Why does the ATPL exam use 30 ft per hPa when the real value is 27?
Because 30 ft per hPa is an exam convention, not a measured value: an average between sea level and 700 hPa. Use it for any height calculation unless the question gives another figure.
A student's question, answered from the rotatestudy 061 and 050 notes.
The real value and the exam value
- The real value. Near the surface, pressure drops about 1 hPa per 27 ft. The figure grows with height, to about 1 hPa per 50 ft near 18 000 ft.
- Where 30 comes from. It is an average between sea level and 700 hPa. The notes trace it to the EASA Learning Objectives (ED Decision 2018/001/R, subject 050, altimetry calculation rules); it already appeared in the former 061 objectives. We have not found a date for the original choice, so we do not give one.
- What to use in the exam. 30 ft per hPa for any height calculation, unless the question explicitly gives another figure. Some questions do give 27 ft per hPa: then use 27. Never substitute a value of your own.
- The calculation. Pressure altitude = indicated altitude + (1013 − QNH) × 30. With QNH below 1013, pressure altitude is higher than indicated altitude. An airfield at 1 000 ft with QNH 983 is 30 hPa below 1013: add 900 ft, for a pressure altitude of 1 900 ft. With 27 you would get 1 810 ft, 90 ft lower. Only the figure the question imposes scores.
- Not the temperature error. Colder air than ISA is denser: the pressure layers compress and the aeroplane flies lower than the altimeter indicates. The notes correct it by 4 ft per 1 000 ft for each degree of deviation from ISA.
Sources: the rotatestudy 061 notes (§ 3.6 and Appendix E) and the 050 notes (§ 1.3). Appendix E of each subject lists the texts it was checked against and the verification date.
The ten free questions below, from 061, include a crosswind component and a true airspeed that starts from a pressure altitude.
Ten free 061 General Navigation questions
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The notes come with 1 month of all 417 questions and the mock exams.
1Using the data below, determine the track error angle.
Planned track: 197°(T)
Actual heading: 213°(T)
Actual in-flight drift: 12° leftShow the answer
Answer C. With 12° of left drift the aircraft actually tracks 213 − 12 = 201°(T). The track error angle is measured from the planned track to the track made good: 201 − 197 = 4°, and since the actual track lies clockwise of the planned one, the error is to the right. Comparing the heading directly with the planned track (16° right) forgets the drift.
2A ground feature is seen on a relative bearing of 315° and, 3 min later, on a relative bearing of 270°. The W/V is calm and the aircraft GS is 180 kts. What is the minimum distance between the aircraft and the ground feature?
Show the answer
Answer B. Relative bearing 315° puts the feature 45° left of the nose; 270° puts it exactly abeam on the left. When the angle off the nose doubles from 45° to 90°, the distance flown equals the distance to the feature. In 3 min at 180 kt the aircraft covers 180 × 3/60 = 9 NM. The feature is abeam at that moment, so 9 NM is also the closest point of approach.
3A pilot is planning a VFR cross-country flight using visual checkpoints. Which of the following would make the most suitable checkpoint?
Show the answer
Answer B. A checkpoint is useful only if it can be found quickly and cannot be confused with anything else, and that depends on contrast with what surrounds it. A feature that blends into its surroundings is hard to spot, one valley among many low hills looks like the next, and small lakes in mountains are numerous and easily mixed up.
4Given the following data for a conic projection chart in the northern hemisphere, calculate the change of longitude:
Great circle track from P to Q, at P: 112°(T)
Great circle track from Q to P, at Q: 316°(T)
Constant of the cone: 0.8Show the answer
Answer D. The track from Q to P at Q is 316°, so the track from P to Q at Q is 316 − 180 = 136°. The straight line changes direction by 136 − 112 = 24°, which is the chart convergency. Chart convergency = change of longitude × constant of the cone, so the change of longitude is 24/0.8 = 30°. 24° is the convergency itself, not the change of longitude.
5What is the Local Mean Time (LMT) at (35°25'S, 110°45'W) when the Coordinated Universal Time (UTC) is 18:34 on 4th September?
Show the answer
Answer D. 110°45' of longitude is 7 h 23 min of time (110.75 x 4 min = 443 min). West of Greenwich, local mean time is behind UTC, so subtract: 18:34 - 7:23 = 11:11 on 4 September. Adding the difference instead gives 01:57 on the 5th, the classic east/west mix-up.
6Using trigonometry and the data below, calculate the crosswind component (XWC) to the nearest knot.
Runway: 32
Wind (magnetic): 020°/25 kt
Variation: 15°EShow the answer
Answer A. Runway designators and the tower wind are both magnetic, so the variation plays no part. Runway 32 is about 320°(M) and a wind from 020° is 60° off the runway axis. Crosswind = 25 × sin 60° = 21.7, about 22 kt. Using the cosine gives the 13 kt headwind component instead, and applying the variation to the wind only distorts the angle.
7Using the information below, what is the TAS?
Pressure altitude: 6000 ft
OAT: +5°C
IAS/CAS: 115 ktShow the answer
Answer D. ISA at 6 000 ft is 15 − 12 = +3 °C, so +5 °C is ISA +2. Pressure at 6 000 ft is about 812 hPa, giving a density ratio σ = (812/1013) × (288/278) ≈ 0.83. TAS = CAS/√σ = 115/0.911 ≈ 126 kt, which is what the airspeed window of a flight computer gives with 6 000 ft set against +5 °C. Multiplying instead of dividing gives about 105 kt, and 115 kt ignores the density correction.
8A pilot is planning a VFR flight that crosses controlled airspace. Because of the airspace restrictions, part of the flight must be flown at low level. Which of the following is a navigation problem that may be met at low level on a VFR flight?
Show the answer
Answer B. The distance to the visible horizon depends on height, so at low level the pilot sees much less of the surrounding country and checkpoints come into view late. Rising ground and trees shorten the view further. Tall features are actually easier to see from low level, and the sensation at low height is one of higher, not lower, speed over the ground.
9Determine the two smallest map scales among the following:
A) 1:500 000
B) 10 cm = 10 NM
C) 5 inch = 10 NM
D) 1:110 000Show the answer
Answer B. Convert everything to a ratio. B: 10 cm = 10 NM = 1 852 000 cm, i.e. 1:185 200. C: 5 inches = 12.7 cm for 1 852 000 cm, about 1:145 800. The smallest scales are those with the largest denominators: A (1:500 000) and B (1:185 200); D is the largest scale of the four.
10What are the UTC and date when it is 23:00 LMT on 21st August at position A (11°00'N, 067°30'W)?
Show the answer
Answer D. Arc to time: 67°30' of longitude is 4 h 30 min. West of Greenwich, local time is behind UTC, so UTC is LMT plus the difference: 23:00 + 4:30 = 03:30 the next day, 22nd August. 18:30 comes from subtracting the longitude, and 03:30 on the 21st forgets that the time has passed midnight.
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