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Question 197-1 : Given.an aircraft is on final approach to runway 32r 322°.the wind velocity reported by the tower is 350°/20 kt..tas on approach is 95 kt..in order to maintain the centre line, the aircraft's heading °m should be ? [ Level reports ]
328°.
Babar350.maximum drift 60/95 x 20 = 12.63.actual drift max drift x sin 350 322 = 12.63 x sin 28 = 5.92°.mainting centre line is 322°+5.92 = 328°.
Question 197-2 : An aircraft takes off from an airport 2 hours before sunset. the pilot flies a track of 090° t , w/v 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. 1944.drift=8° left..set heading 098° 090° + 8° under index, read outbound gs 86 kt..proceed on the same way to finde homeward gs 116 kt....point of safe return psr = 2h x 116 / 86 + 116.point of safe return psr = 1.148 h...1.148 x 86 kt = 98.7 nm.
Question 197-3 : 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 arrival..gso = ground speed out..gsh = ground speed home.
Question 197-4 : Given.required course 045° m.variation is 15°e.w/v is 190° t /30 kt.cas is 120 kt at fl 55 in standard atmosphere..what are the heading °m and gs ?
055° and 147 kt.
.at flight level 55, temperature in standard atmosphere is.15° 2° x 5.5 = 4°c..on 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 kt...now, centre dot on tas 131 kt, under true index put 190° wind direction and mark wind speed 30 kt below, at 101 kt...then, 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 197-5 : Given.airport elevation is 1000 ft..qnh is 988 hpa..what is the approximate airport pressure altitude assume 1 hpa = 30 ft ?
1750 ft.
.1013 988 = 25 hpa..25 hpa x 30 ft = 750 ft..1000 + 750 = 1750 ft.
Question 197-6 : Given.true altitude 9000 ft.oat 32°c.cas 200 kt.the 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 altitude/height changes by 4% for each 10°c temperature deviation from isa...deviation from isa = 15°c 2 x 9 = 3°c. 3°c to 32°c = 29°c. we are in isa 29°c...4% x 9 x 29 = 1044 ft..pressure altitude = 9000 + 1044 = 10044 ft 10000 ft...now, in the airspeed window, set 29°c in front of 10000 ft pressure alitude.. 2523.read cas 200 kt on inner scale and corresponding tas 220 kt on outer scale.
Question 197-7 : Given.course 040° t , tas is 120 kt, wind speed 30 kt..maximum 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°.or.040° 90° = 310°.
Question 197-8 : Given.cas 120 kt, fl 80, oat +20°c..what is the tas ?
141 kt.
.in airspeed window, set tempertaure +20° in front of pressure altitude fl80.on the outside scale, in front of cas 120 kt, you read tas=141 kt.. 2533.cas = ias + correction for position and instrument error...instrument error is an eventual error of the airspeed indicator itself...position error is the error produced from the airflow around the static port wherever that is located and around the probe...at low speed, we consider cas = ias.
Question 197-9 : 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 en/com061 399.jpg..initial 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 197-10 : Given.compass heading 090°, deviation 2°w, variation 12°e, tas 160 kt..whilst maintaining a radial 070° from a vor station, the aircraft flies a ground distance of 14 nm in 6 min..what is the wind °t ?
160°/50 kt.
.first step, search for the ground speed..14 nm in 6 minutes = 14/6 x60 = 140 kt...next step, calculate drift. /com en/com061 484a.jpg.drift x is 18° left 100° 082°...put 160 kt tas on center dot, under true index set 100° heading... /com en/com061 484.jpg..where the ground speed crosses the 18° left drift line, you read on the red scale a wind coming from 160° for 50 kt.
Question 197-11 : Given.m 0.80, oat 50°c, fl 330, gs 490 kt, variation 20°w, magnetic heading 140°, drift is 11° right..calculate the true wind ?
020°/95 kt.
. 2064.tas is 462 kt..magnetic heading 140° variation 20°w = true heading 120°.. 2061.read wind direction and force on the rotative scale.
Question 197-12 : Given pressure altitude 29000 ft, oat 55°c. calculate the density altitude ?
27500 ft.
. /com en/com061 486.jpg..
Question 197-13 : 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. 2056.in front of 150 kt on the inner scale, read the tas 195 kt on the outer scale.
Question 197-14 : Calibrated airspeed cas is indicated airspeed ias corrected for ?
Instrument error and position error.
Question 197-15 : 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.safe endurance 10 h 00..the 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 kt.ground speed home = 480 + 75 = 555 kt..point of safe return psr = 10 x 555 / 390 + 555.point of safe return psr = 5555 / 945.point of safe return psr = 5.878h..distance of the psr from the departure point at a speed of 390 kt.5.878h x 390 = 2292 nm.
