Navigation


Q1: Through which points does the Earth's rotational axis pass? ^t60q1

DE · FR

Answer

C)

Explanation

The Earth's rotational axis is the physical axis around which the planet spins, and it passes through the geographic (true) poles — not the magnetic poles. The geographic poles are fixed points defined by the rotational axis, while the magnetic poles are offset from them and drift over time due to changes in the Earth's molten core.

Q2: Which statement correctly describes the polar axis of the Earth? ^t60q2

DE · FR

Answer

D)

Explanation

The polar axis passes through the geographic poles and is perpendicular (90°) to the plane of the equator by definition. The Earth's axis is indeed tilted 23.5° relative to the plane of its orbit around the sun (the ecliptic), but it is perpendicular to the equatorial plane — those two facts are consistent and not contradictory.

Q3: For navigation systems, which approximate geometrical shape best represents the Earth? ^t60q3

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Answer

B)

Explanation

The Earth is not a perfect sphere — it is slightly flattened at the poles and bulges at the equator due to its rotation. This shape is called an oblate spheroid or ellipsoid. Modern navigation systems (including GPS) use the WGS-84 ellipsoid as the reference model, which accurately accounts for this flattening in coordinate calculations.

Q4: Which of the following statements about a rhumb line is correct? ^t60q4

DE · FR

Answer

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

DE · FR

Answer

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? ^t60q6

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Earth Globe

Answer

B)

Explanation

The equator spans 360 degrees of longitude, and each degree of longitude on the equator equals 60 NM (since 1 NM = 1 arcminute on a great circle). Therefore: 360° x 60 NM = 21,600 NM. In kilometers, the Earth's equatorial circumference is approximately 40,075 km — so option A has the right number but wrong unit. Knowing this relationship (1° = 60 NM on the equator) is fundamental to navigation calculations.

Key Terms

NM = Nautical Mile(s)

Q7: What is the latitude difference between point A (12°53'30''N) and point B (07°34'30''S)? ^t60q7

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Answer

A)

Explanation

When two points are on opposite sides of the equator, the difference in latitude is the sum of their respective latitudes. Here: 12°53'30''N + 07°34'30''S = 20°28'00''. Converting minutes: 53'30'' + 34'30'' = 88'00'' = 1°28'00'', so 12° + 7° + 1°28' = 20°28'00''. Always add latitudes when they are in opposite hemispheres (N and S).

Q8: At what positions are the two polar circles located? ^t60q8

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Answer

D)

Explanation

The Arctic Circle lies at approximately 66.5°N and the Antarctic Circle at 66.5°S — which is 90° - 23.5° = 66.5°, placing them 23.5° away from their respective geographic poles. This 23.5° offset directly corresponds to the axial tilt of the Earth. The Tropics of Cancer and Capricorn (option A) are the ones located 23.5° from the equator.

Q9: Along a meridian, what is the distance between the 48°N and 49°N parallels of latitude? ^t60q9

DE · FR

Answer

C)

Explanation

Along any meridian (line of longitude), 1 degree of latitude always equals 60 nautical miles. This is because meridians are great circles and 1 NM is defined as 1 arcminute of arc along a great circle. The 111 km figure (option A) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations.

Key Terms

NM = Nautical Mile(s)

Q10: Along any line of longitude, what distance corresponds to one degree of latitude? ^t60q10

DE · FR

Answer

D)

Explanation

One degree of latitude = 60 arcminutes, and since 1 NM equals exactly 1 arcminute of latitude along a meridian, 1° of latitude = 60 NM. This relationship holds along any meridian because all meridians are great circles. In SI units, 1° of latitude ≈ 111 km, not 60 km as stated in option C.

Key Terms

NM = Nautical Mile(s)

Q11: Point A lies at exactly 47°50'27''N latitude. Which point is precisely 240 NM north of A? ^t60q11

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Answer

D)

Explanation

Converting 240 NM to degrees of latitude: 240 NM / 60 NM per degree = 4°. Adding 4° to 47°50'27''N gives 51°50'27''N. Moving north increases the latitude value.

Key Terms

NM = Nautical Mile(s)

Q12: Along the equator, what is the distance between the 150°E and 151°E meridians? ^t60q12

DE · FR

Answer

B)

Explanation

On the equator, meridians of longitude are separated by great circle arcs, and 1° of longitude along the equator equals 60 NM — the same as 1° of latitude along any meridian, because the equator is also a great circle. At higher latitudes, the distance between meridians decreases (multiplied by cos(latitude)), but at the equator it is exactly 60 NM per degree.

Key Terms

NM = Nautical Mile(s)

Q13: When two points A and B on the equator are separated by exactly one degree of longitude, what is the great circle distance between them? ^t60q13

DE · FR

Answer

C)

Explanation

The equator itself is a great circle, so the great circle distance between two points on the equator separated by 1° of longitude is simply 60 NM (1° x 60 NM/degree). This is the same principle as measuring along a meridian. Any confusion arises if one tries to calculate using km instead — 1° ≈ 111 km on the equator, but the question asks for NM.

Key Terms

NM = Nautical Mile(s)

Q14: Consider two points A and B on the same parallel of latitude (not the equator). A is at 010°E and B at 020°E. The rhumb line distance between them is always ^t60q14

DE · FR

Answer

D)

Explanation

The rhumb line distance between points on the same parallel of latitude is: 10° x 60 NM x cos(latitude). Since cos(latitude) is always less than 1 for any latitude other than the equator (where it equals exactly 60 NM x 10 = 600 NM), the rhumb line distance is always strictly less than 600 NM. At the equator it would equal 600 NM, but since they are specifically "not on the equator," the distance is always less than 600 NM.

Key Terms

NM = Nautical Mile(s)

Q15: How much time elapses as the sun traverses 20° of longitude? ^t60q15

DE · FR

Answer

B)

Explanation

The Earth rotates 360° in 24 hours, so it rotates 15° per hour, or 1° every 4 minutes. For 20° of longitude: 20 x 4 minutes = 80 minutes = 1 hour 20 minutes. Alternatively: 20° / 15°/h = 1.333 h = 1:20 h. This relationship (15°/hour or 4 min/degree) is essential for time zone calculations and solar noon determination.

Q16: How much time passes as the sun crosses 10° of longitude? ^t60q16

DE · FR

Answer

B)

Explanation

Using the same principle as Q15: the Earth rotates 15° per hour, so 10° corresponds to 10/15 hours = 2/3 hour = 40 minutes = 0:40 h.

Q17: The sun traverses 10° of longitude. What is the corresponding time difference? ^t60q17

DE · FR

Answer

D)

Explanation

This is the same calculation as Q16 but expressed as a decimal fraction of an hour: 10° / 15°/h = 0.6667 h ≈ 0.66 h (40 minutes in decimal hours). Note that Q16 and Q17 appear to ask the same question but expect different answer formats — Q16 expects 0:40 h (40 minutes) while Q17 expects 0.66 h (the decimal equivalent). Both represent the same 40-minute time difference.

Q18: If Central European Summer Time (CEST) is UTC+2, what is the UTC equivalent of 1600 CEST? ^t60q18

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Answer

A)

Explanation

UTC+2 means CEST is 2 hours ahead of UTC. To convert from local time to UTC, subtract the offset: 1600 CEST - 2 hours = 1400 UTC. A simple mnemonic: "to get UTC, subtract the positive offset." This is critical in aviation as all flight plans, ATC communications, and NOTAMs use UTC regardless of local time zone.

Key Terms

ATC = Air Traffic Control

Q19: What is UTC? ^t60q19

DE · FR

Answer

D)

Explanation

Coordinated Universal Time (UTC) is the mandatory time reference for all international aviation operations — flight plans, ATC communications, weather reports (METARs/TAFs), and NOTAMs all use UTC to eliminate confusion from time zone differences. It is not a zonal or local time, and it is not referenced to any geographic location (though it closely tracks Greenwich Mean Time).

Key Terms

ATC = Air Traffic Control

Q20: If Central European Time (CET) is UTC+1, what is the UTC equivalent of 1700 CET? ^t60q20

DE · FR

Answer

C)

Explanation

CET is UTC+1, meaning it is 1 hour ahead of UTC. To convert to UTC, subtract the offset: 1700 CET - 1 hour = 1600 UTC. Switzerland uses CET (UTC+1) in winter and CEST (UTC+2) in summer — knowing the current offset is essential when filing flight plans or reading NOTAMs.

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

DE · FR

Answer

B)

Explanation

The difference in longitude is 016°34' - 013°00' = 3°34' ≈ 3.57°. At 4 minutes per degree, this gives approximately 14.3 minutes ≈ 14 minutes. Vienna is east of Salzburg, so the sun reaches Vienna earlier — both sunrise and sunset occur about 14 minutes earlier in Vienna (as seen in UTC). Local time zones disguise this difference, but in UTC the eastern location always sees solar events first.

Q22: How is "civil twilight" defined? ^t60q22

DE · FR

Answer

A)

Explanation

Civil twilight is the period when the sun's center is between 0° and 6° below the true (geometric) horizon — there is still sufficient natural light for most outdoor activities without artificial lighting. The true horizon (geometric) is used in the formal definition, not the apparent horizon (which is affected by refraction). Nautical twilight uses 12°, and astronomical twilight uses 18° below the true horizon. In aviation regulations, civil twilight often defines the boundary for day/night VFR operations.

Key Terms

VFR = Visual Flight Rules

Q23: Given: WCA: -012°; TH: 125°; MC: 139°; DEV: 002°E. Determine TC, MH, and CH. (2,00 P.) ^t60q23

DE · FR

Answer

B)

Explanation

The heading chain works as follows: TC → (apply WCA) → TH → (apply VAR) → MH → (apply DEV) → CH. Given TH = 125° and WCA = -12°, then TC = TH - WCA = 125° - (-12°) = 137°. For MH: MC = MH + WCA, so MH = MC - WCA = 139° - 12° = 127°. For CH: DEV = 002°E means compass reads 2° high, so CH = MH - DEV = 127° - 2° = 125°. Negative WCA means wind from the right, requiring a left correction in heading.

Key Terms

CH = Compass Heading; DEV = Compass Deviation; MC = Magnetic Course; MH = Magnetic Heading; TC = True Course; TH = True Heading; VAR = Magnetic Variation; WCA = Wind Correction Angle

Q24: Given: TC: 179°; WCA: -12°; VAR: 004° E; DEV: +002°. What are MH and MC? ^t60q24

DE · FR

Answer

A)

Explanation

TH = TC + WCA = 179° + (-12°) = 167°. Then MH = TH - VAR (E is subtracted): MH = 167° - 4° = 163°. For MC: MC = TC - VAR = 179° - 4° = 175°. Alternatively: MC = MH + WCA = 163° + (-12°) = 151° — wait, that doesn't match; MC is measured from magnetic north to the course line, so MC = TC - VAR = 179° - 4° = 175°. East variation is subtracted when converting from True to Magnetic ("East is least").

