JUMBLED NAV QUESTIONS PDF

NAVIGATION QUESTION 1 NAVIGATION QUESTION 1. If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, e...

NAVIGATION QUESTION 1 NAVIGATION QUESTION 1. If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, endurance 3 hours, and actual time of departure is 1105 UTC. The Point of Equal Time (PET) is reached at: a) 1233 UTC b) 1221 UTC c) 1213 UTC 2 If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, endurance 3 hours and actual time of departure is 1105 UTC. The distance from departure to Point of Equal Time (PET) is: a) 147NM b) 203 NM c) 183 NM 3 Find the distance from waypoint 3 (WP 3) to the critical point. Given: distance from WP 3 to WP 4 = 750 NM, TAS out 430 kt, TAS return 425 kt, Tailwind component out 30 kt, head wind component return 40 kt a) 342 NM b) 375 NM c) 408 NM 4 Find the time to the Point of Safe Return (PSR). Given: Maximum useable fuel 15000 kg, Minimum reserve fuel 3500 kg, TAS out 425 kt, Head wind component out 30 kt, TAS return 430 kt, Tailwind component return 20 kt, Average fuel flow 2150 kg/h a) 2 h 51 min b) 3 h 43 min c) 2 h 59 min 5 Given: X = Distance A to point of equal time (PET) between A and BE = Endurance D = Distance A to BO = Groundspeed 'on' H = Groundspeed 'back' The formula for calculating the distance X to point of equal time (PET) is: a) H E x O x HX = ________ O + H b) D x OX = _____ O + H c) D x HX = _____ O + H 6 Given: Course A to B 088° (T)distance 1250 NM Mean TAS 330 kt Mean W/V 340°/60 kt The time from A to the PET between A and B is: a) 1 hour 54 minutes b) 1 hour 42 minutes c) 1 hour 39 minutes 7 Given: Distance X to Y 2700 NM Mach Number 0.75 Temperature -45°C Mean wind component 'on' 10 kt tailwind Mean wind component 'back' 35 kt tailwind. The distance from X to the point of equal time (PET) between X and Y is : a) 1350NM b) 1386 NM c) 1313 NM NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 2 8 Given: Distance A to B 2050 NM Mean groundspeed 'on' 440 kt Mean groundspeed 'back' 540 kt. The distance to the point of equal time (PET) between A and B is: a) 920NM b) 1025 NM c) 1130 NM 9 Given: Distance A to B 3060 NM Mean groundspeed 'out' 440 kt Mean groundspeed 'back' 540 kt Safe Endurance 10 hours The time to the Point of Safe Return (PSR) is: a) 5 hours 30 minutes b) 5 hours 45 minutes c) 3 hours 55 minutes 10 Given the following: D = flight distance X = distance to Point of Equal Time GSo groundspeed out GSr = groundspeed return The correct formula to find distance to Point of Equal Time is: a) X = (D/2) x GSo / (GSo + GSr) b) X = D x GSo / (GSo + GSr) c) X = D x GSr / (GSo + GSr)) 11 Which of the following statements is (are) correct with regard to the advantages of computer flight plans ?1. The computer can file the ATC flight plan.2. Wind data used by the computer is always more up-to-date than that available to the pilot. a) Statement 2 only b) Statement 1 only c) Both statements 12 Which of the following statements is (are) correct with regard to the operation of flight planning computers ?1. The computer can file the ATC flight plan. 2. In the event of inflight re-routing the computer produces a new plan. a) Both statements b) Statement 2 only c) Statement 1 only 13 Which of the following statements is (are) correct with regard to computer flight plans 1. The computer takes account of bad weather on the route and adds extra fuel. 2. The computer calculates alternate fuel sufficient for a missed approach, climb, cruise, descent and approach and landing at the destination alternate. a) Statement 1 only b) Both statements c) Statement 2 only 14 When calculating the fuel required to carry out a given flight, one must take into account: 1 - the wind 2 - foreseeable airborne delays 3 - other weather forecasts 4 - any foreseeable conditions which may delay landing The combination which provides the correct statement is: a) 03-Jan b) 1 - 2 - 3 – 4 c) 2 – 4 NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 3 15 The angle between the plane of the ecliptic and the plane of equator is approximately: a) 23.5° b) 25.3° c) 27.5° 16 Which is the highest latitude listed below at which the sun will rise above the horizon and set every day? a) 72°z b) 68° c) 66° 17 In which two months of the year is the difference between the transit of the 340 Apparent Sun and Mean Sun across the Greenwich Meridian the greatest? a) March and September b) February and November c) June and December 18 What is the highest latitude listed below at which the sun will reach an altitude of 90° above the horizon at some time during the year? a) 66° b) 45° c) 23° 19 Assuming mid-latitudes (40° to 50°N/S). At which time of year is the relationship between the length of day and night, as well as the rate of change of declination of the sun, changing at the greatest rate? a) Summer solstice and winter solstice b) Summer solstice and spring equinox c) Spring equinox and autumn equinox 20 At what approximate date is the earth closest to the sun (perihelion)? a) Beginning of January b) End of March c) Beginning of July 21 At what approximate date is the earth furthest from the sun (aphelion)? a) End of December b) Beginning of July c) Beginning of January 22 Seasons are due to the: a) inclination of the polar axis with the ecliptic plane b) Earth's elliptical orbit around the Sun c) Earth's rotation on its polar axis 23 An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are? a) 45°00'N 169°22W b) 53°20'N 169°22W NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 4 c) 45°00'N 172°38'E 24 The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is: a) 9° b) 7.8° c) 15.6° 25 Given: Waypoint 1. 60°S 030° W Waypoint 2. 60°S 020° W What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025° W? a) 060°11'S b) 060°06'S c) 059°49'S 26 What is the time required to travel along the parallel of latitude 60° N between meridians 010° E and 030° W at a groundspeed of 480 kt? a) 2 HR 30 MIN b) 1 HR 15 MIN c) 1 HR 45 MIN 27 A Rhumb line is: a) any straight line on a Lambert projection b) the shortest distance between two points on a Polyconic projection c) a line on the surface of the earth cutting all meridians at the same angle 28 A great circle track joins position A (59°S 141°W) and B (61°S 148°W). What is the difference between the great circle track at A and B? a) It increases by 6° b) It decreases by 6° c) It increases by 3° 29 What is the longitude of a position 6 NM to the east of 58°42'N 094°00'W? a) 093°54.0'W b) 093°48.5'W c) 093°53.1'W 30 Given: value for the ellipticity of the Earth is 1/297. Earth's semi-major axis, as measured at the equator, equals 6378.4 km. What is the semi-minor axis (km) of the earth at the axis of the Poles? a) 6 356.9 b) 6 378.4 c) 6 367.0 31 Position A is located on the equator at longitude 130°00E. Position B is located 100 NM from A on a bearing of 225°(T).The coordinates of position B are: a) 01°11'S 131°11'E b) 01°11'N 131°11'E c) 01°11'S 128°49'E NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 5 32 In order to fly from position A (10°00'N, 030°00'W) to position B (30°00'N, 050°00'W), maintaining a constant true course, it is necessary to fly: a) the constant average drift route b) a rhumb line track c) the great-circle route d) a straight line plotted on a Lambert chart 33 The rhumb line track between position A (45°00'N, 010°00'W) and position B (48°30'N, 015°00'W) is approximately: a) 330 b) 315 c) 300 34 The diameter of the Earth is approximately: a) 12 700 km b) 6 350 km c) 18 500 km 35 The maximum difference between geocentric and geodetic latitude occurs at about: a) 45° North and South b) 60° North and South c) 90° North and South 36 The great circle distance between position A (59°34.1'N 008°08.4'E) and B (30°25.9'N 171°51.6'W) is: a) 2 700 NM b) 10 800 km c) 5400 NM 37 Given: Position A 45°N, ?°E Position B 45°N, 45°15'E Distance A-B = 280 NMB is to the East of A Required: longitude of position A? a) 38°39'E b) 49°57'E c) 51°51'E 38 If an airplane was to circle around the Earth following parallel 60°N at a ground speed of 480 kt. In order to circle around the Earth along the equator in the same amount of time, it should fly at a ground speed of: a) 550 kt b) 240 kt c) 960 kt 39 An aircraft passes position A (60°00'N 120°00'W) on route to position B (60°00'N 140°30'W). What is the great circle track on departure from A? a) 261° b) 288° c) 279° NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 6 40 An airplane flies from A (59°S 142°W) to B (61°S 148°W) with a TAS of 480 kt. The autopilot is engaged and coupled with an Inertial Navigation System in which AB track is active. On route AB, the true track: a) increases by 5° b) varies by 10° c) decreases by 6° 41 The circumference of the earth is approximately: a) 21600 NM b) 43200 NM c) 5400 NM 42 The Great Circle bearing of 'B' (70°S 060°E), from 'A' (70° S 030° W), is approximately: a) 150°(T)342 b) 135°(T) c) 090°(T) 43 At what approximate latitude is the length of one minute of arc along a meridian equal to one NM (1852 m) correct? a) 90° b) 0° c) 45° 44 An aircraft flies a great circle track from 56° N 070° W to 62° N 110° E. The total distance travelled is? a) 5420NM b) 3720 NM c) 1788 NM 45 Given: A is N55° 000°B is N54° E010°The average true course of the great circle is 100°. The true course of the rhumb line at point A is: a) 096° b) 100° c) 104° 46 The circumference of the parallel of latitude at 60°N is approximately: a) 10 800 NM b) 18 706 NM c) 20 000 NM 47 Given: The coordinates of the heliport at Issy les Moulineaux are:N48°50' E002°16.5'The