### Q151: How many satellites are required for a precise and verified three-dimensional position fix? ^t60q151 - A) Five - B) Two - C) Three - D) Four **Correct: D)** > **Explanation:** A GPS receiver needs signals from at least four satellites to compute a full three-dimensional position fix (latitude, longitude, and altitude). Three satellites would only yield a two-dimensional fix on a known surface, and two satellites are entirely insufficient for any reliable fix. The fourth satellite is essential because it allows the receiver to solve for its own clock error in addition to the three spatial coordinates. Option A (five) refers to the number needed for Receiver Autonomous Integrity Monitoring (RAIM), which adds integrity checking but is not the minimum for a basic 3D fix. ### Q152: Which ground features should be preferred for orientation during visual flight? ^t60q152 - A) Farm tracks and creeks - B) Border lines - C) Power lines - D) Rivers, railroads, highways **Correct: D)** > **Explanation:** For visual navigation, the most reliable ground references are large, prominent linear features such as rivers, railways, and highways, because they are easily visible from altitude, unambiguous, and accurately depicted on aeronautical charts. Option A (farm tracks and creeks) is problematic because these features are too small and numerous to distinguish reliably from the air. Option B (border lines) are political boundaries that have no physical presence visible from the cockpit. Option C (power lines) are extremely difficult to see from a distance and pose a serious collision hazard at low altitude. ### Q153: What is the approximate circumference of the Earth at the equator? See figure (NAV-002) Siehe Anlage 1 ^t60q153 - A) 40000 NM. - B) 12800 km. - C) 21600 NM. - D) 10800 km. **Correct: C)** > **Explanation:** The Earth's equatorial circumference is approximately 21,600 nautical miles, derived from the fundamental navigation relationship: 360 degrees multiplied by 60 NM per degree equals 21,600 NM. This same relationship defines the nautical mile, where one minute of arc along a great circle equals one NM. Option A (40,000 NM) is far too large and confuses nautical miles with kilometres (the metric circumference is approximately 40,075 km). Options B (12,800 km) and D (10,800 km) are both significantly smaller than the actual metric circumference. ### Q154: Given: True course from A to B: 352°. Ground distance: 100 NM. GS: 107 kt. ETD: 0933 UTC. The ETA is… ^t60q154 - A) 1146 UTC. - B) 1029 UTC. - C) 1045 UTC. - D) 1129 UTC. **Correct: B)** > **Explanation:** Flight time equals distance divided by groundspeed: 100 NM / 107 kt = 0.935 hours, which converts to approximately 56 minutes. Adding 56 minutes to the estimated time of departure of 0933 UTC gives an ETA of 1029 UTC. Option A (1146 UTC) overshoots by more than an hour. Option C (1045 UTC) adds 72 minutes, suggesting an incorrect groundspeed of about 83 kt. Option D (1129 UTC) adds nearly two hours to departure time, which is far too long for 100 NM at 107 kt. ### Q155: An aircraft travels 100 km in 56 minutes. The ground speed (GS) equals… ^t60q155 - A) 198 kt. - B) 93 kt - C) 58 km/h - D) 107 km/h. **Correct: D)** > **Explanation:** Groundspeed is calculated as distance divided by time: 100 km / (56/60 hours) = 100 km / 0.933 h = approximately 107 km/h. Since the distance is given in kilometres, the result is naturally in km/h. Option A (198 kt) is far too high and results from an arithmetic error or unit confusion. Option B (93 kt) confuses km/h with knots without proper conversion. Option C (58 km/h) would imply a much longer flight time and appears to be the result of a calculation mistake. ### 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 - A) 435 NM. - B) 693 NM. - C) 375 NM. - D) 202 NM. **Correct: C)** > **Explanation:** Groundspeed equals TAS minus the headwind component: 180 - 25 = 155 kt. Flight time is 2 hours 25 minutes, which equals 2.417 hours. Distance equals groundspeed multiplied by time: 155 kt x 2.417 h = approximately 375 NM. Option A (435 NM) incorrectly uses the full TAS of 180 kt without subtracting the headwind. Option B (693 NM) appears to multiply TAS by an incorrect time value. Option D (202 NM) significantly underestimates the distance. ### Q157: Given: GS 160 kt, TC 177°, wind vector 140°/20 kt. The true heading (TH) equals… ^t60q157 - A) 184°. - B) 173°. - C) 180° - D) 169°. **Correct: B)** > **Explanation:** The wind is blowing from 140 degrees at 20 kt, which is approximately 37 degrees to the left of the true course of 177 degrees, pushing the aircraft to the right of its desired track. To compensate, the pilot must crab into the wind by turning left, reducing the heading. Using the wind correction angle formula, WCA = arcsin(20 x sin37 / 160) = approximately 4 degrees left. True heading therefore equals 177 - 4 = 173 degrees. Option A (184 degrees) incorrectly adds the correction. Options C (180 degrees) and D (169 degrees) reflect incorrect wind correction calculations. ### 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 - A) .+ 5° - B) . - 9° - C) .+ 11° - D) .- 7° **Correct: D)** > **Explanation:** With a true course of 040 degrees and wind from 350 degrees at 30 kt, the wind blows from approximately 50 degrees to the left of the course, pushing the aircraft to the right of track. To correct, the pilot must turn left (negative WCA). The crosswind component is 30 x sin(50) = approximately 23 kt. WCA = -arcsin(23/180) = approximately -7 degrees. The negative sign confirms a leftward correction. Options A (+5 degrees) and C (+11 degrees) incorrectly show a rightward correction, and Option B (-9 degrees) overestimates the correction angle. ### Q159: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The ground speed (GS) equals… ^t60q159 - A) 117 kt. - B) 131 kt. - C) 125 kt. - D) 120 kt. **Correct: C)** > **Explanation:** The wind is from 090 degrees and the aircraft is flying a course of 270 degrees, meaning the aircraft is heading directly into the wind source's direction -- but since "wind from 090" means air moves westward, this is actually a direct tailwind for a westbound course. Groundspeed therefore equals TAS plus tailwind: 100 + 25 = 125 kt. Option A (117 kt) and Option D (120 kt) partially account for the wind but arrive at incorrect values. Option B (131 kt) overestimates the wind contribution. ### Q160: When using GPS for tracking to the next waypoint, a deviation bar with dots is displayed. Which interpretation is correct? ^t60q160 - A) The bar deviation from centre shows track error as angular distance in degrees; full-scale deflection is +-10°. - B) The bar deviation from centre shows track error as absolute distance in NM; full-scale deflection depends on the GPS operating mode. - C) The bar deviation from centre shows track error as angular distance in degrees; full-scale deflection depends on the GPS operating mode. - D) The bar deviation from centre shows track error as absolute distance in NM; full-scale deflection is +-10 NM. **Correct: B)** > **Explanation:** The GPS Course Deviation Indicator (CDI) displays lateral track error as an absolute distance in nautical miles, not as an angular deviation in degrees like a VOR CDI does. The full-scale deflection varies by operating mode: typically plus/minus 5 NM in en-route mode, plus/minus 1 NM in terminal mode, and plus/minus 0.3 NM in approach mode. Options A and C are incorrect because they describe angular deviation, which applies to VOR-based CDIs, not GPS. Option D incorrectly states a fixed full-scale deflection of plus/minus 10 NM regardless of mode. ### Q161: What is the distance from VOR Bruenkendorf (BKD) (53°02'N, 011°33'E) to Pritzwalk (EDBU) (53°11'N, 12°11'E)? See annex (NAV-031) Siehe Anlage 2 ^t60q161 - A) 42 NM - B) 42 km - C) 24 km - D) 24 NM **Correct: D)** > **Explanation:** Using the given coordinates, the latitude difference is 9 minutes (approximately 9 NM north-south), and the longitude difference is 38 minutes. At 53 degrees north, one minute of longitude equals approximately cos(53) = 0.60 NM, so the east-west distance is about 38 x 0.60 = 22.8 NM. Applying the Pythagorean theorem gives the total distance as the square root of (81 + 520) = approximately 24.5 NM, which rounds to 24 NM. Options A and B (42 NM/km) significantly overestimate the distance. Option C (24 km) uses the wrong unit -- aeronautical distances are measured in nautical miles. ### 