Key Terms

DEV = Compass Deviation; MC = Magnetic Course; MH = Magnetic Heading; TC = True Course; TH = True Heading; VAR = Magnetic Variation; WCA = Wind Correction Angle

Q25: The angular difference between the true course and the true heading is known as the ^t60q25

DE · FR

Answer

B)

Explanation

The Wind Correction Angle (WCA) is the angular difference between the true course (the direction of intended track over the ground) and the true heading (the direction the aircraft's nose points). A crosswind requires the pilot to angle the nose into the wind, creating a difference between heading and track — this offset angle is the WCA. It is neither variation (true-to-magnetic difference) nor deviation (magnetic-to-compass difference).

Key Terms

WCA = Wind Correction Angle

Q26: The angular difference between the magnetic course and the true course is called ^t60q26

DE · FR

Answer

C)

Explanation

Magnetic variation (also called declination) is the angle between true north (geographic) and magnetic north at any given location, which creates a difference between the true course and the magnetic course. Variation changes with location and over time as the magnetic poles shift. Deviation is the error introduced by the aircraft's own magnetic field on the compass, affecting the difference between magnetic north and compass north.

Key Terms

WCA = Wind Correction Angle

Q27: How is "magnetic course" (MC) defined? ^t60q27

DE · FR

Answer

D)

Explanation

The magnetic course is the direction of the intended flight path (course line) measured clockwise from magnetic north. It differs from the true course by the local magnetic variation. Pilots use magnetic course because aircraft compasses point to magnetic north, making magnetic references more directly usable for navigation without additional corrections.

Key Terms

MC = Magnetic Course

Q28: How is "True Course" (TC) defined? ^t60q28

DE · FR

Answer

A)

Explanation

The True Course is the angle measured clockwise from true (geographic) north to the intended flight path (course line). It is determined from aeronautical charts, which are oriented to true north. To fly a true course, pilots must apply magnetic variation to get the magnetic course, then apply wind correction angle to get the true heading they must fly.

Key Terms

TC = True Course

Q29: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and VAR? (2,00 P.) ^t60q29

DE · FR

Answer

B)

Explanation

TH = TC + WCA = 183° + 11° = 194°. For variation: VAR is the difference between TC and MC, or equivalently between TH and MH. MH = 198°, TH = 194°, so the difference is 4°. Since MH > TH, magnetic north is east of true north, meaning variation is West (West variation adds to true to get magnetic: MH = TH + VAR, so 198° = 194° + 4°W). Mnemonic: "West is best" — West variation is added going True to Magnetic.

Key Terms

CH = Compass Heading; MC = Magnetic Course; MH = Magnetic Heading; TC = True Course; TH = True Heading; VAR = Magnetic Variation; WCA = Wind Correction Angle

Q30: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. What are TH and DEV? (2,00 P.) ^t60q30

DE · FR

Answer

D)

Explanation

TH = TC + WCA = 183° + 11° = 194°. For deviation: DEV = CH - MH = 200° - 198° = +2°. However, the convention for deviation sign varies — if DEV is defined as what you subtract from CH to get MH, then DEV = -2°. Here CH = 200° > MH = 198°, meaning the compass reads 2° more than magnetic, so DEV = -2° (the compass is deflected eastward, requiring a negative correction). The answer is TH: 194°, DEV: -002°.

Key Terms

CH = Compass Heading; DEV = Compass Deviation; MH = Magnetic Heading; TC = True Course; TH = True Heading; WCA = Wind Correction Angle

Q31: Given: TC: 183°; WCA: +011°; MH: 198°; CH: 200°. Determine VAR and DEV. (2,00 P.) ^t60q31

DE · FR

Answer

B)

Explanation

From Q29: VAR = 4° W (MH 198° > TH 194°, so West variation). From Q30: DEV = -002° (CH 200° > MH 198°, compass reads high, requiring negative deviation correction). The complete heading chain for this problem is: TC 183° → (+11° WCA) → TH 194° → (+4° W VAR) → MH 198° → (+2° DEV) → CH 200°. These three questions (Q29, Q30, Q31) all use the same dataset, testing different parts of the heading conversion chain.

Key Terms

CH = Compass Heading; DEV = Compass Deviation; MH = Magnetic Heading; TC = True Course; TH = True Heading; VAR = Magnetic Variation; WCA = Wind Correction Angle

Q32: At what location does magnetic inclination reach its minimum value? ^t60q32

DE · FR

Answer

C)

Explanation

Magnetic inclination (dip) is the angle at which the Earth's magnetic field lines intersect the horizontal plane. At the magnetic equator (the "aclinic line"), the field lines are horizontal and the dip angle is 0° — the lowest possible value. At the magnetic poles, the field lines are vertical (inclination = 90°). The magnetic equator does not coincide with the geographic equator.

Q33: The angular difference between compass north and magnetic north is referred to as ^t60q33

DE · FR

Answer

B)

Explanation

Deviation is the error in a magnetic compass caused by the aircraft's own magnetic fields (from electrical equipment, metal structure, avionics). It is expressed as the angular difference between magnetic north (what the compass should indicate) and compass north (what it actually indicates). Deviation varies with the aircraft's heading and is recorded on a compass deviation card mounted near the instrument.

Key Terms

WCA = Wind Correction Angle

Q34: What does "compass north" (CN) refer to? ^t60q34

DE · FR

Answer

B)

Explanation

Compass north is the direction the compass needle actually points, which is determined by the combined effect of the Earth's magnetic field AND any local magnetic interference from the aircraft itself. Because of this aircraft-induced deviation, compass north differs from magnetic north. The compass reads this resultant direction, not pure magnetic north — hence the need for a deviation correction card.

Q35: An "isogonal" or "isogonic line" on an aeronautical chart connects all points sharing the same value of ^t60q35

DE · FR

Answer

D)

Explanation

Isogonic lines (also called isogonals) connect all points on Earth that have the same magnetic variation value. They are printed on aeronautical charts so pilots can read the local variation at their position and convert between true and magnetic headings. The agonic line is the special case where variation = 0°. Lines of equal magnetic inclination are called isoclinic lines; lines of equal field intensity are isodynamic lines.

Q36: An "agonic line" on the Earth or on an aeronautical chart connects all points where the ^t60q36

DE · FR

Answer

C)

Explanation

The agonic line is a special isogonic line where magnetic variation equals zero — meaning true north and magnetic north coincide along this line. Aircraft flying along the agonic line need not apply any variation correction; true course equals magnetic course. There are currently two main agonic lines on Earth, passing through North America and through parts of Asia/Australia.

Q37: Which are the official standard units for horizontal distances in aeronautical navigation? ^t60q37

DE · FR

Answer

D)

Explanation

In international aviation, horizontal distances are officially measured in nautical miles (NM) and kilometers (km). The nautical mile is preferred for navigation because it directly relates to the angular measurement system (1 NM = 1 arcminute of latitude). Kilometers are also used, particularly in some countries and on certain charts. Feet and meters are used for vertical distances (altitude/height), not horizontal distance.

Key Terms

NM = Nautical Mile(s)

Q38: How many metres are equivalent to 1000 ft? ^t60q38

DE · FR

Answer

D)

Explanation

1 foot = 0.3048 meters, so 1000 ft = 304.8 m ≈ 300 m. The quick conversion rule is: feet x 0.3 ≈ meters, or equivalently from the exam table: m = ft x 3 / 10. This approximation is accurate enough for practical navigation. For exam purposes: 1000 ft ≈ 300 m, 3000 ft ≈ 900 m, 10,000 ft ≈ 3000 m.

Q39: How many feet correspond to 5500 m? ^t60q39

DE · FR

Answer

D)

Explanation

Using the conversion ft = m x 10 / 3 (from the exam table): 5500 x 10 / 3 = 55000 / 3 ≈ 18,333 ft ≈ 18,000 ft. Alternatively: 1 m ≈ 3.281 ft, so 5500 m x 3.281 ≈ 18,046 ft ≈ 18,000 ft. This altitude is significant in European airspace as it corresponds approximately to FL180 (the base of Class A airspace in some regions).

Key Terms

FL = Flight Level

Q40: What might cause the runway designation at an aerodrome to change (e.g. from runway 06 to runway 07)? ^t60q40

DE · FR

Answer

B)

Explanation

Runway numbers are based on the magnetic heading of the runway, rounded to the nearest 10° and divided by 10. Because the magnetic north pole drifts slowly over time, the local magnetic variation changes — even if the physical runway has not moved, its magnetic bearing changes. When this change is large enough to shift the rounded designation (e.g., from 055° to 065°), the runway is renumbered (from "06" to "07"). Major airports periodically update runway designations for this reason.

Q41: Which flight instrument is affected by electronic devices operated on board the aircraft? ^t60q41

DE · FR

Answer

D)

Explanation

The direct reading (magnetic) compass is sensitive to any magnetic field, including those generated by electrical equipment, avionics, and metal components in the aircraft. This interference is called deviation. Electronic devices that draw current create electromagnetic fields that can deflect the compass needle. That is why pilots are required to record the deviation on a compass card and why compasses are mounted as far from interference sources as possible.

Q42: What are the key characteristics of a Mercator chart? ^t60q42

DE · FR

Answer

A)

Explanation

The Mercator projection is a cylindrical conformal projection where meridians and parallels are straight lines intersecting at right angles. Rhumb lines (constant bearing courses) appear as straight lines — making it useful for constant-heading navigation. However, the scale increases with latitude (Greenland appears as large as Africa) and great circles appear as curved lines. It is not an equal-area projection and is not suitable for high-latitude navigation.

Q43: On a direct Mercator chart, how do rhumb lines and great circles appear? ^t60q43

DE · FR

Answer

D)

Explanation

On a Mercator chart, rhumb lines (constant compass bearing courses) appear as straight lines because the chart is constructed so that meridians are parallel vertical lines and parallels are horizontal lines — any line crossing meridians at a constant angle (a rhumb line) is therefore straight. Great circles, which follow the shortest path on the globe, curve toward the poles when projected onto the Mercator chart and therefore appear as curved lines (bowing toward the nearest pole).

Q44: What are the characteristics of a Lambert conformal chart? ^t60q44

DE · FR

Answer

A)

Explanation

The Lambert Conformal Conic projection is the standard for aeronautical charts (including ICAO charts used in Europe). It is conformal (angles and shapes are preserved locally), nearly true to scale between its two standard parallels, and great circles 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.

Key Terms

ICAO = International Civil Aviation Organization

Q45: The distance between two airports is 220 NM. On an aeronautical chart, a pilot measures 40.7 cm for this distance. What is the chart scale? ^t60q45

DE · FR

Answer

C)

Explanation

Convert 220 NM to centimeters: 220 NM x 1852 m/NM = 407,440 m = 40,744,000 cm. Scale = chart distance / real distance = 40.7 cm / 40,744,000 cm = 1 / 1,000,835 ≈ 1: 1,000,000. The ICAO chart of Switzerland used in the SPL exam is 1:500,000 scale; knowing how to calculate chart scale from measured and actual distances is a standard exam skill.

Key Terms

ICAO = International Civil Aviation Organization; NM = Nautical Mile(s); SPL = Sailplane Pilot Licence

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

DE · FR

![](figures/t60_q46.png)

Answer

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.

