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GENERAL NAVIGATION

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GENERAL NAVIGATION
GENERAL NAVIGATION
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#1. The angle between the plane of the ecliptic and the plane of equator is approximately :

#2. Which is the highest latitude listed below at which the sun will rise above the horizon and set every day?

#3. In which two months of the year is the difference between the transit of the Apparent Sun and Mean Sun across the Greenwich Meridian the greatest?

#4. 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?

#5. 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?

#6. At what approximate date is the earth closest to the sun (perihelion)?

#7. At what approximate date is the earth furthest from the sun (aphelion)?

#8. Seasons are due to the:

#9. 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?

#10. 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:

#11. Given: Waypoint 1. 60°S 030°W Waypoint 2. 60°S 020°WWhat will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W?

#12. 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?

#13. A Rhumb line is :

#14. 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?

#15. What is the longitude of a position 6 NM to the east of 58°42'N 094°00'W?

#16. 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?

#17. 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:

#18. 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:

#19. The rhumb line track between position A (45°00'N, 010°00'W) and position B (48°30'N, 015°00'W) is approximately:

#20. The diameter of the Earth is approximately

#21. The maximum difference between geocentric and geodetic latitude occurs at about:

#22. 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:

#23. Given:Position A 45°N, ?°EPosition B 45°N, 45°15'EDistance A-B = 280 NMB is to the East of ARequired: longitude of position A?

#24. If an aeroplane 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:

#25. 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?

#26. An aeroplane 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:

#27. The circumference of the earth is approximately:

#28. The Great Circle bearing of 'B' (70°S 060°E), from 'A' (70° S 030° W), is approximately:

#29. At what approximate latitude is the length of one minute of arc along a meridian equal to one NM (1852 m) correct?

#30. An aircraft flies a great circle track from 56° N 070° W to 62° N 110° E. The total distance travelled is?

#31. Given :A is N55° 000°B is N54° E010°The average true course of the great circle is 100°.The true course of the rhumbline at point A is:

#32. The circumference of the parallel of latitude at 60°N is approximately

#33. Given:The coordinates of the heliport at Issy les Moulineaux are:N48°50' E002°16.5'The coordinates of the antipodes are :

#34. Given:Position 'A' is N00° E100°, Position 'B' is 240°(T), 200 NM from 'A'.What is the position of 'B'?

#35. The duration of civil twilight is the time:

#36. 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:

#37. . What is the local mean time, position 65°25'N 123°45'W at 2200 UTC?

#38. The Local Mean Time at longitude 095°20'W, at 0000 UTC, is :

#39. . 5 HR 20 MIN 20 SEC corresponds to a longitude difference of:

#40. The main reason that day and night, throughout the year, have different duration, is due to the:

#41. What is the meaning of the term ""standard time"" ?

#42. Civil twilight is defined by :

#43. Given: true track is 348°, drift 17° left, variation 32° W, deviation 4°E. What is the compass heading?

#44. An Agonic line is a line that connects:

#45. Isogonic lines connect positions that have:

#46. Compass deviation is defined as the angle between:

#47. The angle between True North and Magnetic North is called :

#48. Deviation applied to magnetic heading gives:

#49. Isogrives are lines that connect positions that have

#50. The lines on the earth's surface that join points of equal magnetic variation are called

#51. A negative (westerly) magnetic variation signifies that :

#52. The angle between Magnetic North and Compass North is called

#53. The north and south magnetic poles are the only positions on the earth's surface where:

#54. The rhumb-line distance between points A (60°00'N 002°30'E) and B (60°00'N 007°30'W) is:

#55. 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:

#56. The main reason that day and night, throughout the year, have different duration, is due to the:

#57. What is the meaning of the term ""standard time"" ?

#58. Civil twilight is defined by :

#59. Given: true track is 348°, drift 17° left, variation 32° W, deviation 4°E. What is the compass heading?

