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 ?
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An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80 Mean Ground Speed during descent is 271kt What is the minimum rate of descent required ?
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 ?
An aircraft planned to fly from position 'a' to position 'b' distance 480 nm at an average ground speed of 240 kt it departs 'a' at 1000 utc after flying 150 nm along track from 'a' aircraft 2 minutes behind planned time using actual gs experienced what the revised eta at 'b' 150 nm / 240 kt = 0 625 > 37 5 min + 2 min (behind) = 39 5 min > 0 658h 150 nm / 0 658h = 228 kt 480 nm 150 nm = 330 nm 330 nm / 228 kt = 1 447h > 86 8 min 86 8 min + 39 5 min = 126 3 min > 2h 06 min 18 seconds 10 00 + 2h 06 min = 12 06 utc. 
An aircraft 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' aircraft 3 minutes ahead of planned time using actual gs experienced what the revised eta at 'b' 70 nm / 180 kt = 0 389 > 23 3min as we are 3 minute ahead of planned time so our actual time 20 3 min > 0 339h 70 nm / 0 339 = 207 kt 320 nm 70 nm = 250 nm 250 nm / 207 kt = 1 21h > 72 5 min 72 5 min + 20 3 min = 92 8 min > 1h 32 min 48 seconds 12 00 + 1h 33 min = 13 33 utc. 
An aircraft 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' aircraft 1 5 minute behind planned time using actual gs experienced what the revised eta at 'b' 75 nm / 115 kt = 0 652h > 39 min (0 652 x 60) 9 00 + 39 min = 9 39 + 1 5 min = 9 40 5 (we would arrive at 9 39 but we are 1 5 minute behind so +1 5 min ) 40 5 / 60 = 0 675 h (now count speed with revised time) 75 / 0 675 = 111 1 kt (adjusted gs to 1 5 min behind planned of course we are slower as we arrived later) 250 nm 75 nm = 175 nm (the remaining distance to fly) 175 / 111 1 = 1 575 h > 94 5 min (1 575 x 60) (remaining distance divided new gs) 9 40 5 + 1h 34 5 min = 11 15 utc stanley 250/75 = 3 33 3 33 x 1 5min = 5 min 250nm/115kt = 2 17h 2 17 x 60 = 130 min = 2h10min + 5 min = 2h15min > 9 00 +2h15' > 11 15. 
Given distance 'a' to 'b' 475 nm planned gs 315 kt atd actual time departure 1000 utc at 1040 utc a fix obtained at 190 nm along track what gs must be maintained from fix in order to achieve planned eta at 'b' 475 nm / 315 = 1 51 h (1 51 x 60 minutes = 91 minutes or 1h31) estimated time of arrival at 'b' 10h00 + 1h31 = 11h31 fix obtained along track at 10h40 shows a groud speed of 190 nm / 40 minutes = 4 75 nm/minutes 4 75 x 60 minutes = 285 kt instead of 315 kt planned it remains 475 nm 190 nm = 285 nm it remains 51 minutes to achieve distance (10h40 to 11h31) 285 nm / 51 minutes =5 58 nm/minutes 5 6 x 60 = 336 kt. Question 170-8
Given distance 'a' to 'b' 325 nm planned gs 315 kt atd 1130 utc 1205 utc fix obtained 165 nm along track what gs must be maintained from fix in order to achieve planned eta at 'b' 325 nm / 315 kt = 1 03h > 62 min (1 03 x 60) 11 30 + 62 min = 12 32 12 32 12 05 = 27 min > 0 45h (27/60) 325 nm 165 nm = 160 nm (so we need to make 160 nm in 27 min) 160 nm / 0 45 = 355 kt. Question 170-9
Given distance 'a' to 'b' 100 nm fix obtained 40 nm along and 6 nm to left of course what heading alteration must be made to reach 'b' tan 1 (6/40) = 8 53 tan 1 (6/60) = 5 71 total = 14 24° or using formula tke = 6 x 60/40 6 x 60/60 tke = 9° 6° ca = 9° + 6° = 15° (as we are left of course correction to right). Question 170-10
Given distance 'a' to 'b' 90 nm fix obtained 60 nm along and 4 nm to right of course what heading alteration must be made to reach 'b' to calculate heading change at an off course fix to directly reach next waypoint use one in sixty rule (4/60) x 60 = 4° (4/30) x 60 = 8° 8° + 4° = 12° left because we are to right of target course. Question 170-11
Complete line 1 of 'flight navigation log'positions 'a' to 'b' what the hdg° m and eta 2497 Hdg 268° eta 4 utc. . Question 170-12
Complete line 2 of 'flight navigation log' positions 'c' to ' what the hdg° m and eta 2497 Hdg 93° eta 239 utc. . Question 170-13
