Given Distance from departure to destination 340 NMTrue track 320°Wind 160°/40 ktTAS 110 ktWhat is the distance of the PET from the departure point ?
Diploma > registration
Use Route Manual chart NAPOn a direct great circle course from REYKJAVIK 64°10' N 022°00'W to AMSTERDAM 52°32'N 004°50'E the a average true course and b distance are ERR _a_033 510 ?
Given Distance from departure to destination 330 NM Safe Endurance 5 h True Track 170°Wind 140/25TAS 125 ktWhat is the distance of the PSR from the departure point ?
Route manual chart nap the initial magnetic course from a 64°n006°e to c 62°n020°w err _a_033 520 (a) 3 ° (b) 95 nm. img /com_en/com033 520 jpg we are looking initial magnetic course with you protractor aligned on true north you have a true course of 271° add 4°west magnetic variation = 275°. 
The maximum wind velocity °/kt immediately north of tunis 36°n010°e err _a_033 521 (a) 3 ° (b) 95 nm. arrows feathers pennants arrows indicate direction number or pennants and/or feathers correspond to speed example with a 270°/115 kt wind img /com_en/com033 346a jpg pennants correspond to 50 kt feathers correspond to 10 kt half feathers correspond to 5 kt img /com_en/com033 521 jpg . 
Given distance a to b 3060 nmmean groundspeed 'out' 440 ktmean groundspeed 'back' 540 ktsafe endurance 10 hoursthe time to point of safe return psr (a) 3 ° (b) 95 nm. arrows feathers pennants arrows indicate direction number or pennants and/or feathers correspond to speed example with a 270°/115 kt wind img /com_en/com033 346a jpg pennants correspond to 50 kt feathers correspond to 10 kt half feathers correspond to 5 kt img /com_en/com033 521 jpg . 
The approximate mean wind component kt at mach 0 78 along true course 270° at 50°n from 000° to 010°w err _a_033 541 4 kt headwind component. img /com_en/com033 541 jpg (55 kt + 55 kt + 30 kt) / 3 = 46 kt mean wind 240°/46kt headwind component = wind speed x cos (angle between wind the course) headwind component = 46 kt x cos 30° = 40 kt. Question 92-8
Given distance from departure to destination 340 nmgs out 150 ktgs home 120 ktwhat the distance of pet from departure point 4 kt headwind component. ground speed out (gso) = 150 kt ground speed home (gsh) = 120 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 340 x 120 / (150 + 120) distance to pet = 40800 / 270 = 151 nm. Question 92-9
Given distance from departure to destination 470 nmtrue track 237°wind 300/25 kt tas 125 ktwhat the distance of pet from departure point 4 kt headwind component. under index set true track 237° centre dot on tas 125 kt with rotative scale set wind 300°/25 kt you find a left drift of 11° now drift always measured from heading to track turn to set true heading 248° (237° + 11° left drift) under index you now read your ground speed out of 111 kt proceed in same way to find ground speed home of 135 kt (right drift of 9° true heading of 048°) ground speed out (gso) = 111 kt ground speed home (gsh) = 135 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 470 x 135 / (111 + 135) distance to pet = 63450 / 246 = 258 nm (closest answer 256 nm). Question 92-10
From which of following would you expect to find details of search and rescue organisation and procedures sar 4 kt headwind component. under index set true track 237° centre dot on tas 125 kt with rotative scale set wind 300°/25 kt you find a left drift of 11° now drift always measured from heading to track turn to set true heading 248° (237° + 11° left drift) under index you now read your ground speed out of 111 kt proceed in same way to find ground speed home of 135 kt (right drift of 9° true heading of 048°) ground speed out (gso) = 111 kt ground speed home (gsh) = 135 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 470 x 135 / (111 + 135) distance to pet = 63450 / 246 = 258 nm (closest answer 256 nm). Question 92-11
Given distance from departure to destination 2380 nmgs out 420 ktgs home 520 ktwhat the time of pet from departure point 4 kt headwind component. ground speed out = 420 kt ground speed home = 520 kt pet = distance x gsh / (gso + gsh) pet = 2380 x 520 / (420 + 520) pet = 1237600 / 940 = 1316 6 nm time of pet from departure point 1316 6 / 420 = 3 13 h 3 13 x 60 = 188 minutes. Question 92-12
Given distance from departure to destination 338 nm true track 045°wind 225°/35 kt tas 120 ktwhat the distance and time of pet from departure point Distance 2 nm time 46 min. pet point of equal time distance to pet = (d x h)/o + h d = total track distance h = groundspeed home o = groundspeed out d = 338 nm h = 120 ? 35 kt (headwind) = 85 kt o = 120 + 35 kt (tailwind) = 155 kt pet (distance) = 338 x 85 / (155 + 85) distance = 119 7 nm = 120 nm time = (120 / 155 (ground speed out)) x 60 = 46 minutes. Question 92-13
Given distance from departure to destination 1500 nm safe endurance 4 5 h tas 450 kt ground speed out 480 kt ground speed home 410 kt what the time of psr from departure point Distance 2 nm time 46 min. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 480 kt ground speed home = 460 kt point of safe return (psr) = 4 5 x 410 / (480 + 410) point of safe return (psr) = 1845 / 890 point of safe return (psr) = 2 07 h 2 07 x 60 = 124 minutes. Question 92-14