#60. An Agonic line is a line that connects:

#61. Isogonic lines connect positions that have:

#62. Compass deviation is defined as the angle between:

#63. The angle between True North and Magnetic North is called :

#64. Deviation applied to magnetic heading gives:

#65. Isogrives are lines that connect positions that have:

#66. The lines on the earth's surface that join points of equal magnetic variation are called:

#67. A negative (westerly) magnetic variation signifies that :

#68. The angle between Magnetic North and Compass North is called:

#69. The north and south magnetic poles are the only positions on the earth's surface where:

#70. The rhumb-line distance between points A (60°00'N 002°30'E) and B (60°00'N 007°30'W) is:

#71. 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:

#72. 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?

#73. 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:

#74. An aircraft at position 60°N 005°W tracks 090°(T) for 315 km.On completion of the flight the longitude will be:

#75. The 'departure' between positions 60°N 160°E and 60°N 'x' is 900 NM.What is the longitude of 'x'?

#76. An aircraft at latitude 02°20'N tracks 180°(T) for 685 km.On completion of the flight the latitude will be:

#77. An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will its latitude be after 1.5 HR?

#78. An aircraft at latitude 10°North flies south at a groundspeed of 445 km/HR.What will be its latitude after 3 HR?

#79. 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?

#80. 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?

#81. When is the magnetic compass most effective?

#82. . What is the value of the magnetic dip at the magnetic south pole ?

#83. The value of magnetic variation:

#84. . Isogonals converge at the:

#85. A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an:

#86. The horizontal component of the earth's magnetic field:

#87. Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually due to:

#88. The Earth can be considered as being a magnet with the:

#89. Which of the following statements concerning earth magnetism is completely correct?

#90. Which of the following statements concerning the earth's magnetic field is completely correct?

#91. The sensitivity of a direct reading compass varies:

#92. Isogonals are lines of equal :

#93. At a specific location, the value of magnetic variation:

#94. When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compass to:

#95. When decelerating on a westerly heading in the Northern hemisphere, the compass card of a direct reading magnetic compass will turn :

#96. 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:

#97. When turning right from 330°(C) to 040°(C) in the northern hemisphere, the reading of a direct reading magnetic compass will:

#98. When accelerating on an easterly heading in the Northern hemisphere, the compass card of a direct reading magnetic compass will turn :

#99. 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:

#100. When accelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn:

#101. Which of the following statements is correct concerning the effect of turning errors on a direct reading compass?

#102. At the magnetic equator, when accelerating after take off on heading West, a direct reading compass :

#103. Permanent magnetism in aircraft arises chiefly from:

#104. Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that :

#105. In northern hemisphere, during an acceleration in an easterly direction, the magnetic compass will indicate:

#106. The purpose of compass check swing is to:

#107. In a remote indicating compass system the amount of deviation caused by aircraft magnetism and electrical circuits may be minimised by:

#108. A direct reading compass should be swung when:

#109. The direct reading magnetic compass is made aperiodic (dead beat) by:

#110. The annunciator of a remote indicating compass system is used when:

#111. Which one of the following is an advantage of a remote reading compass as compared with a standby compass?

#112. Which of the following is an occasion for carrying out a compass swing on a Direct Reading Compass?

#113. The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane is:

#114. The main reason for usually mounting the detector unit of a remote indicating compass in the wingtip of an aeroplane is to:

#115. The main advantage of a remote indicating compass over a direct reading compass is that it:

#116. 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 :

#117. The nominal scale of a Lambert conformal conic chart is the:

#118. The chart that is generally used for navigation in polar areas is based on a:

#119. A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60° S?

#120. The distance measured between two points on a navigation map is 42 mm (millimetres). The scale of the chart is 1:1 600 000.The actual distance between these two point is approximately:

#121. 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:

#122. On a Lambert conformal conic chart the convergence of the meridians:

#123. A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is:

#124. On a direct Mercator projection, at latitude 45° North, a certain length represents 70 NM. At latitude 30° North, the same length represents approximately:

#125. 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:

#126. On a Mercator chart, the scale:

#127. In a navigation chart a distance of 49 NM is equal to 7 cm. The scale of the chart is approximately:

#128. At 60° N the scale of a direct Mercator chart is 1 : 3 000 000.What is the scale at the equator?

#129. 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?

#130. The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30°S?

