumb lines: curved lines; Great circles: straight lines - C) Rhumb lines: straight lines; Great circles: straight lines - D) Rhumb lines: straight lines; Great circles: curved lines **Correct: D)** > **Explanation:** On a Mercator chart, rhumb lines (constant compass bearing courses) appear as straight lines because the chart is constructed so that meridians are parallel vertical lines and parallels are horizontal lines — any line crossing meridians at a constant angle (a rhumb line) is therefore straight. Great circles, which follow the shortest path on the globe, curve toward the poles when projected onto the Mercator chart and therefore appear as curved lines (bowing toward the nearest pole). ### Q44: What are the characteristics of a Lambert conformal chart? ^t60q44 - A) Conformal and nearly true to scale - B) Conformal and equal-area - C) Rhumb lines depicted as straight lines and conformal - D) Great circles depicted as straight lines and equal-area **Correct: A)** > **Explanation:** The Lambert Conformal Conic projection is the standard for aeronautical charts (including ICAO charts used in Europe). It is conformal (angles and shapes are preserved locally), nearly true to scale between its two standard parallels, and great circles 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. ### Q45: The distance between two airports is 220 NM. On an aeronautical chart, a pilot measures 40.7 cm for this distance. What is the chart scale? ^t60q45 - A) 1 : 2000000. - B) 1 : 250000. - C) 1 : 1000000. - D) 1 : 500000 **Correct: C)** > **Explanation:** Convert 220 NM to centimeters: 220 NM x 1852 m/NM = 407,440 m = 40,744,000 cm. Scale = chart distance / real distance = 40.7 cm / 40,744,000 cm = 1 / 1,000,835 ≈ 1 : 1,000,000. The ICAO chart of Switzerland used in the SPL exam is 1:500,000 scale; knowing how to calculate chart scale from measured and actual distances is a standard exam skill. ### Q46: What is the distance from VOR Bruenkendorf (BKD) (53°02'N, 011°33'E) to Pritzwalk (EDBU) (53°11'N, 12°11'E)? ^t60q46 > *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) 42 km - B) 24 km - C) 42 NM - D) 24 NM **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. ### Q47: On an aeronautical chart, 7.5 cm represents 60.745 NM in reality. What is the chart scale? ^t60q47 - A) 1 : 1500000 - B) 1 : 500000 - C) 1 : 150000 - D) 1 : 1 000000 **Correct: A)** > **Explanation:** Convert 60.745 NM to cm: 60.745 x 1852 m/NM = 112,499 m = 11,249,900 cm. Scale = 7.5 / 11,249,900 ≈ 1 / 1,499,987 ≈ 1 : 1,500,000. This is a less common chart scale — for comparison, the ICAO chart used in Switzerland is 1:500,000 and the German half-million chart (ICAO Karte) is also 1:500,000. ### Q48: A pilot extracts this data from the chart for a short flight from A to B: True course: 245°. Magnetic variation: 7° W. The magnetic course (MC) equals... ^t60q48 - A) 245°. - B) 007°. - C) 252°. - D) 238°. **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°. ### Q49: Given: True course from A to B: 250°. Ground distance: 210 NM. TAS: 130 kt. Headwind component: 15 kt. ETD: 0915 UTC. What is the ETA? (2,00 P.) ^t60q49 - A) 1052 UTC. - B) 1005 UTC. - C) 1115 UTC. - D) 1105 UTC. **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. ### Q50: Given: True course from A to B: 283°. Ground distance: 75 NM. TAS: 105 kt. Headwind component: 12 kt. ETD: 1242 UTC. What is the ETA? ^t60q50 - A) 1356 UTC - B) 1330 UTC - C) 1430 UTC - D) 1320 UTC **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. > Source: Segelflugverband der Schweiz - SFCL_Theorie_Navigation_Version_Schweiz_Uebungen.pdf > Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCL_Theorie_Navigation_Version_Schweiz_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.