Question 197-1 : Given half way between two reporting points the navigation log gives the following information tas 360 kt wv 330°80kt compass heading 237° deviation on this heading 5° variation 19°wwhat is the average ground speed for this leg ? [ Level reports ]

Question 197-2 : Given an aircraft is on final approach to runway 32r 322° the wind velocity reported by the tower is 350°20 kttas on approach is 95 kt in order to maintain the centre line the aircraft's heading °m should be ?

328°.

Babar350 maximum drift 6095 x 20 = 1263actual drift max drift x sin 350 322 = 1263 x sin 28 = 592°mainting centre line is 322°+592 = 328°

Question 197-3 : An aircraft takes off from an airport 2 hours before sunset the pilot flies a track of 090° t wv 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 ?

97 nm.

Resolve this question as a point of safe return question point of safe return psr = endurance x homeward gs outbound gs + homeward gs outbound gs on the computer when you start with a track under index you must first apply drift before read the ground speed 1944drift=8° leftset heading 098° 090° + 8° under index read outbound gs 86 ktproceed on the same way to finde homeward gs 116 ktpoint of safe return psr = 2h x 116 86 + 116 point of safe return psr = 1148 h1148 x 86 kt = 987 nm

Question 197-4 : From the departure point the distance to the point of equal time is ?

Inversely proportional to the sum of ground speed out and ground speed back.

Distance to the point of equal time = d x gsh gso + gsh where d = distance between departure and arrivalgso = ground speed outgsh = ground speed home

Question 197-5 : Given required course 045° m variation is 15°ewv is 190° t 30 ktcas is 120 kt at fl 55 in standard atmospherewhat are the heading °m and gs ?

055° and 147 kt.

At flight level 55 temperature in standard atmosphere is 15° 2° x 55 = 4°con the computer in airspeed window put 4ºc next to fl55 go to cas 120 kt on inner scale and read tas on outer scale 131 ktnow centre dot on tas 131 kt under true index put 190° wind direction and mark wind speed 30 kt below at 101 ktthen rotate to put true track 060° 045° + variation east under true index drift is 9° left it means that on this heading 060° the aircraft would be on a true track of 051°rotate to put true track 060° under drift and note that drift has changed to 10° left as we turn the drift changes and on a heading of 070° we will have a 10° left drift magnetic heading = 070° minus variation east = 055°ground speed is under wind mark 147 kt

Question 197-6 : Given airport elevation is 1000 ftqnh is 988 hpawhat is the approximate airport pressure altitude assume 1 hpa = 30 ft ?

1750 ft.

1013 988 = 25 hpa25 hpa x 30 ft = 750 ft1000 + 750 = 1750 ft

Question 197-7 : Given true altitude 9000 ftoat 32°ccas 200 ktthe true air speed tas is ?

220 kt.

You have first to convert true altitude in pressure altitude you can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isadeviation from isa = 15°c 2 x 9 = 3°c 3°c to 32°c = 29°c we are in isa 29°c4% x 9 x 29 = 1044 ftpressure altitude = 9000 + 1044 = 10044 ft 10000 ft now in the airspeed window set 29°c in front of 10000 ft pressure alitude 2523read cas 200 kt on inner scale and corresponding tas 220 kt on outer scale

Question 197-8 : Given course 040° t tas is 120 kt wind speed 30 ktmaximum drift angle will be obtained for a wind direction of ?

130°.

Maximum drift is obtained when the wind is at a right angle from our course 040° + 90° = 130° or040° 90° = 310°

Question 197-9 : Given cas 120 kt fl 80 oat +20°cwhat is the tas ?

141 kt.

In airspeed window set tempertaure +20° in front of pressure altitude fl80on the outside scale in front of cas 120 kt you read tas=141 kt 2533cas = ias + correction for position and instrument errorinstrument error is an eventual error of the airspeed indicator itselfposition error is the error produced from the airflow around the static port wherever that is located and around the probeat low speed we consider cas = ias

Question 197-10 : Route 'a' 44°n 026°e to 'b' 46°n 024°e forms an angle of 35° with longitude 026°e variation at a is 3°e what is the initial magnetic track from a to b ?

322°.

Draw the situation com encom061 399jpginitial true track from a to b will be 360° 35° = 325°variation 3°e 325° 3° = 322° variation west magnetic best variation east magnetic least

Question 197-11 : Given compass heading 090° deviation 2°w variation 12°e tas 160 ktwhilst maintaining a radial 070° from a vor station the aircraft flies a ground distance of 14 nm in 6 minwhat is the wind °t ?