coordinates of the antipodes are: a) S41°10' W177°43.5' b) S48°50' E177°43.5' c) S48°50' W177°43.5' NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 7 48 Given: Position 'A' is N00° E100°, Position 'B' is 240°(T), 200 NM from 'A'. What is the position of 'B'? a) S01°40' E101°40' b) N01°40' E097°07' c) S01°40' E097°07' 49 The duration of civil twilight is the time: a) agreed by the international aeronautical authorities which is 12 minutes b) between sunset and when the center of the sun is 6° below the true horizon c) needed by the sun to move from the apparent height of 0° to the apparent height of 6° 50 On the 27th of February, at 52°S and 040°E, the sunrise is at 0243 UTC. On the same day, at 52°S and 035°W, the sunrise is at: a) 2143 UTC b) 0243 UTC c) 0743 UTC 51 What is the local mean time, position 65°25'N 123°45'W at 2200 UTC? a) 2200 b) 1345 c) 615 52 The Local Mean Time at longitude 095°20'W, at 0000 UTC, is : a) 1738:40 previous day b) 0621:20 previous day c) 1738:40 same day 53 5 HR 20 MIN 20 SEC corresponds to a longitude difference of: a) 75°00' b) 81°10' c) 80°05' d) 78°45' 54 The main reason that day and night, throughout the year, have different duration, is due to the: a) earth's rotation b) inclination of the ecliptic to the equator c) relative speed of the sun along the ecliptic 55 What is the meaning of the term ""standard time"" ? a) It is the time set by the legal authorities for a country or part of a country b) It is the time zone system applicable only in the USA c) It is an expression for local mean time 56 Civil twilight is defined by: a) sun altitude is 6° below the celestial horizon b) sun altitude is 12° below the celestial horizon c) sun altitude is 18° below the celestial horizon NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 8 57 Given: true track is 348°, drift 17° left, variation 32° W, deviation 4°E. What is the compass heading? a) 007° b) 033° c) 359° 58 An Agonic line is a line that connects: a) points of equal magnetic dip b) positions that have the same variation c) positions that have 0° variation 59 Isogonic lines connect positions that have: a) the same elevation b) 0° variation c) the same variation 60 Compass deviation is defined as the angle between: a) True North and Magnetic North b) Magnetic North and Compass North c) True North and Compass North 61 The angle between True North and Magnetic North is called: a) variation b) deviation c) compass error 62 Deviation applied to magnetic heading gives: a) magnetic course b) true heading c) compass heading 63 Isogrives are lines that connect positions that have: a) the same variation b) the same horizontal magnetic field strength c) the same grivation 64 The lines on the earth's surface that join points of equal magnetic variation are called: a) isotachs b) isogonals c) isogrives 65 A negative (westerly) magnetic variation signifies that: a) True North is East of Magnetic North 344 b) True North is West of Magnetic North c) Compass North is East of Magnetic North 66 The angle between Magnetic North and Compass North is called: a) compass deviation b) compass error c) magnetic variation NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 9 67 The north and south magnetic poles are the only positions on the earth's surface where: a) isogonals converge b) a freely suspended compass needle will stand vertical c) a freely suspended compass needle will stand horizontal 68 The rhumb-line distance between points A (60°00'N 002°30'E) and B (60°00'N 007°30'W) is: a) 300 NM b) 450 NM c) 600 NM 69 An aircraft flies the following rhumb line tracks and distances from position 04°00'N 030°00'W : 600 NM South, then 600 NM East, then 600 NM North, then 600 NM West. The final position of the aircraft is: a) 04°00'N 030°02'W b) 04°00'N 029°58'W c) 04°00'N 030°00'W 70 What is the final position after the following rhumb line tracks and distances have been followed from position 60°00'N 030°00'W? South for 3600 NM, East for 3600 NM, North for 3600 NM, West for 3600 NM. The final position of the aircraft is: a) 59°00'N 060°00'W b) 60°00'N 030°00'E c) 60°00'N 090°00'W 71 An aircraft departing A(N40º 00´ E080º 00´) flies a constant true track of 270º at a ground speed of 120 kt. What are the coordinates of the position reached in 6 HR? a) N40º 00´ E064º 20´ b) N40º 00´ E070º 30´ c) N40º 00´ E060º 00´ 72 A flight is to be made from 'A' 49°S 180°E/W to 'B' 58°S, 180°E/W. The distance in kilometres from 'A' to 'B' is approximately: a) 540 b) 1222 c) 1000 73 An aircraft at position 60°N 005°W tracks 090°(T) for 315 km. On completion of the flight the longitude will be: a) 000°40'E b) 005°15'E c) 002°10'W 74 The 'departure' between positions 60°N 160°E and 60°N 'x' is 900 NM. What is the longitude of 'x'? a) 140°W b) 170°W c) 145°E NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 10 75 An aircraft at latitude 02°20'N tracks 180°(T) for 685 km. On completion of the flight the latitude will be: a) 03°50'S b) 04°10'S c) 04°30'S 76 An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will its latitude be after 1.5 HR? a) 12°15'N b) 02°00'N c) 22°00'N 77 An aircraft at latitude 10°North flies south at a groundspeed of 445 km/HR. What will be its latitude after 3 HR? a) 12°15'S b) 02°00'S c) 22°00'S 78 Given: Position 'A' N60 W020,Position 'B' N60 W021, Position 'C' N59 W020. What are, respectively, the distances from A to B and from A to C? a) 30 NM and 60 NM b) 52 NM and 60 NM c) 60 NM and 30 NM 79 An aircraft is over position HO (55°30'N 060°15'W), where YYR VOR (53°30'N 060°15'W) can be received. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR? a) 208° b) 028° c) 031° 80 When is the magnetic compass most effective? a) About midway between the magnetic poles b) In the region of the magnetic South Pole. c) In the region of the magnetic North Pole. 81 What is the value of the magnetic dip at the magnetic south pole ? a) 45° b) 90° c) 60° 82 The value of magnetic variation: a) has a maximum of 180° b) must be 0° at the magnetic equator c) varies between a maximum of 45° East and 45° West 83 Isogonals converge at the: a) North and South geographic and magnetic poles b) Magnetic equator c) North magnetic pole only NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 11 84 A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an: a) agonic line b) aclinic line c) isogonal 85 The horizontal component of the earth's magnetic field: a) weakens with increasing distance from the magnetic poles b) is approximately the same at magnetic latitudes 50°N and 50°S c) weakens with increasing distance from the nearer magnetic pole 86 Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually due to: a) magnetic pole movement causing numerical values at all locations to increase b) magnetic pole movement causing numerical values at all locations to increase or decrease. c) a reducing field strength causing numerical values at all locations to decrease. 87 The Earth can be considered as being a magnet with the: a) blue pole near the north pole of the earth and the direction of the magnetic force pointing straight up from the earth's surface b) red pole near the north pole of the earth and the direction of the magnetic force pointing straight down to the earth's surface c) blue pole near the north pole of the earth and the direction of the magnetic force pointing straight down to the earth's surface 88 Which of the following statements concerning earth magnetism is completely correct? a) dip An isogonal is a line which connects places with the same magnetic variation, the agonic line is the line of zero magnetic dip b) An isogonal is a line which connects places with the same magnetic variation, the aclinic is the line of zero magnetic c) An isogonal is a line which connects places of equal dip, the aclinic is the line of zero magnetic dip 89 Which of the following statements concerning the earth's magnetic field is completely correct? a) The earth's magnetic field can be classified as transient, semi-permanent or permanent b) At the earth's magnetic equator, the inclination varies depending on whether the geographic equator is north or south of the magnetic equator c) The blue pole of the earth's magnetic field is situated in North Canada 90 The sensitivity of a direct reading compass varies: a) directly with the horizontal component of the earth's magnetic field b) directly with the vertical component of the earth's magnetic field c) inversely with both vertical and horizontal components of the earth's magnetic field NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 12 91 Isogonals are lines of equal: a) pressure. b) compass deviation. c) magnetic variation. 