#131. A Lambert conformal conic projection, with two standard parallels:

#132. The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010°E and 030°W is 30°, is:

#133. 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?

#134. 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:

#135. The chart distance between meridians 10° apart at latitude 65° North is 3.75 inches. The chart scale at this latitude approximates:

#136. On a Lambert conformal conic chart, with two standard parallels, the quoted scale is correct:

#137. 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?

#138. At 47° North the chart distance between meridians 10° apart is 5 inches. The scale of the chart at 47° North approximates:

#139. On a Lambert Conformal Conic chart earth convergency is most accurately represented at the:

#140. On a Transverse Mercator chart, scale is exactly correct along the:

#141. Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2 000 000?

#142. Transverse Mercator projections are used for:

#143. 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 :

#144. 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 :

#145. On a transverse Mercator chart, the scale is exactly correct along the:

#146. On a transverse Mercator chart, with the exception of the Equator, parallels of latitude appear as:

#147. An Oblique Mercator projection is used specifically to produce:

#148. 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 :

#149. 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:

#150. Given:Chart scale is 1 : 1 850 000. The chart distance between two points is 4 centimetres. Earth distance is approximately :

#151. 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:

#152. 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:

#153. 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?

#154. A straight line on a chart 4.89 cm long represents 185 NM.The scale of this chart is approximately :

#155. The scale on a Lambert conformal conic chart :

#156. A direct Mercator graticule is based on a projection that is :

#157. Parallels of latitude on a Direct Mercator chart are :

#158. . A straight line on a Lambert Conformal Projection chart for normal flight planning purposes:

#159. On a Direct Mercator chart, a rhumb line appears as a:

#160. On a Lambert Conformal Conic chart great circles that are not meridians are:

#161. On a Direct Mercator chart a great circle will be represented by a:

#162. The angular difference, on a Lambert conformal conic chart, between the arrival and departure track is equal to:

#163. The parallels on a Lambert Conformal Conic chart are represented by:

#164. Parallels of latitude, except the equator, are:

#165. 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:

#166. On a Direct Mercator chart, meridians are:

#167. On which of the following chart projections is it NOT possible to represent the north or south poles?

#168. Which one of the following, concerning great circles on a Direct Mercator chart, is correct?

#169. On a Lambert conformal conic chart, the distance between parallels of latitude spaced the same number of degrees apart :

#170. 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 ?

#171. Which one of the following describes the appearance of rhumb lines, except meridians, on a Polar Stereographic chart?

#172. What is the value of the convergence factor on a Polar Stereographic chart?

#173. On a Direct Mercator, rhumb lines are:

#174. Contour lines on aeronautical maps and charts connect points :

#175. On a Polar Stereographic chart, the initial great circle course from A 70°N 060°W to B 70°N 060°E is approximately:

#176. 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:

#177. 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:

#178. Given:Magnetic heading 311°Drift angle 10° leftRelative bearing of NDB 270°What is the magnetic bearing of the NDB measured from the aircraft?

#179. 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?

#180. 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?

#181. Given the following:True track: 192°Magnetic variation: 7°EDrift angle: 5° leftWhat is the magnetic heading required to maintain the given track?

#182. Given the following:Magnetic heading: 060°Magnetic variation: 8°WDrift angle: 4° rightWhat is the true track?

#183. Given: True track 180°Drift 8°RCompass heading 195°Deviation -2°Calculate the variation?

#184. Given: True course 300°drift 8°Rvariation 10°Wdeviation -4° Calculate the compass heading?

#185. Given:true track 352°variation 11° Wdeviation is -5°drift 10°R.Calculate the compass heading?

#186. Given:true track 070°variation 30°Wdeviation +1°drift 10°RCalculate the compass heading?

#187. Given: True course from A to B = 090°,TAS = 460 kt, W/V = 360/100kt,Average variation = 10°E,Deviation = -2°.Calculate the compass heading and GS?

#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?

#189. Given: GS = 510 kt.Distance A to B = 43 NMWhat is the time (MIN) from A to B?

#190. Given: GS = 122 kt.Distance from A to B = 985 NM.What is the time from A to B?

#191. Given: GS = 236 kt.Distance from A to B = 354 NMWhat is the time from A to B?

