Atpl bristolgs general navigation exam 4 2023 update

Atpl bristolgs general navigation exam 4 2023 updateAtpl bristolgs general navigation exam 4 2023 update Exam 4, 54 questions; 73 marks; time allowed 2 hours 1. The rhumb-line distance between points A (60°00'N 002°30'E) and B (60º00'N 007°30W) is: (a) 150 NM (b) 300 NM (c) 450 NM (d) 600 NM (1) 2. An aircraft at latitude 02°20'N tracks 180º(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 (d) 09°05'S (2) 3. 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 3600NM, 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 (d) 59°00'N 090°00'W (2) 4. You are flying from A to B in the northern hemisphere. Your initial great circle track is 273°T and conversion angle is 6°. The rhumb line track from A to B is: (a) 279°T (b) 276°T (c) 270°T (d) 267°T (1) 5. An aircraft flies a rhumb line track from A (160°E) to B (170°W) at the same latitude in the Northern hemisphere. Its groundspeed is 300 knots and the flight takes three hours. The latitude of the aircraft is: (a) 30°N (b) 45°N (c) 60°N (d) 75°N (2) 6. What is the approximate date of perihelion, when the Earth is nearest to the Sun ? (a) beginning of July (b) end of December (c) beginning of January (d) end of March (1) 7. The highest latitude at which the sun will be visible above the horizon every day is: (a) 62° (b) 64° (c) 66° (d) 68° (1) 8. An aircraft plans to depart San Juan, Puerto Rico (18°N 066°W) 30 minutes before sunrise on 28th March and will arrive four hours and fifteen minutes later at Santa Maria, Azores (37°N 025°W). If the difference between standard time and UTC is four hours in San Juan and one hour in the Azores and Summer Time is not applicable the local mean time of arrival at Santa Maria will be: (a) 1202 (b) 1208 (c) 1227 (d) 1245 (2) 9. When is the magnetic compass most effective? (a) On the geographic equator (b) In the region of the magnetic North Pole. (c) About midway between the magnetic poles (d) In the region of the magnetic South Pole. (1) 10. The main advantage of a remote indicating compass over a direct reading compass is that it: (a) has less moving parts (b) senses, rather than seeks, the magnetic meridian (c) requires less maintenance (d) is able to magnify the earth's magnetic field in order to attain greater accuracy 11. Scale on a Lambert's conformal chart is: (1) (a) constant over the whole chart (b) constant along a meridian of longitude (c) constant along a parallel of latitude (d) varies with latitude and longitude (1) 12. The most likely use for an Oblique Mercator chart is: (a) charts for intercontinental flights following great circle tracks (b) maps of countries with considerable North/South extent but very little East/West extent (c) trans-polar navigation (d) global depiction of magnetic variation (1) 13. A line is drawn on a Lamberts Conformal chart which follows a parallel of latitude between 007°20'E and 003°30'E and represents 135 nm. The parallel of latitude on which the line is drawn is: (a) 36° (b) 50° (c) 54° (d) 60° (2) 14. On a Transverse Mercator chart a straight line is drawn representing 200 nm which is perpendicular to, and originates from, the central meridian. It is a: (a) small circle (b) rhumb line (c) great circle (d) a curve concave to the prime meridian (1) 15. What is the symbol for an FIR boundary ? (a) (b) (c) (d) ............................ (1) 16. If the rhumb line distance of the 53°N parallel of latitude is measured on a Mercator chart as 133 cm the scale of the chart at 30°S is ? (a) 1: (b) 1: (c) 1: (d) 1: (2) 17. A chart distance represents 120nm at 15°S on a direct Mercator chart. What earth distance does the same chart distance represent at 10°N ? (a) 118.3 nm (b) 124.2 nm (c) 122.3 nm (d) 117.7 nm (2) 18. Where is the scale correct on a Transverse Mercator chart ? (a) at the Standard Parallels (b) along the central meridian and its anti-meridian (c) at the Parallel of Origin (d) scale is constant over the entire chart (1) 19. An aircraft is using a southern hemisphere Lamberts chart with a 'standard' grid (i.e. a grid aligned with the Greenwich meridian) and the Constant of the Cone is given as 0.9. If the aircraft is at 080°W steering a heading of 190°G with 10°S drift its true track will be: (a) 272° (b) 128° (c) 282° (d) 200° (1) 20. Given: TAS = 155 kt, HDG (T) = 216° W/V = 090/60kt. Calculate the Track (ºT) and GS? (a) 226 - 186 kt (b) 224 - 175 kt (c) 222 - 181 kt (d) 231 - 196 kt (1) 21. 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 the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained (b) set heading towards a line feature such as a coastline, motorway, river or railway (c) fly expanding circles until a pinpoint is obtained (d) fly reverse headings and associated timings until the point of departure is regained (1) 22. What is the ISA temperature value at FL 330? (a) -50°C (b) -56°C (c) -66°C (d) -81°C (1) 23. Given: Pressure Altitude 29000 FT , OAT -55°C. Calculate the Density Altitude. (a) 26000 FT (b) 27500 FT (c) 31000 FT (d) 33500 FT (1) 24. 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) 1310 NM (b) 1510 NM (c) 2290 NM (d) 2370 NM (2) 25. 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) 165 NM (b) 170 NM (c) 180 NM (d) 195 NM (2) 26. The distance from departure point to point of equal time is: (a) inversely proportional to groundspeed out and groundspeed back (b) inversely proportional to total distance (c) proportional to cosine of latitude (d) inversely proportional to groundspeed back (1) 27. When flying from A to B, a distance of 120nm, you are 3nm left of track after 30nm. What is the required alteration of heading to make good B ? (a) 8° left (b) 8° right (c) 6° right (d) 4° right (1) 28. An aircraft has a groundspeed of 510 kts and a True Air Speed of 440 kts. If the distance from A to B is 43 nm the time in minutes from A to B will be: (a) 7 (b) 5 (c) 4 (d) 6 (1) 29. If the true track from A to B is 090°, TAS is 460 kts, wind velocity is 360°/100kts, variation is 10°E and deviation is -2°; calculate the compass heading and groundspeed. (a) 069° and 448 kts (b) 068° and 460 kts (c) 078° and 450 kts (d) 070° and 453 kts (2) 30. Given: IAS 120 kts, FL80, COAT +20°. The TAS is: (a) 102 kts (b) 120 kts (c) 132 kts (d) 141 kts (1) 31. An aircraft is flying at FL 140 where the COAT is -5°C. It is flying at an indicated air speed of 260 kts and is experiencing a headwind of 34 knots. When 150 nm from the FIR boundary it is instructed to reduce speed in order to delay arrival at the boundary by 5 minutes. The required reduction in indicated air speed is: (a) 41 kts (b) 33 kts (c) 24 kts (d) 15 kts (2) 32. What is the ratio between the litre and the US gallon ? (a) 1 USG = 4.55 litres (b) 1 litre = 3.78 USG (c) 1 USG = 3.78 litres (d) 1 litre = 4.55 USG (1) Use the attached chart E(LO)1 to answer questions 33 to 42 33. An aircraft is overhead MAC (N55°25.8’W005°39.0’) at 1054 UTC. The aircraft is at FL270, COAT –40°C and the wind is 290°T/50 kts. In order for the aircraft to enter Shanwick Oceanic airspace at 56°N 010°W at 1114 UTC it should fly at a Mach No. of: (a) 0.72 (b) 0.76 (c) 0.80 (d) 0.84 (2) 34. If MAC (N55°25.8’W005°39.0’) 25 nm DME and GOW (N55°25.8’ W004°26.7’) 31 nm DME, the position of the aircraft is: (a) 55°51’N 005°10’W (b) 55°29’N 005°13’W (c) 55°50’N 005°50’W (d) 55°48’N 005°20’W (2) 35. BEL (N54°39.7’W006°13.8’) 085° RMI and 47 nm DME. The position of the aircraft is: (a) 54°30’N 007°30’W (b) 54°20’N 007°26’W (c) 54°35’N 007°33’W (d) 54°33’N 007°00’W (2) 36. An aircraft is overhead TRN (N55°18.8’W004°47.0’) at 2320 UTC; flying with a CAS of 250 kts, at FL180, COAT –15°C, wind 250°T/60kts. Variation is 7°W and the aircraft