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Question 182-1 : Given .tas = 270 kt true hdg = 145° actual true wind = 205°/30kt .calculate the drift angle and gs ? [ Question security ]

6°l 256 kt

Admin .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 exemple 282 6°l - 256 kt

Question 182-2 : Given .tas = 470 kt true heading = 317° wind = 045° t /45 kt .calculate the drift angle and gs ?

5°l 470 kt

Admin .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 exemple 286 5°l - 470 kt

Question 182-3 : Given .tas = 190 kt .true heading = 085° .true wind = 110°/50kt .calculate the drift angle and gs ?

8°l 146 kt

Admin .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 exemple 290 8°l - 146 kt.

Question 182-4 : Given .tas = 132 kt true hdg = 257° true wind = 095°/35 kt .calculate the drift angle and gs ?

4°r 165 kt

Admin .under index set true heading 257° centre dot on tas 132 kt set wind 095°/35kt on the rotative scale . 1453.you read 4 5°right and 164 kt exemple 294 4°r - 165 kt.

Question 182-5 : Given .tas = 370 kt true heading = 181° wind = 095°/35 kt.calculate true track and ground speed ?

186° 370 kt

Admin .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 exemple 298 186° - 370 kt.

Question 182-6 : Given .tas = 125 kt true heading = 355° true wind = 320°/30 kt .calculate the true track and ground speed ?

005° 102 kt

Admin .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 exemple 302 005° - 102 kt.

Question 182-7 : Given .tas = 225 kt .hdg °t = 123° .w/v = 090/60kt .calculate the track °t and gs ?

134° 178 kt

Admin .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° exemple 306 134° - 178 kt.

Question 182-8 : Given .tas = 480 kt true heading = 040° wind = 090°/60 kt .calculate true track and ground speed ?

034° 445 kt

Admin .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 exemple 310 034° - 445 kt.

Question 182-9 : 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 exemple 314 109° - 182 kt.

Question 182-10 : 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 exemple 318 7r - 204 kt.

Question 182-11 : Given .tas = 440 kt true heading = 349° wind = 040/40kt .calculate drift and ground speed ?

4l 415 kt

Admin .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 exemple 322 4l - 415 kt

Question 182-12 : Given tas = 95 kt hdg t = 075° w/v = 310/20kt calculate the drift and gs ?

9r 108 kt

exemple 326 9r - 108 kt

Question 182-13 : Given .tas = 230 kt hdg t = 250° wind = 205/10kt .calculate the drift and gs ?

2r 223 kt

Admin .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 exemple 330 2r - 223 kt.

Question 182-14 : 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 exemple 334 6l - 194 kt

Question 182-15 : Given .tas = 132 kt true heading = 053° wind = 205°/15 kt .calculate the true track and ground speed ?

050° 145 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 exemple 338 050° - 145 kt

Question 182-16 : Given .tas = 90 kt.true heading = 355°.wind = 120/20 kt.calculate true track and ground speed ?

346 102 kt

Admin .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 exemple 342 346 - 102 kt

Question 182-17 : 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 exemple 346 301 - 169 kt.

Question 182-18 : 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 exemple 350 358° - 428 kt.

Question 182-19 : Given .tas = 200 kt track t = 110° w/v = 015/40 kt .calculate the hdg °t and gs ?

099° 199 kt

Admin .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 exemple 354 099° - 199 kt.

Question 182-20 : Given .true hdg = 307° tas = 230 kt track t = 313° gs = 210 kt .calculate the w/v ?

260/30kt

Admin .true heading is 307° true track is 313° our drift is 6° right . 2492.wind 261°/30kt exemple 358 260/30kt.

Question 182-21 : Given .true hdg = 133° tas = 225 kt track t = 144° gs = 206 kt .calculate the w/v ?

075/45kt

Admin .true heading is 133° true track is 144° our drift is 11° right . 2493 exemple 362 075/45kt.

Question 182-22 : Given .true heading = 206° tas = 140 kt true track = 207° gs = 135 kt .calculate the wind ?

180°/05 kt

..true heading is 206° true track is 207° our drift is 1° right . /com en/com061 192 jpg.wind 180°/05 kt exemple 366 180°/05 kt.

Question 182-23 : Given .true heading = 145° tas = 240 kt true track = 150° gs = 210 kt .calculate the wind ?

115°/35 kt

..true heading is 145° true track is 150° our drift is 5° right . /com en/com061 194 jpg.wind 115°/35 kt exemple 370 115°/35 kt.

Question 182-24 : Given .true hdg = 035° tas = 245 kt track t = 046° gs = 220 kt .calculate the w/v ?

340/50kt

Admin .true heading is 035° true track is 046° our drift is 11° right . 2520 exemple 374 340/50kt

Question 182-25 : 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

Admin .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 exemple 378 075°, 39 min.

Question 182-26 : 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

Admin .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 exemple 382 22 kt.

Question 182-27 : 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 exemple 386 10 kt.

Question 182-28 : 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

Admin .wind angle = 100° 063° = 37°.crosswind = windspeed x sin 37° = 20 kt.maximum allowable windspeed = 20 kt / sin 37° = 33 kt exemple 390 33 kt.

Question 182-29 : 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

Admin .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 exemple 394 0736 utc.

Question 182-30 : 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 exemple 398 5 minutes.

Question 182-31 : Given .gs = 122 kt .distance from a to b = 985 nm .what is the time from a to b ?

8 hr 04 min

Admin .985 nm / 122 kt/60 min = 484 minutes 8h04 exemple 402 8 hr 04 min.

Question 182-32 : Given .gs = 435 kt distance from a to b = 1920 nm .what is the time from a to b ?

4 hr 25 min

Admin .1920 nm / 435 kt/60 min = 265 minutes 4h45 exemple 406 4 hr 25 min.

Question 182-33 : Given .gs = 480 kt distance from a to b = 5360 nm .what is the time from a to b ?

11 hr 10 min

Admin .5360 nm / 480 kt/60 min = 670 minutes 11h30 exemple 410 11 hr 10 min.

Question 182-34 : 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 exemple 414 00 hr 59 min.

Question 182-35 : 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 exemple 418 3 hr 12 min.

Question 182-36 : Given .runway direction 083° m .surface wwind 035/35 kt .calculate the effective headwind component ?

24 kt

Admin .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 exemple 422 24 kt.

Question 182-37 : 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

Admin .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 exemple 426 20 kt and 40 kt.

Question 182-38 : Given .runway direction 230° t .surface wind 280° t /40 kt .calculate the effective cross wind component ?

31 kt

Admin .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 exemple 430 31 kt.

Question 182-39 : 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 exemple 434 10 kt.

Question 182-40 : 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

Admin . 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 exemple 438 40 nm.


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