92 At a specific location, the value of magnetic variation: a) depends on the magnetic heading b) depends on the type of compass installed c) varies slowly over time 93 When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compass to: a) lag behind the turning rate of the aircraft b) indicate a turn towards the north c) indicate a turn towards the south 94 When decelerating on a westerly heading in the Northern hemisphere, the compass card of a direct reading magnetic compass will turn: a) anti-clockwise giving an apparent turn towards the south b) clockwise giving an apparent turn toward the south c) clockwise giving an apparent turn towards the north 95 An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If the initial heading was 330°, after 30 seconds of the turn the direct reading magnetic compass should read: a) more than 060° b) 060° c) less than 060° 96 When turning right from 330°(C) to 040°(C) in the northern hemisphere, the reading of a direct reading magnetic compass will: a) under-indicate the turn and liquid swirl will increase the effect b) over-indicate the turn and liquid swirl will decrease the effect c) under-indicate the turn and liquid swirl will decrease the effect 97 When accelerating on an easterly heading in the Northern hemisphere, the compass card of a direct reading magnetic compass will turn: a) clockwise giving an apparent turn toward the south b) clockwise giving an apparent turn toward the north c) anti-clockwise giving an apparent turn toward the north 98 An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135°, after 30 seconds the direct reading magnetic compass should read: a) less than 225° b) 225° c) more than 225° NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 13 99 When accelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn: a) anti-clockwise giving an apparent turn towards the north b) anti-clockwise giving an apparent turn towards the south c) clockwise giving an apparent turn towards the north 100 Which of the following statements is correct concerning the effect of turning errors on a direct reading compass? a) Turning errors are greatest on north/south headings, and are greatest at high latitudes b) Turning errors are greatest on east/west headings, and are least at high latitudes c) Turning errors are greatest on north/south headings, and are least at high latitudes 101 At the magnetic equator, when accelerating after take off on heading West, a DRC: a) indicates the correct heading b) overreads the heading c) underreads the heading 102 Permanent magnetism in aircraft arises chiefly from: a) exposure to the earth's magnetic field during normal operation b) hammering, and the effect of the earth's magnetic field, whilst under construction c) the combined effect of aircraft electrical equipment and the earth's magnetic field 103 Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that: a) on a Westerly heading, a longitudinal acceleration causes an apparent turn to the South b) on an Easterly heading, a longitudinal acceleration causes an apparent turn to the South c) on an Easterly heading, a longitudinal acceleration causes an apparent turn to the North 104 In northern hemisphere, during an acceleration in an easterly direction, the magnetic compass will indicate: a) an increase in heading b) a decrease in heading c) an apparent turn to the South 105 The purpose of compass check swing is to: a) cancel out the vertical component of the earth's magnetic field b) measure the angle between Magnetic North and Compass North c) cancel out the horizontal component of the earth's magnetic field 106 In a remote indicating compass system the amount of deviation caused by aircraft magnetism and electrical circuits may be minimised by: a) positioning the master unit in the center of the aircraft b) the use of repeater cards c) mounting the detector unit in the wingtip NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 14 107 A direct reading compass should be swung when: a) there is a large change in magnetic longitude b) there is a large, and permanent, change in magnetic latitude c) the aircraft is stored for a long period and is frequently moved 108 The direct reading magnetic compass is made aperiodic (dead beat) by: a) using long magnets b) using the lowest acceptable viscosity compass liquid c) keeping the magnetic assembly mass close to the compass point and by using damping wires 109 The annunciator of a remote indicating compass system is used when: a) setting local magnetic variation b) compensating for deviation c) synchronising the magnetic and gyro compass elements 110 Which one of the following is an advantage of a remote reading compass as compared with a standby compass? a) It is lighter than a direct reading compass because it employs, apart from the detector unit, existing aircraft equipment b) It senses the magnetic meridian instead of seeking it, increasing compass sensitivity c) It eliminates the effect of turning and acceleration errors by pendulously suspending the detector unit 111 Which of the following is an occasion for carrying out a compass swing on a Direct Reading Compass? a) Before an aircraft goes on any flight that involves a large change of magnetic latitude b) After an aircraft has passed through a severe electrical storm, or has been struck by lightning c) After any of the aircraft radio equipment has been changed due to unserviceability 112 The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane is: a) to minimise the amount of deviation caused by aircraft magnetism and electrical circuits b) to maximise the units exposure to the earth's magnetic field c) to ensure that the unit is in the most accessible position on the aircraft for ease of maintenance 113 The main reason for usually mounting the detector unit of a remote indicating compass in the wingtip of an aeroplane is to: a) facilitate easy maintenance of the unit and increase its exposure to the Earth's magnetic field b) reduce the amount of deviation caused by aircraft magnetism and electrical circuits c) place it in a position where there is no electrical wiring to cause deviation errors NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 15 114 The main advantage of a remote indicating compass over a direct reading compass 348 is that it: a) has less moving parts b) senses, rather than seeks, the magnetic meridian c) requires less maintenance 115 A chart has the scale 1 : 1 000 000. From A to B on the chart measures 1.5 inches (one inch equals 2.54 centimetres), the distance from A to B in NM is : a) 44.5 b) 38.1 c) 20.6 116 The nominal scale of a Lambert conformal conic chart is the: a) mean scale between pole and equator b) scale at the standard parallels c) mean scale between the parallels of the secant cone 117 The chart that is generally used for navigation in polar areas is based on a: a) Stereographical projection b) Direct Mercator projection c) Gnomonic projection 118 A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60° S? a) 1 : 7 408 000 b) 1 : 1 852 000 c) 1 : 3 208 000 119 The distance measured between two points on a navigation map is 42 mm (millimeters). The scale of the chart is 1:1 600 000. The actual distance between these two point is approximately: a) 36.30 NM b) 370.00 NM c) 67.20 NM 120 The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is: a) 0.39 b) 0.60 c) 0.92 121 On a Lambert conformal conic chart the convergence of the meridians: a) is zero throughout the chart b) is the same as earth convergency at the parallel of origin c) varies as the secant of the latitude 122 A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is: a) 1 : 3 000 000 b) 1 : 6 000 000 NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 16 c) 1 : 5 000 000 123 On a direct Mercator projection, at latitude 45° North, a certain length represents 70 NM. At latitude 30° North, the same length represents approximately: a) 86 NM b) 57 NM c) 70 NM 124 On a direct Mercator projection, the distance measured between two meridians spaced 5° apart at latitude 60°N is 8 cm. The scale of this chart at latitude 60°N is approximately: a) 1 : 7 000 000 b) 1 : 4 750 000 c) 1 : 3 500 000 125 On a Mercator chart, the scale: a) varies as 1/cosine of latitude (1/cosine= secant) b) varies as the sine of the latitude c) is constant throughout the chart 126 In a navigation chart a distance of 49 NM is equal to 7 cm. The scale of the chart is approximately: a) 1 : 1 300 000 b) 1 : 700 000 c) 1 : 130 000 127 At 60° N the scale of a direct Mercator chart is 1 : 3 000 000. What is the scale at the equator? a) 1 : 3 000 000 b) 1 : 6 000 000 c) 1 : 3 500 000 128 What is the chart distance between longitudes 179°E and 175°W on a direct Mercator chart with a scale of 1 : 5 000 000 at the equator? a) 133 mm b) 106 mm c) 167 mm 129 The total length of the 53°N parallel of latitude on a direct Mercator chart is 133cm. What is the approximate scale of the chart at latitude 30°S? a) 1 : 25 000 000 b) 1 : 30 000 000 c) 1 : 18 000 000 130 A Lambert conformal conic projection, with two standard parallels: a) the scale is only correct at parallel of origin b) shows all great circles as straight lines c) the scale is only correct along the standard parallels NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 17 131 The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010°E and 030°W is 30°, is: a) 0.75 b) 0.40 c) 0.50 132 The constant of cone of a Lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctly represented? a) 68°25' b) 66°42' c) 23°18' 133 On a Lambert Conformal chart the distance between meridians 5° apart along latitude 37° North is 9 cm. The scale of the chart at that parallel approximates: a) 1 : 5 000 000 b) 1 : 3 750 000 c) 1 : 2 000 000 134 The chart distance between meridians 10° apart at latitude 65° North is 3.75 inches. The chart scale at this latitude approximates: a) 1 : 6 000 000 b) 1 : 5 000 000 c) 1 : 2 500 000 135 On a Lambert conformal conic chart, with two standard parallels, the quoted scale is correct: a) along the parallel of origin b) in the area between the standard parallels c) along the two standard parallels 136 The convergence factor of a Lambert conformal conic chart is quoted as 0.78535. At what latitude on the chart is earth convergency correctly represented? a) 52°05' b) 51°45' c) 80°39' 137 At 47° North the chart distance between meridians 10° apart is 5 inches. The scale of the chart at 47° North