is routing via airway B2. What is the estimated time of arrival at MULLA ? (a) 2355 UTC (b) 2344 UTC (c) 2336 UTC (d) 2333 UTC (2) 37. If TY (N55°16.4’W008°14.9’) indicates 246° on the RMI and LAY (N55°41.0’W006°14.9’) shows 112° on the RMI the position of the aircraft is: (a) 55°43’N 006°32’W (b) 55°47’N 006°41’W (c) 55°48’N 006°53’W (d) 55°46’N 007°00’W (2) 38. When at FYNER (N56°03.0’W005°07.0’) the DME range from MAC (N55°25.8’W005°39.0’) is: (a) 25 nm (b) 16 nm (c) 41 nm (d) 29 nm (1) 39. TIR (N56°29.6’W006°52.6) and ISY (N55°25.8’W006°15.0’) both show DME range 60 nm. The position of the aircraft is: (a) 55°57’N 007°09’W (b) 56°14’N 007°47’W (c) 55°45’N 008°02’W (d) 55°31’N 007°12’W (2) 40. An aircraft is cleared to route direct from the Shanwick Oceanic CTA at 56N 010W to BEL (N54°39.7’W006°13.8’). If the TAS is 450 kts and the wind is given as Northerly at 70 kts the mean magnetic track to fly will be: (a) 130° (b) 122° (c) 115° (d) 107° (1) 41. If MAC (N55°25.8’W005°39.0’) indicates 124° on the RMI and ISY (N55°41.0’W006°15.0) indicates 80 nm on the DME the position of the aircraft is: (a) 55°52’N 007°11’W (b) 56°15’N 008°25’W (c) 55°58’N 007°09’W (d) 56°19’N 008°20’W (2) 42. An aircraft is routing direct from TRN (N55°18.8’W004°47.0’) to GOW (N55°52.2’W004°26.7’). the Easterly wind is giving seven degrees of drift, variation is 7°W and deviation is –3°. The magnetic heading to fly is: (a) 020° (b) 025° (c) 029° (d) 032° (1) 43. In the B737-400 Flight Management System the CDUs are used during preflight to: (a) manually initialise the IRSs and FMC with dispatch information (b) automatically initialise the IRSs and FMC with dispatch information (c) manually initialise the Flight Director System and FMC with dispatch information (d) manually initialise the IRSs, FMC and Autothrottle with dispatch information (1) 44. What does the INS need for wind calculations ? (a) TAS (b) EAS (c) Mach Number (d) CAS (1) 45. What can be used to enter positions on all INS systems ? (a) Waypoint names (b) geographical co-ordinates (c) hexidecimals (d) ranges and bearings (1) 46. What do you call a system of gyros and accelerometers which is fixed to the aircraft ? (a) Laser (b) Strapdown (c) Fixed (d) Gyro-stabilised platform (1) 47. Alignment of INS and IRS equipment can take place in which of the following modes ? (a) ALIGN and ATT (b) NAV and ALIGN (c) ALIGN only (d) NAV and ATT (1) 48. A pilot turns off the power to his IRS whilst in flight. If he switches the power back on after just a few seconds the effect will be: (a) the IRS can be used providing the position is checked (b) there will be no effect and the IRS can continue to be used (c) the IRS cannot be used and should be shut down (d) the IRS cannot be used to give attitude information but can be used for navigation (1) 49. On an IRS the modes available on the MCP are: (a) OFF – STBY – ALIGN – NAV (b) OFF – STBY – ALIGN – ATT (c) OFF – ALIGN – NAV – ATT (d) OFF – ALIGN – NAV – STBY (1) 50. An IRS is programmed with waypoint 1: N60°00’ W030°00’, waypoint 2: N60°00’ W020°00’, waypoint 3: N60°00’ W010°00’. As the aircraft passes waypoint 2 its track will: (a) decrease by 9° (b) not change (c) increase by 9° (d) increase by 4.5° (2) 51. What is the axis of gyro drift ? (a) vertical and horizontal (b) vertical (c) horizontal (d) longitudinal (1) 52. Which of the following statements is correct concerning Ring Laser Gyros? (a) they do not suffer from ‘lock-in’ and are unaffected by the Earth’s gravitational force (b) they are not necessarily fixed to true north but take a long time to erect (c) they are not necessarily fixed to true north and are quick to erect (d) their alignment will be either with true north or the local vertical (1) 53. If an IRS fails in flight it can provide: (a) attitude and heading information (b) attitude information only (c) attitude and navigation information (d) navigation information but not attitude information (1) 54. 