Key Terms

NM = Nautical Mile(s)

Q47: On an aeronautical chart, 7.5 cm represents 60.745 NM in reality. What is the chart scale? ^t60q47

DE · FR

Answer

A)

Explanation

Convert 60.745 NM to cm: 60.745 x 1852 m/NM = 112,499 m = 11,249,900 cm. Scale = 7.5 / 11,249,900 ≈ 1 / 1,499,987 ≈ 1: 1,500,000. This is a less common chart scale — for comparison, the ICAO chart used in Switzerland is 1:500,000 and the German half-million chart (ICAO Karte) is also 1:500,000.

Key Terms

ICAO = International Civil Aviation Organization; NM = Nautical Mile(s)

Q48: A pilot extracts this data from the chart for a short flight from A to B: True course: 245°. Magnetic variation: 7° W. The magnetic course (MC) equals ^t60q48

DE · FR

Answer

C)

Explanation

When variation is West, magnetic north is west of true north, meaning magnetic bearings are higher (greater) than true bearings. The rule "West is best, East is least" means: West variation → add to True to get Magnetic. MC = TC + VAR(W) = 245° + 7° = 252°. Alternatively: MC = TC - VAR(E), so for West variation (negative East): MC = 245° - (-7°) = 252°.

Key Terms

MC = Magnetic Course; TC = True Course; VAR = Magnetic Variation

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

DE · FR

Answer

D)

Explanation

Ground speed = TAS - headwind = 130 - 15 = 115 kt. Flight time = distance / GS = 210 NM / 115 kt = 1.826 h = 1 h 49.6 min ≈ 1 h 50 min. ETA = ETD + flight time = 0915 + 1:50 = 1105 UTC. This is a standard time/distance/speed calculation. Always compute GS first by applying wind component, then divide distance by GS for time.

Key Terms

ETA = Estimated Time of Arrival; ETD = Estimated Time of Departure; GS = Ground Speed; NM = Nautical Mile(s); TAS = True Airspeed

Q50: Given: True course from A to B: 283°. Ground distance: 75 NM. TAS: 105 kt. Headwind component: 12 kt. ETD: 1242 UTC. What is the ETA? ^t60q50

DE · FR

Answer

B)

Explanation

Ground speed = TAS - headwind = 105 - 12 = 93 kt. Flight time = 75 NM / 93 kt = 0.806 h = 48.4 min ≈ 48 min. ETA = 1242 + 0:48 = 1330 UTC.

Source: Segelflugverband der Schweiz - SFCLTheorieNavigationVersionSchweizUebungen.pdf > Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweizUebungen.pdf.

Key Terms

ETA = Estimated Time of Arrival; ETD = Estimated Time of Departure; GS = Ground Speed; ICAO = International Civil Aviation Organization; NM = Nautical Mile(s); TAS = True Airspeed

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.

Q51: What is the latest time we must land? ^t60q51

DE · FR

Explanation

Swiss VFR regulations define the end of the flying day as 30 minutes after official sunset (or a specified time after evening civil twilight). The landing deadline is looked up in official sunset tables and adjusted for the applicable time zone (MEZ = UTC+1 in winter, MESZ = UTC+2 in summer). June 21 is near the summer solstice, giving the latest sunset of the year; March dates are in standard time (MEZ). Always verify the current eVFG tables, as these values are date and location dependent.

Key Terms

VFR = Visual Flight Rules

Q52: What does the large number 87 near Freiburg on the ICAO chart mean? ^t60q52

DE · FR

Answer

MSA (Minimum Safe Altitude)

Explanation

On the Swiss ICAO 1:500,000 chart, large bold numbers printed near certain cities or waypoints indicate the Minimum Safe Altitude (MSA) in hundreds of feet for that area (so "87" means 8,700 ft MSL). The MSA provides obstacle clearance of at least 300 m (1000 ft) within a defined radius. Pilots use these values for en-route safety altitude planning, especially important in mountainous terrain like the Swiss Jura and Alps.

Key Terms

ICAO = International Civil Aviation Organization; MSA = Minimum Safe Altitude; MSL = Mean Sea Level

Q53: What entry should always be made on the navigation chart before a cross-country flight? ^t60q53

DE · FR

Answer

The TC (True Course)

Explanation

Before a cross-country flight, the pilot should measure and mark the True Course (TC) on the navigation chart using a protractor referenced to the nearest meridian. The TC is the foundation for all subsequent heading calculations: TC → apply variation → MC → apply wind correction → TH → apply deviation → CH. Marking the TC on the chart ensures consistent reference throughout the flight planning process and allows in-flight verification of track.

Key Terms

CH = Compass Heading; MC = Magnetic Course; TC = True Course; TH = True Heading

Q54: How should a final approach over navigationally challenging terrain be conducted? ^t60q54

DE · FR

Answer

Monitor using a time scale, mark known positions on the chart

Explanation

When approaching a destination over navigationally challenging terrain (forests, featureless plains, or complex topography), the pilot should monitor progress using elapsed time against a pre-calculated time scale, and positively identify known landmarks (towns, rivers, roads) and mark them on the chart. This technique — essentially dead reckoning with regular position fixes — prevents the pilot from overflying the destination or becoming lost. In a glider without GPS, time management is critical to ensure arrival with sufficient altitude.

Q55: What does GND mean on the cover page of the gliding chart? ^t60q55

DE · FR

Answer

Upper limit of LS-R for gliders (SF with reduced cloud separation minima)

Explanation

On the Swiss gliding chart cover page, "GND" indicates the lower limit (ground) of certain restricted areas, and the term specifically refers to the upper boundary of LS-R (Luftraum-Segelflug-Reservate) available for gliders operating with reduced cloud separation minima. These zones allow gliders to fly in conditions that would otherwise require instrument flight rules, provided specific weather minima are met. Understanding the legend on the gliding chart cover page is essential for Swiss exam candidates.

Q56: Glider frequencies (ground-to-air, air-to-air, regions)? ^t60q56

DE · FR

Answer

Listed on the gliding chart cover page

Explanation

The Swiss gliding chart cover page contains a complete list of glider frequencies, including ground-to-air and air-to-air communication frequencies organized by region. Common Swiss glider frequencies include 122.300 MHz (universal glider frequency) and regional variants. These must be known before flight as gliders may need to coordinate with each other and with ground stations, especially in busy areas like the Alps or near controlled airspace.

Q57: Military air traffic service hours? ^t60q57

DE · FR

Answer

Gliding chart, bottom right

Explanation

The operating hours of Swiss military airspace and military air traffic services are printed in the lower right corner of the Swiss gliding chart. Military restricted areas (such as those associated with Payerne, Meiringen, and Emmen air bases) may only be active during specific hours, and knowing these hours is critical for planning routes through or near militarily controlled areas. Outside activation times, these areas revert to standard civil airspace classifications.

Q58: Height of the Stockhorn in ft and m? Height of the Stockhorn cable car AGL? ^t60q58

DE · FR

Answer

Stockhorn: 2190 m / 7185 ft; Stockhornbahn AGL: 180 m / 591 ft

Explanation

The Stockhorn (2190 m / 7185 ft MSL) is a prominent peak in the Bernese Prealps visible on the Swiss ICAO chart. Its elevation appears in meters on the chart, and pilots must be able to convert to feet (using ft = m x 10/3: 2190 x 10/3 = 7300 ft, closely matching 7185 ft). The Stockhorn gondola cable (Stockhornbahn) represents an aerial obstacle 180 m AGL — cables and lifts are marked with AGL heights on the gliding chart as they pose significant hazards to low-flying gliders.

Key Terms

AGL = Above Ground Level; ICAO = International Civil Aviation Organization; MSL = Mean Sea Level

Q59: How tall is the tower on the Bantiger (46 58.7 N / 7 31.7 E)? ^t60q59

DE · FR

Answer

188 m / 615 ft

Explanation

The Bantiger tower near Bern is a communication mast shown on the Swiss ICAO and gliding charts at coordinates N46°58.7' / E7°31.7'. Its height is 188 m AGL (615 ft AGL). On the chart, obstacle heights are given in both meters and feet — exam candidates must be able to read the chart and convert between units. Obstacles above 100 m AGL are typically marked with their height and may have obstruction lighting.

Key Terms

AGL = Above Ground Level; ICAO = International Civil Aviation Organization

Q60: How high are you allowed to climb over Egerkingen (32.4 km, 060 from LSZG)? ^t60q60

DE · FR

Answer

Tango Sector status is decisive - if inactive (Basel Info) up to FL100; if active 1750 m or above requires clearance from BSL

Explanation

Egerkingen lies beneath the Tango Sector — a portion of Swiss airspace associated with the Basel/Mulhouse (LFSB/EuroAirport) TMA. When the Tango Sector is inactive (check with Basel Info on the appropriate frequency), the area is uncontrolled airspace up to FL100. When active, the upper limit drops to 1750 m MSL and operations above require a clearance from Basel Approach. This dynamic airspace structure is specific to the Swiss airspace system and requires checking NOTAMs and AIP Switzerland before flight.

Key Terms

AIP = Aeronautical Information Publication; FL = Flight Level; MSL = Mean Sea Level; TMA = Terminal Manoeuvring Area

Q61: What information do we find on the gliding chart for Les Eplatures aerodrome (47 05 N, 6 47.5 E)? ^t60q61

DE · FR

Answer

Gliding chart legend (symbols for controlled vs. uncontrolled fields)

Explanation

Les Eplatures (LSGC) near La Chaux-de-Fonds appears on the Swiss gliding chart with symbols decoded in the chart legend. The legend distinguishes between towered (controlled) and non-towered airfields, glider-specific aerodromes, military fields, and emergency landing strips. Candidates must be able to read the legend and determine the relevant operational information (radio frequencies, runway orientation, airspace class) for any airfield depicted on the chart.

Q62: Usage conditions for LS-R69 T (near Schaffhausen)? ^t60q62

DE · FR

Answer

Gliding chart legend, bottom right. Note: text box on boundary between TMA LSZH 10 (2000 m) and TMA LSZH 3 (1700 m); LSR69 lies in TMA 3

Explanation

LS-R69 is a glider restricted area near Schaffhausen that lies within the Zurich TMA structure. The area overlaps with TMA LSZH 3 (lower limit 1700 m MSL), not TMA LSZH 10 (2000 m) — this distinction is critical because it determines the altitude at which a clearance becomes necessary. Usage conditions are found in the chart legend lower right, and the text boxes on the chart itself clarify which TMA segment applies. Misidentifying the applicable TMA layer could lead to an airspace infringement.

Key Terms

MSL = Mean Sea Level; TMA = Terminal Manoeuvring Area

Q63: Coordinates of Birrfeld aerodrome? ^t60q63

DE · FR

Answer

N 47 26'36'', E 8 14'02''

Explanation

Birrfeld (LSZF) is a glider aerodrome in the canton of Aargau, Switzerland. Reading exact coordinates from the ICAO 1:500,000 chart requires careful use of the latitude and longitude graticule — each degree is divided into minutes, and at this scale, individual minutes of arc are clearly readable. The ability to read and record precise coordinates is tested because pilots may need to report positions to ATC or verify their location against chart features.

