Question 195-1 : Given fl 350 mach 080 oat 55°c calculate the values for tas and local speed of sound lss ? [ Level reports ]

Question 195-2 : Given true heading = 180°tas = 500 ktwv 225° 100 ktcalculate the gs ?

435 kt.

On the computer set 180° under index centre dot on 500 kt under the wind speed 100 kt on the rotating scale 1367you read a groud speed of 435 kt

Question 195-3 : Given true heading = 310° tas = 200 kt gs = 176 kt drift angle 7° rightcalculate the wv ?

270° 33 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 1368you read 270°33kt

Question 195-4 : Given true heading = 090° tas = 200 kt wind = 220°30 ktcalculate the ground speed ?

220 kt.

1369

Question 195-5 : 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 windspeedcrosswind = sin60° x 35 kt = 3031 kt

Question 195-6 : 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°.

Heading311°bearing+270°total581°581° 360° = 221° magnetic bearing of the ndb measured from the aircraft

Question 195-7 : Given the following true track 192° magnetic variation 7°e drift angle 5° leftwhat is the magnetic heading required to maintain the given track ?

190°.

1393use this wonderful table for those questions

Question 195-8 : The angle between the plane of the ecliptic and the plane of equator is approximately ?

235°.

2028

Question 195-9 : Given tas = 485 ktoat = isa +10°cfl 410calculate the mach number ?

0825.

56°c is considered to be the lowest isa temperature and the question states that we are in outside air temperature of isa +10°cat fl410 isa is 56°c isa +10°c will be 46°con the computer 1396by calculation mach = tas x aa = 39square root t° in kelvina = 39 square root 227°k= 58759mach number= 485 58759= 0825

Question 195-10 : At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178° range 135 nmcalculate 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 lajesfor 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° 1398use the latitude scale to find 135 nm

Question 195-11 : At reference1300 utc dr position 37°30'n 021°30'w alter heading port santo ndb 33°03'n 016°23'w tas 450 kt forecast wv 360°30ktcalculate the eta at port santo ndb err a 061 53 ?

1348.

True track is 136°wind 360°30 ktdrift is true heading is ground speed is center dot on 450 kt true index on 136° indicate wind 360°30 kt drift is 3° rightturn to put 133° 136° 3° right drift below true index the groundspeed is under the wind mark 30 kt 470 kt gsplot distance between points and measure along line of longitude com encom061 53jpg6°20' = 633 x 60 nm = 380 nm380 47060 = 485 minutes1300 + 48 minutes 30 secondes = 13h48 30sec

Question 195-12 : 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 ktground speed home = 465 ktpoint of safe return psr = 8 x 465 385 + 465 point of safe return psr = 3720 850point of safe return psr = 43764 hdistance of the psr from the departure point at a speed of 385 kt 43764 h x 385 = 1685 nm

Question 195-13 : Two points a and b are 1000 nm apart tas = 490 kton the flight between a and b the equivalent headwind is 20 kton the return leg between b and a the equivalent tailwind is +40 ktwhat 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 ktground speed home 490 + 40 = 530 ktpet = distance x gsh gso + gsh pet = 1000 x 530 470 + 530 = 530 nm

Question 195-14 : 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 = adtas so gd = gs x ad tas

Question 195-15 : What is the isa temperature value at fl 330 ?

51°c.

15°c 2° x 33 = 51°c

Question 195-16 : Given tas 487kt fl 330 temperature isa + 15calculate the mach number ?

081.

Isa temperature at fl 330 = 15º 2º x 33 = 51ºcdeviation is +15ºc then oat is 51°c + +15°c = 36ºcwith the computer in the airspeed window put the temperature 36ºc in front of 'mach kt ' indexin front of 487 kt on the inner scale you read 081 mach number 1435

Question 195-17 : How many nm would an aircraft travel in 1 minute 45 secondes if gs is 135 kt ?

394 nm.

135 kt 60 minutes = 225 nmminute1 minute 45 secondes = 175 minute225 x 175 = 39375 nm

Question 195-18 : An aircraft travels 100 statute miles in 20 min how long does it take to travel 215 nm ?

50 min.

1 nm = 115 sm215 nm = 215 x 115 = 24725 sm100 sm in 20 minutes = 300 sm in 60 minutes 1h 24725300 = 0824 hour0824 x 60 = 50 min

Question 195-19 : Given tas = 220 ktmagnetic course = 212°wind = 160 ° m 50 ktcalculate 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 195-20 : Given fl250oat 15 °ctas 250 ktcalculate the mach number ?

040.