Question 197-16 : 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 ft...15000 600 = 14400 ft...now, we must correct for temperature.outside temperature is 30°c at fl150..isa at fl150 is 15°c 15 x 2°c = 15°c..we are in isa 15°c...you can use either the computer or with the following rule of thumb, called the '4% rule'.the altitude/height changes by 4% for each 10°c temperature deviation from isa..an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature..any 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 ft..12740 x 0.04 x 1.5 = 764 ft...14400 764 = 13636 ft.
.229° to true index.555 kt to center bore.with a true headin of 229° and a true track of 239°, we have 10° right drift..pencil mark the intersection of the 10° right drift with the ground speed arc 577 kt. 2053.read the wind speed and wind direction 130°/100 kt.
.set heading 090° undex index, center dot on tas 180 kt. 2047.where right drift 5° crosses ground speed arc 180 kt, read wind on the rotating scale 005°/15kt.
.the wind is coming in front of us, our true air speed will be more than 375 kt..set 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. 2048.turn to put heading 215° 255° 40° under true index. 2046.shift the speed arc under the wind speed mark. drift is 6° left and you also notice a tas of 490 kt.. 2045
.center dot on tas 290 kt..true heading 265° under index.put 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 197-24 : An aircraft is flying at fl 200, oat is 0°c..when the actual air pressure on an airfield at msl is placed in the subscale of the altimeter the indicated altitude is 19300 ft..calculate the aircraft's true altitude ?
21 200 ft.
.isa at fl200 is 15°c 2°c x 20 = 25°c..oat is 0°c, we are in isa +25°c...you can use either the computer or with the following rule of thumb, called the '4% rule'.the altitude/height changes by 4% for each 10°c temperature deviation from isa...0.04 x 19300 x 2.5 = 1930 ft..true altitude = 19300 + 1930 = 21230 ft.
Question 197-25 : 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 ft.....at the airfield, isa temperature is 15°c 1.5 x2°c = 12°c..temperature is report to be 20°c, so we are in isa 32°c..temperature correction formula 4° x 14.1 x 32° = 1805 ft....the minimum height above the airfield is 14100 + 1805 ft = 15905 ft..now adding the 1500 ft of the airfield to have an altitude 15905 + 1500 = 17405 ft......note 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 197-26 : Consider the following factors that determine the accuracy of a dead rekoning position.1. the flight time since the last position update..2. the accuracy of the forecasted wind..3. the accuracy of the tas..4. the accuracy of the steered heading...using the list which of the above statements are ?
1, 2, 3 and 4.
Question 197-27 : An aircraft is flying at fl100..the oat = isa 15°c. the qnh given by a station at an elevation 3000 ft is 1035 hpa..calculate the approximate true altitude ?
10 200 ft.
.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...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 ft.10000 + 660 = 10660 ft...we 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 altitude/height changes by 4% for each 10°c temperature deviation from isa.4 ft x 10660 3000 /1000 x 15 = 460 ft.true altitude = 10660 460 = 10200 ft.
Question 197-28 : An aircraft has to fly over a mountain ridge..the highest obstacle, indicated in the navigation chart, has an elevation of 9 800 ft..the qnh, given by a meteorological station at an elevation of 6200 ft, is 1022 hpa..the oat = isa +5°c..calculate the approximate indicated altitude to obtain a ?
11 700 ft.
.9800 ft + 2000 ft = 11800 ft...we 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 altitude/height changes by 4% for each 10°c temperature deviation from isa... /com en/com061 613.jpg....deviation from isa = +5°c..4% x 5.6 x 5 = 112 ft....11800 ft 112 ft = 11688 ft..it is 'minus' 112 ft because air is hotter than standard, true altitude is higher than indicated altitude.
Question 197-29 : An aircraft is flying from a to b a distance of 50 nm..the true course in the flight log is 270°, the forecast wind is 045° t /15 kt and the tas is 120 kt..after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 2.5 nm ahead of the dead ?
17°.
. 1798.with the forecasted wind, we will fly at 130 kt ground speed..at 130 kt and 15 minutes of flight, we will be at 32.5 nm from a..the question states 2.5 nm ahead of the dead reckoning position , so we are at 35 nm from a...use 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 197-30 : An aircraft is flying from a to b a distance of 50 nm..the 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 2.5 nm ahead of the dead ?
17°.