Key Terms

ATC = Air Traffic Control; ICAO = International Civil Aviation Organization

Q64: Coordinates of Montricher aerodrome? ^t60q64

DE · FR

Answer

N 46 35'25'', E 6 24'02''

Explanation

Montricher (LSTR) is a glider airfield in the canton of Vaud, in the French-speaking region of Switzerland. Its coordinates place it on the Swiss Plateau west of Lausanne. Locating it precisely on the ICAO chart and reading the graticule accurately requires practice — at 1:500,000 scale, 1 minute of latitude ≈ 1 NM ≈ 1.85 km, allowing sub-minute precision to be interpolated visually from the grid.

Key Terms

ICAO = International Civil Aviation Organization; NM = Nautical Mile(s)

Q65: Which place is at N 47 07', E 8 00'? ^t60q65

DE · FR

Answer

Willisau

Explanation

Given a set of coordinates, the candidate must locate the point on the Swiss ICAO chart by finding the correct latitude (47°07'N) and longitude (8°00'E) lines and reading the nearest landmark. Willisau is a town in the canton of Lucerne, on the Swiss Plateau. This exercise tests reverse coordinate lookup — starting from numbers and finding the geographic feature, as opposed to the forward direction (finding coordinates from a named place).

Key Terms

ICAO = International Civil Aviation Organization

Q66: Which place is at N 46 11', E 6 16'? ^t60q66

DE · FR

Answer

Annemasse aerodrome

Explanation

These coordinates place the point south of Lake Geneva (Lac Léman) at approximately N46°11' / E6°16', which corresponds to Annemasse aerodrome — a French airfield just across the Swiss-French border near Geneva. This question tests not only chart reading but also awareness that the Swiss ICAO chart extends into neighboring countries (France, Germany, Austria, Italy), and pilots should recognize aerodromes in border regions.

Key Terms

ICAO = International Civil Aviation Organization

Q67: TC from Grenchen aerodrome to Neuchatel aerodrome? ^t60q67

DE · FR

Answer

239

Explanation

To find the true course between two airfields, place a protractor on the chart aligned to the nearest meridian and measure the angle of the straight line connecting the two points. Grenchen (LSZG) is northeast of Neuenburg/Neuchâtel (LSGN), so the course from Grenchen to Neuchâtel runs roughly southwest — approximately 239° true. On the Lambert conformal chart, straight lines closely approximate great circles, and courses are measured from true north at the midpoint meridian.

Key Terms

TC = True Course

Q68: TC from Langenthal aerodrome to Kaegiswil aerodrome? ^t60q68

DE · FR

Answer

132

Explanation

Langenthal (LSPL) is northwest of Kaegiswil (LSPG near Sarnen), so the course from Langenthal to Kaegiswil runs roughly southeast — approximately 132° true. This is measured with a protractor on the ICAO chart, aligned to the meridian passing through or near the midpoint of the route. The course of 132° places the destination to the SE, consistent with Kaegiswil's position in the foothills near Lake Sarnen.

Key Terms

ICAO = International Civil Aviation Organization; TC = True Course

Q69: Distance Laax - Oberalp in km, NM, sm? ^t60q69

DE · FR

Answer

46,3 km / 25 NM / 28,7 sm

Explanation

The distance is measured with a ruler on the 1:500,000 chart and converted using the scale bar. At 1:500,000, 1 cm on the chart = 5 km in reality. Once the distance in km is known, conversion follows: NM = km / 1.852 ≈ km / 2 + 10% (exam formula), and statute miles = km / 1.609. This route runs along the Vorderrhein valley from Laax ski area toward the Oberalp Pass — a classic Swiss glider cross-country segment.

Key Terms

NM = Nautical Mile(s)

Q70: Flight time from Laax 14:52 to Oberalp 15:09? ^t60q70

DE · FR

Answer

17 min

Explanation

Simply subtract departure time from arrival time: 15:09 - 14:52 = 17 minutes. This elapsed flight time, combined with the distance from Q69, gives the speed for Q71. In practice, timing legs of a cross-country flight allows the pilot to verify actual groundspeed against planned groundspeed and detect headwind or tailwind differences from the forecast.

Q71: Speed in km/h, kts, mph? ^t60q71

DE · FR

Answer

163 km/h / 88 kts / 101 mph

Explanation

Ground speed = distance / time = 46.3 km / (17/60) h = 46.3 / 0.2833 = 163.4 km/h ≈ 163 km/h. Converting: kts = km/h / 1.852 ≈ 163 / 2 + 10% ≈ 88 kts; mph = km/h / 1.609 ≈ 101 mph. This three-unit speed result is typical of Swiss navigation exam questions, requiring fluency with all three speed units and their conversion relationships.

Q72: Route LSTB-Buochs-Jungfrau-LSTB: How long in km and NM? ^t60q72

DE · FR

Answer

56+43+59+80 = 238 km / 30+23+32+43 = 128 NM

Explanation

This is a triangular cross-country task measured on the chart: from Bellechasse (LSTB) to Buochs, then to the Jungfrau, and back to Bellechasse. Each leg is measured separately with a ruler on the 1:500,000 chart and the distances summed: 56 + 43 + 59 + 80 = 238 km total. Converting each leg to NM individually then summing (or converting the total: 238 / 1.852 ≈ 128 NM) gives the total task distance used for competition scoring and exam questions.

Key Terms

NM = Nautical Mile(s)

Q73: From Eriswil to Buochs in 18 min - how fast? ^t60q73

DE · FR

Answer

(43 km / 18 min) x 60 = 143 km/h / 77 kts / 89 mph

Explanation

Ground speed = (distance / time) x 60 to convert minutes to hours: (43 km / 18 min) x 60 = 143.3 km/h ≈ 143 km/h. The 43 km distance is taken from the chart measurement for this leg. Converting: kts ≈ 143 / 1.852 ≈ 77 kts; mph ≈ 143 / 1.609 ≈ 89 mph. This type of in-flight speed check — measuring elapsed time between two known points — is how glider pilots monitor actual vs. planned groundspeed during cross-country flights.

Q74: Which airspaces between Bellechasse and Buochs at 1500 m MSL? ^t60q74

DE · FR

Answer

TMA PAY 7 (E), TMA LSZB1 (D - clearance required), LR E MTT, LR E Alps, LS-R15 (if active), TMA LSME 2, CTR LSMA/LSZC (clearances required)

Explanation

This question requires reading all airspace layers on the route between Bellechasse and Buochs at 1500 m MSL, using both the ICAO chart and the gliding chart. Airspace Class D areas (TMA LSZB1, CTR LSMA/LSZC) require an ATC clearance before entry. Airspace Class E areas (TMA PAY 7, LR E MTT, LR E Alpen) are accessible under VFR without clearance but IFR flights have priority. LS-R15 is a glider area that may be active. Systematic left-to-right reading of the chart along the route is the required technique.

Key Terms

ATC = Air Traffic Control; CTR = Control Zone; ICAO = International Civil Aviation Organization; IFR = Instrument Flight Rules; MSL = Mean Sea Level; TMA = Terminal Manoeuvring Area; VFR = Visual Flight Rules

Q75: TC from Jungfrau to Bellechasse? ^t60q75

DE · FR

Answer

308

Explanation

The Jungfrau is located southeast of Bellechasse (LSTB), so the course FROM Jungfrau TO Bellechasse points northwest. A bearing of 308° is northwest of north, consistent with this geometry. The TC is measured with a protractor on the Lambert conformal chart, aligned to the meridian at the midpoint of the route. Note that this is the reciprocal of the course from Bellechasse to Jungfrau (approximately 128°), which confirms 308° is directionally correct.

Key Terms

TC = True Course

Q76: Glide from Jungfrau (4200 m MSL) to Bellechasse with glide ratio 1:30 at 150 km/h - arrival altitude? ^t60q76

DE · FR

Answer

Distance 80 km, altitude loss 2667 m, arrival 1533 m MSL = 1100 m AGL above LSTB (433 m)

Explanation

With a glide ratio of 1:30, the glider covers 30 meters forward for every 1 meter of altitude lost. Height loss over 80 km = 80,000 m / 30 = 2,667 m. Starting at 4200 m MSL: arrival altitude = 4200 - 2667 = 1533 m MSL. Bellechasse (LSTB) elevation is approximately 433 m MSL, so arrival height AGL = 1533 - 433 = 1100 m AGL. This is a classic final glide calculation — comparing arrival altitude with terrain and aerodrome elevation to determine if the glider reaches the destination with sufficient margin.

Key Terms

AGL = Above Ground Level; MSL = Mean Sea Level

Q77: Wind triangle Jungfrau-Bellechasse: TAS 140 km/h, wind 040/15 kts ^t60q77

DE · FR

Answer

GS 137 km/h, WCA 12, TH 320

Explanation

The wind triangle (Winddreieck) is solved graphically or with a mechanical DR calculator: the TC is 308°, TAS is 140 km/h (≈76 kts), and wind is from 040° at 15 kts (≈28 km/h). The wind blows from the NE toward the SW, creating a crosswind component from the right on this NW track. The WCA of +12° (right wind → head left) gives TH = TC + WCA = 308° + 12° = 320°. The headwind component reduces groundspeed from 140 to approximately 137 km/h. These calculations are performed with the mechanical flight computer (e-6B or equivalent) permitted in the Swiss exam.

Key Terms

GS = Ground Speed; TAS = True Airspeed; TC = True Course; TH = True Heading; WCA = Wind Correction Angle

Q78: MH from Jungfrau to Bellechasse (variation 3 E)? ^t60q78

DE · FR

Answer

TH 320 - 3 = MH 317

Explanation

To convert True Heading (TH) to Magnetic Heading (MH), apply the local magnetic variation. With 3° East variation, "East is least" — subtract East variation from True to get Magnetic: MH = TH - VAR(E) = 320° - 3° = 317°. The pilot would set 317° on the directional gyro (aligned to the magnetic compass) to fly this leg. Switzerland has a small easterly variation of about 2-3° in most regions.

Key Terms

MH = Magnetic Heading; TH = True Heading; VAR = Magnetic Variation

Q79: If variation is 25 W - MH? ^t60q79

DE · FR

Answer

TH 320 + 25 = MH 345

Explanation

With 25° West variation, "West is best" — add West variation to True Heading to get Magnetic Heading: MH = TH + VAR(W) = 320° + 25° = 345°. This hypothetical scenario (Switzerland has only ~3° variation, not 25°) is used to test whether candidates understand the direction of correction. West variation increases the magnetic heading number compared to true heading, because magnetic north is west of true north, making all magnetic bearings larger by the amount of variation.

Key Terms

MH = Magnetic Heading; TH = True Heading; VAR = Magnetic Variation

Q80: Transponder codes ^t60q80

DE · FR

| Code | Situation | |------|-----------| | 7000 | VFR in Class E and G airspace | | 7700 | Emergency | | 7600 | Radio failure | | 7500 | Hijack |

Explanation

These four transponder codes are universal ICAO emergency and standard VFR codes, memorized by all pilots. Code 7000 is the standard European VFR squawk in uncontrolled airspace (Class E and G) when no specific code is assigned by ATC. The three emergency codes — 7700 (emergency), 7600 (radio failure), 7500 (unlawful interference/hijack) — are set in order of severity and immediately alert ATC. In Switzerland, 7000 is used in lieu of a specific squawk assignment when flying in uncontrolled airspace outside a TMA or CTR.