approximates: a) 1 : 8 000 000 b) 1 : 6 000 000 c) 1 : 3 000 000 138 On a Lambert Conformal Conic chart earth convergency is most accurately represented at the: a) parallel of origin b) north and south limits of the chart c) standard parallels 139 On a Transverse Mercator chart, scale is exactly correct along the: a) meridian of tangency b) Equator, parallel of origin and prime vertical NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 18 c) datum meridian and meridian perpendicular to it 140 Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2 000 000? a) 130 b) 150 c) 329 141 Transverse Mercator projections are used for: a) radio navigation charts in equatorial areas b) maps of large east/west extent in equatorial areas c) maps of large north/south extent 142 On a Direct Mercator chart at latitude 15°S, a certain length represents a distance of 120 NM on the earth. The same length on the chart will represent on the earth, at latitude 10°N, a distance of : a) 124.2 NM b) 117.7 NM c) 122.3 NM 143 On a Direct Mercator chart at latitude of 45°N, a certain length represents a distance of 90 NM on the earth. The same length on the chart will represent on the earth, at latitude 30°N, a distance of : a) 110 NM b) 73.5 NM c) 78 NM 144 On a transverse Mercator chart, the scale is exactly correct along the: a) meridians of tangency b) equator and parallel of origin c) meridian of tangency and the parallel of latitude perpendicular to it 145 On a transverse Mercator chart, with the exception of the Equator, parallels of latitude appear as a) hyperbolic lines b) straight lines c) ellipses 146 An Oblique Mercator projection is used specifically to produce: a) radio navigational charts in equatorial regions b) charts of the great circle route between two points c) topographical maps of large east/ west extent 147 The two standard parallels of a conical Lambert projection are at N10°40'N and N41°20'. The cone constant of this chart is approximatively: a) 0.66 b) 0.90 c) 0.44 NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 19 148 On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of the chart is approximately: a) 1 : 185 000 b) 1 : 1 000 000 c) 1 : 1 850 000 149 Given: Chart scale is 1 : 1 850 000. The chart distance between two points is 4 centimetres. Earth distance is approximately : a) 74 NM b) 40 NM c) 100 NM 150 On a Mercator chart, at latitude 60°N, the distance measured between W002° and E008° is 20 cm. The scale of this chart at latitude 60°N is approximately: a) 1 : 5 560 000 b) 1 : 278 000 c) 1 : 2 780 000 151 At latitude 60°N the scale of a Mercator projection is 1 : 5 000 000. The length on the chart between 'C' N60° E008° and 'D' N60° W008° is: a) 16.2 cm b) 17.8 cm c) 35.6 cm 152 Assume a Mercator chart. The distance between positions A and B, located on the same parallel and 10° longitude apart, is 6 cm. The scale at the parallel is 1 : 9 260 000. What is the latitude of A and B? a) 30° N or S351 b) 60° N or S c) 0° 153 A straight line on a chart 4.89 cm long represents 185 NM. The scale of this chart is approximately : a) 1 : 7 000 000 b) 1 : 3 500 000 c) 1 : 6 000 000 154 The scale on a Lambert conformal conic chart: a) is constant along a meridian of longitude b) is constant along a parallel of latitude c) is constant across the whole map 155 A direct Mercator graticule is based on a projection that is : a) cylindrical b) conical c) spherical NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 20 156 Parallels of latitude on a Direct Mercator chart are: a) arcs of concentric circles equally spaced b) parallel straight lines equally spaced c) parallel straight lines unequally spaced 157 A straight line on a Lambert Conformal Projection chart for normal flight planning purposes: a) is approximately a Great Circle b) is a Loxodromic line c) is a Rhumb line 158 On a Direct Mercator chart, a rhumb line appears as a: a) spiral curve b) small circle concave to the nearer pole c) straight line 159 On a Lambert Conformal Conic chart great circles that are not meridians are: a) straight lines regardless of distance b) curves concave to the parallel of origin c) curves concave to the pole of projection 160 On a Direct Mercator chart a great circle will be represented by a: a) curve concave to the equator b) complex curve c) curve convex to the equator 161 The angular difference, on a Lambert conformal conic chart, between the arrival and departure track is equal to: a) earth convergence b) map convergence c) conversion angle 162 The parallels on a Lambert Conformal Conic chart are represented by: a) straight lines b) arcs of concentric circles c) parabolic lines 163 Parallels of latitude, except the equator, are: a) both Rhumb lines and Great circles b) Great circles c) Rhumb lines 164 On a Lambert chart (standard parallels 37°N and 65°N), with respect to the straight line drawn on the map between A ( N49° W030°) and B (N48° W040°), the: a) great circle and rhumb line are to the north b) great circle and rhumb line are to the south c) great circle is to the north, the rhumb line is to the south NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 21 165 On a Direct Mercator chart, meridians are: a) parallel, equally spaced, vertical straight lines b) inclined, equally spaced, straight lines that meet at the nearer pole c) parallel, unequally spaced, vertical straight lines 166 On which of the following chart projections is it NOT possible to represent the north or south poles? a) Transverse Mercator b) Lambert's conformal c) Direct Mercator 167 Which one of the following, concerning great circles on a Direct Mercator chart, is correct? a) They approximate to straight lines between the standard parallels b) They are all curves concave to the equator c) With the exception of meridians and the equator, they are curves concave to the equator 168 On a Lambert conformal conic chart, the distance between parallels of latitude spaced the same number of degrees apart: a) is constant between, and expands outside, the standard parallels b) reduces between, and expands outside, the standard parallels c) expands between, and reduces outside, the standard parallels 169 Which one of the following statements is correct concerning the appearance of great circles, with the exception of meridians, on a Polar Stereographic chart whose tangency is at the pole? a) The higher the latitude the closer they approximate to a straight line b) Any straight line is a great circle c) They are complex curves that can be convex and/or concave to the Pole 170 Which one of the following describes the appearance of rhumb lines, except meridians, on a Polar Stereographic chart? a) Curves convex to the Pole b) Ellipses around the Pole c) Curves concave to the Pole 171 What is the value of the convergence factor on a Polar Stereographic chart? a) 0.5 b) 0.866 c) 1.0 172 On a Direct Mercator, rhumb lines are: a) curves concave to the equator b) straight lines c) ellipses NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 22 173 Contour lines on aeronautical maps and charts connect points: a) having the same elevation above sea level b) with the same variation c) having the same longitude 174 On a Polar Stereographic chart, the initial great circle course from A 70°N 060°W to B 70°N 060°E is approximately: a) 030° (T) b) 330° (T) c) 150° (T) 175 On a polar stereographic projection chart showing the South Pole, a straight line joins position A (70°S 065°E) to position B (70°S 025°W).The true course on departure from position A is approximately: a) 250° b) 225° c) 135° 176 Two positions plotted on a polar stereographic chart, A (80°N 000°) and B (70°N 102°W) are joined by a straight line whose highest latitude is reached at 035°W.At point B, the true course is: a) 023° b) 203° c) 247° 177 Given: Magnetic heading 311° Drift angle 10° left Relative bearing of NDB 270° What is the magnetic bearing of the NDB measured from the aircraft? a) 211° b) 208° c) 221° 178 A Lambert conformal conic chart has a constant of the cone of 0.75. The initial course of a straight line track drawn on this chart from A (40°N 050°W) to B is 043°(T) at A, course at B is 055°(T). What is the longitude of B? a) 34°W353 b) 36°W c) 38°W 179 A Lambert conformal conic chart has a constant of the cone of 0.80. A straight line course drawn on this chart from A (53°N 004°W) to B is 080° at A, course at B is 092°(T). What is the longitude of B? a) 009°E b) 009°36'E c) 011°E 180 At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are: a) 090° - 232 kt b) 085° - 226 kt c) 080° - 226 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 23 181 Given: An aircraft is flying a track of 255°(M),2254 UTC, it crosses radial 360° from a VOR station, 2300 UTC, it crosses radial 330° from the same station. At 2300 UTC, the distance between the aircraft and the station is: a) the same as it was at 2254 UTC b) greater than it was at 2254 UTC c) randomly different than it was at 2254 UTC 182 A course of 120°(T) is drawn between 'X' (61°30'N) and 'Y' (58°30'N) on a Lambert Conformal conic chart with a scale of 1 : 1 000 000 at 60°N. The chart distance between 'X' and 'Y' is: a) 66.7 cm b) 33.4 cm c) 38.5 cm 183 Route 'A' (44°N 026°E) to 'B' (46°N 024°E) forms an angle of 35° with longitude 026°E. Average magnetic variation between 'A' and 'B' is 3°E. What is the average magnetic course from 'A' to 'B'? a) 328° b) 322° c) 032° 184 Given: Direct Mercator chart with a scale of 1 : 200 000 at equator, Chart length from 'A' to 'B', in the vicinity of the equator, 11 cm. What is the approximate