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Question 196-1 : Given.tas = 200 kt, track t = 110°, w/v = 015/40 kt..calculate the hdg °t and gs ? [ Level reports ]
099° 199 kt.
.under index, set true track 110°, centre dot on tas 200 kt, with the rotative scale, set wind. 1775.now, drift is always measured from heading to track.turn to set true heading 099° 110° 11° right drift under index, you now read a ground speed of 198 kt... cmarzocchini.the answer is wrong. you have tail wind correct answer using sin and cos and cr3, 098/205... .don't be so confident and read carefully the explanation drift is always measured from heading to track..when you will be on your track, with the correct heading to counteract drift, the wind becomes a headwind.
Question 196-2 : Given.true hdg = 307°, tas = 230 kt, track t = 313°, gs = 210 kt..calculate the w/v ?
260/30kt.
.true heading is 307°, true track is 313° our drift is 6° right.. 2492.wind 261°/30kt.
Question 196-3 : Given.true hdg = 133°, tas = 225 kt, track t = 144°, gs = 206 kt..calculate the w/v ?
075/45kt.
.true heading is 133°, true track is 144° our drift is 11° right.. 2493
...true heading is 145°, true track is 150° our drift is 5° right... /com en/com061 194.jpg..wind 115°/35 kt.
Question 196-6 : Given.true hdg = 035°, tas = 245 kt, track t = 046°, gs = 220 kt..calculate the w/v ?
340/50kt
.true heading is 035°, true track is 046° our drift is 11° right.. 2520
Question 196-7 : Given course required = 085° t , forecast w/v 030/100kt, tas = 470 kt, distance = 265 nm. calculate the true hdg and flight time ?
075°, 39 min.
.tas = 470 kt.true course = 085°.vw = 030°/100kt..drift =.gs =.. a set true track to true index.. b turn the indicator to the wind direction, in this case using the black azimuth graduation the angle being upwind counting anti clockwise... c shift the speed arc corresponding to the true air speed so as to coincide with the wind speed on the indicator... d read the wind correction at the same place. read the ground speed under the center bore from the scal on the axis of the slide... setting.set 85° to true index, set the indicator to 030° on the black azimuth circle being upwind. adjust the speed arc labelled 470 of the diagram slide to the wind speed 10 100 kt of the indicator scale... reading.under the plotted point read the wind correction angle 10°. under the center bore read the ground speed 405 kt.. 1770.then, true heading = true course drift = 085° 10° = 075°..405/60 = 6.75 nm/minutes..265/6.75 = 39 minutes.
Question 196-8 : For a landing on runway 23 227° magnetic surface.wind reported by the atis is 180/30 kt..variation is 13°e..calculate the cross wind component ?
22 kt.
.wind from tower is already corrected for variation. the wind from tower refers to magnetic north...wind angle = 227° 180° = 47°..crosswind = windspeed x sin wind angle..crosswind = 30 kt x sin 47° = 22 kt.
Question 196-9 : 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 ?
10 kt.
.wind from tower atis is recorded by the tower is already corrected for variation. the wind from tower refers to magnetic north..this is a tailwind, our wind angle is = 047°+180° 210° = 17°..tailwind = windspeed x cos 17° = 10 kt..maximum allowable windspeed = 10 kt / cos 17° = 10.46 kt.
Question 196-10 : Given.maximum allowable crosswind component is 20 kt..runway 06, rwy qdm 063° m..wind direction 100° m..calculate the maximum allowable windspeed ?
33 kt.
.wind angle = 100° 063° = 37°..crosswind = windspeed x sin 37° = 20 kt..maximum allowable windspeed = 20 kt / sin 37° = 33 kt.
Question 196-11 : Given.true course a to b = 250°,.distance a to b = 315 nm,.tas = 450 kt..w/v = 200°/60kt..etd a = 0650 utc..what is the eta at b ?
0736 utc.
.set 250° under index, center dot on tas 450 kt and wind 200º/60kt. 2522.drift is 6° right...now, set heading 244° under index, read ground speed 410 kt...315 nm / 410 kt = 0.768 hour = 46 minutes 0.768 x 60...etd at a is 0650 utc + 46 minutes = 0736 utc.
Question 196-12 : Given.gs = 510 kt. distance a to b = 43 nm..what is the time from a to b ?
5 minutes.
.43 nm / 510 kt/60 min = 5 minutes.
Question 196-13 : Given.gs = 122 kt..distance from a to b = 985 nm..what is the time from a to b ?
8 hr 04 min.
.985 nm / 122 kt/60 min = 484 minutes 8h04.
Question 196-14 : Given.gs = 435 kt. distance from a to b = 1920 nm..what is the time from a to b ?
4 hr 25 min.
.1920 nm / 435 kt/60 min = 265 minutes 4h45.
Question 196-15 : Given.gs = 480 kt. distance from a to b = 5360 nm..what is the time from a to b ?