Key Terms

ATC = Air Traffic Control; CTR = Control Zone; ICAO = International Civil Aviation Organization; TMA = Terminal Manoeuvring Area; VFR = Visual Flight Rules

Q81: Unit conversion formulas (exam reference) ^t60q81

DE · FR

| Conversion | Formula | |-----------|---------| | NM from km | km / 2 + 10% | | km from NM | NM × 2 - 10% | | ft from m | m / 3 × 10 | | m from ft | ft × 3 / 10 | | kts from km/h | km/h / 2 + 10% | | km/h from kts | kts × 2 - 10% | | m/s from ft/min | ft/min / 200 | | ft/min from m/s | m/s × 200 |

Key Terms

NM = Nautical Mile(s)

Q82: You are flying below an airspace with a lower limit at FL75, maintaining a 300 m safety margin. Assuming QNH is 1013 hPa, at approximately what altitude are you flying? ^t60q82

DE · FR

Answer

B)

Explanation

FL75 corresponds to 7500 ft at standard pressure (QNH 1013 hPa). 7500 ft × 0.3048 = 2286 m ≈ 2286 m AMSL. Subtracting the safety margin of 300 m: 2286 − 300 = 1986 m. However, the question asks for the flying altitude (below FL75 with 300 m safety margin), which is approximately 2290 m AMSL as the upper limit before applying the margin — corresponding to FL75 converted, which is 2290 m AMSL. Answer B is therefore correct.

Key Terms

AMSL = Above Mean Sea Level; FL = Flight Level; QNH = Pressure adjusted to mean sea level

Q83: A friend departs from France on 6 June (summer time) at 1000 UTC for a cross-country flight toward the Jura. You want to take off from Les Eplatures at the same time. What does your watch show? ^t60q83

DE · FR

Answer

C)

Explanation

In Switzerland on 6 June, summer time is in effect (CEST = UTC+2). To take off at 1000 UTC, your watch must show 1000 + 2h = 1200 LT. France also uses CEST (UTC+2) in summer, so both pilots take off at the same UTC time, but your watches both show 1200 LT.

Q84: Given: TT 220°, WCA -15°, VAR 5°W. What is the MH? ^t60q84

DE · FR

Answer

D)

Explanation

TT (True Track = TC) = 220°, WCA = -15°. TH = TC + WCA = 220° + (-15°) = 205°. With VAR 5°W: MH = TH + VAR (West) = 205° + 5° = 210°. Remember: westerly variation is added to obtain the magnetic heading (West is Best — add). Therefore MH = 210°.

Key Terms

MH = Magnetic Heading; TC = True Course; TH = True Heading; VAR = Magnetic Variation; WCA = Wind Correction Angle

Q85: You intend to follow a TC of 090° from your current position. The wind is a headwind from the right. ^t60q85

DE · FR

Answer

D)

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).

Key Terms

TC = True Course; TH = True Heading; WCA = Wind Correction Angle

Q86: The turning error of a magnetic compass is caused by ^t60q86

DE · FR

Answer

B)

Explanation

The turning error of the magnetic compass is caused by magnetic dip (inclination). When the aircraft turns, the vertical component of the Earth's magnetic field acts on the tilted needle, causing erroneous indications. This error is particularly pronounced at high latitudes where the dip is strong. It manifests during turns passing through magnetic north or south.

Q87: What term describes the deflection of a compass needle caused by electric fields? ^t60q87

DE · FR

Answer

C)

Explanation

The movement of the compass needle caused by electric (or stray magnetic) fields onboard is called deviation. However, the answer key gives C (declination) — which may seem surprising. In this BAZL context, the disturbance of the needle by local electric fields onboard is treated as an additional form of deviation. Note: terminology may vary by source; technically, deviation is caused by the aircraft's own magnetic fields, while electric fields can also disturb the instrument.

Q88: Which statement applies to a chart produced using the Mercator projection (cylinder tangent to the equator)? ^t60q88

DE · FR

Answer

D)

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.

Q89: You measure 12 cm on a 1:200,000 scale chart. What is the actual ground distance? ^t60q89

DE · FR

Answer

B)

Explanation

At a scale of 1:200,000, 1 cm on the chart corresponds to 200,000 cm = 2 km on the ground. Therefore 12 cm on the chart = 12 × 2 km = 24 km on the ground. Simple calculation: actual distance = chart distance × scale denominator = 12 cm × 200,000 = 2,400,000 cm = 24 km.

Q90: Which description matches the information shown on the Swiss ICAO chart for MULHOUSE-HABSHEIM aerodrome (approx. N47°44'/E007°26')? ^t60q90

DE · FR

Answer

C)

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).

Key Terms

AMSL = Above Mean Sea Level; ICAO = International Civil Aviation Organization

Q91: After a thermal flight in the Alps, you glide in a straight line from Erstfeld (46°49'00"N/008°38'00"E) towards Fricktal-Schupfart (47°30'32"N/007°57'00"). You pass through several control zones. On which frequency do you call the third control zone? ^t60q91

DE · FR

Answer

C)

Explanation

Flying a straight line from Erstfeld northwestward to Fricktal-Schupfart, you traverse multiple CTR and TMA sectors visible on the Swiss ICAO 1:500,000 chart. Each controlled airspace sector has its assigned communication frequency printed on the chart. Counting the control zones sequentially along this route, the third one encountered requires contact on 120.425 MHz (option C). The other frequencies listed correspond to different control zones along other routes or in other positions along this route.

Source: Segelflugverband der Schweiz - SFCLTheorieNavigationVersionSchweizUebungen.pdf > Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweizUebungen.pdf

Key Terms

CTR = Control Zone; ICAO = International Civil Aviation Organization; TMA = Terminal Manoeuvring Area

Permitted exam aids: Swiss ICAO chart 1:500,000, Swiss gliding chart, protractor, ruler, mechanical DR computer, compass, non-programmable scientific calculator (TI-30 ECO RS recommended). No alphanumeric or electronic navigation computers are permitted.

Q92: Which geographic features are most useful for orientation during flight? ^t60q92

DE · FR

Answer

B)

Explanation

For visual navigation, major intersections of transport routes — such as motorway junctions, railway branch points, and highway crossings — provide precise, unmistakable position fixes because they appear as distinct point features on both the chart and the ground.

Q93: During flight, you notice that you are drifting to the left. What action do you take to stay on your desired track? ^t60q93

DE · FR

Answer

B)

Explanation

If the aircraft drifts to the left, the wind has a component pushing from the right side of the intended track. To compensate, you increase the heading value (fly a higher heading) so the nose points to the right of the desired track, establishing a crab angle into the wind that offsets the drift.

Q94: During a cross-country flight, you must land at Saanen aerodrome (46°29'11"N/007°14'55"E). On which frequency do you establish radio contact? ^t60q94

DE · FR

Answer

C)

Explanation

Saanen aerodrome (LSGK) uses the frequency 119.430 MHz for aerodrome traffic communications, as indicated on the Swiss ICAO chart and in the Swiss AIP. Before landing at any aerodrome, pilots must consult the chart or AIP to identify the correct radio frequency and establish contact.

Key Terms

AIP = Aeronautical Information Publication; ICAO = International Civil Aviation Organization

Q95: Up to what altitude may you fly a glider over the Oberalppass (146°/52 km from Lucerne) without air traffic control authorisation? ^t60q95

DE · FR

Answer

D)

Explanation

Over the Oberalppass, the Swiss ICAO chart shows that uncontrolled airspace (Class E or G) extends up to 7500 ft AMSL. Below this altitude, VFR flights including gliders may operate without ATC authorisation. Above 7500 ft AMSL, controlled airspace begins and a clearance would be required.

Key Terms

AMSL = Above Mean Sea Level; ATC = Air Traffic Control; ICAO = International Civil Aviation Organization; TMA = Terminal Manoeuvring Area; VFR = Visual Flight Rules

Q96: On the aeronautical chart, north of the Furka Pass (070°/97 km from Sion), there is a red-hatched area marked LS-R8. What does this represent? ^t60q96

DE · FR

Answer

B)

Explanation

The prefix "R" in LS-R8 designates a Restricted area under the Swiss airspace classification system. When a restricted area is active, entry is prohibited unless specific authorisation has been obtained, and pilots must circumnavigate it. Activation status is published via DABS (Daily Airspace Bulletin Switzerland) or available from ATC.

Key Terms

ATC = Air Traffic Control

Q97: The coordinates 46°45'43" N / 006°36'48'' correspond to which aerodrome? ^t60q97

DE · FR

Answer

C)

Explanation

Plotting the coordinates 46 degrees 45 minutes 43 seconds N / 006 degrees 36 minutes 48 seconds E on the Swiss ICAO chart places the position at Motiers aerodrome (LSGM), located in the Val de Travers in the canton of Neuchatel.

Key Terms

ICAO = International Civil Aviation Organization

Q98: After a thermal flight in the Alps, you plan to fly in a straight line from the Gemmi Pass (171°/58 km from Bern Belp) to Grenchen aerodrome. Which magnetic course (MC) do you select? ^t60q98

DE · FR

Answer

D)

Explanation

The Gemmi Pass lies south-southeast of Grenchen, so the true course from Gemmi to Grenchen is roughly north-northwest (approximately 345-350 degrees true). Applying the Swiss magnetic variation of approximately 2-3 degrees East (MC = TC minus easterly variation) yields a magnetic course close to 348 degrees.

Key Terms

MC = Magnetic Course; TC = True Course

Q99: On a cross-country flight from Birrfeld aerodrome (47°26'N, 008°13'E) you turn at Courtelary aerodrome (47°10'N, 007°05'E). On the return leg you land at Grenchen aerodrome (47°10'N, 007°25'E). According to the Swiss gliding chart, the distance flown is ^t60q99

DE · FR

Answer

C)

Explanation

The flight consists of two legs measured on the Swiss gliding chart: Birrfeld to Courtelary (approximately 58 km southwest) and Courtelary to Grenchen (approximately 57 km returning northeast but landing short of Birrfeld). The total distance of both legs is approximately 115 km.

Q100: What onboard equipment does your aircraft need for you to determine your position using a VDF bearing? ^t60q100

DE · FR

Answer

C)

Explanation

VDF (VHF Direction Finding) is a ground-based service in which the station determines the bearing of the aircraft's radio transmission. To use a VDF bearing for position determination, the aircraft needs onboard VOR equipment (VHF omnidirectional range receiver) to interpret and display the bearing information provided by the ground station.

Key Terms

VHF = Very High Frequency

Q101: Which phenomenon is most likely to degrade GPS indications? ^t60q101

DE · FR

Answer

D)

Explanation

GPS signals are microwave transmissions from orbiting satellites that require a clear line of sight between the satellite and the receiver. When flying low in mountainous terrain, surrounding peaks and ridgelines mask portions of the sky, reducing the number of visible satellites and degrading the geometric dilution of precision (GDOP). This can lead to inaccurate position fixes or complete signal loss.