Over london 51°n 000°e/w lowest fl listed which unaffected cat err _a_033 567 Distance 2 nm time 46 min. london located in cat area n°1 this cat area extends from fl240 to fl350 thus two levels are out of this layer fl230 fl360 the question asks the lowest one. Question 92-15
What lowest cloud conditions oktas/ft are forecast 1900 utc at hamburg eddh err _a_033 572 Distance 2 nm time 46 min. london located in cat area n°1 this cat area extends from fl240 to fl350 thus two levels are out of this layer fl230 fl360 the question asks the lowest one. Question 92-16
Given distance from departure to destination 5000 nm safe endurance 10 h tas 450 kt ground speed out 500 kt ground speed home 400 kt what the distance of psr from departure point Distance 2 nm time 46 min. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 500 kt ground speed home = 400 kt point of safe return (psr) = 10 x 400 / (500 + 400) point of safe return (psr) = 4000 / 900 point of safe return (psr) = 4 44 h distance of psr from departure point at a speed of 500 kt 4 44 h x 500 = 2222 nm. Question 92-17
What the earliest time utc if any that thunderstorms are forecast doha otbd err _a_033 603 Distance 2 nm time 46 min. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 500 kt ground speed home = 400 kt point of safe return (psr) = 10 x 400 / (500 + 400) point of safe return (psr) = 4000 / 900 point of safe return (psr) = 4 44 h distance of psr from departure point at a speed of 500 kt 4 44 h x 500 = 2222 nm. Question 92-18
Given maximum landing mass 51300 kgmaximum allowable take off mass 56300 kgdry operating mass 29100 kgload 11700 kgtrip fuel 3000 kgcontingency fuel 215 kgfinal reserve fuel 1250 kgalternate fuel 1300 kgtaxi fuel 200 kgdetermine maximum possible extra fuel that can be uplifted this flight Distance 2 nm time 46 min. Ecqb03 october 2016 29100 + 11700 + 3000 + 215 + 1250 + 1300 ________ 46565 kg today's take off mass maximum possible take off mass = maximum landing mass + trip fuel maximum possible take off mass = 51300 + 3000 = 54300 kg extra fuel = 54300 46565 = 7735 kg. Question 92-19
Which best describes weather if any forecast johannesburg/jan smuts at 0400 utc Distance 2 nm time 46 min. prob30 0305 3000 bcfg bc bancs (patches) fg brouillard (fog). Question 92-20
What minimum visibility forecast taf damascus osdi 1517 Distance 2 nm time 46 min. prob30 0305 3000 bcfg bc bancs (patches) fg brouillard (fog). Question 92-21
Metar egly 301220 24015kt 200v280 8000 ra sct010 bkn025 ovc080 18/15 q0983 tempo 3000 ra bkn008 ovc020= Mean surface wind 24 ° (true) 5 kt varying between 2 ° 28 °. Egly issued at 1220z on 30th surface wind mean 240 deg true 15 kt varying between 200 280 deg prevailing vis 8 km weather light rain cloud 1 2 oktas base 1000 ft 5 7 oktas 2500 ft temperature +18 °c dew point +15°c qnh 983 mb trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft. Question 92-22
For a long distance flight at fl370 long range regime divided into four flight legs with following specifications leg 1 ground distance 2 000 nm headwind component 50 ktleg 2 ground distance 1 000 nm headwind component 30 ktleg 3 ground distance 500 nm tailwind component 70 ktleg 4 ground distance 1 000 nm headwind component 20 kt the total air distance approximately 1518 Mean surface wind 24 ° (true) 5 kt varying between 2 ° 28 °. Egly issued at 1220z on 30th surface wind mean 240 deg true 15 kt varying between 200 280 deg prevailing vis 8 km weather light rain cloud 1 2 oktas base 1000 ft 5 7 oktas 2500 ft temperature +18 °c dew point +15°c qnh 983 mb trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft. Question 92-23
Which best describes maximum intensity of turbulence at fl 290 above london 1519 Mean surface wind 24 ° (true) 5 kt varying between 2 ° 28 °. Egly issued at 1220z on 30th surface wind mean 240 deg true 15 kt varying between 200 280 deg prevailing vis 8 km weather light rain cloud 1 2 oktas base 1000 ft 5 7 oktas 2500 ft temperature +18 °c dew point +15°c qnh 983 mb trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft. Question 92-24
What minimum visibility forecast paris/charles de gaulle at 2100 utc 1521 Mean surface wind 24 ° (true) 5 kt varying between 2 ° 28 °. Egly issued at 1220z on 30th surface wind mean 240 deg true 15 kt varying between 200 280 deg prevailing vis 8 km weather light rain cloud 1 2 oktas base 1000 ft 5 7 oktas 2500 ft temperature +18 °c dew point +15°c qnh 983 mb trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft. Question 92-25
A 'current flight plan' a Filed flight plan with amendments clearance included. Egly issued at 1220z on 30th surface wind mean 240 deg true 15 kt varying between 200 280 deg prevailing vis 8 km weather light rain cloud 1 2 oktas base 1000 ft 5 7 oktas 2500 ft temperature +18 °c dew point +15°c qnh 983 mb trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft. Question 92-26
The maximum permissible take off mass of an aircraft the l wake turbulence category on an atc flight plan Filed flight plan with amendments clearance included. Light aeroplane 7000 kg or less medium aeroplane less than 136000 kg but more than 7000 kg heavy aeroplane 136000 kg or greater for information helicopters produce vortices when in flight there some evidence that per kilogram of gross mass their vortices are more intense than those of fixed wing aircraft. Question 92-27