2485 aviat 617 put temperature 15ºc in front of m kt index in the airspeed windowgo to tas 250 kt on outer scale and read mach number on the inner scaleby calculation mach number = tas lssmach number = 250 39 sqrt 273 15 mach number = 0399

Question 195-21 : During a low level flight 2 parallel roads that are crossed at right angles by an aircraftthe time between these roads can be used to check the aircraft ?

Groundspeed.

Question 195-22 : Given magnetic track = 315° magnetic heading = 301° variation = 5°w tas = 225 ktthe aircraft flies 50 nm in 12 mincalculate the wind °t ?

190°63 kt.

True heading = 296° 301° 5° true course = 310° 315 5° ground speed = 60 x 5012 = 250 kton 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 ktrotate to shift mark under the vertical speed line you read 190°63 kt

Question 195-23 : Given tas = 270 kt true hdg = 270° actual wind 205° t 30ktcalculate 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 195-24 : Given tas = 270 kt true hdg = 145° actual true wind = 205°30ktcalculate 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 1446read the drift and the ground speed 6°l 256 kt

Question 195-25 : Given tas = 470 kt true heading = 317° wind = 045° t 45 ktcalculate 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 1447read drift 5° leftground speed is 470 kt

Question 195-26 : Given tas = 190 kttrue heading = 085°true wind = 110°50ktcalculate 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 1452read drift 8° leftground speed is 147 ktclosest answer 8°l 146 kt

Question 195-27 : Given tas = 132 kt true hdg = 257° true wind = 095°35 ktcalculate the drift angle and gs ?

4°r 165 kt.

Under index set true heading 257° centre dot on tas 132 kt set wind 095°35kt on the rotative scale 1453you read 45°right and 164 kt

Question 195-28 : Given tas = 370 kt true heading = 181° wind = 095°35 ktcalculate true track and ground speed ?

186° 370 kt.

Center dot on tas 370 kttrue heading 181° under index 1454put 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

Question 195-29 : Given tas = 125 kt true heading = 355° true wind = 320°30 ktcalculate the true track and ground speed ?

005° 102 kt.

Center dot on tas 125 kttrue heading 355° under indexput 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 00395 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 wwwjeppesencomdownloadmisccrinstructionspdf page 44 45 true course track and ground speed on top set speed 125 kt over tas indexover 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 circlereading directly up from the wind dot we see that there is a left crosswind component of 17 ktlooking at the outer scale find 17 and opposite it read 8° crab angle 2486since 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 ktlocate 22 on the outer scale and find opposite crab angle of 10° 2487it 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 195-30 : Given tas = 225 kthdg °t = 123°wv = 09060ktcalculate 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 2488rotate to put heading 123° under true index 2489you read a drift of 11° right and a groundspeed of 178 ktadd the drift to your heading to find the true track 134°

Question 195-31 : Given tas = 480 kt true heading = 040° wind = 090°60 ktcalculate true track and ground speed ?

034° 445 kt.

Center dot on tas 480 kttrue heading 040° under index 2026put 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 195-32 : Given tas = 170 kttrue heading = 100°wind = 35030ktcalculate the true track and gs ?

109° 182 kt.

Center dot on tas 170 kttrue heading 100° under indexput 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 encom061 168jpgthe answer is 109° and 182 kt

Question 195-33 : Given tas = 235 kt hdg t = 076° wv = 04040ktcalculate 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 encom061 169jpgread drift 7° rightground speed is 205 kt close enough to answer 204 kt

Question 195-34 : Given tas = 440 kt true heading = 349° wind = 04040ktcalculate drift and ground speed ?

4l 415 kt.

Under index set true heading 349° centre dot on tas 440 kt with the rotative scale set wind 04040ktread drift 4° leftground speed is 415 kt

Question 195-35 : Given tas = 95 kt hdg t = 075° wv = 31020kt calculate the drift and gs ?

9r 108 kt.

Question 195-36 : Given tas = 230 kt hdg t = 250° wind = 20510ktcalculate 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 2490read drift 2° rightground speed is 223 kt

Question 195-37 : Given tas = 205 kt hdg t = 180° wind = 24025ktcalculate 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 encom061 176jpgread drift 6° leftground speed is 194 kt

Question 195-38 : Given tas = 132 kt true heading = 053° wind = 205°15 ktcalculate 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 encom061 178jpgread drift 3° leftground speed is 145 kt

Question 195-39 : Given tas = 90 kttrue heading = 355°wind = 12020 ktcalculate true track and ground speed ?

346 102 kt.

Under index set true heading 355° centre dot on tas 90 kt with the rotative scale set wind 2491read drift 9° leftground speed is 103 kt close enough for the answer

Question 195-40 : Given tas = 155 kt track t = 305° wv = 16018ktcalculate 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 encom061 181jpgnow 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