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 3 (35°S 050°W) to waypoint 4 (35°S 040°W). When the INS indicates longitude 045°W the latitude readout will be: (a) S35°00.0' (b) north of 35°S (c) south of 35°S (d) S34°06.7' (1) Intentionally Blank General Navigation Exam 4, 54 Questions. 73 marks, Time allowed: 2hrs. A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 A B C D 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 A B C D 49 50 51 52 53 54 Intentionally Blank Appendix A to General Navigation Examination 4 Intentionally Blank Intentionally Blank Exam 4: Answers 1. (b) 26. (a) 51. (b) 2. (a) 27. (b) 52. (c) 3. (c) 28. (b) 53. (a) 4. (d) 29. (a) 54. (c) 5. (c) 30. (d) 6. (c) 31. (b) 7. (c) 32. (c) 8. (c) 33. (d) 9. (c) 34. (d) 10. (b) 35. (a) 11. (c) 36. (c) 12. (a) 37. (c) 13. (c) 38. (c) 14. (c) 39. (c) 15. (b) 40. (a) 16. (b) 41. (b) 17. (c) 42. (d) 18. (b) 43. (a) 19. (a) 44. (a) 20. (d) 45. (b) 21. (b) 46. (b) 22. (a) 47. (b) 23. (b) 48. (c) 24. (c) 49. (c) 25. (a) 50. (a) WORKED ANSWERS 1. Change of longitude: 002°30'E 007°30'W 010°00' Departure = C(') x cos lat = 10° x 60 x cos60° = 300 nm Departure formula gives rhumb line distance. 2. 685 km = 369.5 nm = 369.5 ' C = 6°09.5' to south Start at 02°20'N End at 03°49.5'S 4. Great Circle CA GC = 273°T CA = 6° Rhumb line RL = 267°T 5. 160°E 170°W Ch. Long. = 30° x 60 = 1,800' Departure = 3 hrs x 300 kts = 900 nm = C x cos lat Cos Lat = Departure = 900 = 0.5 Ch. Long(') 1,800 Lat. = 60°N 6. Answer (c) 7. 66½°N and 66½°S are the Arctic and Antarctic circles. Above these latitudes there will be days in the winter when the sun will not rise. 8. From Air Almanac sunrise at 18°N 05:58 LMT 28th - :30 Departure time 05:28 LMT 28th 66°W @ 4 min/° 4:24 Departure time 09:52 UTC 28th Flight time 4:15 Arrival time 14:07 UTC 28th 25°W @ 4 min/° 1:40 Arrival time 12:27 LMT 28th 9. Compass will be most effective where H (horizontal component) has maximum strength; at magnetic equator. 10. Detector is not a "North seeking" or "North aligning" system. It does not 'magnify the Earth's field in order to attain greater accuracy' it concentrates the Earth's field in order to attain greater sensitivity. 11. Answer (c) 12. For answer (b) a transverse mercator would be best. For answer (c) a polar stereo or a transverse mercator would do. For answer (d) the global map would be better on a series of Lamberts or some other projection where sheets join easier. 13. Departure = Ch.Long(') x cos Lat. OR Cos Lat. = Departure Ch.Long(') Departure = 135 nm Ch.Long.: = 007°20'E - 003°30'E = 3°50' = 230' Cos Lat = 135 230 = 0.587 = Cos 54° 14. Within approximately 500 nm of the central meridian a great circle is a straight line and the line described is well within that area. 15. Answer (b) 16. On a Mercator chart the rhumb line distance of all parallels of latitude will be the same because the meridians of longitude are straight parallel lines. At 30°S scale = CD = 133 cm . = 133 cm = 133 cm ED 360°x60xcos30 18,706nm 34,674km = 1 . 26,070,521 17. By juggling the Mercator scale formula it can be shown that: EDLat10° = EDLat15° so: EDLat10° = EDLat15° x Cos 10° Cos 10° Cos 15° Cos 15° = 120 x cos 10° cos 15° = 122.3 nm 18. Answer (b) 19. 