Question 199-1 : Given gs = 95 kt distance from a to b = 480 nmwhat is the time from a to b ? [ Level reports ]

Question 199-2 : Given gs = 120 kt distance from a to b = 84 nmwhat is the time from a to b ?

00 hr 42 min.

84 nm 120 kt60 min = 42 minutes

Question 199-3 : Given distance 'a' to 'b' 1973 nmground speed out 430 ktground speed back 385 ktsafe endurance 7 hr 20 minthe distance from 'a' to the point of safe return psr is ?

1490 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 430 ktground speed home = 385 ktpoint of safe return psr = 733 x 385 430 + 385 point of safe return psr = 2822 815point of safe return psr = 346 hdistance of the psr from the departure point at a speed of 430 kt 346 h x 430 = 1489 nm

Question 199-4 : Given distance 'a' to 'b' 2346 nmground speed out 365 ktground speed back 480 ktthe time from 'a' to the point of equal time pet between 'a' and 'b' is ?

219 min.

Ground speed out 365 ktground speed home 480 ktpet = distance x gsh gso + gsh pet = 2346 x 480 365 + 480 = 1332 nm1332 nm 365 kt = 365 h365 h x 60 minutes = 219 minutes

Question 199-5 : Given distance 'q' to 'r' 1760 nmground speed out 435 ktground speed back 385 ktthe time from 'q' to the point of equal time pet between 'q' and 'r' is ?

114 min.

Ground speed out 435 ktground speed home 385 ktpet = distance x gsh gso + gsh pet = 1760 x 385 435 + 385 = 826 nm826 nm 435 kt = 19h19 h x 60 minutes = 114 minutes

Question 199-6 : Given distance 'q' to 'r' 1760 nmground speed out 435 ktground speed back 385 ktsafe endurance 9 hrthe distance from 'q' to the point of safe return psr between 'q' and 'r' is ?

1838 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 435 ktground speed home = 385 ktpoint of safe return psr = 9 x 385 435 + 385 point of safe return psr = 3465 820point of safe return psr = 422 hdistance of the psr from the departure point at a speed of 435 kt 422 h x 435 = 1838 nm

Question 199-7 : An aeroplane is flying at tas 180 kt on a track of 090° the wv is 045° 50kthow far can the aeroplane fly out from its base and return in one hour ?

85 nm.

Centre dot on tas 180 kt rotate to wind direction 045° come down from centre dot for wind speed 50 kt mark end of wind vector at 130 ktrotate to outbound track under heading index note drift14°starboard rotate track to drift note drift now 11°starboard rotate track to drift note drift still 11°starboard our outbound heading will be 079° to track 090° and our ground speed outbound is 141 kt 1382proceed same way to find ground speed homebound you will find 212 kt pnr = t x gso x gsh gso + gsh pnr = 1 x 141 x 212 141 + 212 pnr = 29892 353pnr = 8467 nm

Question 199-8 : An aircraft is maintaining a 52% gradient is at 7 nm from the runway on a flat terrain its height is approximately ?

2210 ft.

1 nm = 6080 ftaircraft is at 7 nm from the runway 7 nm x 6080 ft = 42560 ft42560 x 52100 = 2213 ft

Question 199-9 : An aircraft descends from fl250 to fl100the rate of descent is 1000 ftmin the gs is 360 ktthe flight path angle is ?

16°.

Vertical speed = gradient % * gs1000 = gradient % * 360gradient % = 277gradient % = angle * 10060277 = angle * 10060 > angle = 1662º

Question 199-10 : The outer marker of an ils with a 3° glide slope is located 46 nm from the threshold assuming a glide slope height of 50 ft above the threshold the approximate height of an aircraft passing the outer marker is ?

1450 ft.

46 nm x tan 3° = 024 nm1 nm = approximately 6000 ft024 x 6000 = 1440 ft1440 ft + 50 ft = 1490 ftthe approximate height of the aircraft is 1490 ft close to 1450 ft svandam in a 3° glide slope 1nm = 300 feetthen 46 nm * 300 = 1380 ft1380 ft + 50 ft = 1430 ftthe approximate height of the aircraft is 1430 ft close to 1450 ft johanjog more precise calculation 46 nm*6080 ft1nm = 27968 ft1 60 rule 3º60 = d27968 d = 13984 ft13984 + 50 ft = 14484 ft 1450 ft

Question 199-11 : 730 ftmin equals ?

37 msec.

Question 199-12 : How long will it take to fly 5 nm at a groundspeed of 269 kt ?

1 min 07 sec.

5 26960 =1115 minutes0115 x 60 = 7 secondes1 min 7 sec

Question 199-13 : An aircraft travels 24 statute miles in 47 seconds what is its groundspeed ?

160 kt.

2447 x 3600 = 184 statute miles per hour1 statute mile = 1609 km = 087 nm184 x 087 = 160 kt

Question 199-14 : The icao definition of eta is the ?

Estimated time of arrival at destination.

Question 199-15 : 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 ftmin ?

267 nm.

