Dead Reckoning, Wind & the CX-3

A. 093

C-D-M-V-T, East is added, West subtracted going towards True: compass 095 + 3E = magnetic 098; 098 - 5W = true 093.

The question was recalled without its options; these options were written to it.

D. 7 deg R, 204 kt

The wind comes from 36 deg left of the nose, so it blows the aircraft to the right (starboard drift) and is mostly a headwind. Vector sum: track 083, GS 204 kt, drift 7 deg right. Drift is the angle from heading to track, named by the side the aircraft is blown to.

The question was recalled without its options; these options were written to it.

B. 174 NM

The 30-minute climb covers 189 NAM, so TAS is 378 kt. With a 30 kt headwind GS is 348 kt, and 348 x 0.5 h = 174 NM.

A. 5 deg L

Crosswind = 20 sin 40 = 12.9 kt from the left. WCA = asin(12.9 / 140) = 5.3 deg - steer LEFT, into the wind: heading 085. (1 in 60 check: 12.9 x 60 / 140 = 5.5 deg.)

The question was recalled without its options; these options were written to it.

B. 10 kt

Angle between wind and runway = 20 deg. Crosswind = 30 sin 20 = 10 kt; headwind = 30 cos 20 = 28 kt.

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A. 6 deg right

Track error = 3 x 60/45 = 4 deg; closing angle = 3 x 60/90 = 2 deg. Total correction = 6 deg to the right.

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D. 271 deg and 272 kt

Wind from 55 deg left of the nose: it pushes the track right of heading (to about 271) and takes off about 20 kt: GS 272 kt.

The question was recalled without its options; these options were written to it.

B. 64 NM

Ground distance = air distance + wind effect over the time: 71 - 30 x 14/60 = 71 - 7 = 64 NM.

The question was recalled without its options; these options were written to it.

A. 193Nm

Climb is 36 min = 0.6 h. A 60 kt tailwind adds 60 x 0.6 = 36 NM: 157 + 36 = 193 NM.

D. 406 Nm

GS = 460 + 50 = 510 kt, so the sector takes 450/510 h. Still-air distance = TAS x time = 460 x 0.882 = 406 NAM.

A. 057 / 28 kt

Winds are averaged as VECTORS, not as numbers: resolving both into north and east components and halving gives about 057 / 28 kt. Simply averaging the figures (050 / 30) ignores that the two winds partly cancel and is wrong.

A. the groundspeed is higher and the ground roll longer

The wing needs the same airspeed; with a tailwind the aircraft must also go faster over the ground (TAS + tailwind), so it needs more runway and arrives at a higher groundspeed. That is why a downwind landing in an emergency is fast and long.

B. -61 kt

Wind component here means GS - TAS. With 15 deg drift, GS = 240 cos 15 + along-track wind, so the real along-track wind is 285 - 232 = 53 kt. Reversed: GS = 232 - 53 = 179 kt, component 179 - 240 = -61 kt. Large drift makes the reverse component worse than just changing the sign.

A. 400 NAM and 340 NGM

Air distance = TAS x time = 400 NAM. Ground distance = GS x time = 170 x 2 = 340 NGM.

A. 535 ft/min

ROD = gradient x GS in ft/min = 0.033 x 160 x 6076 / 60 = 535 ft/min (quick: % x GS x 1.013).

The question was recalled without its options; these options were written to it.

A. 1.852 km and 1.609 km

1 NM = 1852 m (one minute of latitude, by definition); 1 SM = 5280 ft = 1609 m. So NM x 1.852 = km, SM x 1.609 = km.

The question was recalled without its options; these options were written to it.

C. 7 deg L

1 in 60: track error = distance off x 60 / distance gone = 7 x 60 / 60 = 7 deg, left of track.

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A. greater

Ground miles = air miles + wind component x time. A tailwind adds ground distance (NGM > NAM); a headwind takes it away. Climb tables give NAM, so correct them for wind before planning top of climb.

B. 336(T), 325 kt

The wind is 5 deg right of the track and almost head-on: crosswind 50 sin 5 = 4 kt needs under 1 deg of correction into wind (right), heading 336; headwind 50 cos 5 = 50 kt, GS = 375 - 50 = 325 kt.

B. 40 and 20 kt

Maximum: crosswind = V sin 60 = 35, V = 40 kt. Minimum: headwind = V cos 60 = 10, V = 20 kt.

The question was recalled without its options; these options were written to it.

B. heading/TAS (air vector), track/GS (ground vector) and the wind velocity

Each side is a VECTOR (direction and speed): the air vector heading/TAS plus the wind vector W/V gives the ground vector track/GS.

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A. 360 / 15 kt

Drift right means the wind blows from the LEFT of the nose (northerly for an easterly heading); GS equal to TAS means no head or tail component, so the wind is square across: 360. Size: 180 sin 5 = 15.7, about 15 kt.

The question was recalled without its options; these options were written to it.

B. track error

Track error = required track vs track made good. Drift = heading vs track made good. Closing angle = correction needed to reach the next point.

The question was recalled without its options; these options were written to it.

A. 7° left

True heading = 255 - 40 = 215. Solving the triangle with GS 375 and wind 235/120 gives TAS about 485 kt and drift 6-7 deg; the wind comes from the right of heading, so drift is left.

C. 192kt,7° right

Wind 350/30 is 58 deg to the left of track 048: crosswind 25 kt, headwind 16 kt. Drift = 7 deg (wind from the left, so right drift); GS = 210 cos 7 - 16 = 192 kt.