#192. Given: GS = 435 kt.Distance from A to B = 1920 NM.What is the time from A to B?

#193. Given: GS = 345 kt.Distance from A to B = 3560 NM.What is the time from A to B?

#194. Given: GS = 480 kt.Distance from A to B = 5360 NM.What is the time from A to B?

#195. Given: GS = 95 kt.Distance from A to B = 480 NM.What is the time from A to B?

#196. Given: GS = 105 kt.Distance from A to B = 103 NM.What is the time from A to B?

#197. Given: GS = 120 kt.Distance from A to B = 84 NM.What is the time from A to B?

#198. The ICAO definition of ETA is the:

#199. 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?

#200. Given:Course 040°(T),TAS is 120 kt,Wind speed 30 kt.Maximum drift angle will be obtained for a wind direction of:

#201. How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?

#202. Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL.What is the endurance?

#203. . What is the ratio between the litre and the US-GAL ?

#204. 265 US-GAL equals? (Specific gravity 0.80)265 US-GAL equals? (Specific gravity 0.80)

#205. 730 FT/MIN equals:

#206. How long will it take to fly 5 NM at a groundspeed of 269 Kt ?

#207. An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed?

#208. An aircraft travels 100 statute miles in 20 MIN, how long does it take to travel 215 NM?

#209. The equivalent of 70 m/sec is approximately:

#210. Given:IAS 120 kt,FL 80,OAT +20°C. What is the TAS?

#211. 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:

#212. For a given track the:Wind component = +45 ktDrift angle = 15° leftTAS = 240 ktWhat is the wind component on the reverse track?

#213. Given:Magnetic heading = 255°VAR = 40°WGS = 375 ktW/V = 235°(T) / 120 ktCalculate the drift angle?

#214. Given:True Heading = 180° TAS = 500 ktW/V 225° / 100 ktCalculate the GS?

#215. Given:True heading = 310° TAS = 200 ktGS = 176 ktDrift angle 7° right.Calculate the W/V?

#216. Given:True Heading = 090° TAS = 180 ktGS = 180 ktDrift 5° rightCalculate the W/ V?

#217. Given:True Heading = 090°TAS = 200 ktW/V = 220° / 30 kt.Calculate the GS?

#218. 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?

#219. 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)?

#220. 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?

#221. 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:

#222. The reported surface wind from the Control Tower is 240°/35 kt. Runway 30 (300°). What is the cross-wind component?

#223. Given:TAS = 132 kt,True HDG = 257°W/V = 095°(T)/35kt.Calculate the drift angle and GS?

#224. Given:TAS = 270 kt,True HDG = 270°,Actual wind 205°(T)/30kt,Calculate the drift angle and GS?

#225. Given:TAS = 270 kt,True HDG = 145°,Actual wind = 205°(T)/30kt.Calculate the drift angle and GS?

#226. Given:TAS = 470 kt,True HDG = 317°W/V = 045°(T)/45ktCalculate the drift angle and GS?

#227. Given:TAS = 140 kt,True HDG = 302°,W/V = 045°(T)/45ktCalculate the drift angle and GS?

#228. Given:TAS = 290 kt,True HDG = 171°,W/V = 310°(T)/30ktCalculate the drift angle and GS?

#229. Given:TAS = 485 kt,True HDG = 226°,W/V = 110°(T)/95kt.Calculate the drift angle and GS?

#230. Given:TAS = 235 kt,HDG (T) = 076°W/V = 040/40kt.Calculate the drift angle and GS?

#231. Given: TAS = 440 kt,HDG (T) = 349°W/V = 040/40kt.Calculate the drift and GS?

#232. Given: TAS = 465 kt,HDG (T) = 124°,W/V = 170/80kt.Calculate the drift and GS?

#233. Given: TAS = 95 kt,HDG (T) = 075°,W/V = 310/20kt.Calculate the drift and GS?

#234. Given: TAS = 140 kt,HDG (T) = 005°,W/V = 265/25kt.Calculate the drift and GS?

#235. Given: TAS = 190 kt,HDG (T) = 355°,W/V = 165/25kt.Calculate the drift and GS?