080°W True North Grid North Grid North Convergence = C x n = 80° x 0.9 = 72° Hdg. 190°G Drift 10°Stbd Track200°G Conv. 72° Track272°T 20. On CRP5 or similar: Centre dot on TAS, 155 kts, rotate to put wind direction, 090° under heading index and put wind mark 60 kts below centre dot at 95 kts. Rotate to put heading, 216° under heading index. Note at wind mark drift, 15°Stbd, will give track 231° and groundspeed is 190 kts. 21. Answer (b) 22. +15° - (1.98° x 33) = +15° - 65.34° = -50.34°C 23. On CRP5 or similar: In AIRSPEED window put temperature, -55°C, next to pressure altitude, 29,000 feet; go to DENSITY ALTITUDE window and note that the index is half way between answers (a) and (b) ! Bummer ! We must calculate: At 29,000 feet ISA = +15° -(2° x 29) = -43°C OAT = -55°C Dev. = -12°C Density altitude is 120 feet per degree of ISA deviation from pressure altitude (lower if colder). Density altitude = 120x12 = 1,440 feet lower 29,000 - 1,440 = 27,560 feet 24. Time to PSR = E x H Where: E = fuel endurance = 10 hours (O +H) O = g/s out = 480-90 = 390 kts H = g/s home= 480+75= 555 kts = 10 x 555 (O + H) = 945 945 = 5.87 hours at 390 kts (g/s out) = 2,290 nm 25. Distance to ETP = D x H Where: D = total distance = 360 nm (O + H) O = g/s out = 180 - 15 = 165 kts H = g/s home=180 + 15= 195 kts = 360 x 195 (O + H) = 360 360 = 195 nm from A = 360 - 195 = 165 nm from B 26. From formula in question 26 above it can be seen that the distance from departure point to point of equal time varies: directly with D (total distance) directly with H (groundspeed home) inversely with (O + H) ( g/s out and g/s home) 28. CRP5 or similar: Put 60 on inner main scale next to groundspeed, 510 kts, on outer main scale. Go to distance, 43 nm, on outer main scale and read time next to it, 5 minutes, on inner main scale. Or, calculator: 43 x 60 = 5 minutes 510 29. CRP5 or similar: Centre dot on TAS, 460 kts, rotate to wind direction, 360°, put wind mark 100 kts under centre dot at 360 kts. Rotate to put true track, 090°, under heading index; note drift 13°Stbd and rotate track 090° to drift 13°Stbd. Drift is still 13°Stbd so required heading, under heading index, is 077°T and groundspeed, under the wind mark, is 448 kts. C D M V T 069° -2°(W) 067° 10°E 077° 30. CRP5 or similar: In AIRSPEED window put COAT, +20°C, next to pressure altitude, FL80. IAS is virtually CAS (same as RAS) so go to 120 kts on inner main scale and read TAS next to it on outer scale;141 kts. 31. On CRP5 or similar In AIRSPEED window put COAT, -5°C, next to pressure altitude, FL140. IAS is virtually CAS (same as RAS) so go to 260 kts on inner main scale and read next to it 327 kts; move COMP.CORR. index 0.25 of a division; go back to 260 kts on the inner scale and read TAS next to it on outer scale; 325 kts. TAS 325 kts Wind comp. -34 kts Groundspeed 291 kts At present groundspeed, 291 kts, 150 nm will take 30.9 minutes. To cover 150 nm in 35.9 minutes will require groundspeed 250 kts. This is a reduction in groundspeed of 41 kts so TAS must reduce by 41 kts to 284 kts. On CRP5 or similar In AIRSPEED window put COAT, -5°C, next to pressure altitude, FL140. Go to required TAS on outer scale and read required CAS on inner scale; 227 kts. IAS must reduce from 260 kts to 227 kts = 33 kts reduction. 32. On CRP5 or similar: Put 1 (10) on inner main scale next to US GAL index on outer scale. Go to LTR index on outer scale and read on inner scale number of litres in 1 US gallon = 3.8. 33. On chart note that track from MAC to 56°N 010°W is 291°M and distance is 151 nm. Aircraft must cover 151 nm in 20 minutes (between 10:54 and 11:14 UTC) which will require a groundspeed of 453 kts. The wind from 290°T is virtually on the nose therefore: required groundspeed 453 kts headwind component -50 kts required TAS 503 kts On CRP5 or similar: In AIRSPEED window put COAT -40°C next to MACH NO. INDEX. Go to required TAS, 503 kts, on outer main scale and read Mach No. next to it on inner main scale; 0.845. 34. On chart; with compasses and, using latitude scale, arc off ranges of 25 nm from MAC and 31 nm from GOW. They will cross in two places but only one of them, TABIT is offered. 