Q102: Given: MC 225 degrees, magnetic declination (variation) 5 degrees E. What is the TC? ^t60q102

DE · FR

Answer

D)

Explanation

True Course (TC) is calculated from Magnetic Course (MC) by accounting for magnetic declination. With easterly variation, magnetic north lies east of true north, so MC is larger than TC. The formula is TC = MC minus East variation: 225 degrees minus 5 degrees = 220 degrees.

Key Terms

MC = Magnetic Course; TC = True Course

Q103: In poor visibility, you fly from Gruyeres (222°/46 km from Bern) towards Lausanne (051°/52 km from Geneva). Which true course (TC) do you select? ^t60q103

DE · FR

Answer

D)

Explanation

Using the radial and distance references to plot both positions on the Swiss ICAO chart — Gruyeres at 222 degrees/46 km from Bern and Lausanne at 051 degrees/52 km from Geneva — and measuring the true course between them with a protractor yields approximately 261 degrees (roughly west-southwest).

Key Terms

ICAO = International Civil Aviation Organization; TC = True Course

Q104: You want to determine your position using a VDF bearing, but the controller reports the signals are too weak for assessment. What is the likely reason? ^t60q104

DE · FR

Answer

C)

Explanation

VDF operates on VHF frequencies, which propagate in a quasi-optical (line-of-sight) manner. If the aircraft is flying too low, the curvature of the Earth or intervening terrain blocks the signal path between the aircraft and the ground station, resulting in weak or undetectable signals.

Key Terms

VHF = Very High Frequency

Q105: What does the term "agonic line" mean? ^t60q105

DE · FR

Answer

A)

Explanation

The agonic line is a specific isogonic line along which the magnetic declination (variation) is exactly zero degrees — meaning true north and magnetic north are aligned. Along this line, a magnetic compass points directly to geographic north without any correction needed.

Q106: What is 4572 m expressed in feet? ^t60q106

DE · FR

Answer

B)

Explanation

To convert metres to feet, multiply by the conversion factor 3.2808 (since 1 metre = 3.2808 feet). Calculating: 4572 m multiplied by 3.2808 = 15,000 ft. This is a standard altitude conversion that aviation pilots should be able to perform quickly.

Q107: Which of the following statements is correct? ^t60q107

DE · FR

Answer

D)

Explanation

Lines of longitude (meridians) converge toward the poles, so the distance between two degrees of longitude is greatest at the equator (60 NM or 111 km) and decreases to zero at the poles, following the cosine of the latitude. This is a fundamental property of the spherical coordinate system.

Key Terms

NM = Nautical Mile(s)

Q108: Which value must you mark on the navigation chart before a cross-country flight? ^t60q108

DE · FR

Answer

C)

Explanation

On a navigation chart, the course line is drawn relative to the chart's grid, which is oriented to geographic (true) north. Therefore, the value measured and marked on the chart is the True Course (TC) — the angle between true north and the intended track line. Magnetic heading (option B), true heading (option A), and compass heading (option D) all incorporate corrections for wind, magnetic variation, or compass deviation that are calculated separately during flight planning, not drawn on the chart itself.

Key Terms

CH = Compass Heading; MH = Magnetic Heading; TC = True Course; TH = True Heading

Q109: In flight, you notice a drift to the right. How do you correct? ^t60q109

DE · FR

Answer

C)

Explanation

If the aircraft drifts to the right, the wind has a component pushing from the left side. To counteract this drift and maintain the desired track, you must turn into the wind by increasing the heading value (turning the nose further to the right to establish a crab angle into the wind component).

Q110: Up to what maximum altitude may you fly a glider over Lenzburg (255°/28 km from Zurich) without notification or authorisation? ^t60q110

DE · FR

Answer

D)

Explanation

Lenzburg lies beneath the Zurich TMA structure. According to the Swiss ICAO chart, the lowest TMA sector in this area has its floor at 1700 m AMSL. Below this altitude, the airspace is uncontrolled (Class E or G), and gliders may fly without ATC notification or authorisation. Above 1700 m AMSL, you enter controlled airspace requiring a clearance.

Key Terms

AMSL = Above Mean Sea Level; ATC = Air Traffic Control; ICAO = International Civil Aviation Organization; TMA = Terminal Manoeuvring Area

Q111: How does the map grid appear in a Lambert (normal conic) projection? ^t60q111

DE · FR

Answer

C)

Explanation

In a Lambert conformal conic projection, the cone is placed over the globe so that meridians project as straight lines converging toward the apex (the pole), while parallels of latitude appear as concentric arcs (parallel curves) centered on the pole. This projection preserves angles (conformality), making it ideal for aeronautical charts.

Q112: You depart from Bern on 10 June (summer time) at 1030 LT. The flight duration is 80 minutes. At what UTC time do you land? ^t60q112

DE · FR

Answer

D)

Explanation

On 10 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure at 1030 LT (CEST) equals 0830 UTC. Adding 80 minutes of flight time: 0830 + 0080 = 0950 UTC.

Q113: What are the coordinates of Bellechasse aerodrome (285°/28 km from Bern)? ^t60q113

DE · FR

Answer

D)

Explanation

Bellechasse aerodrome (LSGE) is located west-northwest of Bern, near the town of Bellechasse in the canton of Fribourg. Plotting the position at 285 degrees/28 km from Bern on the Swiss ICAO chart yields coordinates of approximately 46 degrees 59 minutes N / 007 degrees 08 minutes E.

Key Terms

ICAO = International Civil Aviation Organization

Q114: During a cross-country flight, "POOR GPS COVERAGE" appears on the screen. What could be the cause? ^t60q114

DE · FR

Answer

C)

Explanation

The "POOR GPS COVERAGE" message indicates that the receiver cannot track enough satellites with adequate geometry for a reliable position fix. The most common cause during cross-country glider flights is terrain masking — flying in deep valleys or near steep mountain faces that block satellite signals from view.

Q115: The magnetic compass of an aircraft is affected by metallic parts and electrical equipment. What is this influence called? ^t60q115

DE · FR

Answer

C)

Explanation

Deviation is the error in a magnetic compass caused by local magnetic fields from the aircraft's own metallic structure, electrical wiring, and electronic equipment. It varies with heading and is recorded on a deviation card in the cockpit.

Q116: You plan a cross-country flight Courtelary (315°/43 km from Bern-Belp) - Dittingen (192°/18 km from Basel-Mulhouse) - Birrfeld (265°/24 km from Zurich) - Courtelary. What is the total distance? ^t60q116

DE · FR

Answer

D)

Explanation

This is a closed triangular cross-country route with three legs: Courtelary to Dittingen, Dittingen to Birrfeld, and Birrfeld back to Courtelary. Each position is plotted on the Swiss ICAO 1:500,000 chart using the given radial/distance references, and the leg distances are measured with a ruler. The sum of all three legs yields approximately 189 km.

Key Terms

ICAO = International Civil Aviation Organization

Q117: Your GPS displays heights in metres, but you need feet. Can you change this? ^t60q117

DE · FR

Answer

B)

Explanation

Modern aviation GPS units allow pilots to change the display units (metres, feet, kilometres, nautical miles, etc.) through the device's settings menu (SETTING MODE). This is a simple user-accessible configuration change that does not require any maintenance intervention.

Q118: On a map, 5 cm correspond to a distance of 10 km. What is the scale? ^t60q118

DE · FR

Answer

D)

Explanation

To determine map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio of map distance to real distance is 5 cm to 1,000,000 cm, which simplifies to 1 cm representing 200,000 cm, giving a scale of 1:200,000.

Q119: During a long approach over a difficult navigation area, which method is most effective? ^t60q119

DE · FR

Answer

C)

Explanation

Over a difficult navigation area during a long approach, the most effective technique is to use time-based dead reckoning: monitor elapsed time with a time ruler (marking planned time checkpoints along the route) and confirm your position by identifying ground features as they appear, marking each verified position on the map. This combines time estimation with visual confirmation for maximum accuracy.

Q120: If you are south of the Montreux - Thun - Lucerne - Rapperswil line, on which frequency do you communicate with other glider pilots? ^t60q120

DE · FR

Answer

C)

Explanation

In Switzerland, glider-to-glider communication frequencies are divided geographically. South of the Montreux-Thun-Lucerne-Rapperswil line, the designated common glider frequency is 122.475 MHz. This frequency is used for traffic awareness, thermal information sharing, and safety communication among glider pilots operating in the southern Swiss Alps and surrounding areas. The other listed frequencies are either assigned to the northern sector or serve different aviation purposes.

Q121: What does the designation LS-R6, shown as a red hatched area north of Grindelwald (127°/52 km from Bern), mean? ^t60q121

DE · FR

Answer

D)

Explanation

LS-R6 is a restricted area (the "R" stands for Restricted in Swiss airspace classification). When active, entry is prohibited for all aircraft except helicopter emergency medical service (EMS) flights, which are exempted due to their life-saving mission.

Q122: How do you find the magnetic declination (variation) values for a given location? ^t60q122

DE · FR

Answer

D)

Explanation

Magnetic declination (variation) is found by reading the isogonic lines printed on aeronautical charts such as the Swiss ICAO 1:500,000 chart. Isogonic lines connect points of equal magnetic declination and are updated periodically to reflect the slow drift of Earth's magnetic field.

Key Terms

ICAO = International Civil Aviation Organization

Q123: In flight, you notice a drift to the left. How do you correct? ^t60q123

DE · FR

Answer

B)

Explanation

If the aircraft drifts to the left, the wind is pushing it from the right side of the flight path. To correct, the pilot must turn into the wind by increasing the heading value (turning right). This applies a wind correction angle that offsets the crosswind component. Turning left (option A) or decreasing the heading (option C) would worsen the drift. Flying faster (option D) reduces drift angle slightly but does not correct it — proper heading adjustment is the correct technique.

Q124: What does the indication GND on the cover of the gliding chart (top left, approximately 15 NM west of St Gallen-Altenrhein, 088°/75 km from Zurich-Kloten) mean? ^t60q124

DE · FR

Answer

D)

Explanation

The GND designation on the Swiss gliding chart indicates that reduced cloud separation distances are permitted inside the designated zones outside military flying service hours. When the military is not active, glider pilots benefit from relaxed minima in these areas.

Key Terms

NM = Nautical Mile(s)

Q125: Given: TC 180 degrees, MC 200 degrees. What is the magnetic declination (variation)? ^t60q125

DE · FR

Answer

C)

Explanation

Magnetic declination (variation) is the difference between True Course (TC) and Magnetic Course (MC), calculated as: Variation = TC - MC = 180° - 200° = -20°. A negative value indicates West declination, so the answer is 20°W. The mnemonic "variation west, magnetic best" (magnetic heading is greater) confirms this: when MC is greater than TC, variation is West.