Complete line 3 of 'flight navigation log' positions 'e' to 'f' what the hdg° m and eta 2497 Hdg 5° eta 2 5 utc. . Question 170-14
Complete line 4 of 'flight navigation log' positions 'g' to 'h' what the hdg° m and eta 2497 Hdg 344° eta 336 utc. . Question 170-15
Complete line 5 of 'flight navigation log' positions 'j' to 'k' what the hdg° m and eta 2497 Hdg 337° eta 422 utc. . Question 170-16
Complete line 6 of 'flight navigation log' positions ' to 'm' what the hdg° m and eta 2497 Hdg 75° eta 5 2 utc. you have to find tas set 55°c in airspeed window (next to mkt ) read on outer main scale in front of 0 84 a tas of 485 kt below center dot set tas 485 kt under index set true course 070° with rotative scale set wind (020°/60kt) read below 60 kt a 6° right drift true heading = 070° 6° = 064° gs 441 kt after having set true heading on computer magnetic heading = 064° + 11° variation = 075° (495/441)x60 = 67 minutes. Question 170-17
Given tas = 197 kt true course = 240° w/v = 180/30kt descent initiated at fl 220 and completed at fl 40 distance to be covered during descent 39 nm what the approximate rate of descent Hdg 75° eta 5 2 utc. using flight computer you will get gs = 182 kt distance = rate x time 39 nm = 182 x time t = 39/182 = 0 214h > 12 8 min (total time of descent) 22000 ft 4000 ft = 18000 ft (height to be flown in descent) 18000 / 12 8 min = 1406 ft/min. Question 170-18
Given ils glide path angle = 3 5° ground speed = 150 kt what the approximate rate of descent Hdg 75° eta 5 2 utc. 1 in 60 rule a rule of thumb (3 5 x 150 x 100)/60 = 875 ft/min. Question 170-19
Given aircraft height 2500 ft ils gp angle 3° at what approximate distance from thr can you expect to capture gp Hdg 75° eta 5 2 utc. distance= height x 60 / angle° distance= 2500x60/3° distance= 50000 ft then divide 6080 (as 1nm=6080 ft). Question 170-20
An island appears 60° to left of centre line on an airborne weather radar display what the true bearing of aircraft from island if at time of observation aircraft was on a magnetic heading mh of 276° with magnetic variation 10°e Hdg 75° eta 5 2 utc. 276°(mh) + 10°e = 286°(th) 286° 60° = 226° (th to island to get from island simply reverse it) 226° 180° = 046°. Question 170-21
An island appears 45° to right of centre line on an airborne weather radar display what the true bearing of aircraft from island if at time of observation aircraft was on a magnetic heading mh of 215° with magnetic variation 21°w Hdg 75° eta 5 2 utc. 215°(mh) 21°w = 194°(th) 194° + 45° = 239° (th to island to get from island simply reverse it) 239° 180° = 059°. Question 170-22
An island appears 30° to right of centre line on an airborne weather radar display what the true bearing of aircraft from island if at time of observation aircraft was on a magnetic heading mh of 355° with magnetic variation var 15°e Hdg 75° eta 5 2 utc. 355°(mh) + 15°e = 010°(th) 010° + 30° = 040° (th to island to get from island simply reverse it) 040° + 180° = 220°. Question 170-23
An island appears 30° to left of centre line on an airborne weather radar display what the true bearing of aircraft from island if at time of observation aircraft was on a magnetic heading mh of 020° with magnetic variation var 25°w Hdg 75° eta 5 2 utc. 020°(mh) 25°w = 355°(th) island on left so 355° 30° = 325° true bearing from island so 325° 180° = 145°. Question 170-24
Given an aircraft flying a track of 255° m 2254 utc it crosses radial 360° from a vor station 2300 utc it crosses radial 330° from same station at 2300 utc distance between aircraft and station The same as it was at 2254 utc. this an isosceles triangle the first angle at 22 54 255° 180° = 75° the second angle at vor 30° the last one = 180° (75°+30°) = 75°. Question 170-25