35. The RMI shows the magnetic bearing from the aircraft to the station. Plot the reciprocal (085° + 180°) = 265°M from BEL and, with compasses, arc off range 47 nm using latitude scale for distance; where they cross is where you are. 36. On chart note that the airway alignment is 213°M, variation is shown as 7°W (for this information you may need to refer to your own copy of the chart which you will have with you in the exam) so the true track is 206°T Also from the chart the distance to run to MULLA is shown as 9 + 20 + 32 + 14 = 75 nm. On CRP5 or similar. In AIRSPEED window put COAT -15°C next to FL180; go to CAS 250 kts on inner main scale and read on outer main scale 333 kts; move COMP.CORR. index 0.33 of a division and go back to CAS 250 kts to read off TAS 330 kts. Put centre dot over TAS 330 kts, rotate to put wind direction, 250°T, under heading index and put mark 60 kts below centre dot at 270 kts. Rotate to put track, 206°T, under heading index and note drift 9°Port; rotate to put track, 206°T, under drift, 9°Port. Note drift now 8°Port; rotate to put track 206°T under drift 8°Port. Drift is still 8°Port so heading 214°T will be required to track 206°T (213°M). Read groundspeed under wind mark, 283 kts. 75 nm at groundspeed 283 kts takes 15.9 minutes over head TRN 23:20 UTC ETA MULLA 23:35.9 UTC 37. RMI shows magnetic bearing to the station. Plot the reciprocal bearings from the stations (taking care to align protractor with magnetic north) and where they cross is where you are: RMI bearing TY 246°M LAY 112°M -180° +180° Bearings to plot: 066°M 292°M 38. From the chart:: 25 + 16 = 41 nm 39. With compasses arc off 60 nm (using latitude scale) from TIR and ISY. These arcs cross in two places but only one of them is offered as an answer. 40. On the chart draw in the track from 56°N 010°W to BEL. The question asks for the mean magnetic track so it should be measured close to the mid-point by either: (a) centre protractor on track and align with magnetic north indicator of any nearby VOR or NDB to measure magnetic track directly. or (b) centre protractor on track and align with meridian of longitude to measure true track and then apply variation to calculate magnetic track. 41. RMI shows magnetic bearing to the station. Plot the reciprocal bearing from MAC (taking care to align protractor with magnetic north) and with compasses arc off 80 nm using the latitude scale for distance. Where they cross is where you are: RMI bearing MAC 124°M +180° Bearing to plot: 304°M 42. From the chart the track from TRN to GOW is 025°M. The easterly wind will give 7° port drift so the required heading will be 032°M 43. Answer (a) 44. To calculate the wind required data: heading: calculated by INS track: calculated by INS groundspeed: calculated by INS TAS: input from ADC 45. All systems will accept latitude and longitude positions; some will accept other ways of defining positions. 46. Answer (b) 47. Normally alignment takes place and is completed in the ALIGN mode but if NAV is selected and the aircraft is not moved the INS or IRS will go through the alignment sequence before switching to the NAV mode. 48. Once the IRS power supply is switched off it has lost all alignment references. It may be possible to subsequently use the system for attitude information but not for navigation so the only correct answer offered is (c). 49. With the laser gyros in the IRS there is no STBY mode. 50. At waypoint 2: track 1 to 2 greater than 090°T track 2 to 3 less than 090°T track has decreased by 2 x conversion angle which is convergency = 10° x sin60° = 8.7° 51. Drift is in the horizontal plane about the vertical axis. 52. Laser gyros do suffer from 'lock-in' (that's why there are dither motors fitted), they are quick to 'erect' (i.e. be ready for use) and they are aligned with the aircraft (not true north or the local vertical). 53. Depending on the nature of the failure the IRS could provide attitude and heading information (but not navigation information); answer (a), or absolutely nothing (not an option on offer). 54.