.on the computer, set 180° under index, centre dot on 500 kt, under the wind speed 100 kt on the rotating scale.. 1367.you read a groud speed of 435 kt.
.use low speed scale on the computer, set 310° under index, centre dot on 200 kt, where right drift 7° crosses ground speed arc 176 kt, read the wind on the rotating scale. 1368.you read 270°/33kt.
Question 195-3 : Given.true heading = 090°, tas = 200 kt, wind = 220°/30 kt..calculate the ground speed ?
220 kt.
. 1369
Question 195-4 : The reported surface wind from the control tower is 240°/35 kt. runway 30 300°..what is the cross wind component ?
30 kt.
.angle between the wind and the direction of the runway 300° 240° = 60°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin60° x 35 kt = 30.31 kt.
Question 195-5 : Given magnetic heading 311°, drift angle 10° left, relative bearing of ndb 270°. what is the magnetic bearing of the ndb measured from the aircraft ?
221°.
.heading.....311°..bearing....+270°..total.......581°..581° 360° = 221° magnetic bearing of the ndb measured from the aircraft.
Question 195-6 : Given the following.true track 192°, magnetic variation 7°e, drift angle 5° left..what is the magnetic heading required to maintain the given track ?
190°.
. 1393.use this wonderful table for those questions.
Question 195-7 : The angle between the plane of the ecliptic and the plane of equator is approximately ?
23.5°.
. 2028
Question 195-8 : Given.tas = 485 kt.oat = isa +10°c..fl 410..calculate the mach number ?
0.825
. 56°c is considered to be the lowest isa temperature, and the question states that we are in outside air temperature of isa +10°c...at fl410, isa is 56°c, isa +10°c will be 46°c...on the computer. 1396..by calculation.mach = tas x a.a = 39square root t° in kelvin...a = 39 square root 227°k= 587,59.mach number= 485/ 587,59= 0,825.
Question 195-9 : At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178°, range 135 nm..calculate the aircraft position at 1215 utc. 2484 ?
40°55'n 027°55'w.
.rmi reads 178°, it is the magnetic direction to reach the station vor tacan at lajes...for a vor, we must apply the variation at the station, on the chart, variation line is 15°w..178° variation west, magnetic best , so 'minus' 15° 178° 15° = 163°...from lajes, 163° + 180° = 343°.. 1398..use the latitude scale to find 135 nm.
Question 195-10 : At reference..1300 utc dr position 37°30'n 021°30'w alter heading port santo ndb 33°03'n 016°23'w.tas 450 kt, forecast w/v 360°/30kt..calculate the eta at port santo ndb.. err a 061 53 ?
1348
.true track is 136°..wind 360°/30 kt..drift is .true heading is ..ground speed is ...center dot on 450 kt, true index on 136°, indicate wind 360°/30 kt drift is 3° right...turn to put 133° 136° 3° right drift below true index, the groundspeed is under the wind mark 30 kt 470 kt gs....plot distance between points and measure along line of longitude.. /com en/com061 53.jpg..6°20' = 6.33 x 60 nm = 380 nm...380 / 470/60 = 48.5 minutes....1300 + 48 minutes 30 secondes = 13h48 30sec.
Question 195-11 : For a distance of 1860 nm between q and r, a ground speed 'out' of 385 kt, a ground speed 'home' of 465 kt and an endurance of 8 hours excluding reserves the distance from q to the point of safe return psr is ?
1685 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 385 kt.ground speed home = 465 kt..point of safe return psr = 8 x 465 / 385 + 465.point of safe return psr = 3720 / 850.point of safe return psr = 4.3764 h..distance of the psr from the departure point at a speed of 385 kt.4.3764 h x 385 = 1685 nm.
Question 195-12 : Two points a and b are 1000 nm apart. tas = 490 kt..on the flight between a and b the equivalent headwind is 20 kt..on the return leg between b and a, the equivalent tailwind is +40 kt..what distance from a, along the route a to b, is the the point of equal time pet ?
530 nm.
.ground speed out 490 20 = 470 kt.ground speed home 490 + 40 = 530 kt..pet = distance x gsh / gso + gsh.pet = 1000 x 530 / 470 + 530 = 530 nm.
Question 195-13 : Given ad = air distance, gd = ground distance, tas = true airspeed, gs = groundspeed. which of the following is the correct formula to calculate ground distance gd gone ?
Gd = ad x gs /tas
.gd = gs x t but t = ad/tas so gd = gs x ad /tas.