#236. Given: TAS = 230 kt,HDG (T) = 250°,W/V = 205/10kt.Calculate the drift and GS?

#237. Given: TAS = 205 kt,HDG (T) = 180°,W/V = 240/25kt.Calculate the drift and GS?

#238. Given: TAS = 250 kt,HDG (T) = 029°,W/V = 035/45kt.Calculate the drift and GS?

#239. Given: TAS = 132 kt,HDG (T) = 053°,W/V = 205/15kt.Calculate the Track (°T) and GS?

#240. 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?

#241. 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?

#242. Given: Maximum allowable crosswind component is 20 kt. Runway 06, RWY QDM 063°(M). Wind direction 100°(M)Calculate the maximum allowable windspeed?

#243. Given:TAS = 472 kt,True HDG = 005°,W/V = 110°(T)/50kt.Calculate the drift angle and GS?

#244. Given:TAS = 190 kt,True HDG = 085°,W/V = 110°(T)/50kt.Calculate the drift angle and GS?

#245. Given: TAS = 220 kt,Magnetic course = 212 º, W/V 160 º(M)/ 50kt,Calculate the GS?

#246. 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)?

#247. Given:TAS = 370 kt,True HDG = 181°,W/V = 095°(T)/35kt.Calculate the true track and GS?

#248. Given:TAS = 375 kt,True HDG = 124°,W/V = 130°(T)/55kt.Calculate the true track and GS?

#249. Given:TAS = 125 kt,True HDG = 355°,W/V = 320°(T)/30kt.Calculate the true track and GS?

#250. Given: TAS = 198 kt,HDG (°T) = 180,W/V = 359/25.Calculate the Track(°T) and GS?

#251. Given: TAS = 135 kt,HDG (°T) = 278,W/V = 140/20ktCalculate the Track (°T) and GS?

#252. Given: TAS = 225 kt,HDG (°T) = 123°,W/V = 090/60kt.Calculate the Track (°T) and GS?

#253. Given: TAS = 480 kt,HDG (°T) = 040°,W/V = 090/60kt.Calculate the Track (°T) and GS?

#254. Given: TAS = 155 kt,HDG (T) = 216°,W/V = 090/60kt.Caslculate the Track (°T) and GS?

#255. Given: TAS = 90 kt,HDG (T) = 355°,W/V = 120/20kt.Calculate the Track (°T) and GS?

#256. Given: TAS = 485 kt,HDG (T) = 168°,W/V = 130/75kt.Calculate the Track (°T) and GS?

#257. Given: TAS = 155 kt,Track (T) = 305°,W/V = 160/18kt.Calculate the HDG (°T) and GS?

#258. Given: TAS = 130 kt,Track (T) = 003°,W/V = 190/40kt.Calculate the HDG (°T) and GS?

#259. Given: TAS = 227 kt,Track (T) = 316°, W/V = 205/15kt.Calculate the HDG (°T) and GS?

#260. Given: TAS = 465 kt,Track (T) = 007°,W/V = 300/80kt.Calculate the HDG (°T) and GS?

#261. Given: TAS = 200 kt,Track (T) = 073°,W/V = 210/20kt.Calculate the HDG (°T) and GS?

#262. Given: TAS = 200 kt,Track (T) = 110°,W/V = 015/40kt.Calculate the HDG (°T) and GS?

#263. Given: TAS = 270 kt,Track (T) = 260°,W/V = 275/30kt.Calculate the HDG (°T) and GS?

#264. Given: True HDG = 307°,TAS = 230 kt,Track (T) = 313°,GS = 210 kt.Calculate the W/V?

#265. Given:True HDG = 233°,TAS = 480 kt,Track (T) = 240°,GS = 523 kt.Calculate the W/V?

#266. Given: True HDG = 133°,TAS = 225 kt,Track (T) = 144°,GS = 206 kt.Calculate the W/V?

#267. Given: True HDG = 074°,TAS = 230 kt,Track (T) = 066°,GS = 242 kt.Calculate the W/V?

#268. Given: True HDG = 206°,TAS = 140 kt,Track (T) = 207°,GS = 135 kt.Calculate the W/V?