How many hours in advance of eobt estimated off block time should a atc flight plan be filed in case of flights into areas subject to air traffic flow management atfm Filed flight plan with amendments clearance included. Atfm (air traffic flow management) defined as 'a function established with objective of contributing to a safe orderly expeditious flow of air traffic ensuring that atc capacity utilised to maximum extent possible that traffic volume compatible with capacities declared the appropriate air traffic service provider in case of atfm (air traffic flow management) flight plan should be filed at least three hours in advance of eobt. Question 92-28
Which of following statements are correct with regard to advantages of computer flight plans 1 the computer can file atc flight plan 2 wind data used the computer always more up to date than that available to pilot Filed flight plan with amendments clearance included. Flight planning computers can generate both atc flight plan operational flight plan (by taking in account forecast wind take off mass traffic load temperature etc ) but since it based on forecast data while in flight weather info available to pilot will always be more accurate than anything stored in a computer. Question 92-29
In ats flight plan item 10 'standard equipment' considered to be err _a_033 112 Vhf rtf adf vor ils. img /com_en/com033 112 png icao doc 4444. Question 92-30
In an atc flight plan item 15 in order to define a position as a bearing and distance from a vor group of figures should consist of 1522 Vor ident magnetic bearing distance in nautical miles. Icao doc4444 pans atm . Question 92-31
When an atc flight plan has been submitted a controlled flight flight plan should be amended or cancelled in event of off block time being delayed Vor ident magnetic bearing distance in nautical miles. Icao doc4444 pans atm . Question 92-32
When completing an atc flight plan an elapsed time item 16 of 1 hour 55 minutes should be entered Vor ident magnetic bearing distance in nautical miles. Icao doc4444 pans atm . Question 92-33
When a pilot fills in an atc flight plan he must indicate wake turbulence category this category a function of which mass Maximum certified take off mass. Icao doc4444 pans atm . Question 92-34
Which of following statements are correct with regard to operation of flight planning computers 1 the computer can file atc flight plan 2 in event of in flight re routing computer produces a new plan Maximum certified take off mass. flight planning computers can generate both atc flight plan operational flight plan (by taking in account forecast wind take off mass traffic load temperature etc ) but since it based on forecast data while in flight weather info available to pilot will always be more accurate than anything stored in a computer the computer can not produce a new 'vali flight plan. Question 92-35
From options given below select those flights which require flight plan notification 1 any public transport flight2 any ifr flight3 any flight which to be carried out in regions which are designated to ease provision of alerting service or operations of search and rescue4 any cross border flights5 any flight which involves overflying water Maximum certified take off mass. a flight plan required if you are expecting to use atc services if you are going to cross borders. Question 92-36
In ats flight plan a non scheduled flight which of following letters should be entered in item 8 type of flight 1522 Maximum certified take off mass. Icao doc4444 pans atm . Question 92-37
In atc flight plan item 15 a cruising speed of 470 knots will be entered as Maximum certified take off mass. Icao doc4444 pans atm . Question 92-38
In event that selcal prescribed an appropriate authority in which section of ats flight plan will selcal code be entered 1522 Maximum certified take off mass. Icao doc4444 pans atm . Question 92-39
When an atc flight plan submitted a flight outside designated ats routes points included in item 15 route should not normally be at intervals of more than 3 minutes flying time or 37 km. Icao doc4444 pans atm . Question 92-40
On an atc flight plan an aircraft indicated as 'h' 'heavy' Is of highest wake turbulence category. icao mandates separation minima based upon wake vortex categories that are in turn based upon maximum take off weight (mtow) of aircraft these minima are categorised are as follows light mtow of 7000 kilograms or less medium mtow of greater than 7000 kilograms but less than 136000 kilograms heavy mtow of 136000 kilograms or greater atpl examination icao based there no 'super' wake turbulence category airbus a380 but a recommandation inviting pilots to use expression 'super' immediately after their call sign in initial radiotelephony contact between ats units the wake turbulence category 'j' airbus a380 only a guidance pending.
Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.
3639 Free Training Exam