11 hr 10 min.
.5360 nm / 480 kt/60 min = 670 minutes 11h30.
Question 196-16 : Given.gs = 105 kt. distance from a to b = 103 nm..what is the time from a to b ?
00 hr 59 min.
.103 nm / 105 kt/60 min = 59 minutes.
Question 196-17 : Given.gs = 135 kt. distance from a to b = 433 nm..what is the time from a to b ?
3 hr 12 min.
.433 nm / 135 kt/60 min = 192 minutes 3h32.
Question 196-18 : Given.runway direction 083° m.surface wwind 035/35 kt..calculate the effective headwind component ?
24 kt.
.angle between the wind and the direction of the runway 083° 035° = 48°..effective headwind = cos of the angle between the wind and the direction of the runway x windspeed..effective headwind = cos 48° x 35 kt = 23.42 kt.
Question 196-19 : 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 ?
20 kt and 40 kt.
.crosswind = 35 kt maximum.35 = x. sin 60.x = 35 / sin 60 = 40 kt..headwind = 10 kt minimum.10 = x. cos 60.x = 10 / cos 60 = 20 kt.
Question 196-20 : Given.runway direction 230° t.surface wind 280° t /40 kt..calculate the effective cross wind component ?
31 kt.
.angle between the wind and the direction of the runway 280° 230° = 50°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin50° x 40 kt = 30.64 kt.
Question 196-21 : Given.runway direction 210° m , surface w/v 230° m /30 kt..calculate the cross wind component ?
10 kt.
Angle between the wind and the direction of the runway 230° 210° = 20°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin20° x 30 kt = 10.26 kt.
Question 196-22 : An aircraft obtains a relative bearing of 315° from an ndb at 08h30. at 08h40 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 08h40 ?
40 nm.
. 2521.you have an isoceles triangle and the angles are 45°. 45° and 90°...the hypotonuse is the distance from the 0830 position to the ndb, the two equal sides are. the distance travelled between 0830 and 0840..and. the distance from the 0840 position to the ndb..in 10 minutes at 240 kt the aircraft will travel 40 nm so this is also the distance from the 0830 position and the ndb...you can also use the 1 in 60 rule.315° 270° = 45°.240 kt / 60 min = 4° per minute.10 min x 4° = 40 nm.
Question 196-23 : The equivalent of 70 m/sec is approximately ?
136 kt.
.1 nm = 0.5 m/s..70 / 0.5 = 140 kt closest to 136 kt than 145 kt.....if you want to find the exact answer.70 m/s x 3600 secondes = 252000 m/h.252000 m/h = 252 km/h.252 / 1.852 = 136 kt.
Question 196-24 : Given.runway direction 305° m ,surface w/v 260° m /30 kt..calculate the cross wind component ?
21 kt.
.angle between the wind and the direction of the runway 305° 260° = 45°...crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin45° x 30 kt = 21.2 kt.
Question 196-25 : 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 ?
6° right.
Question 196-26 : 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 ?
1000 km.
.you are travelling south along the greenwich anti meridian, from 49°s to 58°s which is a 9° change of latitude..9° x 60 nm = 540 nm.540 nm x 1.852 = 1000 km.
Question 196-27 : 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' ?
8° right.
Question 196-28 : An aircraft was over 'a' at 1435 hours flying direct to 'b'. given.distance 'a' to 'b' 2900 nm.true airspeed 470 kt.mean wind component 'out' +55 kt.mean wind component 'back' 75 kt..the eta for reaching the point of equal time pet between 'a' and 'b' is ?
1657.
Ground speed out 470 + 55 = 525 kt.ground speed home 470 75 = 395 kt..pet = d x gsh / gso + gsh.pet = 2900 x 395 / 525 + 395 = 1245 nm...1245 nm / 525 kt = 2.37h..2.37 x 60 minutes = 142 minutes 2h42minutes..14h35 + 2h42 = 16h57.
Question 196-29 : Given.distance 'a' to 'b' 2484 nm.groundspeed 'out' 420 kt.groundspeed 'back' 500 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?
193 minutes.
.ground speed out 420 kt.ground speed home 500 kt..pet = distance x gsh / gso + gsh.pet = 2484 x 500 / 420 + 500 = 1350 nm...1350 nm / 420 kt = 3.21 h..3.21 h x 60 minutes = 193 minutes.
Question 196-30 : Given.distance 'a' to 'b' 2484 nm.ground speed out 420 kt.ground speed home 500 kt.safe endurance 08 h 30 minutes..the distance from 'a' to the point of safe return psr is ?
1940 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 420 kt.ground speed home = 500 kt..point of safe return psr = 8.5 x 500 / 420 + 500.point of safe return psr = 4250 / 920.point of safe return psr = 4.62 h..distance of the psr from the departure point at a speed of 420 kt.4.62 h x 420 = 1940 nm.