B. 320° / 60 kt

Track = 090 + 5 = 095. Ground vector minus air vector gives wind towards 140 at 59 kt, i.e. from 320 at about 60 kt.

B. 20kt and 40 kt

Minimum wind = 10 / cos 60 = 20 kt; maximum = 35 / sin 60 = 40 kt.

A. 6°L-487 kt

The wind 110/50 blows towards 290, 105 deg to the right of the heading 005. Adding the wind vector to the air vector (472 kt on 005) gives a ground vector of 487 kt on 359: drift 6 deg left, GS 487 kt.

A. 24kt

Angle between wind and runway = 83 - 35 = 48 deg: headwind = 35 x cos 48 = 23.4, about 24 kt.

C. 13 Min, 271, 26 nm, 2.5 Gal

Descent 10,500 to 2,700 ft (1,000 ft above the 1,700 ft airport) = 7,800 ft at 600 ft/min = 13 min. TAS about 135 kt, headwind component about 12 kt gives GS about 123 kt: 26 NM. Drift about 12 deg: heading 263 + 12 = 275 T, 270 M, 271 C. Fuel 11.5 x 13/60 = 2.5 gal.

B. 1045 Hrs, 437.5 nm, 408.7 nm.

The correct answer is: 1045 Hrs, 437.5 nm, 408.7 nm..

A. 30 kt

Wind 240 against runway 300 is 60 deg off: crosswind = 35 x sin 60 = 30 kt.

A. 50 min

100 SM in 20 min = 300 SM/h = 261 kt. 215 NM / 261 kt = 0.82 h = about 50 min.

A. 37 min

Wind 65 deg off the nose from the left: headwind 40 cos 65 = 17 kt, crosswind 36 kt (4.4 deg drift). GS = 470 - 17 = 452 kt. Time = 275 / 452 = 0.61 h = 37 min.

A. 130°

Drift is greatest when the wind is at 90 deg to the track: wind from 040 + 90 = 130 (or 310). Maximum drift = arcsin (30/120) = 14.5 deg.

B. 130

For a FIXED TRACK the crosswind component, and so the drift, is greatest when the wind blows at 90 deg to the track: 040 + 90 = 130 (or 310). Drift = asin(30/120) = 14.5 deg.

C. WCA -2° , GS 396 Kts

Wind 340 is 37 deg left of track 017: crosswind 18 kt gives WCA about -2 (heading left); headwind 24 kt gives GS 396 kt.

D. 4.98 %

Gradient = height gained / distance covered. In one minute the aircraft covers 198 x 6076 / 60 = 20,050 ft and climbs 1000 ft: 1000 / 20,050 = 4.99 %. Rule of thumb: ROC (ft/min) = gradient % x GS (kt) x 1.013.

The question was recalled without its options; these options were written to it.

A. 0.000,000,001

A nanosecond is 10^-9 s = 0.000 000 001 s.

Our independent check found that more than one option can be argued for, as worded in the recall. The key gives the answer most references support.

A. 136 kt

1 kt = 0.514 m/s, so 70 / 0.514 = 136 kt (70 m/s x 3.6 = 252 km/h).

The question was recalled without its options; these options were written to it.

A. 530 ft/min

ROD = GS x 101 x tan 3 = 100 x 101 x 0.0524 = 530 ft/min (quick: GS x 5 = 500).

A. 480 ft/min

ROD = GS x 101 x tan 3 = 90 x 101 x 0.0524 = 478 ft/min (quick: GS x 5 = 450).

The question was recalled without its options; these options were written to it.

A. 10 kt

Convert the wind to magnetic first: 160 T + 10 W = 170 M. Angle to the runway 30 deg: crosswind 20 sin 30 = 10 kt (headwind 17 kt). Using the true wind directly would give 20 deg and 7 kt.

A. 87

27,000 ft to lose x 3 NM / 1000 ft = 81 NM, plus the 6 NM: begin about 87 NM out.

B. decrease

The glide path fixes the height lost per mile. At a lower groundspeed the miles pass more slowly, so fewer feet per minute are needed - a headwind on final means a smaller ROD.

A. 630 ft/min

ROD = GS x 101.3 x tan(angle) = 142 x 101.3 x 0.0437 = 628 ft/min. (For 3 deg the shortcut is GS x 5; for 2.5 deg it is GS x 4.4.)

The question was recalled without its options; these options were written to it.

A. 020°/95 kts,

LSS at -50 C = 38.95 x sqrt 223 = 582 kt, so TAS = 0.80 x 582 = 465 kt. TH = 140 - 20 = 120; track = 131. Wind = ground vector - air vector gives 020/95.

C. 186, 370 kt

The wind blows from 86 deg left of the nose - almost a pure crosswind from the left: drift 35/370 x 57 = 5 deg right, track 186; with no head or tail component the GS stays about 370 kt.

The question was recalled without its options; these options were written to it.

A. 1 h 10 min, 1150 UTC

ETE = 210 / 180 = 1.167 h = 1 h 10 min. ETA = 1040 + 1:10 = 1150 UTC.

A. 83 NM

Height to lose 25,000 ft / 1500 = 16.7 min. Distance = 300 x 16.7 / 60 = 83 NM. (Shortcut: 3 NM per 1000 ft at 300 kt / 1500 ft/min gives 75 NM - close, but work it when the options are this far apart.)

C. 098

Track = heading + starboard drift (or - port drift): 090 + 8 = 098. The lubber line shows heading; the track line shows where the aircraft actually goes.