The distance between two waypoints 200 nm to calculate compass heading pilot used 2°e magnetic variation instead of 2°w assuming that forecast w/v applied what will off track distance be at second waypoint The same as it was at 2254 utc. for each degree of error that you have at every 60 nm of travel you will be 1 nm off track you have 200/60 3 33 nm off track each degree of error total error 4° (from 2°e to 2°w) 4° x 3 33 nm = 13 33 nm using goniometric functions tan4° = ? / 200 ? = tan4° x 200 ? = 13 99 nm. Question 170-26
Given eta to cross a meridian 2100 utcgs 441 kttas 491 ktat 2010 utc atc requests a speed reduction to cross meridian at 2105 utc the reduction to tas will be approximately The same as it was at 2254 utc. at 20h10 airplane at 50 minutes of meridian with a speed of 441 kt thus at a distance of 367 5 nm ((441/60) x 50) now it has to travel 367 5 nm in 55 minutes (367 5 / 55) x 60 = 401 kt tas reduction is 441 401 = 40 kt. Question 170-27
The flight log gives following data 'true track drift true heading magnetic variation magnetic heading compass deviation compass heading' the right solution in same order 9° 3°l 22° 2°e 2 ° +4° 6°. use this wonderful table those questions. Question 170-28
At 0020 utc an aircraft crossing 310° radial at 40 nm of a vor/dme station at 0035 utc radial 040° and dme distance 40 nm magnetic variation zero the true track and ground speed are 9° 3°l 22° 2°e 2 ° +4° 6°. draw situation it a isosceles triangle with (at least) two equal sides the angle at vor 90° the other two angles are 45° at 00 20 bearing from aircraft to vor 310° 180° = 130° track is 130° 45° = 085° it not mandatory to calculate groundspeed but you can use pythagoras 40² + 40² = (distance between position at 00 20 postion at 00 35)² distance = 56 567 nm 56 nm covered in 15 minutes (56 567/15)x60 = 226 kt. Question 170-29
Given tas 120 ktata 'x' 1232 utceta 'y' 1247 utcata 'y' 1250 utcwhat eta 'z' 2500 9° 3°l 22° 2°e 2 ° +4° 6°. x to y = 30 nm in 18 minutes (from 12h32 to 12h50) = 100 kt ground speed y to z = 20 nm at 100 kt = 12 minutes 12h50 + 12 minutes = 13h02. Question 170-30
Given fl120 oat isa standard cas 200 kt track 222° m heading 215° m variation 15°w time to fly 105 nm 21 min what the wind 9° 3°l 22° 2°e 2 ° +4° 6°. oat at fl120 is 15° (2° x 12) = 9°c magnetic heading 215° variation 15°w = true heading 200° our magnetic track 222° minus 15°w = true track 207° (105 nm / 21 min) x 60 = 300 kt ground speed on computer under centre dot set tas 239 kt under index set true heading 200° mark point where drift 7°right crosses ground speed 300 kt wind is 050°/70 kt. Question 170-31
A useful method of a pilot resolving during a visual flight any uncertainty in aircraft's position to maintain visual contact with ground and Set heading towards a line feature such as a coastline motorway river or railway. oat at fl120 is 15° (2° x 12) = 9°c magnetic heading 215° variation 15°w = true heading 200° our magnetic track 222° minus 15°w = true track 207° (105 nm / 21 min) x 60 = 300 kt ground speed on computer under centre dot set tas 239 kt under index set true heading 200° mark point where drift 7°right crosses ground speed 300 kt wind is 050°/70 kt. Question 170-32
An aircraft descending down a 6% slope whilst maintaining a ground speed of 300 kt the rate of descent of aircraft approximately Set heading towards a line feature such as a coastline motorway river or railway. 1 nm = 6080 ft 6% ==> 0 06 (300 kt x 6080 ft / 60 min) x 0 06 = 1824 ft/min. Question 170-33
An aircraft flying according flight log at annex after 15 minutes of flying with planned tas and true heading aircraft 3 nm north of intended track and 2 5 nm ahead of dead reckoning position to reach destination b from this position true heading should be 2516 Set heading towards a line feature such as a coastline motorway river or railway. 15 min / 60 = 0 25h 0 25 x 130 kt (gs) = 32 5 nm 32 5 nm + 2 5 nm (ahead of dr) = 35 nm 50 nm (distance) 35 nm = 15 nm tke = 3 x 60/35 correction angle = 3 x 60/15 tke = 5° 12° ca = 5° + 12° = 17° as we are north of 275° to get back we need to fly more to south therefore minus 17° so 275° 17° = 258°. Question 170-34