#269. Given:True HDG = 054°,TAS = 450 kt,Track (T) = 059°,GS = 416 kt.Calculate the W/V?

#270. Given: True HDG = 145°,TAS = 240 kt,Track (T) = 150°,GS = 210 kt.Calculate the W/V?

#271. Given:True HDG = 002°,TAS = 130 kt,Track (T) = 353°,GS = 132 kt.Calculate the W/V?

#272. Given: True HDG = 035°,TAS = 245 kt,Track (T) = 046°,GS = 220 kt.Calculate the W/V?

#273. Given: course required = 085° (T),Forecast W/V 030/100kt,TAS = 470 kt,Distance = 265 NM.Calculate the true HDG and flight time?

#274. Given:Runway direction 083°(M),Surface W/V 035/35kt.Calculate the effective headwind component?

#275. 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?

#276. Given: Runway direction 230°(T), Surface W/V 280°(T)/40 kt. Calculate the effective cross-wind component?

#277. Given:Runway direction 210°(M),Surface W/V 230°(M)/30kt. Calculate the crosswind component? 13 kt

#278. Given:Runway direction 305°(M),Surface W/V 260°(M)/30 kt. Calculate the crosswind component?

#279. Given:Magnetic track = 075°,HDG = 066°(M),VAR = 11°E,TAS = 275 ktAircraft flies 48 NM in 10 MIN.Calculate the true W/V °?

#280. 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?

#281. 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 :

#282. Given:FL120, 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?

#283. . 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

#284. An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain, its height is approximately

#285. Given:FL 350,Mach 0.80,OAT -55°C. Calculate the values for TAS and local speed of sound (LSS)?

#286. Given: Pressure Altitude 29000 FT, OAT -55°C. Calculate the Density Altitude?

#287. Given:TAS = 485 kt,OAT = ISA +10°C,FL 410.Calculate the Mach Number?

#288. What is the ISA temperature value at FL 330?

#289. Given: TAS 487kt, FL 330, Temperature ISA + 15. Calculate the MACH Number?

#290. Given: FL250, OAT -15 ºC, TAS 250 kt.Calculate the Mach No.?

#291. Given:Airport elevation is 1000 ft. QNH is 988 hPa.What is the approximate airport pressure altitude?(Assume 1 hPa = 27 FT)

#292. Given :True altitude 9000 FT,OAT -32°C,CAS 200 kt.What is the TAS?

#293. Given:Aircraft at FL 150 overhead an airportElevation 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)

#294. 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 :

#295. Given:A polar stereographic chart whose grid is aligned with the zero meridian. Grid track 344°, Longitude 115°00'W,Calculate the true course?

#296. 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:

#297. 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)?

#298. Given:AD = Air distance GD = Ground distanceTAS = True AirspeedGS = GroundspeedWhich of the following is the correct formula to calculate ground distance (GD) gone?

#299. An aircraft was over 'A' at 1435 hours flying direct to 'B'.Given:Distance 'A' to 'B' 2900 NMTrue airspeed 470 ktMean wind component 'out' +55 ktMean wind component 'back' -75 ktThe ETA for reaching the Point of Equal Time (PET) between 'A' and 'B' is:

#300. 61.4.7.0 (4483) An aircraft was over 'A' at 1435 hours flying direct to 'B'.Given:Distance 'A' to 'B' 2900 NMTrue airspeed 470 ktMean wind component 'out' +55 ktMean wind component 'back' -75 ktSafe endurance 9 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is:

#301. Given:Distance 'A' to 'B' 2484 NMGroundspeed 'out' 420 ktGroundspeed 'back' 500 ktThe time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is:

#302. Given:Distance 'A' to 'B' 2484 NMMean groundspeed 'out' 420 ktMean groundspeed 'back' 500 ktSafe endurance 08 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is:

#303. An aircraft was over 'Q' at 1320 hours flying direct to 'R'.Given:Distance 'Q' to 'R' 3016 NMTrue airspeed 480 ktMean wind component 'out' -90 ktMean wind component 'back' +75 ktThe ETA for reaching the Point of Equal Time (PET) between 'Q' and 'R' is:

#304. 3016 NMTrue airspeed 480 ktMean wind component 'out' -90 ktMean wind component 'back' +75 ktSafe endurance 10:00 HR The distance from 'Q' to the Point of Safe Return (PSR) 'Q' is:

#305. Given:Distance 'A' to 'B' 1973 NMGroundspeed 'out' 430 ktGroundspeed 'back' 385 ktThe time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is:

#306. Given:Distance 'A' to 'B' 1973 NMGroundspeed 'out' 430 ktGroundspeed 'back' 385 ktSafe endurance 7 HR 20 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is:

#307. Given:Distance 'A' to 'B' 2346 NMGroundspeed 'out' 365 ktGroundspeed 'back' 480 ktThe time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is:

#308. Given:Distance 'A' to 'B' 2346 NMGroundspeed 'out' 365 ktGroundspeed 'back' 480 ktSafe endurance 8 HR 30 MIN The time from 'A' to the Point of Safe Return (PSR) 'A' is:

#309. Given:Distance 'Q' to 'R' 1760 NMGroundspeed 'out' 435 ktGroundspeed 'back' 385 ktThe time from 'Q' to the Point of Equal Time (PET) between 'Q' and 'R' is:

#310. Given:Distance 'Q' to 'R' 1760 NMGroundspeed 'out' 435 ktGroundspeed 'back' 385 ktSafe endurance 9 HR The distance from 'Q' to the Point of Safe Return (PSR) between 'Q' and 'R' is:

#311. Given:Distance 'A' to 'B' 3623 NMGroundspeed 'out' 370 ktGroundspeed 'back' 300 ktThe time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is:

#312. 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:

#313. From the departure point, the distance to the point of equal time is :

#314. 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?

#315. A ground feature appears 30° to the left of the centre 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:

#316. 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:

#317. An island appears 30° to the left of the centre line on an airborne weather radar

#318. 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:

#319. 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?

#320. 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?

#321. 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:

#322. An island appears 60° 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 (MH) of 276° with the magnetic variation (VAR) 10°E?

#323. An island appears 45° to the right 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 (MH) of 215° with the magnetic variation (VAR) 21°W?

#324. An island appears 30° to the right 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 (MH) of 355° with the magnetic variation (VAR) 15°E?

#325. 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 (MH) of 020° with the magnetic variation (VAR) 25°W?

#326. An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt.The rate of descent of the aircraft is approximately:

#327. 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?

#328. 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:

#329. . 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:

#330. 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:

#331. 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?

#332. 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:

#333. 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:

#334. An aircraft at FL330 is rerquired 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?

#335. 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?

#336. An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL80. The mean GS for the descent is 340 kt. What is the minimum rate of descent required?

#337. What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS?

#338. 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?

#339. Given:ILS GP angle = 3.5 DEG,GS = 150 kt.What is the approximate rate of descent?

#340. Given:aircraft height 2500 FT,ILS GP angle 3°.At what approximate distance from THR can you expect to capture the GP?

#341. A pilot receives the following signals from a VOR DME station: radial 180°+/- 1°, distance = 200 NM. What is the approximate error?

#342. 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:

#343. 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:

#344. 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?

#345. 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:

#346. 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?

#347. 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?

#348. 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'?

#349. 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'?

#350. 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'?

#351. 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'?

#352. 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'?

#353. 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'?

#354. 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'?

#355. 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'?

#356. 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:

#357. 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?

#358. 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?

#359. The flight log gives the following data :""True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compass heading""The right solution, in the same order, is :

#360. The purpose of the Flight Management System (FMS), as for example installed in the B737-400, is to provide:

#361. Which component of the B737-400 Flight Management System (FMS) is used to enter flight plan routeing and performance parameters?

#362. What indication, if any, is given in the B737-400 Flight Management System if radio updating is not available?

#363. What is the validity period of the 'permanent' data base of aeronautical information stored in the FMC In the B737-400 Flight Management System?

#364. In the B737-400 Flight Management System the CDUs are used during preflight to:

#365. How is the radio position determined by the FMC in the B737-400 Electronic Flight Instrument System?