160°50 kt.

First step search for the ground speed 14 nm in 6 minutes = 146 x60 = 140 ktnext step calculate drift com encom061 484ajpgdrift x is 18° left 100° 082° put 160 kt tas on center dot under true index set 100° heading com encom061 484jpgwhere the ground speed crosses the 18° left drift line you read on the red scale a wind coming from 160° for 50 kt

Question 197-12 : Given m 080 oat 50°c fl 330 gs 490 kt variation 20°w magnetic heading 140° drift is 11° rightcalculate the true wind ?

020°95 kt.

2064tas is 462 ktmagnetic heading 140° variation 20°w = true heading 120° 2061read wind direction and force on the rotative scale

Question 197-13 : Given pressure altitude 29000 ft oat 55°c calculate the density altitude ?

27500 ft.

com encom061 486jpg

Question 197-14 : An aircraft is flying at fl180 and the outside air temperature is 30°c if the cas is 150 kt what is the tas ?

195 kt.

In airspeed window put 30°c next to fl180 2056in front of 150 kt on the inner scale read the tas 195 kt on the outer scale

Question 197-15 : Calibrated airspeed cas is indicated airspeed ias corrected for ?

Instrument error and position error.

Question 197-16 : 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 ktsafe endurance 10 h 00the distance from 'q' to the point of safe return psr 'q' is ?

2290 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 480 90 = 390 ktground speed home = 480 + 75 = 555 ktpoint of safe return psr = 10 x 555 390 + 555 point of safe return psr = 5555 945point of safe return psr = 5878hdistance of the psr from the departure point at a speed of 390 kt 5878h x 390 = 2292 nm

Question 197-17 : An aircraft is flying at fl150 with an outside air temperature of 30° above an airport where the elevation is 1660 ft and the qnh is 993 hpa calculate the true altitude assume 30 ft = 1 hpa ?

13 660 ft.

You have to turn your altimeter subscale setting knob counterclockwise from 1013 to 993 indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft15000 600 = 14400 ftnow we must correct for temperature outside temperature is 30°c at fl150isa at fl150 is 15°c 15 x 2°c = 15°cwe are in isa 15°cyou can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isaan altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperatureany temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight therefore 14400 1660 = 12740 ft12740 x 004 x 15 = 764 ft14400 764 = 13636 ft

Question 197-18 : Given true track 239°true heading 229°tas 555 ktgs 577 ktcalculate the wind velocity ?

130°100 kt.

229° to true index555 kt to center borewith a true headin of 229° and a true track of 239° we have 10° right driftpencil mark the intersection of the 10° right drift with the ground speed arc 577 kt 2053read the wind speed and wind direction 130°100 kt

Question 197-19 : Given true track 245°drift 5° rightvariation 3°ecompass heading 242°calculate the deviation ?

5°w.

2050use this wonderful table for those questions

Question 197-20 : Given true heading 090°tas 180 ktgs 180 ktdrift 5° rightthe wind is ?

005° 15 kt.

Set heading 090° undex index center dot on tas 180 kt 2047where right drift 5° crosses ground speed arc 180 kt read wind on the rotating scale 005°15kt

Question 197-21 : Given magnetic heading = 255°variation = 40°wgs = 375 ktwv = 235° t 120 ktcalculate the drift angle ?

6° left.

The wind is coming in front of us our true air speed will be more than 375 ktset wind index under true index and set 400 kt for example in center dot mark wind at 120 kt below the center dot on the rotating indicator 2048turn to put heading 215° 255° 40° under true index 2046shift the speed arc under the wind speed mark drift is 6° left and you also notice a tas of 490 kt 2045

Question 197-22 : Given true track = 095° tas = 160 kt true heading = 087° gs = 130 ktcalculate the wind ?

057°36 kt.

Question 197-23 : Given true track 245°drift 5° rightvariation 3°ecompass heading 242°calculate the magnetic heading ?

237°.

2043use this wonderful table for those questionsnote compass heading 242° is given for nothing

Question 197-24 : Given heading 265°tas 290 ktwind 210°35 ktcalculate track and groundspeed ?

271° and 272 kt.