An island observed to be 30° to right of nose of aircraft the aircraft heading 290° m variation 10° e the bearing ° t from aircraft to island Set heading towards a line feature such as a coastline motorway river or railway. magnetic heading 290° variation east magnetic least magnetic heading = true heading 10°e true heading = magnetic heading + 10°e = 300° the true bearing from aircraft to island is 300° + 30°right = 330°. Question 170-35
An aircraft follows a radial to a vor/dme station at 10 00 dme reads 120 nm at 10 03 dme reads 105 nm the estimated time overhead vor/dme station Set heading towards a line feature such as a coastline motorway river or railway. 15 nm in 3 minutes = 5 nm per minute distance to station 105 nm 105 / 5 nm/min = 21 minutes 10 03 + 00 21 = 10 24. Question 170-36
You are departing from an airport which has an elevation of 2000 ft the qnh 1013 hpa 10 nm away there a waypoint you are required to pass at an altitude of 7500 ft given a groundspeed of 100 kt what the minimum rate of climb Set heading towards a line feature such as a coastline motorway river or railway. total climb = 7500 ft 2000 ft = 5500 ft 10 nm at 100 kt = 10 nm/(100/60) = 6 minutes 5500 ft / 6 min = 916 6 ft/min. Question 170-37
At reference or see europe low altitude enroute chart e lo 1aan aircraft flying from inverness vordme n57°32 6' w004°02 5 to aberdeen vordme n57°18 6' w002°16 0' at 1000 utc fix of aircraft determined vordme inverness radial = 114dme distance = 20 5 nm at 1006 utc fix of aircraft determined vordme aberdeen radial = 294dme distance = 10 5 nm what the average gs of aircraft between 1000 utc and 1006 utc 2503 Set heading towards a line feature such as a coastline motorway river or railway. the distance between vor's 59 nm 20 5 nm + 10 5 nm = 31 nm 59 nm 31 nm = 28 nm (we need to make 28 nm in 6 minutes so ) 28 / 0 1 = 280 kt (6 minutes = 0 1h). Question 170-38
The distance between point of departure and destination 340 nm and wind velocity in whole area 100°/25 kt tas 140kt true track 135° and safe endurance 3h and 10 min how long will it take to reach point of safe return Set heading towards a line feature such as a coastline motorway river or railway. psr = e x h / o + h (e endurance h gs home o gs out) using flight computer you will get 20 kt headwind so gsout = 120 kt gshome = 160 kt) psr = 190 x 160 kt/120 + 160 (3h 10min > 190 min) psr = 108 6 min > 1h 49 min. Question 170-39
You are tracking 200° radial inbound to a vor and your true heading 010° at vor you then track 090° radial outbound and are showing a heading of 080°m the variation +5° and tas 240 kt what the wind °t has affected aircraft Set heading towards a line feature such as a coastline motorway river or railway. 200° radial inbound = magnetic track 020° variation +5° = true track 025° true heading = 010° drift = 15° right 090° radial outbound = magnetic track 090° variation +5° = true track 095° magnetic heading 080° variation +5° = true heading 085° drift = 10° right on computer centre dot on tas 240 kt put true heading 010° under index mark a line down 15° right drift line rotate to put true heading 085° under index mark a line down 10° right drift line the point where these two lines intersect the end of wind vector rotate to position it under centre dot read wind 310°/65 kt. Question 170-40
An aircraft flying according flight log at annex after 15 minutes of flying with planned tas and true heading aircraft 3 nm north of intended track and 2 5 nm ahead of dead reckoning position to reach destination b from this position true heading should be 2504 Set heading towards a line feature such as a coastline motorway river or railway. 15 min / 60 = 0 25h 0 25 x 130 kt (gs) = 32 5 nm 32 5 nm + 2 5nm (ahead of dr) = 35 nm 50 nm(dist ) 35 nm = 15 nm tke = 3 x 60/35 3 x 60/15 tke = 5° 12° ca = 5° + 12° = 17° as we are north of 095° to get back we need to fly more to south therefore add 17° so 095 + 17° = 112°.
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