15000 ft 3000 ftmin = 5 minutes5 min 60 min = 00833 hour00833 x 320 = 2667 nm

Question 199-16 : An island appears 30° to the left of the centre line on an airborne weather radar display what is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation 12°w ?

054°.

Magnetic heading 276ºvariation 12ºwtrue heading 264ºisland bearing 30ºlefttrue bearing of the island from the aircraft 234ºtrue bearing of the aircraft from the island 234°+ 180º = 054°

Question 199-17 : 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 ?

960 ftmin.

37000 ft 13000 ft = 24000 ft120 nm 288kt = 0417h > 25 min 0417 x 60 24000 ft 25 min = 960 ftmin

Question 199-18 : An aircraft at fl310 m083 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 ?

M074.

1745set temperature 30°c in airspeed windowin front of 83 mach 83 inner scale on the outer scale you read tas = 503 kt 360503 x 60 = 43 minuteswe are at 43 minutes from the reporting point 360 x 60 = 48 minutes = 360 x 60 48 = 450 ktin airspeed window set outside temperature 30°c in front of mach index go to 450 kt on the outer scale and you read 74 mach 074 on the inner scale

Question 199-19 : 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 ktwhen the relative bearing was 280° the distance and true bearing of the aircraft from the feature was ?

30 nm and 240°.

5 minutes at 360 kt = 360 60 x 5 = 30 nm 1415magnetic heading 165°variation 25°wtrue heading = 140°true track= 140° + 10° = 150°we have an isosceles triangle and in an isosceles triangle two sides are equal in lengthrelative bearing 280°true bearing of the feature from the aircraft = 140 + 280 360 = 060°true bearing of the aircraft from the feature = 060 + 180 = 240°

Question 199-20 : An aircraft at fl350 is required to descend to cross a dme facility at fl80 maximum rate of descent is 1800 ftmin and mean gs for descent is 276 kt the minimum range from the dme at which descent should start is ?

69 nm.

35000 ft 8000 ft = 27000 ft27000 ft 1800 ftmin = 15 min > 025h 15 60 025 x 276 kt = 69 nm

Question 199-21 : 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 plannedassuming flight conditions do not change when 100 nm from the reporting point ias should be reduced to ?

159 kt.

Use nav computer to find tas 1390ias 200 kt = tas 240 ktground speed = 240 kt + 30 kt = 270 kt100 nm at gs 270 kt = 22 minutes100 nm in 27 minutes = 100 27 = 37 nm per minute37 x 60 = 222 kt222 kt 30 kt wind = 192 kttas 192 kt ==> with nav computer ==> 159 kt ias

Question 199-22 : An aircraft at fl350 is required to cross a vordme 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 ?

1340 ftmin.

24000 ft to descend33560 = 5583 nmmin1005583 = 18 minutes24000 18 min = 1340 ftmin

Question 199-23 : An aircraft at fl370 m086 oat 44°c headwind component 110 kt is required to reduce speed in order to cross a reporting point 5 minutes later than planned if the speed reduction were to be made 420 nm from the reporting point what mach number is required ?

M081.

2494tas is 503 kt ground speed is 503 110 = 393 kt420 nm at 393 kt = 107 h 107 x 60 = 64 minutes atc asks you to cross a rportin poitn 5 minutes later in 64 + 5 minutes69 60 = 115 h420 115 = 365 kt 2495365 kt + 110 kt = 475 kt = m 081

Question 199-24 : An aircraft at fl390 is required to descend to cross a dme facility at fl70 maximum rate of descent is 2500 ftmin mean gs during descent is 248 kt what is the minimum range from the dme at which descent should commence ?

53 nm.

39000 7000 = 32000 ft32000 2500 = 128 minutes248 kt 60 minutes = 414 nmminute128 x 414 = 53 nm

Question 199-25 : An aircraft at fl370 is required to commence descent when 100 nm from a dme facility and to cross the station at fl120if the mean gs during the descent is 396 kt the minimum rate of descent required is approximately ?

1650 ftmin.

37000 12000 = 25000 ft396 kt 60 minutes = 66 nmminute100 nm 66 = 1515 minutes before fly over dme25000 15 = 1650 ftmin

Question 199-26 : 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 minutes later than planned assuming that flight conditions do not change when 150 nm from the reporting point the aircraft should reduce ias by ?

20 kt.

Given ias cas 210 kt fl140 and oat 5°con the computer we find tas = 262 ktwe can now find the groundspeed 262 35 = 227 kt150 nm at 227 kt = 396 minuteswe must reduce speed to cross a point 5 minutes later 150 nm in 396 + 5 minutes = 202 ktwe must reduce ground speed by 227 202 = 25 kta tas of 262 25 = 237 kt is requirednow back to the computer tas 237 kt fl140 and oat 5°c new ias is 190 ktwe should reduce ias by 210 190 = 20 kt

Question 199-27 : At 0422 an aircraft at fl370 gs 320kt is on the direct track to vor 'x' 185 nm distantthe aircraft is required to cross vor 'x' at fl80 for a mean rate of descent of 1800 ftmin at a mean gs of 232 kt the latest time at which to commence descent is ?