distance from 'A' to 'B'? a) 21 NM b) 12 NM c) 22 NM d) 14 NM 185 Given the following: Magnetic heading: 060° Magnetic variation: 8°W Drift angle: 4° right What is the true track? a) 048° b) 064° c) 056° 186 Given: True track 180° Drift 8°R Compass heading 195° Deviation -2° Calculate the variation? a) 21°W b) 25°W c) 5°W 187 Given: True course 300° drift 8° R variation 10° W deviation -4° Calculate the compass heading? a) 294° b) 322° c) 306° NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 24 188 Given: True course A to B = 250° Distance A to B = 315 NMTAS = 450 kt. W/V = 200°/60kt. ETD A = 0650 UTC. What is the ETA at B? a) 0730 UTC b) 0736 UTC c) 0810 UTC 189 Given: GS = 510 kt. Distance A to B = 43 NM What is the time (MIN) from A to B? a) 6 b) 4 c) 5 d) 7 190 The ICAO definition of ETA is the: a) estimated time of arrival at destination b) actual time of arrival at a point or fix c) estimated time of arrival at an en-route point or fix 191 Given: Required course 045°(M),Variation is 15°E, W/V is 190°(T)/30 kt, CAS is 120 kt at FL 55 in standard atmosphere. What are the heading (°M) and GS? a) 036° and 151 kt b) 055° and 147 kt c) 052° and 154 kt 192 Given: Course 040°(T), TAS is 120 kt, Wind speed 30 kt. Maximum drift angle will be obtained for a wind direction of: a) 130° b) 145° c) 115° 193 How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt? a) 2.36 b) 3.94 c) 3.25 194 Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance? a) 3 HR 53 MIN b) 3 HR 12 MIN c) 4 HR 32 MIN 195 What is the ratio between the litre and the US-GAL? a) 1 US-GAL equals 4.55 litres b) 1 litre equals 3.78 US-GAL c) 1 US-GAL equals 3.78 litres 196 265 US-GAL equals? (Specific gravity 0.80) a) 862 kg b) 803 kg c) 895 kg NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 25 197 730 FT/MIN equals: a) 3.7 m/sec b) 5.2 m/sec c) 1.6 m/sec 198 How long will it take to fly 5 NM at a groundspeed of 269 Kt ? a) 2 MIN 30 SEC b) 1 MIN 55 SEC c) 1 MIN 07 SEC 199 An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed? a) 209 kt b) 183 kt c) 160 kt 200 An aircraft travels 100 statute miles in 20 MIN, how long does it take to travel 215 NM? a) 100 MIN b) 50 MIN c) 90 MIN 201 The equivalent of 70 m/sec is approximately: a) 136 kt b) 145 kt c) 210 kt 202 Given: IAS 120 kt, FL 80, OAT +20°C. What is the TAS? a) 102 kt b) 141 kt c) 120 kt 203 An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30 kt. The GS and drift angle are: a) 192 kt, 7° left b) 200 kt, 3.5° right c) 192 kt, 7° right 204 Given: Magnetic heading = 255°VAR = 40°WGS = 375 ktW/V = 235°(T) / 120 kt Calculate the drift angle? a) 9° left b) 7° right c) 7° left 205 Given: True Heading = 180° TAS = 500 ktW/V 225° / 100 kt Calculate the GS? a) 600 kt b) 435 kt c) 535 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 26 206 Given: True heading = 310° TAS = 200 kt GS = 176 kt Drift angle 7° right. Calculate the W/V? a) 270° / 33 kt b) 360° / 33 kt364 c) 090° / 33 kt 207 Given: True Heading = 090° TAS = 180 kt GS = 180 kt Drift 5° right Calculate the W/ V? a) 010° / 15 kt b) 190° / 15 kt c) 360° / 15 kt 208 Given: M 0.80, OAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right. Calculate the true W/V? a) 020°/95 kt b) 025°/47 kt c) 200°/95 kt 209 Given: Compass Heading 090°, Deviation 2°W, Variation 12°E, TAS 160 kt. Whilst maintaining a radial 070° from a VOR station, the aircraft flies a ground distance of 14 NM in 6 MIN. What is the W/V °(T)? a) 340°/98 kt b) 340°/25 kt c) 160°/50 kt 210 An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045° / 50kt.How far can the aeroplane fly out from its base and return in one hour? a) 85 NM b) 88 NM c) 56 NM 211 The following information is displayed on an Inertial Navigation System: GS 520 kt, True HDG 090°, Drift angle 5° right, TAS 480 kt. SAT (static air temperature) -51°C.The W/V being experienced is: a) 320° / 60 kt b) 225° / 60 kt c) 220° / 60 kt 212 The reported surface wind from the Control Tower is 240°/35 kt. Runway 30 (300°). What is the cross-wind component? a) 24 kt b) 30 kt c) 27 kt 213 Given: TAS = 190 kt, HDG (T) = 355°,W/V = 165/25kt. Calculate the drift and GS? a) 1R - 175 kt b) 1L - 225 kt c) 1L - 215 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 27 214 Given: TAS = 250 kt, HDG (T) = 029°,W/V = 035/45kt.Calculate the drift and GS? a) 1R - 205 kt b) 1L - 205 kt c) 1L - 265 kt 215 Given: TAS = 132 kt, HDG (T) = 053°,W/V = 205/15kt.Calculate the Track (°T) and GS? a) 050 - 145 kt b) 057 - 144 kt c) 052 - 143 kt 216 For a landing on runway 23 (227° magnetic) surface W/V reported by the ATIS is 180/30kt. VAR is 13°E. Calculate the cross wind component? a) 15 kt b) 26 kt c) 22 kt 217 Given: Maximum allowable tailwind component for landing 10 kt. Planned runway 05 (047° magnetic). The direction of the surface wind reported by ATIS 210°. Variation is 17°E. Calculate the maximum allowable windspeed that can be accepted without exceeding the tailwind limit? a) 8 kt b) 18 kt c) 11 kt 218 Given: Maximum allowable crosswind component is 20 kt. Runway 06, RWY QDM 063°(M). Wind direction 100°(M)Calculate the maximum allowable windspeed? a) 33 kt b) 31 kt c) 26 kt 219 Given: TAS = 220 kt, Magnetic course = 212 º, W/V 160 º(M)/ 50kt, Calculate the GS? a) 290 kt b) 186 kt c) 246 kt 220 Given: Magnetic track = 315 º, HDG = 301 º(M),VAR = 5ºW, TAS = 225 kt, The aircraft flies 50 NM in 12 MIN. Calculate the W/V(°T)? a) 190 º/63 kt b) 355 º/15 kt c) 195 º/61 kt 221 Given: TAS = 225 kt, HDG (°T) = 123°, W/V = 090/60kt. Calculate the Track (°T) and GS? a) 120 - 190 kt b) 134 - 188 kt c) 134 - 178 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 28 222 Given: TAS = 227 kt, Track (T) = 316°, W/V = 205/15kt. Calculate the HDG (°T) and GS? a) 311 - 230 kt b) 312 - 232 kt c) 313 - 235 kt 223 Given: True HDG = 054°, TAS = 450 kt, Track (T) = 059°, GS = 416 kt. Calculate the W/V? a) 010/50kt b) 005/50kt c) 010/55kt 224 Given: True HDG = 035°, TAS = 245 kt, Track (T) = 046°, GS = 220 kt. Calculate the W/V? a) 335/45kt b) 340/50kt c) 335/55kt 225 Given: course required = 085° (T),Forecast W/V 030/100kt, TAS = 470 kt, Distance = 265 NM. Calculate the true HDG and flight time? a) 076°, 34 MIN b) 075°, 39 MIN c) 096°, 29 MIN 226 Given: Runway direction 083°(M),Surface W/V 035/35kt. Calculate the effective headwind component? a) 31 kt b) 27 kt c) 24 kt 227 Given: For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-wind limitation of 35 kt. The angle between the wind direction and the runway is 60°, Calculate the minimum and maximum allowable wind speeds? a) 12 kt and 38 kt b) 20 kt and 40 kt c) 15 kt and 43 kt 228 Given: Runway direction 230°(T), Surface W/V 280°(T)/40 kt. Calculate the effective cross-wind component? a) 31 kt b) 36 kt c) 21 kt 229 Given: Runway direction 210°(M), Surface W/V 230°(M)/30kt. Calculate the crosswind component? a) 19 kt b) 10 kt c) 16 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 29 230 Given: Runway direction 305°(M), Surface W/V 260°(M)/30 kt. Calculate the crosswind component? a) 27 kt b) 24 kt c) 21 kt 231 Given: Magnetic track = 075°, HDG = 066°(M), VAR = 11°E, TAS = 275 kt Aircraft flies 48 NM in 10 MIN. Calculate the true W/V °? a) 210°/15 kt b) 320°/50 kt c) 340°/45 kt 232 Given: Magnetic track = 210°, Magnetic HDG = 215°, VAR = 15°E, TAS = 360 kt, Aircraft flies 64 NM in 12 MIN. Calculate the true W/V? a) 265°/50 kt b) 195°/50 kt c) 235°/50 kt 233 Given: An aircraft is on final approach to runway 32R (322°), The wind velocity reported by the tower is 350°/20 kt., TAS on approach is 95 kt. In order to maintain the centre line, the aircraft's heading (°M) should be: a) 316° b) 322° c) 328° 234 Given: FL 120, OAT is ISA standard, CAS is 200 kt, Track is 222°(M),Heading is 215° (M), Variation is 15°W. Time to fly 105 NM is 21 MIN. What is the W/V? a) 050°(T) / 70 kt. b) 040°(T) / 105 kt. c) 055°(T) / 105 kt. 235 A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position is to maintain visual contact with the ground and: a) fly expanding circles until a pinpoint is obtained b) fly the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained c) set heading towards a line feature such as a coastline, motorway, river or railway 236 An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain, its height is approximately: a) 680 FT b) 2210 FT c) 1890 FT d) 3640 FT 237 Given: FL 350, Mach 0.80, OAT -55°C. Calculate the values for TAS and local speed of sound (LSS)? a) 490 kt, LSS 461 kt b) 237 kt, LSS 296 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 30 c) 461 kt , LSS 576 kt 238 Given: Pressure Altitude 29000 FT, OAT -55°C. Calculate the Density Altitude? a) 33500 FT b) 31000 FT c) 27500 FT 239 Given: TAS = 485 kt, OAT = ISA +10°C, FL 410. Calculate the Mach Number? a) 0.90 b) 0.825 c) 0.85 240 What is the ISA temperature value at FL 330? a) -56°C b) -50°C c) -66°C 241 Given: TAS 487kt, FL 330, Temperature ISA + 15. Calculate the MACH Number? a) 0.76 b) 0.84 c) 0.81 242 Given: FL250, OAT -15 ºC, TAS 250 kt. Calculate the Mach No.? a) 0.42 b) 0.40 c) 0.44 243 Given: Airport elevation is 1000 ft. QNH is 988 hPa. What is the approximate airport pressure altitude? (Assume 1 hPa = 27 FT) a) 680 b) 320 FT c) 1680 FT 244 Given: True altitude 9000 FT, OAT -32°C, CAS 200 kt. What is the TAS? a) 200 kt b) 215 kt c) 220 kt 245 Given: Aircraft at FL 150 overhead an airport Elevation of airport 720 FT. QNH is 1003 hPa.OAT at FL150 -5°C. What is the true altitude of the aircraft?