Question 195-14 : What is the isa temperature value at fl 330 ?
51°c
15°c 2° x 33 = 51°c.
Question 195-15 : Given.tas 487kt, fl 330, temperature isa + 15..calculate the mach number ?
0.81
.isa temperature at fl 330 = 15º 2º x 33 = 51ºc...deviation is +15ºc then oat is 51°c + +15°c = 36ºc...with the computer.in the airspeed window put the temperature 36ºc in front of 'mach kt ' index..in front of 487 kt, on the inner scale, you read 0.81 mach number.. 1435
Question 195-16 : How many nm would an aircraft travel in 1 minute 45 secondes if gs is 135 kt ?
Question 195-17 : An aircraft travels 100 statute miles in 20 min, how long does it take to travel 215 nm ?
50 min.
.1 nm = 1.15 sm..215 nm = 215 x 1.15 = 247.25 sm..100 sm in 20 minutes = 300 sm in 60 minutes 1h...247.25/300 = 0.824 hour..0.824 x 60 = 50 min.
Question 195-18 : Given.tas = 220 kt.magnetic course = 212°.wind = 160 ° m / 50 kt.calculate the gs ?
186 kt.
.the given wind has a headwind component so the groundspeed is going to be less than the given tas and all but one of the answers are more.
Question 195-19 : Given.fl250.oat 15 °c.tas 250 kt.calculate the mach number ?
0.40
. 2485.. aviat 617.put temperature 15ºc in front of m kt index, in the airspeed window...go to tas 250 kt on outer scale and read mach number on the inner scale...by calculation..mach number = tas / lss..mach number = 250 / 39 sqrt 273 15..mach number = 0.399.
Question 195-20 : During a low level flight 2 parallel roads that are crossed at right angles by an aircraft..the time between these roads can be used to check the aircraft ?
Groundspeed.
Question 195-21 : Given.magnetic track = 315°, magnetic heading = 301°, variation = 5°w, tas = 225 kt..the aircraft flies 50 nm in 12 min..calculate the wind °t ?
190°/63 kt.
.true heading = 296° 301° 5°..true course = 310° 315 5°..ground speed = 60 x 50/12 = 250 kt...on nav computer.set tas 225 kt on center dot, under true index set true heading 296°...mark where drift 14º right crosses ground speed 250 kt...rotate to shift mark under the vertical speed line, you read 190°/63 kt.
Question 195-22 : Given.tas = 270 kt, true hdg = 270°, actual wind 205° t /30kt..calculate the drift angle and gs ?
6r 259kt
.under index, set true heading 270°, centre dot on tas, 270 kt, with the rotative scale, set wind 205°/30 kt.read the drift and the ground speed 6°r 259 kt.
Question 195-23 : Given.tas = 270 kt, true hdg = 145°, actual true wind = 205°/30kt..calculate the drift angle and gs ?
6°l 256 kt
.under index, set true track 145°, centre dot on tas, 270 kt, with the rotative scale, set wind. 1446.read the drift and the ground speed 6°l 256 kt.
Question 195-24 : Given.tas = 470 kt, true heading = 317°, wind = 045° t /45 kt..calculate the drift angle and gs ?
5°l 470 kt
.under index, set true heading 317°, centre dot on tas, 470 kt, with the rotative scale, set wind. 1447.read drift 5° left..ground speed is 470 kt.
Question 195-25 : Given.tas = 190 kt,.true heading = 085°,.true wind = 110°/50kt..calculate the drift angle and gs ?
8°l 146 kt.
.under index, set true heading 085°, centre dot on tas, 190 kt, with the rotative scale, set wind. 1452.read drift 8° left..ground speed is 147 kt..closest answer 8°l 146 kt.
.center dot on tas 370 kt..true heading 181° under index. 1454.put wind direction under the red compass rose, under 35 kt, your drift is 5° right, giving a track of 186° and a groundspeed under the wind mark of 370 kt.