Key Terms

MC = Magnetic Course; TC = True Course

Q126: During a triangle flight Grenchen (350°/31 km from Bern-Belp) - Kagiswil (090°/57 km from Bern-Belp) - Buttwil (221°/28 km from Zurich-Kloten) - Grenchen, on the return from Buttwil you must land at Langenthal (032°/35 km from Bern-Belp). What is the straight-line distance flown? ^t60q126

DE · FR

Answer

D)

Explanation

The total distance is the sum of the individual legs: Grenchen to Kagiswil, Kagiswil to Buttwil, and Buttwil to Langenthal (since the pilot diverted instead of returning to Grenchen). Measuring these legs on the 1:500,000 ICAO chart using the given radial/distance references from Bern-Belp and Zurich-Kloten yields a total of approximately 178 km.

Key Terms

ICAO = International Civil Aviation Organization

Q127: South of Gruyeres aerodrome there is a zone designated LS-D7. What is this? ^t60q127

DE · FR

Answer

A)

Explanation

The prefix "D" in LS-D7 designates a Danger zone under the Swiss airspace classification system. The upper limit of this zone is 9000 ft AMSL (above mean sea level).

Key Terms

AGL = Above Ground Level; AMSL = Above Mean Sea Level

Q128: On a map, 4 cm correspond to 10 km. What is the scale? ^t60q128

DE · FR

Answer

D)

Explanation

To find the map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio is 4 cm on the map to 1,000,000 cm in reality, so 1 cm represents 250,000 cm, giving a scale of 1:250,000.

Q129: Up to what altitude does the Locarno CTR (352°/18 km from Lugano-Agno) extend? ^t60q129

DE · FR

Answer

D)

Explanation

The Locarno CTR (Control Zone) extends from the surface up to 3,950 ft AMSL (above mean sea level), as published on the Swiss aeronautical charts.

Key Terms

AGL = Above Ground Level; AMSL = Above Mean Sea Level; CTR = Control Zone; FL = Flight Level

Q130: You are above Fraubrunnen (north of Bern-Belp airport), N47°05'/E007°32', at 4500 ft AMSL. Your height above the ground is approximately 3000 ft. In which airspace are you? ^t60q130

DE · FR

Answer

C)

Explanation

At Fraubrunnen (north of Bern-Belp) at 4500 ft AMSL, the aircraft is below the BERN 2 TMA, which begins at 5500 ft AMSL in this area, and above the Bern CTR, which only extends to a lower altitude. This places the aircraft in Class E airspace.

Key Terms

AMSL = Above Mean Sea Level; CTR = Control Zone; TMA = Terminal Manoeuvring Area

Q131: Your GPS displays distances in NM, but you need km for your calculations. Can you change this? ^t60q131

DE · FR

Answer

C)

Explanation

Modern aviation GPS units allow the pilot to change distance display units (NM to km or vice versa) through the device's SETTING MODE menu. This is a simple user preference and requires no technical workshop intervention.

Key Terms

NM = Nautical Mile(s)

Q132: You depart from Bern on 5 June (summer time) at 0945 UTC for a glider flight lasting 45 minutes. At what local time do you land? ^t60q132

DE · FR

Answer

B)

Explanation

On 5 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure is at 0945 UTC, and the flight lasts 45 minutes, so landing occurs at 0945 + 0045 = 1030 UTC. Converting to local time: 1030 UTC + 2 hours = 1230 CEST. However, the correct answer given is B (1130 LT), which would correspond to UTC+1 conversion. This suggests the question intends standard CET (UTC+1) or uses a different convention.

Q133: 54 NM correspond to: ^t60q133

DE · FR

Answer

C)

Explanation

The conversion factor is 1 NM = 1.852 km. Therefore 54 NM x 1.852 km/NM = 100.008 km, which rounds to 100.00 km.

Key Terms

NM = Nautical Mile(s)

Q134: Which statement about GPS is correct? ^t60q134

DE · FR

Answer

B)

Explanation

GPS is highly accurate for position determination, but satellite signals can be disrupted by terrain shading, atmospheric conditions, or intentional interference. Pilots must always cross-check GPS position against visual ground references.

Q135: What is meant by an "isogonic line"? ^t60q135

DE · FR

Answer

C)

Explanation

An isogonic line connects all points on a chart that have the same magnetic declination (variation). These lines are printed on aeronautical charts to help pilots convert between true and magnetic bearings.

Q136: In poor visibility, you fly from the Saentis (110°/65 km from Zurich-Kloten) towards Amlikon (075°/40 km from Zurich-Kloten). Which true course (TC) do you select? ^t60q136

DE · FR

Answer

C)

Explanation

Plotting both positions relative to Zurich-Kloten on the chart, the Saentis lies to the southeast (110°/65 km) and Amlikon to the east-northeast (075°/40 km). The route from Saentis to Amlikon heads northwest, yielding a true course of approximately 328°.

Key Terms

TC = True Course

Q137: What onboard equipment must your glider have for you to determine your position using a VDF bearing? ^t60q137

DE · FR

Answer

C)

Explanation

VDF (VHF Direction Finding) works by having a ground station take a bearing on the pilot's radio transmission. The only equipment the aircraft needs is a standard VHF radio communication system — the pilot transmits, and the ground station determines the direction.

Key Terms

ELT = Emergency Locator Transmitter; VHF = Very High Frequency

Q138: How does the map grid appear in a normal cylindrical projection (Mercator projection)? ^t60q138

DE · FR

Answer

C)

Explanation

In a Mercator (normal cylindrical) projection, both meridians and parallels appear as straight lines that intersect at right angles, forming a rectangular grid. Meridians are evenly spaced vertical lines and parallels are horizontal lines (though their spacing increases toward the poles).

Q139: Up to what maximum altitude may you fly a glider over Burgdorf (035°/19 km from Bern-Belp) without notification or authorisation? ^t60q139

DE · FR

Answer

D)

Explanation

Above Burgdorf, the lower boundary of the Bern TMA is at 1700 m AMSL. Below this altitude, a glider may fly freely without notification or authorization in Class E or G airspace.

Key Terms

AGL = Above Ground Level; AMSL = Above Mean Sea Level; TMA = Terminal Manoeuvring Area

Q140: What is the name of the location at coordinates 46°29' N / 007°15' E? ^t60q140

DE · FR

Answer

C)

Explanation

The coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome, which serves the Gstaad area in the Bernese Oberland.

Q141: What is meant by the "geographic longitude" of a location? ^t60q141

DE · FR

Answer

D)

Explanation

Geographic longitude is the angular distance measured east or west from the Prime Meridian (0° at Greenwich) to the local meridian passing through the given location, expressed in degrees (0° to 180°E or W).

Q142: The term 'magnetic course' (MC) is defined as ^t60q142

DE · FR

Answer

D)

Explanation

Magnetic Course (MC) is defined as the angle measured clockwise from magnetic north to the intended course line over the ground. It is the course referenced to the Earth's magnetic field rather than to true (geographic) north.

Key Terms

MC = Magnetic Course; TC = True Course

Q143: An aircraft is flying at FL 75 with an outside air temperature (OAT) of -9°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q143

DE · FR

Answer

C)

Explanation

True altitude accounts for non-standard temperature effects on pressure altitude. ISA temperature at approximately 6500 ft is about +2°C (15° - 2°/1000 ft x 6.5). With OAT of -9°C, the air is approximately 11°C colder than ISA. Cold air is denser, meaning pressure levels are compressed closer to the ground, so the aircraft is actually lower than the altimeter indicates. Using the correction of roughly 4 ft per 1°C per 1000 ft: 11°C x 4 x 6.5 = approximately 286 ft below QNH altitude, yielding about 6250 ft true altitude.

Key Terms

FL = Flight Level; ISA = International Standard Atmosphere; QNH = Pressure adjusted to mean sea level

Q144: An aircraft flies at a pressure altitude of 7000 ft with OAT +11°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q144

DE · FR

Answer

A)

Explanation

At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +11°C is about 9-10°C warmer than ISA. In warmer-than-standard air, the atmosphere is expanded, so the aircraft sits higher than the altimeter indicates. Applying the temperature correction (approximately +10°C x 4 ft/°C/1000 ft x 6.5 = +260 ft) to the QNH altitude gives approximately 6500 + 250 = 6750 ft true altitude.

Key Terms

ISA = International Standard Atmosphere; QNH = Pressure adjusted to mean sea level

Q145: An aircraft flies at a pressure altitude of 7000 ft with OAT +21°C. The QNH altitude is 6500 ft. The true altitude equals ^t60q145

DE · FR

Answer

A)

Explanation

At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +21°C means the air is about 19-20°C warmer than standard. Warm air expands, placing the aircraft significantly higher than indicated. The correction is approximately +20°C x 4 ft/°C/1000 ft x 6.5 = +520 ft, yielding about 6500 + 500 = 7000 ft true altitude. This large warm correction brings the true altitude up to match the pressure altitude.

Key Terms

ISA = International Standard Atmosphere; QNH = Pressure adjusted to mean sea level

Q146: Given: True course: 255°. TAS: 100 kt. Wind: 200°/10 kt. The true heading equals ^t60q146

DE · FR

Answer

D)

Explanation

With TC 255° and wind from 200°, the wind comes from approximately 55° to the left of the course line. This crosswind pushes the aircraft to the right of track. To compensate, the pilot must crab into the wind (turn left), reducing the heading below the course value. The wind correction angle is approximately sin^-1(10 x sin55° / 100) = sin^-1(0.082) = about 5°. True heading = 255° - 5° = 250°.

Key Terms

TAS = True Airspeed; TC = True Course

Q147: Given: True course: 165°. TAS: 90 kt. Wind: 130°/20 kt. Distance: 153 NM. The true heading equals ^t60q147

DE · FR

Answer

D)

Explanation

The wind from 130° on a 165° course comes from approximately 35° to the left of the nose, pushing the aircraft right of track. The pilot must crab left to compensate. WCA = sin^-1(20 x sin35° / 90) = sin^-1(0.127) = approximately 7°. True heading = 165° - 7° = 158°.

Key Terms

NM = Nautical Mile(s); TAS = True Airspeed; WCA = Wind Correction Angle

Q148: An aircraft follows a true course (TC) of 040° at a constant TAS of 180 kt. The wind vector is 350°/30 kt. The groundspeed (GS) equals ^t60q148

DE · FR

Answer

D)

Explanation

With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left-front. The headwind component is 30 x cos50° = approximately 19 kt, and the crosswind component is 30 x sin50° = approximately 23 kt. The wind correction angle is about 7°, and the groundspeed is calculated from the navigation triangle as TAS minus the effective headwind component, approximately 180 - 21 = 159 kt.

Key Terms

GS = Ground Speed; TAS = True Airspeed; TC = True Course

Q149: Given: True course: 120°. TAS: 120 kt. Wind: 150°/12 kt. The WCA equals ^t60q149

DE · FR

Answer

C)

Explanation

With TC 120° and wind from 150°, the wind comes from 30° to the right of and behind the course line. This pushes the aircraft to the left of track, requiring the pilot to crab to the right. WCA = sin^-1(12 x sin30° / 120) = sin^-1(6/120) = sin^-1(0.05) = approximately 3° to the right.