(Assume 1 hPa = 27 FT) a) 15 840 FT b) 15 280 FT c) 14 160 FT 246 An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 FT, QFE = 963 hPa, temperature = 32°C).Five minutes later, passing 5 000 FT on QFE, the second altimeter set on 1 013 hPa will indicate approximately: a) 6 800 FT b) 6 400 FT NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 31 c) 6 000 FT 247 Given: A polar stereographic chart whose grid is aligned with the zero meridian. Grid track 344°, Longitude 115°00'W, Calculate the true course? a) 099° b) 229° c) 279° 248 For a distance of 1860 NM between Q and R, a ground speed ""out"" of 385 kt, a ground speed ""back"" of 465 kt and an endurance of 8 HR (excluding reserves) the distance from Q to the point of safe return (PSR) is: a) 930 NM b) 1532 NM c) 1685 NM 249 Two points A and B are 1000 NM apart. TAS = 490 kt. On the flight between A and B the equivalent headwind is -20 kt. On the return leg between B and A, the equivalent headwind is +40 kt. What distance from A, along the route A to B, is the the Point of Equal Time (PET)? a) 455 NM b) 470 NM c) 530 NM 250 Given: AD = Air distance GD = Ground distance TAS = True Airspeed GS = Groundspeed Which of the following is the correct formula to calculate ground distance (GD) gone? a) GD = (AD X GS)/TAS b) GD = (AD - TAS)/TAS c) GD = AD X (GS -TAS)/GS 251 An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B' 2900 NM True airspeed 470 kt Mean wind component 'out' +55 kt Mean wind component 'back' -75 kt The ETA for reaching the Point of Equal Time (PET) between 'A' and 'B' is: a) 1846 b) 1744 c) 1657 252 An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B' 2900 NM True airspeed 470 kt Mean wind component 'out' +55 kt Mean wind component 'back' -75 kt Safe endurance 9 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is: a) 1759 NM b) 1611 NM c) 2141 NM 253 Given: Distance 'A' to 'B' 2484 NM Groundspeed 'out' 420 kt Groundspeed 'back' 500 kt The time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: a) 173 MIN b) 163 MIN NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 32 c) 193 MIN 254 Given: Distance 'A' to 'B' 2484 NM Mean groundspeed 'out' 420 kt Mean groundspeed 'back' 500 kt Safe endurance 08 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is: a) 1736 NM b) 1908 NM c) 1940 NM 255 An aircraft was over 'Q' at 1320 hours flying direct to 'R'. Given: Distance 'Q' to 'R' 3016 NM True airspeed 480 kt Mean wind component 'out' -90 kt Mean wind component 'back' +75 kt The ETA for reaching the Point of Equal Time (PET) between 'Q' and 'R' is: a) 1756 b) 1752 c) 1820 256 An aircraft was over 'Q' at 1320 hours flying direct to 'R'. Given: Distance 'Q' to 'R' 3016 NM True airspeed 480 kt Mean wind component 'out' -90 kt Mean wind component 'back' +75 kt Safe endurance 10:00 HR The distance from 'Q' to the Point of Safe Return (PSR) 'Q' is: a) 2370 NM b) 2290 NM c) 1310 NM 257 Given: Distance 'A' to 'B' 1973 NM Groundspeed 'out' 430 kt Groundspeed 'back' 385 kt The time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: a) 130 MIN b) 145 MIN c) 162 MIN 258 Given: Distance 'A' to 'B' 1973 NM Groundspeed 'out' 430 kt Groundspeed 'back' 385 kt Safe endurance 7 HR 20 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is: a) 1664 NM b) 1490 NM c) 1698 NM 259 Given: Distance 'A' to 'B' 2346 NM Groundspeed 'out' 365 kt Groundspeed 'back' 480 kt The time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: a) 197 MIN b) 290 MIN c) 219 MIN 260 An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a track of 090°(T), W/V 130°/ 20 kt, TAS 100 kt. In order to return to the point of departure before sunset, the furthest distance which may be travelled is: a) 115 NM b) 97 NM c) 105 NM NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 33 261 From the departure point, the distance to the point of equal time is : a) inversely proportional to the sum of ground speed out and ground speed back b) proportional to the sum of ground speed out and ground speed back c) inversely proportional to the total distance to go 262 Given: Distance A to B is 360 NM. Wind component A - B is -15 kt, Wind component B - A is +15 kt, TAS is 180 kt. What is the distance from the equal-time-point to B? a) 180 NM b) 195 NM c) 165 NM 263 A ground feature appears 30° to the left of the center line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the feature is: a) 160° b) 220° c) 310° 264 During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft: a) track b) position c) groundspeed 265 An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation 12°W? a) 318° b) 054° c) 234° 266 A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°(M), variation 25° W, drift 10° Right and GS 360 kt. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was: a) 30 NM and 240° b) 40 NM and 110° c) 40 NM and 290° 267 An island is observed by weather radar to be 15° to the left. The aircraft heading is 120°(M) and the magnetic variation 17°W. What is the true bearing of the aircraft from the island? a) 302° b) 268° c) 088° NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 34 268 A ground feature was observed on a relative bearing of 315° and 3 MIN later on a relative bearing of 270°. The W/V is calm, aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature? a) 3 NM b) 12 NM c) 9 NM 269 An island is observed to be 15° to the left. The aircraft heading is 120°(M), variation 17°(W).The bearing °(T) from the aircraft to the island is: a) 122 b) 88 c) 268 270 An island appears 60° to the left of the center line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 276° with the magnetic variation (VAR) 10°E? a) 226° b) 086° c) 046° 271 An island appears 45° to the right of the center line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 215° with the magnetic variation (VAR) 21°W? a) 239° b) 101° c) 059° 272 An island appears 30° to the right of the center line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 355° with the magnetic variation (VAR) 15° E? a) 220° b) 130° c) 160° 273 An island appears 30° to the left of the center line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 020° with the magnetic variation (VAR) 25° W? a) 195° b) 145° c) 205° NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 35 274 An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt. The rate of descent of the aircraft is approximately: a) 650 FT/MIN b) 6500 FT/MIN c) 4500 FT/MIN 275 Assuming zero wind, what distance will be covered by an aircraft descending 15000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN? a) 38.4 NM b) 19.2 NM c) 26.7 NM 276 An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facility at FL130. If the mean GS for the descent is 288 kt, the minimum rate of descent required is: a) 960 FT/MIN b) 860 FT/MIN c) 890 FT/MIN 277 An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 FT/MIN and mean GS for descent is 276 kt. The minimum range from the DME at which descent should start is: a) 49 NM b) 79 NM c) 69 NM 278 An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kt, the minimum rate of descent required is: a) 1390 FT/MIN b) 1340 FT/MIN c) 1240 FT/MIN 279 An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2500 FT/MIN, mean GS during descent is 248 kt. What is the minimum range from the DME at which descent should commence? a) 53 NM b) 58 NM c) 63 NM 280 An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL120. If the mean GS during the descent is 396 kt, the minimum rate of descent required is approximately: a) 1000 FT/MIN b) 2400 FT/MIN c) 1650 FT/MIN NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 36 281 At 0422 an aircraft at FL370, GS 320kt, is on the direct track to VOR 'X' 185 NM distant. The aircraft is required to cross VOR 'X' at FL80. For a mean rate of descent of 1800 FT/MIN at a mean GS of 232 kt, the latest time at which to commence descent is: a) 445 b) 448 c) 451 282 An aircraft at FL330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required? a) 1650 FT/MIN b) 1950 FT/MIN c) 1750 FT/MIN 283 An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271kt. What is the minimum rate of descent required? a) 2000 FT/MIN b) 1900 FT/MIN c) 1700 FT/MIN 284 What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS? a) Mach number increases, TAS increases b) Mach number remains constant, TAS increases c) Mach number decreases, TAS decreases 285 Given: TAS = 197 kt, True course = 240°,W/V = 180/30kt. Descent is initiated at FL 220 and completed at FL 40. Distance to be covered during descent is 39 NM. What is the approximate rate of descent? a) 800 FT/MIN b) 1400 FT/MIN c) 950 FT/MIN 286 Given: ILS GP angle = 3.5 DEG, GS = 150 kt. What is the approximate rate of descent? a) 700 FT/MIN b) 1000 FT/MIN c) 900 FT/MIN 287 Given: aircraft height 2500 FT, ILS GP angle 3°. At what approximate distance from THR can you expect to capture the GP? a) 13.1 NM b) 7.0 NM c) 8.3 NM 288 A pilot receives the following signals from a VOR DME station: radial 180°+/- 1°, distance = 200 NM. What is the approximate error? a) +/- 1 NM b) +/- 3.5 NM c) +/- 2 NM NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 37 289 An aircraft at FL310, M0.83, temperature -30°C, is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged, when 360 NM from the reporting point Mach Number should be reduced to: a) M0.74 b) M0.76 c) M0.78 290 An aircraft at FL120, IAS 200kt, OAT -5° and wind component +30kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to: a) 159 kt b) 165 kt c) 169 kt 291 An aircraft at FL370, M0.86, OAT -44°C, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. If the speed reduction were to be made 420 NM from the reporting point, what Mach