Center dot on tas 290 kttrue heading 265° under indexput wind direction under the red compass rose under 35 kt your drift is 6° right giving a track of 271° and a groundspeed under the wind mark of 272 kt 2036

Question 197-25 : An aircraft is flying at fl 200 oat is 0°cwhen the actual air pressure on an airfield at msl is placed in the subscale of the altimeter the indicated altitude is 19300 ftcalculate the aircraft's true altitude ?

21 200 ft.

Isa at fl200 is 15°c 2°c x 20 = 25°coat is 0°c we are in isa +25°cyou can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isa004 x 19300 x 25 = 1930 fttrue altitude = 19300 + 1930 = 21230 ft

Question 197-26 : An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft the qnh at the nearest airfield is 991 hpa the elevation is 1500 ft and the temperature is 20°c calculate the minimum altitude required ?

17 400 ft.

The aircraft must be at a true height above the airfield of 13600 + 2000 1500 = 14100 ftat the airfield isa temperature is 15°c 15 x2°c = 12°ctemperature is report to be 20°c so we are in isa 32°ctemperature correction formula 4° x 141 x 32° = 1805 ftthe minimum height above the airfield is 14100 + 1805 ft = 15905 ftnow adding the 1500 ft of the airfield to have an altitude 15905 + 1500 = 17405 ftnote qnh is calculated from the qfe reduced to mean sea level msl assuming isa conditions there is no temperature error between airfield elevation qfe and mean sea level msl

Question 197-27 : Consider the following factors that determine the accuracy of a dead rekoning position 1 the flight time since the last position update2 the accuracy of the forecasted wind3 the accuracy of the tas4 the accuracy of the steered headingusing the list which of the above statements are correct ?

1 2 3 and 4.

Question 197-28 : An aircraft is flying at fl100the oat = isa 15°c the qnh given by a station at an elevation 3000 ft is 1035 hpacalculate the approximate true altitude ?

10 200 ft.

Note 061 general navigation learning objectives states for questions involving height calculation 30 fthpa is to be used unless another figure is specified in the question you have to turn your altimeter subscale setting knob clockwise from 1013 to 1035 indicated altitude will be increased by 22 hpa x 30 ft = 660 ft10000 + 660 = 10660 ftwe now have to correct the temperature above the station you can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isa 4 ft x 10660 3000 1000 x 15 = 460 fttrue altitude = 10660 460 = 10200 ft

Question 197-29 : An aircraft has to fly over a mountain ridgethe highest obstacle indicated in the navigation chart has an elevation of 9 800 ftthe qnh given by a meteorological station at an elevation of 6200 ft is 1022 hpathe oat = isa +5°ccalculate the approximate indicated altitude to obtain a clearance of 2000 ?

11 700 ft.

9800 ft + 2000 ft = 11800 ftwe have to correct the temperature above the qfe datum you can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isa com encom061 613jpgdeviation from isa = +5°c4% x 56 x 5 = 112 ft11800 ft 112 ft = 11688 ftit is 'minus' 112 ft because air is hotter than standard true altitude is higher than indicated altitude

Question 197-30 : An aircraft is flying from a to b a distance of 50 nmthe true course in the flight log is 270° the forecast wind is 045° t 15 kt and the tas is 120 ktafter 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 25 nm ahead of the dead ?

17°.

1798with the forecasted wind we will fly at 130 kt ground speedat 130 kt and 15 minutes of flight we will be at 325 nm from athe question states 25 nm ahead of the dead reckoning position so we are at 35 nm from ause the one in sixty rule track error angle from a = 3 nm x 60 35 nm = 5° it's the drift to applied in order to correct the wind track error angle to join b from our current position = 3 nm x 60 15 nm = 12° to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17°

Question 197-31 : An aircraft is flying from a to b a distance of 50 nmthe true course in the flight log is 090° the forecast wind is 225° t 15kt and the tas is 120 kt after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm north of the intended track and 25 nm ahead of the dead ?

17°.

Draw the exercice com encom061 623jpgwithout wind at 120 kt and 15 minutes of flight we are at 325 nm from athe question states 25 nm ahead of the dead reckoning position so we are at 35 nm from ause the one in sixty rule track error angle from a = 3 nm x 60 35 nm = 5° it's the drift to applied in order to correct the wind track error angle to join b from our current position = 3 nm x 60 15 nm = 12° to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17°

Question 197-32 : An aircraft is flying from a to bthe true course according to the flight log is 090° the estimated wind is 225° t 15 kt and the tas is 120 ktafter 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 25 nm ahead of the dead reckoning ?