.draw the exercice.. /com en/com061 623.jpg..without wind, at 120 kt and 15 minutes of flight, we are at 32.5 nm from a..the question states 2.5 nm ahead of the dead reckoning position , so we are at 35 nm from a...use 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..the true course according to the flight log is 090°, the estimated wind is 225° t /15 kt and the tas is 120 kt..after 15 minutes of flying with the planned tas and true heading, the aircraft is 3 nm south of the intended track and 2.5 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 index...make a wind mark on centre line 15 kt below centre dot at 105 kt...rotate to set 090° true track under true index...wind mark has moved to 5° left drift...rotate to lined up 090° with 5° left drift...wind mark has stayed at 5° left drift, you find.true heading 095°...ground speed 130 kt... 130/60 x 15 min = 32.5 nm.actual aircraft position is 2.5 nm ahead dead reckoning position, at 32.5 + 2.5 = 35 nm...track error angle = distance off track x 60 / distance along track..track error angle = 3 x 60 / 35..track error angle = 180 / 35 = 5°.
Question 197-32 : 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 same..what 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 en/com061 627.jpg..read the wind 360°/17 kt.
Question 197-33 : An aircraft is flying from salco to berry head on magnetic track 007°, tas 445 kt..the wind is 050° t /40 kt..variation 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 ?
21 nm.
.first step, find the aircraft magnetic heading. /com en/com061 631.jpg..calculate the drift between our true track 002° and the true wind 050°/40 kt with your computer, the drift is 4° left, you have to apply a 4°right wind angle correction and also a ground speed of 415 kt....from the aircraft, at 1000 utc, the locator is at 60° to the left 011° to 311° = 60°...from the aircraft, at 1003 utc, the locator is at 105° to the left 011° to 266° = 105°....we have an isosceles triangle, and in an isosceles triangle, two sides are equal in length...in 45° in 3 minutes..415 kt/60 = 6.92 nm/min..3 min x 6.92 = 20.76 nm.
Question 197-34 : An aircraft is flying at fl250, oat = 45°c. the qnh, given by a station at msl, is 993.2 hpa..calculate the approximate true altitude ?
23400 ft.
.you have to turn your altimeter subscale setting knob counterclockwise, from 1013.2 to 993.2.indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft..25000 ft 600 ft = 24400 ft..you can use either the computer or with the following rule of thumb, called the '4% rule'.the altitude/height changes by 4% for each 10°c temperature deviation from isa...deviation from isa = 15° 2 x 25 = 35°c.. 35°c to 45°c = 10°c. we are in isa 10°c..0.04 x 24400 x 1 = 976 ft..true altitude = 24400 976 = 23424 ft...keep in mind that air is colder than standard, thus, the air column is contracted, our true altitude is lower than our indicated altitude.
Question 197-35 : A vor is situated at position n55°26', w005°42'..the variation at the vor is 9°w..the position of the aircraft is n60°00'n, w010°00'..the variation at the aircraft position is 11°w..the initial true track angle of the great circle from the aircraft position to the vor is 101.5°..which radial is the ?
294°.
. /com en/com061 638.jpg..first step, apply convergency.convergency = difference of longitude x sin mean latitude..convergency = 10°w 5°42'w x sin 60 +55°26' /2.convergency = 4.35° x sin 57.5 = 3.67°..second step, find true track vor.true track at vor = 101.5° + convergency = 101.5 + 3.67° = around 105° t..third step, we must apply variation at the vor.105 + 9°w = 114° magnetic...last step, we are looking for a radial.114° + 180° = 294°.
Question 197-36 : 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°w..the wind is ?
320°/50 kt.
.following radial 200° inbound magnetic track 020° with a magnetic heading of 010°, we have a 10° right drift...our true heading is 005° we have to applied the 5°west magnetic variation since it is a vor..after having overfly the vor, we fly outbound on radial 090° magnetic track 090°. our magnetic heading is 080°, so our true heading is 075°...on the computer, set 240 kt under the center dot, and 005° below true index, draw a line along the right 10° drift line..rotate and put 075° below true index, draw a line along the right 10° drift line...rotate to bring back the intersection of the lines under the central wind line. 1788
Question 197-37 : 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 ?
1
. /com en/com061 644.jpg..point 1 is always located on the right side of the radials.
Question 197-38 : 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 hpa..calculate the clearance above a mountain ridge with an elevation of 20410 ft ?
4 200 ft.
Question 197-39 : Given.an aircraft is flying at fl100, oat = isa 15°c..the 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 ft...10000 + 513 = 10513 ft...now, we must correct for temperature.we are in isa 15°c...you can use either the computer or with the following rule of thumb, called the '4% rule'.the altitude/height changes by 4% for each 10°c temperature deviation from isa..an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature..any 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 ft..10613 x 0.04 x 1.5 = 637 ft...10513 637 = 9876 ft closest answer is 9900 ft.
Question 197-40 : The accuracy of the, manually calculated, dead reckoning position of an aircraft is, among other things, affected by ?
The accuracy of the forecasted wind.
.dead reckoning is the process of calculating one's current position by using a previously determined position, or fix, and advancing that position based upon known or estimated speeds over elapsed time, and course. you have to take the wind into account, and the more accurate the wind information is, the more accurate the manually calculated position will be.
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