Number is required? a) M0.73 b) M0.81 c) M0.75 292 An aircraft at FL140, IAS 210 kt, OAT -5°C and wind component minus 35 kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change, when 150 NM from the reporting point the IAS should be reduced by: a) 25 kt b) 20 kt c) 30 kt 293 An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing from the same position is 270°. Assuming no drift and a GS of 240 kt, what is the approximate range from the NDB at 0840? a) 40 NM b) 50 NM c) 60 NM 294 The distance between positions A and B is 180 NM. An aircraft departs position A and after having travelled 60 NM, its position is pinpointed 4 NM left of the intended track. Assuming no change in wind velocity, what alteration of heading must be made in order to arrive at position B? a) 2° Left b) 8° Right c) 6° Right NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 38 295 Given: Distance A to B = 120 NM, After 30 NM aircraft is 3 NM to the left of course. What heading alteration should be made in order to arrive at point 'B'? a) 6° right b) 8° right c) 4° right 296 An aircraft is planned to fly from position 'A' to position 'B', distance 480 NM at an average GS of 240 kt. It departs 'A' at 1000 UTC. After flying 150 NM along track from 'A', the aircraft is 2 MIN behind planned time. Using the actual GS experienced, what is the revised ETA at 'B'? a) 1206 b) 1203 c) 1153 297 An aircraft is planned to fly from position 'A' to position 'B', distance 320 NM, at an average GS of 180 kt. It departs 'A' at 1200 UTC. After flying 70 NM along track from 'A', the aircraft is 3 MIN ahead of planned time. Using the actual GS experienced, what is the revised ETA at 'B'? a) 1347 UTC b) 1401 UTC c) 1333 UTC 298 An aircraft is planned to fly from position 'A' to position 'B', distance 250 NM at an average GS of 115 kt. It departs 'A' at 0900 UTC. After flying 75 NM along track from 'A', the aircraft is 1.5 MIN behind planned time. Using the actual GS experienced, what is the revised ETA at 'B'? a) 1044 UTC b) 1110 UTC c) 1115 UTC 299 Given: Distance 'A' to 'B' is 475 NM, Planned GS 315 kt, ATD 1000 UTC, 1040 UTC - fix obtained 190 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at 'B'? a) 300 kt b) 360 kt. c) 340 kt 300 Given: Distance 'A' to 'B' is 325 NM, Planned GS 315 kt, ATD 1130 UTC, 1205 UTC - fix obtained 165 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at 'B'? a) 355 kt b) 375 kt c) 395 kt NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 39 301 Given: Distance 'A' to 'B' is 100 NM, Fix obtained 40 NM along and 6 NM to the left of course. What heading alteration must be made to reach 'B'? a) 9° Right b) 15° Right c) 6° Right 302 Given: Distance 'A' to 'B' is 90 NM, Fix obtained 60 NM along and 4 NM to the right of course. What heading alteration must be made to reach 'B'? a) 12° Left b) 16° Left c) 4° Left 303 Given : ETA to cross a meridian is 2100 UTCGS is 441 kt TAS is 491 kt At 2010 UTC, ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately: a) 75 kt b) 90 kt c) 40 kt 304 The distance between two waypoints is 200 NM, To calculate compass heading, the pilot used 2°E magnetic variation instead of 2°W. Assuming that the forecast W/V applied, what will the off track distance be at the second waypoint? a) 7 NM b) 14 NM c) 0 NM 305 Given: Half way between two reporting points the navigation log gives the following information: TAS 360 kt, W/V 330°/80kt, Compass heading 237°, Deviation on this heading -5°, Variation 19°W. What is the average ground speed for this leg? a) 354 kt b) 403 kt c) 373 kt 306 The purpose of the Flight Management System (FMS), as for example installed in the B737-400, is to provide: a) continuous automatic navigation guidance as well as manual performance management b) manual navigation guidance and automatic performance management c) continuous automatic navigation guidance and performance management 307 Which component of the B737-400 Flight Management System (FMS) is used to 377 enter flight plan routing and performance parameters? a) Flight Management Computer b) Multi-Function Control Display Unit c) Inertial Reference System NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 40 308 What indication, if any, is given in the B737-400 Flight Management System if radio updating is not available? a) A warning message is displayed on the EHSI and MFDU b) A warning message is displayed on the IRS displays c) A warning message is displayed on the Flight Director System 309 What is the validity period of the 'permanent' data base of aeronautical information stored in the FMC In the B737-400 Flight Management System? a) 3 calendar months b) one calendar month c) 28 days 310 In the B737-400 Flight Management System the CDUs are used during preflight to: a) manually initialize the Flight Director System and FMC with dispatch information b) automatically initialize the IRSs and FMC with dispatch information c) manually initialize the IRSs and FMC with dispatch information 311 How is the radio position determined by the FMC in the B737-400 Electronic Flight Instrument System? a) DME/DME or VOR/DME b) DME/DME c) DME ranges and/ or VOR/ADF bearings 312 In which of the following situations is the FMC present position of a B737-400 Electronic Flight Instrument System likely to be least accurate? a) Just after take-off b) At top of climb c) At top of descent 313 What are, in order of highest priority followed by lowest, the two levels of message produced by the CDU of the B737-400 Electronic Flight Instrument System? a) Alerting and Advisory b) Urgent and Routine c) Priority and Alerting 314 Which of the following can all be stored as five letter waypoint identifiers through the CDU of a B737-400 Electronic Flight Instrument System? a) Airway names, navaid identifiers, airport names, waypoint code numbers b) Waypoint names, navaid identifiers, runway numbers, airport ICAO identifiers c) Waypoint names, navaid frequencies, runway codes, airport ICAO identifiers 315 Which of the following lists all the methods that can be used to enter 'Created Waypoints' into the CDU of a B737-400 Electronic Flight Instrument System? a) Identifier bearing/distance, place bearing/place bearing, latitude and longitude, waypoint name b) Identifier bearing/distance, place bearing/place distance, along/across-track displacement, latitude and longitude c) Identifier bearing/distance, place bearing/place bearing, along-track displacement, latitude and longitude NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 41 316 Which FMC/CDU page normally appears on initial power application to the B737-400 Electronic Flight Instrument System? a) INITIAL b) IDENT c) POS INIT 317 Which of the following lists the first three pages of the FMC/CDU normally used to enter data on initial start-up of the B737-400 Electronic Flight Instrument System? a) IDENT - POS INIT - RTE b) POS INIT - RTE - IDENT c) IDENT - RTE - DEPARTURE 318 With reference to inertial navigation systems, a TAS input is: a) not required b) required to provide a W/V read out c) required for Polar navigation 319 The platform of an inertial navigation system (INS) is maintained at right angles to the local vertical by applying corrections for the effects of: a) aircraft maneuvers, earth rotation, transport wander and Coriolis b) gyroscopic inertia, earth rotation and real drift 378 c) vertical velocities, earth precession, centrifugal forces and transport drift 320 Some inertial reference and navigation systems are known as ""strapdown"". This means that: a) the gyroscopes and accelerometers become part of the unit's fixture to the aircraft structure b) only the gyros, and not the accelerometers, become part of the unit's fixture to the aircraft structure c) gyros and accelerometers are mounted on a stabilised platform in the aircraft 321 In order to maintain an accurate vertical using a pendulous system, an aircraft inertial platform incorporates a device: a) without damping and a period of 84.4 SEC b) without damping and a period of 84.4 MIN c) with damping and a period of 84.4 MIN 322 The term drift refers to the wander of the axis of a gyro in: a) the vertical and horizontal plane b) the vertical plane c) the horizontal plane 323 The resultant of the first integration from the north/south accelerometer of an inertial navigation system (INS) in the NAV MODE is: a) groundspeed b) velocity along the local meridian c) change latitude NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 42 324 Double integration of the output from the east/west accelerometer of an inertial navigation system (INS) in the NAV MODE give: a) distance east/west b) vehicle longitude c) distance north/south 325 In an Inertial Navigation System (INS), Ground Speed (GS) is calculated: a) from TAS and W/V from RNAV data b) from TAS and W/V from Air Data Computer (ADC) c) by integrating measured acceleration 326 One of the errors inherent in a ring laser gyroscope occurs at low input rotation rates tending towards zero when a phenomenon known as 'lock-in' is experienced. What is the name of the technique, effected by means of a piezo-electric motor, that is used to correct this error? a) dither b) cavity rotation c) zero drop 327 The resultant of the first integration of the output from the east/west accelerometer of an inertial navigation system (INS) in NAV MODE is: a) vehicle longitude b) change of longitude c) velocity along the local parallel of latitude 328 Which of the following lists, which compares an Inertial Reference System that utilises Ring Laser Gyroscopes (RLG) instead of