Key Terms

TAS = True Airspeed; TC = True Course; WCA = Wind Correction Angle

Q150: The distance from 'A' to 'B' is 120 NM. At 55 NM from 'A' the pilot finds a deviation of 7 NM to the right. What approximate course change is needed to reach 'B' directly? ^t60q150

DE · FR

Answer

D)

Explanation

Using the 1:60 rule, the opening angle (track error from A) is (7/55) x 60 = approximately 7.6° or about 8°. The remaining distance to B is 120 - 55 = 65 NM, so the closing angle to reach B is (7/65) x 60 = approximately 6.5° or about 6°. The total course correction needed is the sum of both angles: 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left).

Key Terms

NM = Nautical Mile(s)

Q151: How many satellites are required for a precise and verified three-dimensional position fix? ^t60q151

DE · FR

Answer

D)

Explanation

A GPS receiver needs signals from at least four satellites for a three-dimensional position fix (latitude, longitude, and altitude). Three satellites would provide only a two-dimensional fix, and the fourth is needed to solve for the receiver's clock error in addition to three spatial coordinates.

Q152: Which ground features should be preferred for orientation during visual flight? ^t60q152

DE · FR

Answer

D)

Explanation

Rivers, railroads, and highways are the preferred visual navigation references because they are large, prominent linear features that are easily identifiable from altitude and accurately depicted on aeronautical charts.

Q153: What is the approximate circumference of the Earth at the equator? ^t60q153

DE · FR

![](figures/t60_q153.png)

Answer

C)

Explanation

The Earth's equatorial circumference is approximately 21,600 NM. This derives from the fundamental navigation relationship: 360° of longitude x 60 NM per degree = 21,600 NM, since one nautical mile equals one minute of arc on a great circle. In metric terms, the circumference is about 40,075 km, but that does not match any of the other options correctly.

Key Terms

NM = Nautical Mile(s)

Q154: Given: True course from A to B: 352°. Ground distance: 100 NM. GS: 107 kt. ETD: 0933 UTC. The ETA is ^t60q154

DE · FR

Answer

B)

Explanation

Flight time equals distance divided by groundspeed: 100 NM / 107 kt = 0.935 hours = 56 minutes. Adding 56 minutes to the ETD of 0933 UTC gives 0933 + 0056 = 1029 UTC.

Key Terms

ETA = Estimated Time of Arrival; ETD = Estimated Time of Departure; GS = Ground Speed; NM = Nautical Mile(s)

Q155: An aircraft travels 100 km in 56 minutes. The ground speed (GS) equals ^t60q155

DE · FR

Answer

D)

Explanation

Groundspeed = distance / time = 100 km / (56/60 hours) = 100 x (60/56) = 107.1 km/h. Since the distance is given in kilometres, the result is naturally in km/h.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s)

Q156: An aircraft flies with TAS 180 kt and a headwind component of 25 kt for 2 hours and 25 minutes. The distance flown equals ^t60q156

DE · FR

Answer

C)

Explanation

Groundspeed = TAS minus headwind = 180 - 25 = 155 kt. Flight time = 2 hours 25 minutes = 2.417 hours. Distance = GS x time = 155 x 2.417 = 374.6 NM, approximately 375 NM.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s); TAS = True Airspeed

Q157: Given: GS 160 kt, TC 177°, wind vector 140°/20 kt. The true heading (TH) equals ^t60q157

DE · FR

Answer

B)

Explanation

The wind from 140° on a 177° true course comes from approximately 37° to the left of the course, pushing the aircraft to the right. The pilot must crab left to compensate. WCA = sin^-1(20 x sin37° / 160) = sin^-1(12/160) = sin^-1(0.075) = approximately 4°. True heading = 177° - 4° = 173°.

Key Terms

GS = Ground Speed; TC = True Course; TH = True Heading; WCA = Wind Correction Angle

Q158: An aircraft follows TC 040° at a constant TAS of 180 kt. The wind vector is 350°/30 kt. The wind correction angle (WCA) equals ^t60q158

DE · FR

Answer

D)

Explanation

With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left side. The crosswind component = 30 x sin50° = approximately 23 kt pushes the aircraft to the right of track. To maintain course, the pilot crabs left (negative WCA). WCA = -sin^-1(23/180) = -sin^-1(0.128) = approximately -7°.

Key Terms

TAS = True Airspeed; TC = True Course; WCA = Wind Correction Angle

Q159: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The ground speed (GS) equals ^t60q159

DE · FR

Answer

C)

Explanation

The aircraft flies on TC 270° (westbound) and the wind blows from 090° (east). Since the wind comes from directly behind the aircraft, it is a pure tailwind. Groundspeed = TAS + tailwind = 100 + 25 = 125 kt. There is no crosswind component, so no wind correction angle is needed.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s); TAS = True Airspeed; TC = True Course

Q160: When using GPS for tracking to the next waypoint, a deviation bar with dots is displayed. Which interpretation is correct? ^t60q160

DE · FR

Answer

B)

Explanation

The GPS CDI (Course Deviation Indicator) displays lateral track error as an absolute distance in nautical miles, not as angular degrees like a VOR CDI. The full-scale deflection varies by operating mode: typically +/-5 NM in en-route mode, +/-1 NM in terminal mode, and +/-0.3 NM in approach mode.

Key Terms

NM = Nautical Mile(s)

Q161: What is the distance from VOR Bruenkendorf (BKD) (53°02'N, 011°33'E) to Pritzwalk (EDBU) (53°11'N, 12°11'E)? ^t60q161

DE · FR

![](figures/t60_q161.png)

Answer

D)

Explanation

Using the coordinates: latitude difference = 9' (= 9 NM north-south). Longitude difference = 38'; at latitude 53°N, 1 minute of longitude = cos(53°) NM = approximately 0.60 NM, giving 38 x 0.60 = 22.8 NM east-west. Total distance = sqrt(9^2 + 22.8^2) = sqrt(81 + 520) = sqrt(601) = approximately 24.5 NM, rounded to 24 NM.

Key Terms

NM = Nautical Mile(s)

Q162: An aircraft flies with TAS 120 kt and experiences 35 kt tailwind. How much time is needed for a distance of 185 NM? ^t60q162

DE · FR

Answer

C)

Explanation

Groundspeed = TAS + tailwind = 120 + 35 = 155 kt. Flight time = distance / GS = 185 / 155 = 1.194 hours = 1 hour 12 minutes.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s); TAS = True Airspeed

Q163: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The flight time equals ^t60q163

DE · FR

Answer

C)

Explanation

Flying on TC 270° with wind from 090° means the wind is a direct tailwind (blowing from directly behind). GS = TAS + tailwind = 100 + 25 = 125 kt. Flight time = 100 NM / 125 kt = 0.80 hours = 48 minutes.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s); TAS = True Airspeed; TC = True Course

Q164: Which answer completes the flight plan (marked cells)? (3,00 P.) ^t60q164

DE · FR

![](figures/t60_q164.png)

Answer

D)

Explanation

The flight plan conversion chain proceeds from True Course through wind correction to True Heading (TH), then applying magnetic variation to get Magnetic Heading (MH), and finally accounting for compass deviation for Magnetic Course (MC). The values TH 185°, MH 184°, and MC 178° are consistent with the sequential application of a small wind correction angle, a 1° easterly variation, and compass deviation.

Key Terms

MC = Magnetic Course; MH = Magnetic Heading; TC = True Course; TH = True Heading

Q165: What is meant by the term "terrestrial navigation"? ^t60q165

DE · FR

Answer

B)

Explanation

Terrestrial navigation (also known as pilotage or map reading) is the technique of orienting the aircraft by visually identifying ground features — towns, rivers, roads, railways, lakes — and matching them to the aeronautical chart.

Q166: What flight time is required for a distance of 236 NM at a ground speed of 134 kt? ^t60q166

DE · FR

Answer

C)

Explanation

Flight time = distance / groundspeed = 236 NM / 134 kt = 1.761 hours. Converting the decimal fraction: 0.761 x 60 = 45.7 minutes, approximately 46 minutes, giving a total of 1 hour 46 minutes.

Key Terms

GS = Ground Speed; NM = Nautical Mile(s)

Q167: What is the true course (TC) from Uelzen (EDVU) (52°59'N, 10°28'E) to Neustadt (EDAN) (53°22'N, 011°37'E)? ^t60q167

DE · FR

![](figures/t60_q167.png)

Answer

D)

Explanation

Neustadt lies to the north-northeast of Uelzen (higher latitude and further east). Plotting the route from Uelzen to Neustadt on the chart yields a northeast heading of approximately 061°.

Key Terms

TC = True Course

Q168: What does the 1:60 rule mean? ^t60q168

DE · FR

Answer

C)

Explanation

The 1:60 rule is a mental math shortcut stating that at a distance of 60 NM, a 1° track error produces approximately 1 NM of lateral offset. Mathematically, this works because the arc length of 1° on a 60 NM radius circle is 2 x pi x 60 / 360 = approximately 1.047 NM, close enough to 1 NM for practical navigation.

Key Terms

NM = Nautical Mile(s)

Q169: An aircraft follows TC 220° at a constant TAS of 220 kt. The wind vector is 270°/50 kt. The ground speed (GS) equals ^t60q169

DE · FR

Answer

C)

Explanation

With TC 220° and wind from 270°, the wind angle is 50° from the right-front of the aircraft. The headwind component = 50 x cos50° = approximately 32 kt, and the crosswind component = 50 x sin50° = approximately 38 kt. Using the navigation wind triangle, the groundspeed works out to approximately 185 kt after accounting for both the headwind reduction and the crab angle.

Key Terms

GS = Ground Speed; TAS = True Airspeed; TC = True Course

Q170: An aeroplane has a heading of 090°. The distance to fly is 90 NM. After 45 NM the aeroplane is 4.5 NM north of the planned flight path. What corrected heading is needed to reach the destination directly? ^t60q170

DE · FR

Answer

C)

Explanation

Applying the 1:60 rule: the opening angle (track error) = (4.5 / 45) x 60 = 6° off track to the north. The remaining distance is 90 - 45 = 45 NM. The closing angle to reach the destination = (4.5 / 45) x 60 = 6°. Total correction = opening angle + closing angle = 6° + 6° = 12° to the right (south), since the aircraft has drifted north of track.

Key Terms

NM = Nautical Mile(s)

Q171: What is the distance from Neustadt (EDAN) (53°22'N, 011°37'E) to Uelzen (EDVU) (52°59'N, 10°28'E)? ^t60q171

DE · FR

![](figures/t60_q171.png)

Answer

A)

Explanation

From the coordinates: latitude difference = 23' (= 23 NM north-south). Longitude difference = 69'; at approximately 53°N latitude, 1' of longitude = cos(53°) = 0.602 NM, so 69 x 0.602 = 41.5 NM east-west. Total distance = sqrt(23^2 + 41.5^2) = sqrt(529 + 1722) = sqrt(2251) = approximately 47 NM, rounded to 46 NM on the chart.

Key Terms

NM = Nautical Mile(s)

Q172: What does the term terrestrial navigation mean? ^t60q172

DE · FR

Answer

B)

Explanation

Terrestrial navigation is the method of navigating by visually identifying ground features such as roads, rivers, railways, towns, and lakes, and matching them to an aeronautical chart. It is the primary VFR navigation technique and sometimes called pilotage or map reading.

Key Terms

VFR = Visual Flight Rules