5°r.

Frist step find the ground speed place centre dot on 120 kt tas place 225° wind direction under true indexmake a wind mark on centre line 15 kt below centre dot at 105 kt rotate to set 090° true track under true indexwind mark has moved to 5° left driftrotate to lined up 090° with 5° left driftwind mark has stayed at 5° left drift you find true heading 095°ground speed 130 kt 13060 x 15 min = 325 nmactual aircraft position is 25 nm ahead dead reckoning position at 325 + 25 = 35 nmtrack error angle = distance off track x 60 distance along tracktrack error angle = 3 x 60 35track error angle = 180 35 = 5°

Question 197-33 : On a true heading of 090° the aircraft experiences drift of 5°right on a true heading of 180° the aircraft experiences no drift on both headings the tas is 200 kt and it is assumed that the wind is the samewhat is the experienced wind speed and direction ?

360°17 kt.

Since on a true heading of 180° there is no drift the wind is coming from 180° or 360° under index set true heading 090° centre dot on tas 200 kt with the rotative scale set the 5°right drift com encom061 627jpgread the wind 360°17 kt

Question 197-34 : An aircraft is flying from salco to berry head on magnetic track 007° tas 445 kt the wind is 050° t 40 ktvariation 5°w deviation +2°at 1000 utc the rb of locator py is 311°at 1003 utc the rb of locator py is 266°calculate the distance of the aircraft from locator py at 1003 utc ?

Question 197-35 : An aircraft is flying at fl250 oat = 45°c the qnh given by a station at msl is 9932 hpacalculate the approximate true altitude ?

23400 ft.

You have to turn your altimeter subscale setting knob counterclockwise from 10132 to 9932 indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft25000 ft 600 ft = 24400 ftyou can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isadeviation from isa = 15° 2 x 25 = 35°c 35°c to 45°c = 10°c we are in isa 10°c004 x 24400 x 1 = 976 fttrue altitude = 24400 976 = 23424 ftkeep in mind that air is colder than standard thus the air column is contracted our true altitude is lower than our indicated altitude

Question 197-36 : A vor is situated at position n55°26' w005°42' the variation at the vor is 9°wthe position of the aircraft is n60°00'n w010°00' the variation at the aircraft position is 11°wthe initial true track angle of the great circle from the aircraft position to the vor is 1015°which radial is the aircraft on ?

294°.

com encom061 638jpgfirst step apply convergency convergency = difference of longitude x sin mean latitude convergency = 10°w 5°42'w x sin 60 +55°26' 2 convergency = 435° x sin 575 = 367°second step find true track vor true track at vor = 1015° + convergency = 1015 + 367° = around 105° t third step we must apply variation at the vor 105 + 9°w = 114° magneticlast step we are looking for a radial 114° + 180° = 294°

Question 197-37 : An aircraft tracks radial 200° inbound to a vor station with a magnetic heading of 010°after being overhead the vor station the aircraft tracks radial 090° outbound with a mh of 080°the tas is 240 kt and the magnetic variation in the area is 5°wthe wind is ?

Question 197-38 : The fix of the aircraft position is determined by radials from three vor stations the measurements contain small random errors known systematic errors and unknown systematic errors the measured radials are corrected for known systematic errors and are plotted on a navigation chart the result is ?

Question 197-39 : An aircraft flies at fl 250 with an oat of 45°c the qnh given by a meteorological station with an elevation of 2830 ft is 1033 hpacalculate the clearance above a mountain ridge with an elevation of 20410 ft ?

4 200 ft.

Question 197-40 : Given an aircraft is flying at fl100 oat = isa 15°cthe qnh given by a meteorological station with an elevation of 100 ft below msl is 1032 hpa 1 hpa = 27 ft calculate the approximate true altitude of this aircraft ?

9900 ft.

You have to turn your altimeter subscale setting knob clockwise from 1013 to 1032 indicated altitude will be increased by 19 hpa x 27 ft = 513 ft10000 + 513 = 10513 ftnow we must correct for temperature we are in isa 15°cyou can use either the computer or with the following rule of thumb called the '4% rule' the altitudeheight changes by 4% for each 10°c temperature deviation from isaan altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperatureany temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight therefore 10513 100 = 10613 ft10613 x 004 x 15 = 637 ft10513 637 = 9876 ft closest answer is 9900 ft