conventional gyroscopes, is completely correct? a) There is little or no 'spin up' time and it is insensitive to gravitational ('g') forces b) The platform is kept stable relative to the earth mathematically rather than mechanically but it has a longer 'spin up' time c) It does not suffer from 'lock in' error and it is insensitive to gravitational ('g') forces 329 The principle of 'Schuler Tuning' as applied to the operation of Inertial Navigation Systems/ Inertial Reference Systems is applicable to: a) both gyro-stabilised platform and 'strapdown' systems b) only gyro-stabilised systems c) both gyro-stabilised and laser gyro systems but only when operating in the non 'strapdown' mode 330 What additional information is required to be input to an Inertial Navigation System (INS) in order to obtain an W/V readout? a) Altitude and OAT b) IAS c) TAS NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 43 331 What is the name given to an Inertial Reference System (IRS) which has the gyros and accelerometers as part of the unit's fixture to the aircraft structure? a) Strapdown b) Rigid c) Solid state 332 During initial alignment an inertial navigation system is north aligned by inputs from: a) computer matching of measured gravity magnitude to gravity magnitude of initial alignment b) the aircraft remote reading compass system c) horizontal accelerometers and the east gyro 333 During the initial alignment of an inertial navigation system (INS) the equipment: a) will accept a 10° error in initial latitude but will not accept a 10° error in initial longitude b) will not accept a 10° error in initial latitude but will accept a 10° error in initial longitude c) will not accept a 10° error in initial latitude or initial longitude 334 Which of the following statement is correct concerning gyro-compassing of an inertial navigation system (INS)? a) Gyro-compassing of an INS is not possible in flight because it cannot differentiate between movement induced and misalignment induced accelerations. b) Gyro-compassing of an INS is possible in flight because it can differentiate between movement induced and misalignment induced accelerations. c) Gyro-compassing of an INS is possible in flight because it cannot differentiate between movement induced and misalignment induced accelerations. 335 Which of the following statements concerning the loss of alignment by an Inertial Reference System (IRS) in flight is correct? a) The mode selector has to be rotated to ATT then back through ALIGN to NAV in order to obtain an in-flight realignment b) The IRS has to be coupled to the remaining serviceable system and a realignment carried out in flight c) The navigation mode, including present position and ground speed outputs, is inoperative for the remainder of the flight 336 The alignment time, at mid-latitudes, for an Inertial Reference System using laser ring gyros is approximately: a) 2 MIN b) 20 MIN c) 10 MIN 337 Which of the following statements concerning the alignment procedure for Inertial Navigation Systems(INS)/Inertial Reference Systems (IRS) at mid-latitudes is correct? a) INS/IRS can only be aligned in the ALIGN mode b) INS/IRS can be aligned in either the ALIGN or NAV mode c) INS/IRS can be aligned in either the ALIGN or ATT mode NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 44 338 A pilot accidently turning OFF the INS in flight, and then turns it back ON a few moments later. Following this incident: a) it can only be used for attitude reference b) no useful information can be obtained from the INS c) everything returns to normal and is usable 339 The azimuth gyro of an inertial unit has a drift of 0.01°/HR. After a flight of 12 HR with a ground speed of 500 kt, the error on the aeroplane position is approximately: a) 12 NM b) 1 NM c) 6 NM 340 The drift of the azimuth gyro on an inertial unit induces an error in the position given by this unit. ""t"" being the elapsed time. The total error is: a) proportional to t/2 b) proportional to the square of time, t² c) proportional to t 341 With reference to an inertial navigation system (INS), the initial great circle track between computer inserted waypoints will be displayed when the control display unit (CDU) is selected to: a) TK/GS b) HDG/DA c) DSRTK/STS 342 Gyro-compassing of an inertial reference system (IRS) is accomplished with the mode selector switched to: a) ATT/REF b) STBY c) ALIGN 343 Which of the following correctly lists the order of available selections of the Mode Selector switches of an inertial reference system (IRS) mode panel? a) OFF - ALIGN - NAV - ATT b) OFF - ON - ALIGN - NAV c) OFF - STBY - ALIGN - NAV 344 ATT Mode of the Inertial Reference System (IRS) is a back-up mode providing: a) only attitude and heading information b) only attitude information c) navigation information 345 Which of the following statements concerning the operation of an Inertial Navigation System (INS)/Inertial Reference System (IRS) is correct? a) NAV mode must be selected prior to the loading of passengers and/or freight b) NAV mode must be selected on the runway just prior to take-off c) NAV mode must be selected prior to movement of the aircraft off the gate NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 45 346 Which of the following statements concerning the aircraft positions indicated on a triple fit Inertial Navigation System (INS)/ Inertial Reference System (IRS) on the CDU is correct? a) The positions will be the same because they are an average of three different positions b) The positions are likely to differ because they are calculated from different sources c) The positions will only differ if one of the systems has been decoupled because of a detected malfunction 347 Waypoints can be entered in an INS memory in different formats. In which of the following formats can waypoints be entered into all INSs? a) hexadecimal b) bearing and distance c) geographic coordinates 348 An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. The following two points have been entered in the INS computer: WPT 1: 60°N 030°WWPT 2: 60°N 020°W When 025°W is passed the latitude shown on the display unit of the inertial navigation system will be: a) 59°49.0'N b) 60°00.0'N c) 60°05.7'N 349 An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertial system. The coordinates of A (45°S 010°W) and B (45°S 030°W) have been entered. The true course of the aircraft on its arrival at B, to the nearest degree, is: a) 284° b) 277° c) 263° 350 As the INS position of the departure aerodrome, coordinates 35°32.7'N 139°46.3'W are input instead of 35°32.7'N 139°46.3'E. When the aircraft subsequently passes point 52°N 180°W, the longitude value shown on the INS will be: a) 099° 32.6'E b) 099° 32.6'W c) 080° 27.4'E 351 The following points are entered into an inertial navigation system (INS).WPT 1: 60°N 30°WWPT 2: 60°N 20°WWPT 3: 60°N 10°W The inertial navigation system is connected to the automatic pilot on route (1-2-3).The track change when passing WPT 2 will be approximately: a) a 9° increase b) zero c) a 9° decrease NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 46 352 The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS) and the aircraft is flying from waypoint No. 2 (60°00'S 070°00'W) to No. 3 (60°00'S 080°00'W).Comparing the initial track (°T) at 070°00'W and the final track (°T) at 080°00'W, the difference between them is that the initial track is approximately: a) 9° less than the final one b) 5° greater than the final one c) 9° greater than the final one 353 The automatic flight control system is coupled to the guidance outputs from an inertial navigation system. Which pair of latitudes will give the greatest difference between initial track read-out and the average true course given, in each case, a difference of longitude of 10°? a) 60°N to 60°N b) 60°N to 50°N c) 30°S to 30°N 354 The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS).The aircraft is flying between inserted waypoints No. 3 (55°00'N 020°00'W) and No.4 (55°00'N 030°00'W). With DSRTK/STS selected on the CDU, to the nearest whole degree, the initial track read- out from waypoint No. 3 will be: a) 266° b) 278° c) 274° 355 Which of the following statements concerning the position indicated on the Inertial Reference System (IRS) display is correct? a) It is not updated once the IRS mode is set to NAV b) It is constantly updated from information obtained by the FMC c) It is updated when 'go-around' is selected on take-off 356 What is the source of magnetic variation information in a Flight Management System (FMS)? a) The FMS calculates MH and MT from the FMC position b) The main directional gyro which is coupled to the magnetic sensor (flux valve) positioned in the wingtip c) Magnetic variation information is stored in each IRS memory, it is applied to the true heading calculated by the respective IRS 357 Where and when are the IRS positions updated? a) IRS positions are updated by pressing the 'Take-off/ Go-around' button at the start of the take - off roll b) During flight IRS positions are automatically updated by the FMC c) Only on the ground during the alignment procedure NAVRANG SINGH E-mail: [email protected] M: 9818282209 NAVIGATION QUESTION 47 358 The sensors of an INS measure: a) acceleration b) velocity c) the horizontal component of the earth's rotation d) precession 359 Given the following: True track: 192° Magnetic variation: 7°E Drift angle: 5° left What is the magnetic heading required to maintain the given track? a) 194° b) 190° c) 204° NAVRANG SINGH E-mail: [email protected] M: 9818282209