.center dot on tas 125 kt..true heading 355° under index.put wind direction under the red compass rose, under 30 kt your drift is 10° right, giving a track of 005° and a groundspeed under the wind mark of 102 kt.. 1457.. cmarzocchini.all this questions doesnt work with cr3, wich is used for us in spain. be carefull about this. ie in this question, just in this one, the correct answer using cr3 is 003/95. i only did this comment in this question, but, take it into account for the reminder questions... .feel free to download the cr3 instructions here.http //www.jeppesen.com/download/misc/crinstructions.pdf.. page 44 45 true course track and ground speed .on top, set speed 125 kt over tas index..over tc index set 355°...locate the wind dot by finding the 320° line on the green scale and making the point where this line intersects the green 30 kt circle...reading directly up from the wind dot we see that there is a left crosswind component of 17 kt..looking at the outer scale, find 17 and opposite it read 8° crab angle.. 2486.since the wind is from the left, the true heading must be left of the true course. therefore rotate the top disc 8° to the left counter clockwise , now the tc index points to 003°...looking directly above the wind dot after the above move, you now find that the crosswind component has changed to 22 kt instead of 17 kt...locate 22 on the outer scale and find opposite crab angle of 10°.. 2487.it now appears that the first crab angle of 8° was 2° too less. therefore, add 2° of the previous tc adjustment, making a true course reading of 005°..you can read a headwind of 23kt, therefore 125 kt 23 kt = 102 kt ground speed.
Question 195-29 : Given.tas = 225 kt..hdg °t = 123°..w/v = 090/60kt..calculate the track °t and gs ?
134° 178 kt.
.put 225 kt in center dot, under true index set wind direction 090°. mark wind on centre line at 165 kt 60 kt below centre dot.. 2488.rotate to put heading 123° under true index. 2489.you read a drift of 11° right and a groundspeed of 178 kt..add the drift to your heading to find the true track 134°.
.center dot on tas 480 kt..true heading 040° under index. 2026.put wind direction under the red compass rose, under 60 kt, your drift is 6° left, giving a track of 034° and a groundspeed under the wind mark of 445 kt.
Question 195-31 : Given.tas = 170 kt.true heading = 100°.wind = 350/30kt..calculate the true track and gs ?
109° 182 kt.
...center dot on tas 170 kt..true heading 100° under index.put wind direction under the red compass rose, under 30 kt, your drift is 9° right, giving a track of 109° and a groundspeed under the wind mark of 182 kt... /com en/com061 168.jpg.the answer is 109° and 182 kt.
Question 195-32 : Given. tas = 235 kt, hdg t = 076°, w/v = 040/40kt..calculate the drift angle and gs ?
7r 204 kt.
...under index, set true heading 076°, centre dot on tas, 235 kt, with the rotative scale, set wind.. /com en/com061 169.jpg..read drift 7° right..ground speed is 205 kt close enough to answer 204 kt.
.under index, set true heading 349°, centre dot on tas, 440 kt, with the rotative scale, set wind 040/40kt...read drift 4° left..ground speed is 415 kt.
Question 195-34 : Given tas = 95 kt, hdg t = 075°, w/v = 310/20kt. calculate the drift and gs ?
9r 108 kt
Question 195-35 : Given.tas = 230 kt, hdg t = 250°, wind = 205/10kt..calculate the drift and gs ?
2r 223 kt.
.under index, set true heading 250°, centre dot on tas, 230 kt, with the rotative scale, set wind. 2490.read drift 2° right..ground speed is 223 kt.
Question 195-36 : Given.tas = 205 kt, hdg t = 180°, wind = 240/25kt..calculate the drift and gs ?
6l 194 kt
...under index, set true heading 180°, centre dot on tas, 205 kt, with the rotative scale, set wind.. /com en/com061 176.jpg..read drift 6° left..ground speed is 194 kt.
...under index, set true heading 053°, centre dot on tas, 132 kt, with the rotative scale, set wind. /com en/com061 178.jpg..read drift 3° left..ground speed is 145 kt.
.under index, set true heading 355°, centre dot on tas, 90 kt, with the rotative scale, set wind. 2491.read drift 9° left..ground speed is 103 kt close enough for the answer.
Question 195-39 : Given.tas = 155 kt, track t = 305°, w/v = 160/18kt..calculate the hdg °t and gs ?
301 169 kt.
...under index, set true track 305°, centre dot on tas, 155 kt, with the rotative scale, set wind.. /com en/com061 181.jpg..now, drift is always measured from heading to track.turn to set true heading 301° 305° 4° right drift under index, you now read a ground speed of 169 kt.
Question 195-40 : Given.tas = 465 kt, track t = 007°, w/v = 300/80kt..calculate the hdg °t and gs ?
358° 428 kt.
...under index, set true track 007°, centre dot on tas, 465 kt, with the rotative scale, set wind.. /com en/com061 184.jpg..now, drift is always measured from heading to track.turn to set true heading 358° 007° 9° right drift under index, you now read a ground speed of 428 kt.
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