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 - A) 2 h 11 min - B) 0 h 50 min - C) 1 h 12 min - D) 1 h 32 min **Correct: C)** > **Explanation:** With a 35 kt tailwind, the groundspeed is TAS plus tailwind: 120 + 35 = 155 kt. Flight time equals distance divided by groundspeed: 185 NM / 155 kt = 1.194 hours, which converts to approximately 1 hour and 12 minutes. Option A (2 h 11 min) appears to ignore the tailwind entirely and use a much lower speed. Option B (50 min) would require a groundspeed of about 222 kt, which is far too high. Option D (1 h 32 min) corresponds to using only the TAS of 120 kt without adding the tailwind benefit. ### Q163: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The flight time equals… ^t60q163 - A) 62 Min. - B) 37 Min. - C) 48 Min. - D) 84 Min. **Correct: C)** > **Explanation:** Wind from 090 degrees on a 270-degree course is a direct tailwind, giving a groundspeed of 100 + 25 = 125 kt. Flight time equals 100 NM / 125 kt = 0.8 hours = 48 minutes. Option A (62 min) would correspond to a groundspeed of about 97 kt, suggesting headwind instead of tailwind. Option B (37 min) implies a groundspeed of roughly 162 kt, which is far too high. Option D (84 min) treats the tailwind as a headwind, yielding GS = 75 kt. ### Q164: Which answer completes the flight plan (marked cells)? See annex (NAV-014) (3,00 P.) Siehe Anlage 3 ^t60q164 - A) TH: 185°. MH: 185°. MC: 180°. - B) TH: 173°. MH: 174°. MC: 178°. - C) TH: 173°. MH: 184°. MC: 178°. - D) TH: 185°. MH: 184°. MC: 178°. **Correct: D)** > **Explanation:** This flight plan completion requires the correct sequential application of navigation corrections. Starting from the true course, the wind correction angle is applied to obtain the true heading (TH: 185 degrees). Then the local magnetic variation is subtracted to convert to magnetic heading (MH: 184 degrees). Finally, compass deviation is applied to arrive at the magnetic compass heading (MC: 178 degrees). Option A incorrectly shows MH equal to TH, ignoring variation. Option B starts with a wrong true heading. Option C has an inconsistent jump between TH and MH that does not match any standard variation value. ### Q165: What is meant by the term "terrestrial navigation"? ^t60q165 - A) Orientation by instrument readings during visual flight - B) Orientation by ground features during visual flight - C) Orientation by GPS during visual flight - D) Orientation by ground celestial objects during visual flight **Correct: B)** > **Explanation:** Terrestrial navigation, also known as pilotage or map reading, is the method of orienting and navigating an aircraft by visually identifying ground features such as roads, rivers, towns, and railways and matching them to an aeronautical chart. Option A describes instrument-based navigation, which is a separate discipline. Option C describes satellite navigation using GPS technology. Option D confusingly combines "ground" with "celestial objects" -- celestial navigation uses stars and the sun, not ground features, and is an entirely different method. ### Q166: What flight time is required for a distance of 236 NM at a ground speed of 134 kt? ^t60q166 - A) 0:46 h - B) 0:34 h - C) 1:46 h - D) 1:34 h **Correct: C)** > **Explanation:** Flight time equals distance divided by groundspeed: 236 NM / 134 kt = 1.761 hours. Converting the decimal fraction to minutes: 0.761 x 60 = approximately 46 minutes, giving a total flight time of 1 hour 46 minutes. Option A (0:46) drops the whole hour, suggesting the calculation stopped at the fractional part. Option B (0:34) is far too short. Option D (1:34) underestimates the minutes portion, which would correspond to a groundspeed of about 150 kt. ### 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)? See annex (NAV-031) Siehe Anlage 2 ^t60q167 - A) 235° - B) 241° - C) 055° - D) 061° **Correct: D)** > **Explanation:** Neustadt (EDAN) lies to the north-east of Uelzen (EDVU) -- it is both further north (53 degrees 22 minutes vs 52 degrees 59 minutes) and further east (011 degrees 37 minutes vs 010 degrees 28 minutes). The true course from Uelzen to Neustadt therefore points north-east, approximately 061 degrees. Option B (241 degrees) is the reciprocal course from Neustadt back to Uelzen. Option