04h45.

Fl370 to fl80 = 29000 ft29000 1800 ftmin = 161 minutes during the descent the aircraft will cover 23260 x 161 = 6225 nmdistance before top of descent tod 184 6225 = 12175 nmtime before tod 12175 32060 = 228 minutesthe latest time to commence descent is 04h22 + 23 minutes = 04h45

Question 199-28 : An aircraft at fl330 is required to commence descent when 65 nm from a vor and to cross the vor at fl100 the mean gs during the descent is 330 kt what is the minimum rate of descent required ?

1950 ft min.

33000 10000 = 23000 ft330 kt 60 minutes = 55 nmminute65 nm 55 = 118 minutes before fly over dme23000 118 = 1950 ftmin

Question 199-29 : 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 ?

1900 ft min.

50 nm 271 kt = 0185h > 11 min29000 ft 8000 ft = 21000 ft21000 ft 11 min = 1900 ftmin

Question 199-30 : 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 ?

1800 ftmin.

Fl350 fl080 = 27000 ft85 nm 340 kt = 025 h 15 minutes 27000 ft 15 min = 1800 ftmin

Question 199-31 : An aircraft is 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' the aircraft is 2 minutes behind planned timeusing the actual gs experienced what is the revised eta at 'b' ?

12 06 utc.

150 nm 240 kt = 0625 > 375 min + 2 min behind = 395 min > 0658h150 nm 0658h = 228 kt480 nm 150 nm = 330 nm330 nm 228 kt = 1447h > 868 min868 min + 395 min = 1263 min > 2h 06 min 18 seconds10 00 + 2h 06 min = 12 06 utc

Question 199-32 : 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 minutes ahead of planned timeusing the actual gs experienced what is the revised eta at 'b' ?

13 33 utc.

70 nm 180 kt = 0389 > 233minas we are 3 minute ahead of planned time so our actual time is 203 min > 0339h70 nm 0339 = 207 kt320 nm 70 nm = 250 nm250 nm 207 kt = 121h > 725 min725 min + 203 min = 928 min > 1h 32 min 48 seconds12 00 + 1h 33 min = 13 33 utc

Question 199-33 : 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 15 minute behind planned timeusing the actual gs experienced what is the revised eta at 'b' ?

11 15 utc.

75 nm 115 kt = 0652h > 39 min 0652 x 60 9 00 + 39 min = 9 39 + 15 min = 9 405 we would arrive at 9 39 but we are 15 minute behind so +15 min 405 60 = 0675 h now count the speed with revised time 75 0675 = 1111 kt adjusted gs to 15 min behind planned of course we are slower as we arrived later 250 nm 75 nm = 175 nm the remaining distance to fly 175 1111 = 1575 h > 945 min 1575 x 60 remaining distance divided by new gs 9 405 + 1h 345 min = 11 15 utc stanley 25075 = 333333 x 15min = 5 min250nm115kt = 217h217 x 60 = 130 min = 2h10min + 5 min = 2h15min > 9 00 +2h15' > 11 15

Question 199-34 : Given distance 'a' to 'b' is 475 nm planned gs 315 kt atd actual time departure 1000 utcat 1040 utc a fix is obtained at 190 nm along trackwhat gs must be maintained from the fix in order to achieve planned eta at 'b' ?

340 kt.

Question 199-35 : Given distance 'a' to 'b' is 325 nm planned gs 315 kt atd 1130 utc 1205 utc fix obtained 165 nm along trackwhat gs must be maintained from the fix in order to achieve planned eta at 'b' ?

355 kt.

325 nm 315 kt = 103h > 62 min 103 x 60 11 30 + 62 min = 12 3212 32 12 05 = 27 min > 045h 2760 325 nm 165 nm = 160 nm so we need to make 160 nm in 27 min 160 nm 045 = 355 kt

Question 199-36 : 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' ?

15° right.

Tan 1 640 = 853tan 1 660 = 571total = 1424°or using formula tke = 6 x 6040 and 6 x 6060tke = 9° and 6°ca = 9° + 6° = 15° as we are left of the course correction is to the right

Question 199-37 : Given distance 'a' to 'b' is 90 nm fix obtained 60 nm along and 4 nm to the right of coursewhat heading alteration must be made to reach 'b' ?

12° left.

To calculate the heading change at an off course fix to directly reach the next waypoint use the one in sixty rule 460 x 60 = 4° 430 x 60 = 8°8° + 4° = 12° and left because we are to the right of the target course

Question 199-38 : Complete line 1 of the 'flight navigation log'positions 'a' to 'b' what is the hdg° m and eta 2497 ?

Hdg 268° eta 1114 utc.

2496

Question 199-39 : Complete line 2 of the 'flight navigation log' positions 'c' to 'd' what is the hdg° m and eta 2497 ?

Hdg 193° eta 1239 utc.

2496

Question 199-40 : Complete line 3 of the 'flight navigation log' positions 'e' to 'f' what is the hdg° m and eta 2497 ?

Hdg 105° eta 1205 utc.

2496