#366. 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

#367. 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?

#368. Which of the following can all be stored as five letter waypoint identifiers through the CDU of a B737-400 Electronic Flight Instrument System?

#369. 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?

#370. Which FMC/CDU page normally appears on initial power application to the B737-400 Electronic Flight Instrument System?

#371. 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?

#372. With reference to inertial navigation systems, a TAS input is:

#373. The platform of an inertial navigation system (INS) is maintained at right angles to the local vertical by applying corrections for the effects of:

#374. Some inertial reference and navigation systems are known as ""strapdown"". This means that:

#375. In order to maintain an accurate vertical using a pendulous system, an aircraft inertial platform incorporates a device:

#376. The term drift refers to the wander of the axis of a gyro in:

#377. The resultant of the first integration from the north/south accelerometer of an inertial navigation system (INS) in the NAV MODE is:

#378. Double integration of the output from the east/west accelerometer of an inertial navigation system (INS) in the NAV MODE give:

#379. n an Inertial Navigation System (INS), Ground Speed (GS) is calculated:

#380. 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?

#381. The resultant of the first integration of the output from the east/west accelerometer of an inertial navigation system (INS) in NAV MODE is:

#382. Which of the following lists, which compares an Inertial Reference System that utilises Ring Laser Gyroscopes (RLG) instead of conventional gyroscopes, is completely correct?

#383. The principle of 'Schuler Tuning' as applied to the operation of Inertial Navigation Systems/ Inertial Reference Systems is applicable to:

#384. What additional information is required to be input to an Inertial Navigation System (INS) in order to obtain an W/V readout?

#385. 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?

#386. During initial alignment an inertial navigation system is north aligned by inputs from:

#387. During the initial alignment of an inertial navigation system (INS) the equipment:

#388. Which of the following statement is correct concerning gyro-compassing of an inertial navigation system (INS)?

#389. Which of the following statements concerning the loss of alignment by an Inertial Reference System (IRS) in flight is correct?

#390. The alignment time, at mid-latitudes, for an Inertial Reference System using laser ring gyros is approximately:

#391. Which of the following statements concerning the alignment procedure for Inertial Navigation Systems(INS)/Inertial Reference Systems (IRS) at mid-latitudes is correct?

#392. A pilot accidently turning OFF the INS in flight, and then turns it back ON a few moments later. Following this incident:

#393. 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 :

#394. 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:

#395. 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:

#396. Gyrocompassing of an inertial reference system (IRS) is accomplished with the mode selector switched to:

#397. Which of the following correctly lists the order of available selections of the Mode Selector switches of an inertial reference system (IRS) mode panel?

#398. ATT Mode of the Inertial Reference System (IRS) is a back-up mode providing

#399. Which of the following statements concerning the operation of an Inertial Navigation System (INS)/Inertial Reference System (IRS) is correct?

#400. 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?

#401. Waypoints can be entered in an INS memory in different formats.In which of the following formats can waypoints be entered into all INSs?

#402. An aircraft equipped with an Inertial Navigation System (INS) flies with INS 1 coupled with autopilot 1. Both inertial navigation systems are navigating from way-point A to B. The inertial systems' Central Display Units (CDU) sho shows:- XTK on INS 1 = 0- XTK on INS 2 = 8L(XTK = cross track)From this information it can be deduced that:

#403. 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°WWhen 025°W is passed the latitude shown on the display unit of the inertial navigation system will be:

#404. 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:

#405. 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:

#406. 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°WThe inertial navigation system is connected to the automatic pilot on route (1-2-3).The track change when passing WPT 2 will be approximately:

#407. 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:

#408. 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°?

#409. 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:

#410. Which of the following statements concerning the position indicated on the Inertial Reference System (IRS) display is correct?

#411. What is the source of magnetic variation information in a Flight Management System (FMS)?

#412. Where and when are the IRS positions updated?

#413. The sensors of an INS measure:

#414. What is the approximate maximum theoretical range at which an aircraft at FL130 could receive information from a VDF facility which is sited 1024 FT above MSL?

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GENERAL NAVIGATION
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