A (235 degrees) is also a south-westerly bearing and therefore in the wrong direction. Option C (055 degrees) is close but does not match the precise bearing plotted on the chart. ### Q168: What does the 1:60 rule mean? ^t60q168 - A) 10 NM lateral offset at 1° drift after 60 NM - B) 60 NM lateral offset at 1° drift after 1 NM - C) 1 NM lateral offset at 1° drift after 60 NM - D) 6 NM lateral offset at 1° drift after 10 NM **Correct: C)** > **Explanation:** The 1:60 rule is a practical navigation shortcut stating that 1 degree of track error produces a lateral offset of 1 NM after flying 60 NM. This works because the arc length of 1 degree on a circle of radius 60 NM is approximately 1.047 NM, close enough to 1 NM for practical use. The rule allows pilots to quickly estimate track corrections in flight without complex calculations. Option A overstates the offset by a factor of 10. Option B reverses the distance and offset values entirely. Option D uses a non-standard ratio that does not follow from the geometry. ### 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 - A) 135 kt. - B) 170 kt. - C) 185 kt. - D) 255 kt. **Correct: C)** > **Explanation:** With a true course of 220 degrees and wind from 270 degrees at 50 kt, the angle between the wind direction and the course is 50 degrees. The headwind component is 50 x cos(50) = approximately 32 kt, while the crosswind component is 50 x sin(50) = approximately 38 kt. Solving the wind triangle yields a groundspeed of approximately 185 kt. Option A (135 kt) underestimates the groundspeed significantly. Option B (170 kt) is slightly too low. Option D (255 kt) incorrectly adds the wind to TAS as if it were a pure tailwind, which it is not given the 50-degree angle. ### 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 - A) 9° to the right - B) 6° to the right - C) 12° to the right - D) 18° to the right **Correct: C)** > **Explanation:** Using the 1:60 rule, the track error angle is (4.5 NM / 45 NM) x 60 = 6 degrees. Since the aircraft is north of the planned track while heading east (090 degrees), it must correct to the right (southward). The closing angle to reach the destination over the remaining 45 NM is also (4.5 NM / 45 NM) x 60 = 6 degrees. The total heading correction is the sum of the opening and closing angles: 6 + 6 = 12 degrees to the right. Option B (6 degrees) accounts for only one component. Option A (9 degrees) and Option D (18 degrees) result from incorrect calculations. ### Q171: What is the distance from Neustadt (EDAN) (53°22'N, 011°37'E) to Uelzen (EDVU) (52°59'N, 10°28'E)? See annex (NAV-031) Siehe Anlage 2 ^t60q171 - A) 46 NM - B) 78 km - C) 78 km - D) 46 km **Correct: A)** > **Explanation:** The latitude difference between the two points is 23 minutes (approximately 23 NM north-south). The longitude difference is 69 minutes, and at approximately 53 degrees north, one minute of longitude equals about 0.60 NM, giving an east-west distance of 69 x 0.60 = approximately 41 NM. The total distance is the square root of (23 squared + 41 squared) = the square root of (529 + 1681) = approximately 47 NM, which rounds to 46 NM. Options B and C both show 78 km, which would convert to about 42 NM and does not match the calculation. Option D (46 km) uses the wrong unit -- it would equal only about 25 NM. ### Q172: What does the term terrestrial navigation mean? ^t60q172 - A) Orientation by GPS during visual flight - B) Orientation by ground features during visual flight - C) Orientation by instrument readings during visual flight - D) Orientation by ground celestial objects during visual flight **Correct: B)** > **Explanation:** Terrestrial navigation refers to the practice of navigating by visually identifying features on the Earth's surface -- such as roads, rivers, railways, towns, and lakes -- and correlating them with an aeronautical chart. It is the most fundamental VFR navigation skill, sometimes called pilotage or map reading. Option A (GPS) describes satellite-based navigation, an entirely different technology. Option C (instrument readings) describes instrument navigation. Option D incorrectly merges terrestrial and celestial concepts; celestial navigation uses astronomical bodies like stars, not ground features.