Question 196-31 : An aircraft was over 'q' at 1320 hours flying direct to 'r'. given.distance 'q' to 'r' 3016 nm.true airspeed 480 kt.mean wind component 'out' 90 kt.mean wind component 'back' +75 kt..endurance 10 h.what is the eta at the point of equal time pet ?
1752.
.ground speed out 480 90 = 390 kt.ground speed home 480 + 75 = 555 kt..pet = d x gsh / gso + gsh.pet = 3016 x 555 / 390 + 555 = 1771 nm...1771 nm / 390 kt = 4.54 h..4.54 x 60 minutes = 272 minutes 4h32minutes..13h40 + 4h32 = 17h52.
Question 196-32 : Given.distance 'a' to 'b' 1973 nm.groundspeed 'out' 430 kt.groundspeed 'back' 385 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?
130 minutes.
.ground speed out 430 kt.ground speed home 385 kt..pet = distance x gsh / gso + gsh.pet = 1973 x 385 / 430 + 385 = 932 nm...932 nm / 430 kt = 2.16 h..2.16 h x 60 minutes = 129.6 minutes.
Question 196-33 : Given.distance 'a' to 'b' 2346 nm.ground speed out 365 kt.ground speed back 480 kt.safe endurance 8 h 30 minutes.the time from 'a' to the point of safe return is ?
290 minutes.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 365 kt.ground speed home = 480 kt..point of safe return psr = 8.5 x 480 / 365 + 480.point of safe return psr = 4080 / 845.point of safe return psr = 4.83 h..the time from 'a' to the point of safe return is 4.83 x 60 minutes = 290 minutes.
Question 196-34 : Given.distance 'a' to 'b' 3623 nm.groundspeed 'out' 370 kt.groundspeed 'back' 300 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?
263 minutes.
.ground speed out 370 kt.ground speed home 300 kt..pet = distance x gsh / gso + gsh.pet = 3623 x 300 / 370 + 300 = 1622 nm...1622 nm / 370 kt = 4.38h..4.38 h x 60 minutes = 262.8 minutes.
.48 nm in 10 minutes > gs = 288 kt..on the computer.under index set true heading 077° and in center dot, tas 275 kt...our true track is 086°, so drift is 9° right...mark the point where the 9° right drift crosses the ground speed 288 kt...on the rotating scale, you can read a wind of 335°/45kt. 1742
.64 nm in 12 minutes > gs = 320 kt..on the computer.under index set true heading 230° and over center dot, tas 360 kt...our true track is 225°, so drift is 5° left...mark the point where the 5° left drift crosses the ground speed 320 kt...on the rotating scale, you can read a wind of 265°/50 kt. 2531
Question 196-37 : Given.aircraft at fl150 overhead an airport..elevation 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 ?
15 300 ft.
.at fl150, isa = 15°c 2°c x 15 = 15°c.oat is 5°c, we are in air mass 10°c warmer than isa...changing subscale from 1013 hpa to 1003 hpa, means that indicated altitude on the altimeter will decrease by 270 ft 10 hpa.15000 270 = 14730 ft...temperature correction.4 x 15 x 10 = 600 ft...14730 + 600 = 15330 ft 'plus' 600 ft because air mass is warmer than isa... maxscail.how do you find 4x15x10=600ft, is it a formula.. .yes, this is the rule of thumb formula, called the '4% rule'.the altitude/height changes by 4% for each 10°c temperature deviation from isa..this is an official formula given by easa for altitude/height calculations at the exam.
Question 196-38 : An aircraft takes off from the aerodrome of brioude altitude 1483 ft, qfe = 963 hpa, temperature = 32°c. five minutes later, passing 5000 ft on qfe, the second altimeter set on 1013 hpa will indicate approximately ?
6 500 ft.
.difference between 963 hpa and 1013 hpa is 50 hpa. 50 hpa x 30 ft/hpa = 1500 ft...5000 + 1500 = 6500 ft...your altimeter indicates your pressure altitude, not your true altitude, this is the reason why we do not correct the temperature we only want to know the reading of the altimeter...note 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa is to be used unless another figure is specified in the question.
Question 196-39 : 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 ?
165 nm.
Babar350.e*o*h / o+h....e endurance..o gs outnbound..h gs inbound....e = distance * o...eoh/ o+h = 2,18*165*195/ 360 is giving distance from etp from a 195 nm...the question is etp from b so 360 195 = 165 nm.
Question 196-40 : 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 ?
403 kt.
.first step, find the true heading..237° 5° deviation on this heading = magnetic heading 232°..232° 19° variation west = true heading 213°..put 360 kt in center dot, under index set true heading 213°, on the rotating scale, set wind 330°/80kt.. 2532.you read a ground speed of 403 kt..the answer will be more accurate on a real computer such as the aviat 617 for example.
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